Mathematics

New Saraswati House (India) Pvt. Ltd.New Delhi-110002 (INDIA)

��������� ��

�������� ���� ��

��������

��� ������

Surender VermaM.Sc. (Maths), B.Ed

Delhi Public School,Dwarka, New Delhi

ByP.W

- 288 pages / 01-06-2016

Second Floor, MGM Tower, 19 Ansari Road, Daryaganj, New Delhi-110002 (India) Phone : +91-11-43556600Fax : +91-11-43556688E-mail : delhi@saraswatihouse.comWebsite : www.saraswatihouse.comCIN : U22110DL2013PTC262320Import-Export Licence No. 0513086293

Branches:

• Ahmedabad (079) 22160722 • Bengaluru (080) 26619880, 26676396 • Bhopal +91-7554003654 • Chennai (044) 28416531 • Dehradun 09837452852• Guwahati (0361) 2457198• Hyderabad (040) 42615566 • Jaipur (0141) 4006022 • Jalandhar (0181) 4642600, 4643600 • Kochi (0484) 4033369 • Kolkata (033) 40042314 • Lucknow (0522) 4062517 • Mumbai (022) 28737050, 28737090 • Patna (0612) 2570403 • Ranchi (0651) 2244654

Revised edition 2017

Published by: New Saraswati House (India) Pvt. Ltd.19 Ansari Road, Daryaganj, New Delhi-110002 (India)

The moral rights of the author has been asserted.

©Reserved with the Publishers

All rights reserved under the Copyright Act. No part of this publication may be reproduced, transcribed, transmitted, stored in a retrieval system or translated into any language or computer, in any form or by any means, electronic, mechanical, magnetic, optical, chemical, manual, photocopy or otherwise without the prior permission of the copyright owner. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages.

This book is meant for educational and learning purposes. The author(s) of the book has/have taken all reasonable care to ensure that the contents of the book do not violate any copyright or other intellectual property rights of any person in any manner whatsoever. In the event the author(s) has/have been unable to track any source and if any copyright has been inadvertently infringed, please notify the publisher in writing for any corrective action.

��������

1. Real Numbers

� Worksheets (1 to 6) ..................................................................................................... 6

• Assessment Sheets (1 and 2) ................................................................................... 11

• Chapter Test ............................................................................................................. 13

2. Polynomials

� Worksheets (10 to 15) ...............................................................................................15

• Assessment Sheets (3 and 4) ...................................................................................23

• Chapter Test ............................................................................................................. 26

3. Pair of Linear Equations in Two Variables

� Worksheets (18 to 29) ...............................................................................................29

• Assessment Sheets (5 and 6) ...................................................................................45

• Chapter Test ............................................................................................................. 48

4. Triangles

� Worksheets (33 to 45) ...............................................................................................51

• Assessment Sheets (7 and 8) ...................................................................................68

• Chapter Test ............................................................................................................. 71

5. Introduction to Trigonometry

� Worksheets (50 to 58) ...............................................................................................73

• Assessment Sheets (9 and 10) ................................................................................. 84

• Chapter Test ............................................................................................................. 86

– 3 –

6. Statistics

� Worksheets (62 to 65) ...............................................................................................88

• Assessment Sheet 11 ................................................................................................ 93

• Chapter Test ............................................................................................................. 97

PRACTICE PAPERS (1 to 5) ..................................................................................... 99

– 4 –

��������� ��

���������������

���

������������

������� �������������

����������

6 �� � � � � � � � � �

�������� �

1. (C) We know that the factors of a prime are1 and the prime itself only.Therefore, the common factor of p and qwill be 1 only. Hence, HCF (p, q) = 1.

2. (A) As prime factors of 1005 are:1005 = 5 × 3 × 67.∴ 7 is not a prime factor of 1005.

3.3

4 3 3125 5 1

= = = 0.0625. 162 5 16×5

Clearly, the decimal form of 4 3125

.2 5 termi-

nates after four places.

4. Terminating

Hint: 24 192= = 0.192

125 1000.

5. LCM =First number × Second number

HCF

=96 × 404

4= 24 × 404 = 9696.

6. (i) 660 (ii) 330Hint: Going in opposite direction to thefactor tree, we obtain2 × 165 = 330 (ii) and 2 × 330 = 660 (i).

7. HCF = 3; LCM = 420Hint: 12 = 22 × 3; 15 = 3 × 5; 21 = 3 × 7.

8. (i) Terminating

Hint: 1 3543 543

=250 2 × 5

.

(ii) Non-terminating repeating

Hint: 2 19 1 1

= =108 12 2 × 3

.

9. Hint: Let 5 – 2 3 = ab

; b ≠ 0

⇒ 3 = 5 –2b a

b

As RHS of this equation is rational, butLHS is irrational so a contradiction.

10. Let a be any odd positive integer and b = 4.By Euclid’s lemma there exist integers qand r such that

a = 4q + r, 0 ≤ r < 4... a = 4q or 4q + 1 or 4q + 2 or 4q + 3.Therefore, for a to be odd, we have to take

a = 4q +1 or 4q +3.

11. The maximum capacity (in kg) of a bag willbe the HCF of 490, 588 and 882. Let us findout the required HCF by prime factorisationmethod.

490 = 2 × 5 × 72

588 = 22 × 3 × 72

882 = 2 × 32 × 72

∴ HCF = 2 × 72 = 98

Thus, the maximum capacity of a bag is98 kg.

�������� �

1. (A) HCF (p, q) = 1 ⇒ p and q are coprime.

If p and q are coprime with q ≠ 0 and pq

is

a rational number, then q has only 2 and 5as prime factors.

i.e., q = 2m × 5n where, m and n are non-negative integers.

2. (B) Going to opposite direction to thefactor tree, we obtain

3 × 7 = 21 (ii) and 2 × 21 = 42 (i).

3. 2 = 1.414... and 3 = 1.732...

Therefore, we can take 1.5 = 32

as 2 < 32

< 3 .

��������

REAL NUMBERS

7��� ����� �

4. Required number =23×1449

161

=1449

7 = 207.

5. Hint: As 12576 > 4052... 12576 = 4052 × 3 + 420Further 4052 = 420 × 9 + 272Further 420 = 272 × 1 + 148Further 272 = 148 × 1 + 124Further 148 = 124 × 1 + 24Further 124 = 24 × 5 + 4Further 24 = 4 × 6 + 0.In the last equation, remainder is zero.Hence, the required HCF = 4.

6. First given number is composite as5 × 3 × 11 + 11 = 11 (15 + 1) = 11 × 16

= 11 × 2 × 8But second given number is prime as5 × 7 + 7 × 3 + 3 = 35 + 21 + 3 = 59.

7. No. Prime factors of 6n will be of type 2n × 3n.As it doesn't have 5 as a prime factor, so 6n

can't end with the digit 5.

8. Hint: Let a be any positive integer... a = 3q or 3q + 1 or 3q + 2... a2 = 9q2 = 3m; m = 3q2

or a2 = (3q +1)2 = 3m + 1, m = q (3q + 2)or a2 = (3q + 2)2 = 3m + 1, m = 3q2 + 4q + 1.

9. We represent 6, 72 and 120 in their primefactors.

6 = 2 × 372 = 23 × 32

120 = 23 × 3 × 5Now, HCF = 2 × 3 = 6And LCM = 23 × 32 × 5 = 360.

10. Hint: Let 2 5�

= x, a rational number

⇒ 2 = x + 5Squaring both sides, we get

2 = x2 + 5 + 2x 5

⇒ 5 =2– – 32

xx

RHS of this last equation is rational, butLHS is irrational which is a contradiction.

11. Length = 6 m 30 cm = 630 cmBreadth = 5 m 85 cm = 585 cmHeight = 3 m 60 cm = 360 cmThe required length (in cm) of the tape willbe the Highest Common Factor (HCF) ofthe numbers 630, 585 and 360.Let us find out the HCF.

630 = 2 × 32 × 5 × 7585 = 32 × 5 × 13360 = 23 × 32 × 5

∴ HCF = 32 × 5 = 45Hence, the length of the tape will be 45 cm.

�������� –

1. (C) 4 3 4 443 43 × 5 215

= = 2 × 5 (2 × 5) 10

= 0.0215

Hence, the number terminates after fourplaces of decimal.

2. (A) � �� � � � � �2 2

2 3 2 3 2 3� � � �

= 2 – 3 = –1.– 1 is a rational number.

3. 128 = 27; 240 = 24 × 3 × 5.Now, HCF (128, 240) = 24 = 16.

4. 232

Hint: First number =HCF × LCM

Second number.

5. No.Hint: Prime factors of 15n does not contain2p × 5q in factor, p, q being positive integers.

6. Rational number = 0.27Irrational number = 0.26010010001... .

7. (i) 145 29 8 232

= × = = 0.232625 125 8 1000

.

(ii) 7 125 875

× = = 0.087580 125 10000

.

8. Let us assume, to the contrary that 2 isrational. We can take integers a and b ≠ 0such that

3 = ab

, where a and b are coprime.

⇒ 3b2 = a2

⇒ a2 is divisible by 3

8 �� � � � � � � � � �

⇒ a is divisible by 3 ...(i)We can write a = 3c for some integer c⇒ a2 = 9c2

⇒ 3b2 = 9c2 (... a2 = 3b2)⇒ b2 = 3c2

⇒ b2 is divisible by 3⇒ b is divisible by 3 ...(ii)From (i) and (ii), we observe that a and bhave atleast 3 as a common factor. But thiscontradicts the fact that a and b are co-prime. This means that our assumption isnot correct.

Hence, 3 is an irrational number.

9. As: 1032 = 408 × 2 + 216 ...(i)408 = 216 × 1 + 192 ...(ii)216 = 192 × 1 + 24 ...(iii)192 = 24 × 8 + 0 ...(iv)

⇒ HCF = 24∴ From (iii)⇒ 24 = 216 – 192

= 216 – [408 – 216] [Use (ii)]= 2 × 216 – 408= 2[1032 – 2 × 408] – 408

[Use (i)]⇒ 24 = 1032 × 2 – 5 × 408⇒ m = 2.

10. Hint: Let x be any positive integer.Then it is of the form 3q or 3q + 1 or 3q + 2.If x = 3q, then

x3 = (3q)3 = 9m; m = 3q3

If x = 3q + 1, thenx3 = (3q + 1)3

= 9m + 1; m = q(3q2 + 3q + 1).If x = 3q + 2, then

x3 = (3q + 2)3

= 9m + 8; m = q (3q2 + 6q + 4).

11. The maximum number of columns must bethe highest common factor (HCF) of 616and 32. Let us find out the HCF by themethod of Euclid's division lemma.Since 616 > 32, we apply division lemmato 616 and 32, to get

616 = 32 × 19 + 8

Since the remainder 8 �� 0, we apply thedivision lemma to 32 and 8, to get

32 = 8 × 4 + 0The remainder has now become zero, soour procedure stops. Since the divisor atthis stage is 8, the HCF of 616 and 32 is 8Hence, the maximum number of columnsis 8.

��������– �

1. (B) Non-terminating repeating.Hint: Denominator is not in the exact formof 2m × 5n, where m, n are non-negativeintegers.

2. (C) 0 ≤ r < b.

3. � �� � � � � �2 2

6 5 6 5 6 5� � � �

= 6 – 5 = 1 = Rational number.4. Terminating decimal form as denominator

4 of 1074

is of the form 2n × 5m.

Here n = 2, m = 0

5. (i) 1001 (ii) 91Hint: 7 × 13 = (ii) and (ii) × 11 = (i).

6. Let us represent each of the numbers 30,72 and 432 as a product of primes. 30 = 2 × 3 × 5 72 = 23 × 32

432 = 24 × 33

Now, HCF = 2 × 3 = 6and LCM = 24 × 33 × 5 = 2160.

7. Here, 396 > 82.∴ 396 = 82 × 4 + 68Further 82 = 68 × 1 + 14Further 68 = 14 × 4 + 12Further 14 = 12 × 1 + 2Further 12 = 2 × 6 + 0In the last equation, the remainder is zeroand the divisor is 2.Hence, the required HCF = 2.

8. Hint: Let 3 + 2 5 = ab

; b ≠ 0

⇒– 32

a bb

= 5 = Rational

9��� ����� �

Which is a contradiction as 5 is an irra-tional number.

9. (i) The given fraction can be written as

4 3 443 43 × 5

= = 0.02152 ·5 10

Hence, the given number terminates afterfour places of decimal.

(ii) The given fraction can be written as

4

5 5 5

359 2 × 359=

2 × 5 2 × 5 =

5744100000

= 0.05744Hence, the given number terminates afterfive places of decimal.

10. The required number of students will bethe highest common factor (HCF) of 312,260 and 156. Let us find out the HCF bythe method of prime factorisation.

312 = 23 × 3 × 13

260 = 22 × 5 × 13

156 = 22 × 3 × 13

∴ HCF = 22 × 13 = 52Number of buses required

= Total number of students

Number of students in one bus

= 312 + 260 + 156

= 1452

Thus, the maximum number of students ina bus and number of buses required are52 and 14 respectively.

11. Hint: Let x = any positive integerx = 5m, 5m + 1, 5m + 2, 5m + 3 or 5m + 4Now take square of all these form.

�������� – �

1. (C) Let the quotient is m when n2 – 1 isdivided by 8.∴ n2 – 1 = 8 × m⇒ n2 – 1 = 0, 8, 16, 24, 32, ... .

⇒ n2 = 1, 9, 17, 25, 33, ...

⇒ n = ± 1, ± 3, ± 17 ± 5, ± 33 , ...∴ n = An odd integer is the right

Answer.

2. Hint: HCF (65, 117) = 13Now, 65m – 117 = 13.∴ m = 2 will satisfy this equation.

3. Hint: LCM of 18, 24, 30, 42 = 2520∴ Required number = 2520 + 1 = 2521.

4. Prime factors of numbers 1 to 10 are:1 = 1; 2 = 1 × 2; 3 = 1 × 3; 4 = 1 × 22

5 = 1 × 5; 6 = 1 × 2 × 3; 7 = 1 × 7;8 = 1 × 23; 9 = 1 × 32; 10 = 1 × 2 × 5Now,LCM = 1 × 23 × 32 × 5 × 7

= 8 × 9 × 5 × 7 = 2520 is requirednumber.

5. 2.

Hint: 5 3

5 3

�

�

= 2x – 15

⇒ 4 – 15 = 2x – 15

⇒ x = 2, which is a rational number.6. Hint: Any odd positive integer will be

type of 4q + 1 or 4q + 3∴ (4q + 1)2 = 16q2 + 8q + 1

= 8 (2q2 + q) + 1= 8n + 1

Also, (4q + 3)2 = 16q2 + 24q + 9= 8 (2q2 + 3q + 1) + 1= 8n + 1.

7. 35 cmHint: Find HCF.

8. Hint: Let 5 3 2�

=ab

where a, b are integers and b ≠ 0Squaring on both sides,

5 18 6 10� � =2

2ab

⇒2

223a

b�

= 6 10

⇒�

2 2

2

23

6

b a

b= 10 ... a contradiction.

10 �� � � � � � � � � �

9. (i) Terminating (ii) Terminating.

10. The required number of burfis will be thehighest common factor of 420 and 130.Let us find out the HCF using Euclid'sdivision lemma.It is clear that 420 > 130. We apply Divisionlemma to 420 and 130, to get

420 = 130 × 3 + 30Since the remainder 30 �� 0, so we applyDivision lemma to 130 and 30, to get

130 = 30 × 4 + 10Again the remainder 10 �� 0, so we applyDivision lemma to 30 and 10, to get

30 = 10 × 3 + 0Now, the remainder is zero. So the HCF of420 and 130 is the divisor at the last stagethat is 10.Hence, the required number of burfis is 10.

11. Let n = 3q, 3q + 1 or 3q + 2.Case I. If n = 3q, then

n = 3q divisible by 3n + 2 = 3q + 2 ⇒ Not divisible by 3n + 4 = 3q + 4 = 3(q +1) + 1

⇒ Not divisible by 3.Case II. If n = 3q + 1 then only

n + 2 = 3q + 1 + 2 = 3q + 3= 3(q + 1) is divisible by 3.

Case III. If n = 3q + 2 then onlyn + 4 = 3q + 6 = 3(q + 2)

is divisible by 3.

�������� – �

1. (C) 3825 = 52 × 32 × 17So, 11 is not a prime factor of 3825.

2. (C) As p and p + 1 are two consecutivenatural numbers, HCF = 1 andLCM = p (p + 1).

3. Hint: The given number is 51 17

or 1500 500

∴ Denominator = 500 = 22 × 53

Clearly, the denominator is exactly in theform 2m × 5n, where m and n are non-negative integers; so the given numberhas a terminating decimal expansion.

4. 1800Hint: ��� 8 = 23; 9 = 32; 25 = 52

∴ HCF (8, 9, 25) = 1

LCM (8, 9, 25) = 1800.

5. – 19

Hint: HCF (210, 55) = 5∴ 210 × 5 + 55y = 5⇒ 55y = 5 – 1050

⇒ y = 104555

�

= – 19.

6. Irrational

Hint: 2 3

2 3

�

�

=3

x

⇒ 7 – 4 3 = 3

x

⇒ 7 3 – 12 = x = Irrational.

7. Rational Number = 0.55Irrational number = 0.5477477747... .

8. 15Hint: HCF (1380, 1455, 1620) = 15.

9. (i) 0.052. (ii) 5.8352.10. We know that any positive integer is either

of the form 3q, 3q + 1 or 3q + 2 for someinteger q.Now, three cases arise.Case I. When p = 3q,p + 2 = 3q + 2 and p + 4 = 3q + 4Here, p = 3q is exactly divisible by 3

p + 2 = 3q + 2 leaves 2 as remain-der when it is divided by 3

p + 4 = 3q + 4 or 3 (q + 1) + 1 leaves1 as remainder when it isdivided by 3.

Case II. When p = 3q + 1, p + 2 = 3q + 3 and p + 4 = 3q + 5

Here, p = 3q + 1 leaves 1 as remainder when it is divided by 3

p + 2 = 3q + 3 or 3 (q + 1) is exactly divisible by 3

p + 4 = 3q + 5 or 3(q + 1) + 2 leaves 2 as remainder when it is divided by 3.

11��� ����� �

Case III. When p = 3q + 2, p + 2 = 3q + 4and p + 4 = 3q + 6Here, p = 3q + 2 leaves 2 as remainder whenit is divided by 3.p + 2 = 3q + 4 or 3(q + 1) + 1 leaves 1 asremainder when it is divided by 3p + 4 = 3q + 6 or 3(q + 2) is exactly divisibleby 3.Hence, in all the cases, one and only onenumber out of p, p + 2 and p + 4 is divisibleby 3, where p is any positive integer.

ORAny positive odd integer is type of 2q + 1where q is a whole number.∴ (2q + 1)2 = 4q2 + 4q + 1 = 4q (q + 1) + 1

...(i)Now, q(q + 1) is either 0 or evenSo it is 2m, where m is some number.∴ From (i) ⇒ (2q + 1)2 = 8m + 1.

11. Since, height of each stack is the same,therefore, the number of books in each stackis equal to the HCF of 96, 240 and 336.Let us find their HCF

96 = 24 × 2 × 3240 = 24 × 3 × 5336 = 24 × 3 × 7

So, HCF = 24 × 3 = 48.Now, number of stacks of English books

=9648

= 2

Number of stacks of Hindi books

= 24048

= 5

Number of stacks of Mathematics books

= 33648

= 7.

�������������� – �

1. (D) The denominator of each fraction in theoptions (A), (B) and (C) can be expressed inthe form 2n 5m, where m, n being non-negative integer.

2. (A) Let x be any positive integer then it isof the form 3q or 3q + 1 or 3q + 2. So, x2

canbe written in the form 3m or 3m + 1.

3. HCF × LCM = Product of the two numbers⇒ 40 × 252 × p = 2520 × 6600

⇒ p =2520×6600

40× 252= 1650.

4. No; because HCF must divide LCM and hereHCF = 18 which doesn't divide LCM whichis 380.

5. True, If the number 3n ends with the digit 0,then its prime factorisation contains theprimes 2 and 5. But by the FundamentalTheorem of Arithmetic, there is no prime otherthan 3 in the factorisation of 3n.

6. The required number would be the HCF of967 – 7 = 960 and 2060 – 12 = 2048.Let us find the HCF of 960 and 2048 byusing Euclid’s algorithm.Since 2048 > 960∴ 2048 = 960 × 2 + 128

960 = 128 × 7 + 64128 = 64 × 2 + 0

Since the remainder becomes zero and thedivisor at this stage is 64, the HCF of 960and 2048 is 64.Hence, the required number is 64.

7.

Clearly, 456 = 23 × 3 × 19and 360 = 23 × 32 × 5∴ HCF = 23 × 3 = 24

Hence LCM =456×360

24= 6840.

8. (i) Time taken by Ram to complete one cycle= 180 seconds.

Time taken by Shyam to complete one cycle= 150 seconds.

∴ Consider LCM of 180 and 150.

12 �� � � � � � � � � �

∴ 180 = 22 × 5 × 32

150 = 2 × 52 × 3

∴ LCM of 180 and 150 = 22 × 52 × 32

= 4 × 25 × 9

= 900 seconds

=90060

= 15 minutes

∴ They both will again meet after 15 minutes.(ii) Since they started at 6 a.m. and they will be

meeting again after 15 minutes.∴ The time will be 6:15 a.m.

(iii) L.C.M. of real numbers.(iv) Since Ram and Shyam go for morning walk

daily. So, it depicts their discipline andhealth consciousness.

9. Let a be any odd positive integer. Then, it isof the form 6p + 1, 6p + 3 or 6p + 5.Here, three cases arise.Case I: When a = 6p + 1,

∴ a2 = 36p2 + 12p + 1= 6p(6p + 2) + 1 = 6q + 1,

where q = p(6p + 2).Case II: When a = 6p + 3,

∴ a2 = 36p2 + 36p + 9= 36p2 + 36p + 6 + 3= 6(6p2 + 6p + 1) + 3= 6q + 3,

where q = 6p2 + 6p + 1.Case III: When a = 6p + 5,

∴ a2 = 36p2 + 60p + 25= 36p2 + 60p + 24 + 1= 6(6p2 + 10p + 4) + 1= 6q + 1,

where q = 6p2 + 10p + 4.Hence, a is of the form 6q + 1 or 6q + 3.

�������������� – �

1. (D) 145871250

= 11.6696.

Clearly, the decimal expansion terminatesafter four decimal places.

2. (C) LCM (p, q) = x3 y2 z3.

3. HCF × LCM = Product of the two numbers.

⇒ 9 × LCM = 306 × 657

⇒ LCM = 306×657

9 = 22338.

4. As given number can be written as 2525which is product of prime numbers: 5 × 5 ×101. Hence it is a composite number.

5. The maximum number out of 3, 5, 15, 25,75 is 75. Therefore, the HCF of 525 and 3000is 75.

6. The denominator of 2575000

or 51410000

is 10000.

10000 = 104 = 24 × 54

Further,2575000

= 451410

= 0.0514.

7. Let x = 2p + 1 and y = 2q + 1∴ x2 + y2 = (2p + 1)2 + (2q + 1)2

= 4p2 + 4p + 1 + 4q2 + 4q + 1= 4(p2 + p + q2 + q) + 2= S + T

where S = 4(p2 + p + q2 + q) and T = 2S is divisible by 4 and so an even integer.T is not divisible by 4 but an even integer.Therefore, S + T is even, as sum of any twoeven number is even, and not divisible by 4.

8. Let us assume the contrary that 5 is arational number.We can take coprime a and b (say) such that

5 =ab

; b ≠ 0

⇒ b 5 = a

Square both the sides to get5b2 = a2

⇒ a2 is divisible by 5⇒ a is divisible by 5 because if square of anumber is divisible by a prime, then thenumber is divisible by the prime.Let us take some integer c such that

a = 5cSquaring both sides, we get

a2 = 25c2

13��� ����� �

Substitute a2 = 25c2 in 5b2 = a2 to get

5b2 = 25c2

b2 = 5c2

⇒ b2 is divisible by 5⇒ b is divisible by 5Therefore, both a and b are divisible by 5.This contradicts the fact that a and b arecoprime that is a and b have no commonfactor.∴ Our assumption is false.So, we conclude that 5 is an irrationalnumber.

9. (i) First we will find HCF of 732 and 942 byusing Euclid’s lemma:

∴ 942 = 732 × 1 + 210732 = 210 × 3 + 102210 = 102 × 2 + 6102 = 6 × 17 + 0

∴ The last divisor = 6∴ HCF = 6∴ The least number of students in which

732 apples can be distributed such that

each student will get 6 apples = 732

6= 122.

Similarly, the least number of students inwhich 942 oranges can be distributed suchthat each student will get 6 oranges

= 9426

= 157.

∴ Total least number of students required =122 + 157 = 279.(ii) HCF of two real numbers.

(iii) Harmony and love.

���������

1. (D) Since 32844 = 2 × 2 × 3 × 7 × 17 × 23

So, 11 is not prime factor of 32844.2. (C)

As 8q is even and 6 is even, 8q + 6 is even.

3. ... LCM =306 × 1314

18 = 22338.

4. Yes.2 × 3 × 5 × 13 × 17 + 13

= 13 × (2 × 3 × 5 × 17 + 1)

= 13 × 511

= a composite number.

5. � �2

2 – 9 = 2 – 2 18 9�

= 11 – 2 18

= irrational.

6. No.Hint: Prime factors of 9n will be type of

32n, i.e., × × ×3 3 ... 3 Even no.

of times.

�����������

7. ... 0.56125 =56125

100000=

449800

=449

32 25�

= 5 2449

2 5�... 2n × 5m = 25 × 52

∴ n = 5, m = 2.8. 120 = 23 × 3 × 5

105 = 3 × 5 × 7

150 = 2 × 3 × 52

∴ HCF = 3 × 5 = 15

And LCM = 23 × 3 × 52 × 7

= 8 × 3 × 25 × 7

= 4200.9. Hint:

Let 2 – 3 3 = x, where x is rational.

⇒ � �2

2 – 3 3 = x2

⇒ 2 + 27 – 6 6 = x2

⇒ 29 – x2 = 6 6

⇒229 –

6x

= 6 .

Since 6 is not a perfect square. So 6 isalways irrational.

∴ It's a contradiction.

14 �� � � � � � � � � �

10. We know that any positive integer is of theform 3q or 3q + 1 or 3q + 2.

Case I: n = 3q

⇒ n3 = (3q)3 = 9 × 3q3 = 9m

⇒ n3 + 1 = 9m + 1, where m = 3q3.

Case II: n = 3q + 1⇒ n3 = (3q + 1)3

= 27q3 + 1 + 27q2 + 9q

= 9q (3q2 + 3q + 1) + 1= 9m + 1

⇒ n3 + 1 = 9m + 2, wherem = q(3q2 + 3q + 1).

Case III: n = 3q + 2⇒ n3 = (3q + 2)3

= 27q3 + 8 + 54q2 + 36q

n3 + 1 = 27q3 + 54q2 + 36q + 9= 9(3q3 + 6q2 + 4q + 1)= 9m,

where m = 3q3 + 6q2 + 4q + 1.

Hence, n3 + 1 can be expressed in the form9m, 9m + 1 or 9m + 2, for some integer m.

11. (i) We will find HCF of 96 and 112 by usingEuclid’s lemma:∴ 112 = 96 × 1 + 16and 96 = 16 × 6 + 0⇒ the last divisor = 16∴ HCF = 16∴ The minimum number of boxes required

for apples = 9616

= 6

and the minimum number of boxes

required for oranges = 11216

= 7

∴ Total minimum number of boxes required= 7 + 6 = 13.(ii) Concept used is HCF of two real

numbers using Euclid’s lemma.(iii) By distributing fruits in orphanage his

kindness and concern towards theneedful has been reflected.

��

15���� ��� ��

8. Solving α + β = 3 and α – β = – 1,we get α = 1, β = 2∴ Polynomial is x2 – (α + β) x + αβ⇒ p(x) = x2 – 3x + 2.

9. According to the division algorithm,p(x) = g(x) × q(x) + r(x)

⇒ x3 – 3x2 + x + 2 = g(x) × (x – 2) + (– 2x + 4)(As given in question)

⇒ g(x) =3 2 – 3 + 3 – 2

– 2x x x

x

To find g(x), we proceed as given below.

Thus, g(x) = x2 – x + 1.

10.13

� ; 32

Hint: 6x2 – 7x – 3= 6x2 – 9x + 2x – 3= 3x(2x – 3) + 1(2x – 3)= (2x – 3) (3x + 1)2x – 3 = 0 gives

∴ x = 32

3x + 1 = 0 gives x = –13

∴ α + β = 1

3�

+32

=76

=– ba

∴ α . β = –13

. 32

= –12

= ca

.

��������

POLYNOMIALS

�������� ��

1. (C) Since the given graph of y = p(x) cutsx-axis at three points, so the number ofzeroes of p(x) are 3.

2. Sum of zeroes = � � �

� �

( 5) 53 3

ba

Product of zeroes = �

13

ca

3. – 1

Hint: ���� �

� �� � ��

.

4. Let one zero be α, then the other one will be1�

.

∴ α .1�

= –15k

⇒ k = – 15.

5. Sum of zeroes (S) =2 3

–43

�

=3 – 84 3

= 5–

4 3

Product of zeroes (P) =2 3

– ×43

=1

–2

Now, required polynomial will be

x2 – Sx + P, i.e., x2 + 54 3

x1

–2

or 4 3 x2 + 5x – 2 3 .

6. Let f (x) = 2x2 + 2ax + 5x + 10If x + a is a factor of f (x), then f (– a) = 0Therefore, 2a2 – 2a2 – 5a + 10 = 0⇒ a = 2.

7. x3 – 4x2 + x + 6Hint: If the zeroes are �����and γ of a cubicalpolynomial, then the polynomial will be(x – α) (x – β) (x – γ)= (x – 3) (x – 2) (x + 1) = x3 – 4x2 + x + 6.

16 �� � � � � � � ��

5. p = 2Hint: (2)3 – 3(2)2 + 3(2) – p = 0⇒ 8 – 12 + 6 – p = 0⇒ 2 – p = 0∴ p = 2.

6. Let � and ��be the two zeroes off(x) = ax2 + 2x + 3a

Then, ������= – 2a

and ���= 3aa

����

According to the question,

– 2a = 3

⇒ a = – 23 .

7. Let the third zero be �, then

sum of the zeroes = – 2

3coefficient of coefficient of

xx

⇒ 2 + 3 + � = –– 61

⇒ � = 1Hence, the third zero is 1.

8. Let us divide 6x4 + 8x3 + 17x2 + 21x + 7 by3x2 + 4x + 1.

Clearly, the remainder is x + 2.Now, ax + b = x + 2Comparing the coefficients of like powersof x both the sides, we obtain

a = 1, b = 2.

9. We know:Dividend = (Divisor × Quotient) + Remainder⇒ 4x3 – 8x2 + 8x + 1 = g(x) × (2x – 1) + x + 3⇒ g(x) × (2x – 1) = 4x3 – 8x2 + 7x – 2

⇒ g(x) = 3 24 – 8 7 – 2

2 – 1x x x

x�

11. Let p(x) = x4 + x3 – 34x2 – 4x + 120Given zeroes of p(x) are 2 and – 2∴ (x – 2) (x + 2) = x4 – 4 is a factor of p(x).We divide p(x) by x2 – 4,

x2 + – 30x

x x x3 2– 30 – 4 +120

x4 + – 34 – 4 + 120x x x3 2

x x4 2– 4–

+

–

+

–

+

– 4x x3

– 30 + 20x2 1

– 30 + 20x2 1

x2 – 4

0

∴ p(x) = (x2 – 4) (x2 + x – 30)∴ Other zeroes of p(x) are given by

x2 + x – 30 = 0⇒ x2 + 6x – 5x – 30 = 0⇒ x(x + 6) – 5(x + 6) = 0⇒ (x – 5) (x + 6) = 0

x = 5, – 6Hence, all the zeroes are 2, – 2, 5 and – 6.

�������� ��

1. (D) Let us take option (D)p(x) = (x2 – 2) – (x2 + 3x) = – 3x – 2This is a linear polynomial.

2. � p(x) = 2x2 – 2x + 1∴ Sum of zeroes = 1

Product of zeroes =12

.

3. Let � = 5 and � = – 5, then the quadraticpolynomial will be x2 – (�����)x + ��or x2 – 25.

4. p(x) = 4x2 – 4x + 1= 4x2 – 2x – 2x + 1= 2x (2x – 1) – 1(2x – 1)= (2x – 1) (2x – 1)

For zeroes, 2x – 1 = 0 and 2x – 1 = 0

∴ x = 12

,12

.

17���� ��� ��

Now,

2 1x –

2 – 3 + 2x x2

4 – 8 + 7 – 2x x x3 2

4 – 2x x3 2

– 6 + 7 – 22x x

– 6 + 32x x

4 – 2

4 – 2

x

x

+–

+ –

+–

0

Hence, g(x) = 2x2 – 3x + 2.

10. 3 and 1

Hint: x2 – 3x – x + 3 = (x – 3 ) (x – 1)

For zeroes, x – 3 = 0 and x – 1 = 0

⇒ x = 3 , 1

Now, sum of zeroes = 3 + 1

= – 2Coefficient of

Coefficient of xx

And product of zeroes = 3

= 2Constant term

Coefficient of x.

11. (i) To find the number of sweets which wasdistributed among the slum children, wedivide the total number of sweets by numberof children Mr. Vinod has. Remainder thusobtained is the required number of sweets.

x 62 + + 8x

6 – 16 – 12 + 21x x x3 2

x4 + 2x x x3 2– 13 – 12 + 21x x x4 3 2– 4 + 3– –

–

–

+

+–

–

x x2 – 4 + 3

6 – 24 + 18x x x3 2

8 – 30 + 21x x2

8 – 32 + 24x x2

2 – 3x+

Hence, the number of sweets which wasdistributed amoung the slum children was2x – 3.

(ii) Helping one another, fair division.

�������� – �

1. (C) Sum of zeroes = – (– 5)

13

� �� �� �

= 15

Product of zeroes =

3213

= 92

.

2. (A) Let the zeroes be α, β, γ. Then αβγ = –1c

If γ = – 1, then αβ = c ...(i)Further, (– 1)3 + a (– 1)2 + b (– 1) + c = 0⇒ – 1 + a – b + c = 0⇒ c = b – a + 1 ...(ii)From equations (i) and (ii), we have

αβ = b – a + 1.3. Sum of zeroes = 6

⇒ 6 = –– 3

1k

∴ k =63

= 2.

4. Let one zero be α, then the other one will

be 1�

.

So, α .1�

= 24

4a

a �

a2 – 4a + 4 = 0⇒ (a – 2)2 = 0⇒ a = 2.

5. Given polynomial is: f (x) = x2 – px – 2p – c

∴ α + β = pand α . β = – 2p – c∴ (α + 2) (β + 2) = αβ + 2 (α + β) + 4

= – 2p – c + 2p + 4= (4 – c).

18 �� � � � � � � � � �

6. λ = 6Hint: (α + β)2 = (α − β)2 + 4αβ.

7. x = –1 or 3; f(x) = x2 – 2x – 3Hint: x = – 1 or 3,∴ Sum of zeroes = 2Product of zeroes = – 3

∴ p(x) = x2 – (α + β)x + αβ= x2 – 2x – 3.

8. x2 – x – 474

Hint: f (x) = {x2 – (sum of zeroes) x + (product of zeroes)}

9. The number which to be subtracted is theremainder when 4x4 + 2x3 – 8x2 + 3x – 7 isdivided by 2x2 + x – 2. To find the remainder,we proceed as following.

2 2x + x –2

2 – 2x2

4 + 2 – 8 + 3 – 7x x x x4 3 2

4 + 2 – 4x x x4 3 2

– + 3 – 7x x24

– 2 4x x +24

5 11x –

––

–+ + –

+

Hence, 5x – 11 must be subtracted from4x4 + 2x3 – 8x2 + 3x – 7 so that it becomesexactly divisible by 2x2 + x – 2.

10. g(x) = x2 + 2x + 1

Hint: p(x) = g(x) × q(x) + r(x)

⇒ g(x) = ( ) – ( )

( )p x r x

q x

where, p(x) = 3x3 + x2 + 2x + 5

q(x) = 3x – 5

and r(x) = 9x + 10.

11. Since x = 53

and x = – 53

are zeroes of

p(x) = 3x4 + 6x3 – 2x2 – 10x – 5, so p(x) is

divisible by 5 5 – +

3 3� �� �� �� �� �� �� �� �� �� �� �� �

x x , i.e., x2 – 53 .

Here, other two zeroes of p(x) are the twozeroes of quotient 3x2 + 6x + 3Put 3x2 + 6x + 3 = 0⇒ 3(x + 1)2 = 0⇒ x = – 1 and x = – 1

Hence, all the zeroes of p(x) are 53

,

– 53

, – 1 and – 1.

��������–�

1. (B) Sum of zeroes = – 99 = –ve Product of zeroes = 127 = +veIf the sum of both zeroes is negative, thenthe zeroes would be either both negative orone negative and other one positive. If theproduct of both the zeroes is positive, thenthe zeroes would be either both positive orboth negative.Consequently, we obtain that both thezeroes are negative.

2.3 11

– , –2 2

Hint: Given polynomial can be written as:

p(x) = 2x2 + 3x – 11

Sum of zeroes =–ba

Product of zeroes =ca .

3. We know that the degree of the remainderis less than the degree of divisor or does'texist.

19���� ��� ��

Here, degree of the divisor is 3, therefore,the possible degree of the remainderaccording to the options can be any out of0, 1 and 2.

4. k = 0.

Hint: Substitute x = – 2 in x2 + 2x + k = 0.

5. Since������are the zeroes of x2 + px + q,�then ���+�����– p; �����q

Now, + = = –pq

α +β1 1α β αβ

and × = = q1 1 1 1α β αβ

So the polynomial having zeroes and� �

�� �

will be

q(x) = x2 – + ×x� � � ��

� � � �

� � � �� �� �� �� �� �� �� � � �

= 2 1+ +

px x

q q

or q(x) = qx2 + px + 1.

6. g(x) = x2 + 2x + 7.Hint: Divide x3 + 3x – 14 by x – 2.

7. One example is:p(x) = 3x2 – 3x + 12.

g(x) = x2 – x + 4

∴ q(x) = 3

r(x) = 0.

8. 1 1, –

7 7

Hint: For zeroes: 21x2 – 3 = 0

x2 = 17

∴ x = 17

� .

9. Since a = 2 is a zero of a3 – 3a2 – 10a + 24,therefore a3 – 3a2 – 10a + 24 is divisible bya – 2. Further the obtained quotient willprovide the other two zeroes.

a2 – a – 12= (a – 4) (a + 3)

For other zeroes, put a – 4 = 0 and a + 3 = 0a = – 3, 4

Thus, the other two zeroes are – 3 and 4.

10. g(x) = x + 1.

Hint: Applying division algorithm, we get

x4 + 1 = g(x) × (x3 – x2 + x – 1) + 2

⇒ g(x) =4

3 2– 1

– + – 1x

x x x

= � �� �� �

� �� �

2

2

+1 – 1 +1

– 1 +1

x x x

x x

= x + 1.

11.2

2– 2b acc

Hint: �

� �2 21 1

=2 2

2 2� ��

� �

= � �2

2 22��� � ��

� �=

2

2– 2b ac

c.

ORLet us divide x4 + 2x3 + 8x2 + 12x + 18 by x2 + 5.

20 �� � � � � � � ��

���������62 – 4 × 4 = 20 ⇒���–����= 2 5±Thus, the difference of zeroes is 2 5± .

5.� ��

� �=� ��

��

2 2

=� �

2 2� �� � ��

��

= 25 126� =

136

.

6. x2 – 1 = (x + 1) (x – 1)∴ x = – 1 or 1, both will satisfy with thegiven polynomial.∴ We get, p + q + r + s + t = 0 ...(i)

and p – q + r – s + t = 0 ...(ii)From (ii),

p + r + t = q + s2(q + s) = 0 ⇒ q + s = 0 [From (i)]

∴ p + r + t = q + s = 0.

7. No.Hint: Divide q(x) by g(x). If the remainderobtained is zero, then the g(x) is a factor ofq(x) otherwise not.

8. a = 1, b = 7Hint: Put remainder = 0 and equate coefficientof x in the remainder and constant termwith zero.

9. According to division algorithm, p(x) = g(x) × q(x) + r(x)

(i) p(x) = 6x2 + 3x + 2, g(x) = 3q(x) = 2x2 + x, r(x) = 2

(ii) p(x) = 8x3 + 6x2 – x + 7, g(x) = 2x2 + 1q(x) = 4x + 3, r(x) = – 5x + 4

(iii) p(x) = 9x2 + 6x + 5, g(x) = 3x + 2,q(x) = 3x, r(x) = 5.

10. Given quadratic polynomial is

5 5 x2 + 30x + 8 5

= 5 5 x2 + 30x + 8 5

= 5 5 x2 + 20x + 10x + 8 5

= 5x ( )5 4x + + 2 5 ( )5 4x +

= ( )+5 2 5x ( )+5 4x

Clearly, the remainder is 2x + 3.Now, px + q = 2x + 3Comparing the coefficients of like powers ofx both the sides, we get

p = 2, q = 3.

�������� – ��

1. (C) Let each of two equal roots be α.

Then α + α = –ba and α .α = c

a

⇒ 2

–2ba

� �� �� �

=ca ⇒ a(4ac – b2) = 0

But a ≠ 0So, b2 – 4ac = 0⇒ b2 = 4acThis last equation holds iff a and c havesame sign.

2. (A) ��������32

�������12

∴ (��–��)2 = (��+��)2 – 4��

� 94

– 2 = 14

⇒ ��������–�� � 12

±

∴ ���� 12

� ����1 or ������� ����12

∴ � � �52

� � � � �� or � � � ��

���� �� 52

�

Hence, the required polynomial can be

x2 – 53

2� ��� �� �

x +52

× 3, i.e., x2 – 112

x + 152

.

3. p(x) = x2– (sum of zeroes) x + product ofzeroes

= x2 – (12

+ 2)x + 12

× 2

= x2 – 52

x + 1

4. Let zeroes be � and �.�����������������Using���–���������������–������we get

21���� ��� ��

To find its zeroes, put 5 2 5x � = 0 and

5 x + 4 = 0.

⇒ x = –25

and x =– 4

5

i.e., x = –2 5

5 and x = –

4 55

So, sum of zeroes =– 2 5

5 –

4 55

= –6 5

5And product of zeroes

= 2 5 4 5– × –5 5

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 85

.

Also, sum of zeroes = – 2Coefficient of

Coefficient of x

x

= – 305 5

= – 6 55

And product of zeroes = 2Constant term

Coefficient of x

= 8 55 5

=85

.

Hence verified.OR

q(x) = 3x2 – 2x + 1

Hint: Let S =1 11 1

� � � ��

� � � �

P =1 11 1

� �� � � �� �� �� �� � � �

∴ Required polynomial q(x) = x2 – Sx + P.

11. As 32

and – 32

are the zeroes of the given

quadratic polynomial, so 3–

2x

⎛ ⎞⎜ ⎟⎝ ⎠

and

32

x⎛ ⎞

+⎜ ⎟⎝ ⎠

will be the factors of that. Conse-

quently,3

–2

x⎛ ⎞⎜ ⎟⎝ ⎠

×32

x⎛ ⎞

+⎜ ⎟⎝ ⎠

, i.e., � �� �� �

2 3–

2x

must be the factor of that. Let us divide

2x4 – 10x3 + 5x2 + 15x – 12 by 2 3–

2x .

2 – 10 + 8x x2

– 10 + 8 + 15 12x x x –3 2

2 – 10 + 5 + 15 – 12x x x x4 3 2

2 – 3x x4 2

3

2

+

–

–

+

x2 –

– 10 + 15x x3

8 12x –2

–

8 12x –2

+

0

Now, 2x4 – 10x3 + 5x2 + 15x – 12

= 2 3–

2x� �

� �� �(2x2 – 10x + 8)

By splitting –10x, we factorise 2x2 – 10x + 8as (x – 4) (2x – 2). So, its zeroes are givenby x = 4 and x = 1.

Therefore, all zeroes of the given poly-

nomial are 32

, – 32

, 1 and 4.

�������� – ��

1. (D) Let zeroes be ��and ���then(��– �)2 = 144

⇒ ��– � ��± 12 ...(i)��+ � ��– p ...(ii)

��� ��45 ...(iii)Also, we have

(��– �)2 = (��+ �)2 – 4��⇒ 144 = p2 – 180⇒ p = ± 18.

2. 1 – cHint: f (x) = x2 – px – (p + c)(α + 1) (β + 1) = αβ + (α + β) + 1.

3.pr

Hint: � � � � �������� � �

�� �� �� ���.

4. Let the given linear polynomial bey = ax + b ...(i)

This passes through points (1, –1), (2, 1) and

3, 0

2� �� �� �

22 �� � � � � � � ��

� – 1 = a + b ...(ii)1 = 2a + b ...(iii)

0 =32

a + b ...(iv)

Solving equations (ii) and (iii), we get a = 2,b = – 3 which satisfy to equation (iv).Consequently, using equation (i), we get

y = 2x – 3∴ Polynomial is p(x) = 2x – 3

Since p(x) = 0 if x = 32

⇒ x = 32

is zero of p(x).

5. Let us divide ax3 + bx – c by x2 + bx + c bythe long division method.

ax ab–

– + ( –abx b – ac) x c2

ax bx c3 + –

ax + abx + acx3 2

–

+

– –

++

x bx c2 + +

– –abx ab x – abc2 2

( + ) –ab b – ac x + abc c2

Put remainder = 0⇒ (ab2 + b – ac)x + (abc – c) = 0⇒ ab2 + b – ac = 0 and abc – c = 0Consider abc – c = 0 ⇒ (ab – 1) c = 0⇒ ab = 1 or c = 0. Hence, ab = 1.

6. Hint: Let f (x) = x3 – mx2 – 2npx + np2

(x – p) is a factor of p(x)⇒ f (x) = 0 at x = p.⇒ p3 – p2m – p2n = 0⇒ p2 [(p – (m + n)] = 0⇒ p = m + n since p ≠ 0.

7. x3 – 4x2 + x + 6

Hint: The required cubic polynomial is givenby (x – 3) (x – 2) (x + 1) or x3 – 4x2 + x + 6This is the required polynomial.

8. – 2, 3, 4Hint: α + β + γ = 5

αβ + βγ + αγ = – 2αβγ = – 24

Let αβ = 12

∴ γ = – 2∴ α + β = 7⇒ (α – β)2 = 1⇒ α – β = ± 1∴ α – β = 1 or α – β = – 1Solving α + β = 7 and α – β = 1, we get

α = 4, β = 3And solving α + β = 7 and α – β = – 1we get α = 3, β = 4.

9. f(x) would become exactly divisible by g(x)if the remainder is subtracted from f(x).Let us divide f(x) by g(x) to get the remainder.

x 62 + + 8x

6 – 16 – 12 + 21x x x3 2

x4 + 2x x x3 2– 13 – 12 + 21

x x x4 3 2– 4 + 3– –

–

–

+

+–

–

x x2 – 4 + 3

6 – 24 + 18x x x3 2

8 – 30 + 21x x2

8 – 32 + 24x x2

2 – 3x

+

Hence, we should subtract 2x – 3 from f(x).

10. If 2 ± 3 are zeroes of p(x), then x – ( )2 3+

and x – ( )2 – 3 are factors of p(x).

Consequently ( ){ }– 2 3x + ( ){ }– 2 – 3x

i.e., (x – 2)2 – 3, i.e., x2 – 4x + 1 is factor ofp(x).Further,

x2 – 2 – 35x

– 2 – 27 + 138 – 35x x x3 2

x x x x4 3 2– 6 – 26 + 138 – 35

x x x4 3 2– 4 +– –

+

+

+

–

–

+

x x2 – 4 + 1

– 2 + 8 – 2x x x3 2

– 35 + 140 – 35x x2

– 35 + 140 – 35x x2

0

+

23���� ��� ��

Clearly x2 – 2x – 35 is a factor of p(x)⇒ (x – 7)(x + 5) is a factor of p(x)⇒ x – 7 and x + 5 are factors of p(x)⇒ x – 7 = 0 and x + 5 = 0 give other zeroes

of p(x)⇒ x = 7 and x = – 5 are other zeroes of p(x).Hence, 7 and – 5 are required zeroes.

11. Hint: 2 2 4 4

2 2 2 2+

+ =α β α β β α α β

2 2 2 2

2 2+ – 2 – 2

={(α β) αβ} α β

α β .

OR

Given polynomial is:f (x) = pqx2 + (q2 – pr)x – qr

= pqx2 + (q2 – pr)x – qr= pqx2 + q2x – prx – qr

= qx(px + q) – r(px + q)= (px + q)(qx – r)

px + q = 0 and qx – r = 0 provide the zeroes

of f (x). So zeroes are –qp and r

q .

Sum of zeroes = –qp +

rq =

2–pr qpq

= – 2

Coefficient of

Coefficient of

x

x

Product of zeroes = – ×q rp q = –

qrpq

= 2Constant term

Coefficient of x.

�������������� – �

1. (C) Let each of two equal roots be α.

Then α + α = –ba

and α .α =ca

⇒ 2

–2ba

⎛ ⎞⎜ ⎟⎝ ⎠

=ca ⇒ a(4ac – b2) = 0

But a ≠ 0.

So, b2 – 4ac = 0⇒ b2 = 4acThis last equation holds iff a and c havesame sign.

2. Required quadratic polynomial

= x2 – (sum of zeroes)x + product of zeroes

= x2 – 2 3x – 5 3 .

3. p(x)= x2 – ax – (a + 1)At x = – 1, p(x) = (–1)2 – a( –1) – (a + 1)

= 1 + a – a – 1 = 0q(x) = ax2 – x – (a + 1)

At x = – 1, q(x) = a( –1)2 – ( –1) – (a + 1)= a + 1 – a – 1 = 0

Therefore, x + 1 is the common factor of p(x)and q(x).

4. Correct,

f (x) = x2 – p(x +1) – c= x2 – px – (c + p)∴ α + β = p; αβ= – (c + p)Now, (α + 1) (β + 1)= αβ + (α + β) + 1

= – (c + p) + p + 1= – c – p + p + 1= 1 – c.

5. Let f(x) = 6x3 + 2 x2 – 10x – 4 2

As 2 is a zero of f (x), (x – 2 ) is a factorof f (x).Let us divide f(x) by ( )– 2x .

∴ f (x) = ( )– 2x � �� �26 7 2 4x x

= ( )– 2x � �� � �26 3 2 4 2 4x x x

= ( )– 2x � � � �� �3 2 4 2 1x x

24 �� � � � � � � ��

Hence, � � � �� �3 2 4 2 1x x gives x = –2 2

3

or x = –12

Therefore, other two zeroes are –2 2

3 and

–2

2.

6. p(y) = y2 + 3 5

2y – 5

= 12

(2y2 + 3 5 y – 10)

= 12

(2y2 + 4 5 y – 5y – 10)

= 12

[2y(y + 2 5 ) – 5(y + 2 5 )]

= 12

(y + 2 5 ) (2y – 5 )

y + 2 5 = 0 and 2y – 5 = 0 give the required

zeroes, that are – 2 5 and 52

.

7. α and β are zeroes of f (x) = x2 – x – 2

Sum of zeroes = α + β = – –11

= 1 ...(i)

Product of zeroes = αβ = –21

= – 2 ...(ii)

∴ (2α + 1) + (2β + 1) = 2(α + β) + 2

= 2(1) + 2 [Using (i)]

= 4 ...(iii)

And (2α + 1) (2β + 1) = 4αβ + 2α + 2β + 1

= 4αβ + 2 (α + β) + 1

= 4 (– 2) + 2 (1) + 1

[Using (i) and (ii)]

= – 5 ...(iv)Now, required polynomial can be given by

x2 – {(2α + 1) + (2β + 1)}x + (2α + 1)(2β + 1)

i.e., x2 – 4x – 5. [Using (iii) and (iv)]

8. Let us divide p(x) by 2x2 – 5.

3 + 4 + 52 xx

x x ax b8 + 10 + +3 2

6 + 8 5x a b4 – + +x x x3 2

6 – 15x x4 2

–

+

+

+

–

–

8 – 20x x3

10 + (20 + )x a x + b2

10 – 25x2

2 – 5x2

(20 + ) + 25a x + b

Here, remainder is (20 + a)x + b + 25.If the polynomial p(x) is exactly divisible by2x2 – 5, the remainder must be zero.∴ (20 + a)x + (b + 25) = 0Comparing the coefficients of like powers ofx between both the sides, we have20 + a = 0 and 25 + b = 0⇒ a = – 20 and b = – 25.

�������������� – �

1. (C) Sum of zeroes = – 3 2

–3

= 2

Product of zeroes = 13

.

2. At x = 2, p(x) = 0, i.e., p(2) = 0∴ a (2)2 – 3 × 2 (a – 1) – 1 = 0⇒ 4a – 6a + 6 – 1 = 0

⇒ a = 52

.

3. Sum of zeroes = α + β = 5Product of zeroes = αβ = 4

Now,1α

+1β

– 2αβ =α + βαβ

– 2αβ

= 54

– 2 × 4

=27

–4

.

4. Using division algorithm, we haveg(x) × (x – 2) – 2x + 4 = x3 – 3x2 + x + 2

25���� ��� ��

⇒ g(x) =�

3 2– 3 3 – 2– 2

x x xx

Here, at x = 2,

x3 – 3x2 + 3x – 2

= 8 – 12 + 6 – 2 = 0∴ x3 – 3x2 + 3x – 2

= (x – 2) (x2 – x + 1)

∴ g(x) = 2( – 2)( – 1)

( – 2)x x x

x�

⇒ g(x) = x2 – x + 1.

5. Given s = 2 and p = –32

The required polynomial is given byk[x2 – sx + p]

i.e., k � �� �� �

2 3– 2 –

2x x , where k is any real

number.

Let us find zeroes of this polynomial.

k(x2 – 2 x – 32

) =2k

(2x2 – 2 2 x – 3)

=2k

( 2x – 3)( 2 x + 1)

2 x – 3 = 0 and 2 x + 1 = 0 provides thezeroes.

Hence 3 2

2 and

– 22

are the required

zeroes.

6. Let f(x) = �24 3 5 – 2 3x x

= 24 3 8 – 3 – 2 3x x x�

= 4 ( 3 2) – 3 ( 3 2)x x x� �

= ( 3 2)(4 – 3)x x�

To find zeroes of f (x), put

3 2 0x � � and 4 – 3 0x �

⇒ – 2 – 2 3

33x � � and

34

x �

Thus, the zeroes are α = –2 3

3and β = 3

4 Sum of zeroes = α + β

= – 2 33

+3

4=

– 5 312

=5 3

–4 3×

= – 5

4 3

= – 2Coefficient of Coefficient of

xx

Product of zeroes = αβ = –2 3

3.

34

= – 2 34 3

= 2

Constant termCoefficient of x

.

Hence verified.

7. (i) Let y = p(x)∴ y = – x2 + x + 6

The table for some values of x and theircorresponding values of y is given by

x – 2 – 1 0 1 2 3

y 0 4 6 6 4 0

Let us draw the graph of p(x) using thistable.

26 �� � � � � � � � � �

From the graph, it is clear that the zeroes ofp(x) are – 2 and 3.

(ii) Let y = p(x)

∴ y = x3 – 4x

The table for some values of x and theircorresponding values of y is given by

x – 2 – 1 0 1 2

y 0 3 0 – 3 0

Let us draw the graph of p(x) by using thistable.

From the graph, it is clear that the zeroes of p(x) are – 2, 0 and 2 .

8. (i) First we divide x4 + x3 + 8x2 +ax + b byx2 + 1 as follows:

� �

� � � � �

�

� �

� � �

�

� �

� � �

�

� �

� � �

2

2 4 3 2

4 2

3 2

3

2

2

7

1 8

7

7 ( 1)7 7

( 1) ( 7)

x x

x x x x ax b

x x

x x ax bx x

x a x bx

a x b

Since, x4 + x3 + 8x2 + ax + b is divisible by x2 + 1,therefore remainder = 0i.e.,(a – 1)x + (b – 7) = 0 or (a – 1)x + (b – 7) = 0.x + 0

Equating the corresponding terms, we have

a – 1 = 0 and b – 7 = 0

i.e., a = 1 and b = 7

(ii) Common good, Social responsibility.

����������

1. (B) Let p(x) = x 72

x� ��� �� �∴ Zeroes are given by

x = 0 and 72

x� ��� �� � = 0.

Hence zeroes are 0 and – 72

.

2. ��α + β = – 52

, αβ = 12

∴ α + β + αβ = – 2.

3. p(x) = x2 – (α + β)x + αβ= x2 + x – 2.

4. ⇒ Let p(x) = 2x3 + 4x2 + 5x + 7Now, p(x) = g(x) × 2x + (7 – 5x)

g(x) = ( ) – (7 – 5 )

2p x x

x

=� � � �

3 22 4 5 7 – 7 52

x x x xx

= x2 + 2x + 5.

5. 6 5 9, –

5 4

Hint: α + β = – 2

Coefficient of

Coefficient of

x

x

αβ = 2Constant term

Coefficient of x

6.–13

Hint: α = – β

α + β = 0 ⇒–ba

= 0.

27� �� ��� � �

7. f (x) = ax3 + bx2 + cx + d

g(x) = ax2 + bx + c

q(x) = x

r(x) = d.8. If α, β and γ are the zeroes of a cubic

polynomial f(x), thenf(x) = x3 – (α + β + γ) x2

+ (αβ + βγ + γα) x – αβγHere, α + β + γ = 4, αβ + βγ + γα = 1

and αβγ = – 6

∴ f(x) = x3 – 4x2 + x + 6.9. We have

24 3 5 – 2 3x x+ = ( )( )3 2 4 – 3x x+

So, the value of �24 3 5 – 2 3x x is zero

when, 3 2x + = 0 or 4 – 3x = 0,

i.e., when x =– 2

3 or x =

34

.

Therefore, the zeroes of 24 3 + 5 – 2 3x x

are – 2

3 and 3

4.

Now, sum of zeroes– 2

3+

34

=– 5

4 3

= 2

– Coefficient of

Coefficient of

x

x

Product of zeroes = � �� �

�� � � �� � � �

– 2 343

= – 2 3

4 3

= 2Constant term

Coefficient of x.

10. (i) Let p(x) = Total Relief Fund

g(x) = Number of families whoreceived Relief Fund

q(x) = Amount each family received

r(x) = Amount left after distribution

When the polynomial p(x) is divided by apolynomial g(x) such that q(x) and r(x) are

respectively the quotient and the remainder, thedivision algorithm is

p(x) = g(x) . q(x) + r(x) ...(i)

According to the question,

p(x) = 3x3 + x2 + 2x + 5

q(x) = 3x – 5

r(x) = 9x + 10

Substituting these values of p(x), q(x) and r(x) inthe equation (i), we get

3x3 + x2 + 2x + 5 = g(x) . (3x – 5) + 9x + 10

⇒ (3x – 5) g(x) = 3x3 + x2 + 2x + 5 – 9x – 10

= 3x3 + x2 – 7x – 5

⇒ g(x) =3 23 7 5

3 5x x x

x� � �

�

To find g(x), we proceed as following:

2

3 2

3 2

2

2

2 13 5 3 7 5

3 5

6 7 56 10

3 53 5

0

x xx x x x

x x

x xx x

xx

� �

� � � �

�

� �

� �

�

� �

�

�

� �

Thus, g(x) = x2 + 2x + 1.

(ii) Common good, Accountability, socialresponsibility.

11. Since –1

3and

13

are zeroes.

Therefore, 1 1–

3 3x x

⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

will be a factor

of p(x), i.e., 2 1–

3x is a factor of p(x).

28 �� � � � � � � ��

3 – 15 + 182 xx

– 15 + 18 + 5 – 6x x x3 2

3 – 15 + 17 + 5 – 6x x x x4 3 2

3x – x4 2

–

+

–

+

–

+

– 15 + 5x x3

x18 – 62

x18 – 62

0

x2 –31

Here, 3x2 – 15x + 18 = x2 – 5x + 6 = (x – 3) (x – 2)

Other zeroes are given by x – 3 = 0 andx – 2 = 0.So, other zeroes are 3 and 2.

Hence all the zeroes are 1

–3

, 13

, 3 and 2.

���

29���� ����� �� � � ��� ��

��������

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

�������� ��

1. (B) Since (3, a) lies on the line 2x – 3y = 5,therefore, (3, a) must satisfy this equation.∴ 2 (3) – 3 (a) = 5

⇒ 3a = 1 ⇒ a = 13

.

2. k = 6

Hint: – 32=

– 9k.

3. k = 6

Hint: The condition of inconsistency of twoequations a1x + b1y = c1 and a2x + b2y = c2 is

given by �1 1 1

2 2 2=

a b ca b c

.

4. Adding the given equations, we get80x + 80y = 240

or x + y = 3 ...(i)Subtracting given first equation from otherone, we get

6x – 6y = – 6or x – y = –1 ...(ii)Solving equations (i) and (ii), we obtain

x = 1, y = 2.

5. x = 3, y = 2

Hint: Let �

1x y

= u, 1

x y�= v.

∴ Given equations become10u + 2v = 4 and 15u – 5v = – 2.

6. False.Let us substitute c = 40, The given equationsbecome

x – 2y = 8or 5x – 10y = 40

Here,15

= – 2–10

=840

⇒ The equations represent a pair ofcoincident lines.

⇒ The equations have infinitely manysolutions at c = 40 and no solutions atc ≠ 40.

⇒ For no value of c, the given pair has aunique solution.

7. The given equations are4(2x + 3y) = 9 + 7y

and 3x + 2y = 4or 8x + 5y –9 = 0

3x + 2y – 4 = 0By cross-multiplication, we have

− 20 + 18x

= 32 + 27

y�

�

= 1

16 15�

⇒– 2x = –

– 5y = 1

1

x = – 2 and y = 5Hence, x = – 2, y = 5 is the solution of thegiven system of equations.

8. To draw a line, we need atleast twosolutions of its corresponding equation.x + 3y = 6; at x = 0, y = 2 and x = 3, y = 1.So, two solutions of x + 3y = 6 are:

x 0 3

y 2 1

2x – 3y = 12; at x = 0, y = – 4 and at x = 6, y = 0So, two solutions of 2x – 3y = 12 are:

Now, we draw the graph of given systemof equations by using their correspondingsolutions obtained in the above tables.

x 0 6

y – 4 0

30 �� � � � � � �� ��

2. x – 5y = 5.

(2, k) lies on it.

∴ 2 – 5(k) = 5 ⇒ 5(k) = – 3

⇒ k = 3–

5.

3. Condition for parallel lines is

1

2

aa

= 1

2

bb

≠ 1

2

cc

⇒ 13

=– 2k

≠ – 3– 1

⇒ k = – 6.

4. The given lines to be coincident, if

12 I

k = 3

IIk

= – ( – 3)

–III

kk

Taking I and II, we havek2 = 36 ⇒ k = ± 6. ...(i)Taking II and III, we havek2 – 3k = 3k ⇒ k(k – 6) = 0⇒ k = 0 or 6 ... (ii)Using (i) and (ii), we obtain

k = 6.

5. x = 5, y = 2

Hint: Adding the given equations,we get 2x + y = 12 ...(i)Subtracting the given equations,we get 3x + y = 17 ...(ii)

6. Yes.Applying the condition

1

2

aa = 1

2

bb

= 1

2

cc ,

we have

13

= 26

= – 3– 9

That is true.Therefore, the pair of equations is consistentwith infinitely many solutions.

From the graph, the two lines intersect they-axis at (0, 2) and (0, – 4).

9. Let the fixed charges and change per kmbe � x and � y respectively.

x + 10y = 105 ...(i)x + 25y = 255 ...(ii)

Subtracting equation (i) from equation (ii),we get

15y = 150y = 10 ...(iii)

From equations (i) and (iii), we getx = 5

Now, the fare for travelling a distance of35 km

= x + 35y= 5 + 35 × 10= � 355.

Fixed charge = � 5Charge per km = � 10Total charge for 35 km = � 355.

�������� �

1. (D) As y = 2 and y = 7,both represent straight lines parallel tox-axis∴ y = 2 and y = 7 are parallel lines.Hence, the given pair of equations has nosolution.

31���� ����� �� � � ��� ��

7. x = 25552

, y =615104

Hint: The system:9x – 10y + 15 = 0 5x + 6y – 60 = 0

By cross-multiplication, we have 1

= =600 90 540 75 54 + 50

yx �

� � �

.

8. For equation 3x + y – 2 = 0,

For equation 2x – 3y – 5 = 0,

As the lines corresponding to the givenequations intersect each other at (1, – 1),the required solution is x = 1, y = – 1.

9. Let the man's starting salary and fixedannually increment be x and y respectively.According to the question,

x + 4 y = 15000 ...(i)x + 10 y = 18000 ...(ii)

Equations (i) and (ii) form the required pairof linear equations. Let us solve this pair.Subtracting equation (i) from equation (ii),we get

6y = 3000 ⇒ y = 500Substituting y = 500 in equation (ii),we get

x = 13000Hence, starting salary was � 13000 andannual increment was � 500.The value imbibe by the man are:consistency, hard work and sincerity

�������� ��

1. (D) Line x = a is parallel to y-axis and theline y = b is parallel to x-axis. These linesintersect each other at (a, b).

2. As the lines are intersecting each other,3a

≠2–1

⇒ a ≠ – 32

.

3. 3x – y – 5 = 0 and 6x – 2y – k = 0 have nosolution⇒ These equations represent a pair of

parallel lines.

⇒ 36

= –1– 2

≠ – 5– k

⇒ k ≠ – 10.

4. No.For infinitely many solutions, the followingcondition must be satisfied.

λ2

=36

= 7

–14

But, here 36

≠ –714

as 12

≠ 1

–2

Hence, no value of ‘λ’ provides the pair ofinfinitely many solutions.

5. The given system of equations can be writtenas

ax + by – (a – b) = 0bx – ay – (a + b) = 0

By cross-multiplication,

– ( + ) – ( – ) ( )

xb a b a a b

i

= –

– ( + ) + ( – ) ( )

ya a b b a b

ii

= 2 21

– – ( )

a biii

x 0 1y 2 – 1

x – 2 1y – 3 – 1

32 �� � � � � � �� ��

Taking (i) and (iii), we get

x =2 2

2 2

– – –

– –

ab b a ab

a b

+ = 1

Taking (ii) and (iii), we get

y = –2 2

2 2

– – –

– –

a ab ab b

a b

+ = – 1

Hence x = 1, y = – 1 is the solution of thegiven system of equations.

6. x = 6, y = – 4, m = 0

Hint: Take 1x = u and 1

y= v.

7. No; (6, 0) , (4, 0)

Hint: For x + 3y = 6

For 3x + 9y = 12

Let us draw the graph of lines using thetables obtained above.

In the graph, lines are parallel. So, the pairof equations is not consistent.The lines intersect the x–axis at (4, 0) and(6, 0).

8. (i) Let l = length of the rectangle b = breadth of the rectangleAccording to question,

(l + 7)(b – 3) = lb ...(i)

(l – 7)(b + 5) = lb ...(ii)

From equation (i),

lb + 7b – 3l – 21 = lb⇒ 7b – 3l = 21 ...(iii)

From equation (ii),lb – 7b + 5l – 35 = lb

⇒ – 7b + 5l = 35 ...(iv)Adding equations (iii) and (iv), we get

2l = 56 ⇒ l = 28 mPutting the value of l in equation (iii), we get

b = 15 m.∴ l = 28 m, b = 15 m.

(ii) Solution of system of linear equations in twovariables.

(iii) Love for environment and human beings.

�������� – ��

1. (D) Let unit's and ten's digit be x and yrespectively.

x + y = 9 ...(i)10y + x + 27 = 10x + y ...(ii)

Solving equations (i) and (ii), we have x = 6, y = 3Hence, the required number is 3 × 10 + 6,that is, 36.

2. Given equation is 5(x – y) = 3

⇒ 5x – 5y – 3 = 0

Let a2 = 10, b2 = 10 and c2 = 6

For coincident; 1 1 1

2 2 2

a b ca b c

� �

Hence 1 1

2 2

5 1 5 1;

10 2 10 2a ba b

� � � �

and 1

2

3 16 2

cc

� �

∴ � �

1 1 1

2 2 2

a b ca b c

So required equation which can coincide is10x – 10y – 6 = 0

3. p = 6

Hint: 3 5

=10p

⇒ p = 6

x 0 3

y 2 1

x 1 4

y 1 0

33���� ����� �� � � ��� ��

Note: At p = 6, the given system has bothzero and non-zero solutions.

4. a = 5, b = 1Hint: According to the condition of infinitelymany solutions, we reaches at

+ 2 – 21= =

2 3 7a b a b

.

5. x = 1, y = 1Hint: Simplifying the given linearequations, we have

7 2 8 7– = 5, + = 15

y x y x

Now take 1 x

= u, 1 y

= v; and solve.

6. x = 4 –

5a b

a, y =

− + 45

a bb

Hint: – 3 (2 + ) + 2 ( + 2 )

xb a b b a b

= –

– 2 (2 + ) + 3 ( + 2 )y

a a b a a b

= 1

2 2 – 3 3a b a b� �

Take first and third terms as well as secondand third terms and solve.

7. a = 7, b = 3Hint: For infinitely many solutions,

24

=( )( )

– – 4– – 1

aa

= 2 15 1bb�

�

Take12

= 41

aa�

�

and 12

= 2 15 1bb�

�

.

8. Table for values of x and y as regardingequation 3x + y – 5 = 0 is

Similarly table for equation 2x – y – 5 = 0 is

Let us draw the graph of lines using thetables obtained above.

The lines intersect y-axis at (0, 5) and(0, – 5).

9. (i) Total distance = 300 kmLet speed of train = x km/hand speed of bus = y km/h

As speed =distance

time

∴ time =distance

speed

According to question,

60x

+240

y= 4 hours ...(i)

and100

x +

200y

= 4 hours and

10 minutes. ...(ii)From equation (i),

15x

+ 60y

= 1 ...(iii)

From equation (ii),

100x

+200

y=

256

hours

⇒ 4x

+8y

= 16

⇒1x

+2y

=1

24...(iv)

∴ We will solve equations (iii) and (iv) byelimination method.Applying (iii) – 15 × (iv), we get:

x 0 1

y 5 2

x 0 1

y – 5 – 3

34 �� � � � � � �� ��

15x

+ 60y = 1

15x

+ 30y

=1524

– – –

30y = 1 –

1524

⇒ y30

=924

⇒ y = 30 24

9�

= 10 × 8 = 80 km/h

Putting the value of y in equation (iv), we get,1x

+ 2

80=

124

⇒1x

= 1

24 –

140

=40 2424 40

�

�

= 16

24 40�

= 1

60 ⇒ x = 60 km/h

(ii) Solution of system of linear equations intwo variables.

(iii) By opting for public transport it depicts thatshe is a responsible citizen, so herresponsibility and rationality have beendepicted here.

�������� – ��

1. (D) The condition to be coincident for linesax + by + c = 0 and dx + ey + f = 0 is givenby

ad

= be

= cf

⇒ ae = bd; bf = ce.

Note: Two lines are coincident if both theequations follow the condition of infinitelymany solutions.

2. (A) For no solutions,

12k

=3k

≠ –( – 2)

–k

k⇒ k = ± 6

If k = 6

6 3 6 – 2 4 2

12 6 6 6 3� � � � (True)

If k = – 6

– 6 – 83 412 – 6 – 6 3

� � � (True)

∴ Required value of k, can be 6 or – 6.

3. Let the required equation be ax + by + c = 0.

Then,2

a=

– 3b

≠ – 5c

⇒2

a=

– 3b

= k (say)

⇒ a = 2 k, b = – 3 k, k ≠ –5c

any real

number

Then, 2 kx – 3 ky + c = 0

⇒ 2 x – 3 y + ck

= 0

Putting k = – c, we have

⇒ 2 x – 3 y = 1.

4. For infinite number of solutions, we have

+2

p q =

– 3–( – 3)p q�

= –7

–(4 )p q�

On solving,+2

p q=

– 3–( – 3)p q�

and

+– 3

–( – 3)p q=

+– 7

–(4 )p q,

we obtain p = – 5, q = – 1.

5. x = 1, y = 2Hint: Adding and subtracting the given twoequations, we have

x + y = 3 ...(i)and x – y = – 1 ...(ii)Now, solve equations (i) and (ii).

6. x = a2, y = b2

Hint: Given system of linear equations maybe written as

bx + ay – ab (a + b) = 0b2x + a2y – 2 a2b2 = 0

Solve these two equations by the method ofcross-multiplication.

7. Let the two digits number be 10x + y.Since ten's digit exceeds twice the unit'sdigit by 2

35���� ����� �� � � ��� ��

∴ x = 2y + 2⇒ x – 2y – 2 = 0 ...(i)Since the number obtained by inter-changing the digits, i.e., 10y + x is 5 morethan three times the sum of the digits.... 10y + x = 3 (x + y) + 5⇒ 2x – 7y + 5 = 0 ...(ii)On solving equations (i) and (ii), we obtain x = 8 and y = 3... 10x + y = 83Hence, the required two-digit number is 83.

8. Tables for equations 3x + y – 11 = 0 andx – y – 1 = 0 are respectively

and

Let us draw the graph.

From the graph, it is clear that the linesintersect each other at a point A(3, 2). Sothe solution is x = 3, y = 2.

The line 3x + y – 11 = 0 intersects they-axis at B(0, 11) and the line x – y – 1 = 0intersects the y-axis at C (0, – 1). Draw the

perpendicular AM from A on the y-axis tointersect it at M.Now, in Δ ABC,base BC = 11 + 1 = 12 units,height AM = 3 units.

... ar(ΔABC) =12

× base × height

=12

× 12 × 3 = 18 sq. units

Hence, x = 3, y = 2; area = 18 sq. units.

9. Speed of boat = 6 km/hr,Speed of stream = 2 km/hrHint: Let the speed of boat in still water =x km/h and the speed of stream = y km/h

12 40+ = 8

x – y x + y ...(i)

DistanceUsing Time

Speed� �� ��� �� �

16 32+ = 8

x – y x + y ...(ii)

Put x – y = u, x + y = v and solve further tofind x and y.

OR

Let each boy receives � x and the number ofboys be y. Then sum of money which isdistributed is � xy.

Had there been 10 boys more, each wouldhave received a rupee less,

... (y + 10) (x – 1) = xy

⇒ 10x – y = 10 ...(i)

Had there been 15 boys fewer, each wouldhave received � 3 more,

... (y – 15) (x + 3) = xy

⇒ 5x – y = –15 ...(ii)Solving (i) and (ii), we get

x = 5 and y = 40

... xy = 200

Hence, sum of money = � 200

And number of boys = 40.

x 3 4

y 2 – 1

x 0 4

y – 1 3

36 �� � � � � � �� ��

– =

– 3 ( + 2 ) + 2 ( – 2 ) ( )

ya b a b

ii

1=

( + 2 ) (2 + ) – (2 – ) ( – 2 ) ( )

a b a b a b a biii

Taking (i) and (iii), 5 – 2

=10 b a

xab

Again taking (ii) and (iii), +1010

a by

ab�

Thus, 5 – 2

=10 b a

xab

, + 10=

10 a b

yab

is the

solution of the given system of equations.

8. Speed of rowing = 6 km/hr,

Speed of current = 4 km/hr

Hint: 2 × (x + y) = 20

[Time × Speed = Distance]

2 × (x – y) = 4Where, x = speed of rowing and,

y = speed of current.

OR

Let fare from A to B and from A to C be� x and � y respectively.According to the given conditions,

2x + 3y = 795 ...(i)

3x + 5y = 1300 ...(ii)

Solving eqn. (i) and (ii), we obtain

x = 75, y = 215

Hence, fares from A to B is � 75 and from Ato C is � 215.

9. Let us make the table for the values of x andcorresponding values of y to the equation

2x + y – 8 = 0

x 2 4

y 4 0

�������� – ��

1. (B) As the given equations are homogeneousso only solution will be x = 0, y = 0.

2. (C) x = 4, y = 9

Hint: Put 1

x = u ,

1 y

= v and solve.

3. Here, � �– 32 9

6 – 9 – 5

... Lines are parallel.

4. The given equations have a unique solution

⇒��������� al

≠ bm

⇒��am ≠ bl.

5. The given equation can be written as6ax + 6by = 3a + 2b ...(i)

and 6bx – 6ay = 3b – 2a ...(ii)Multiplying equation (i) by a and (ii) by band adding the results, we have

6(a2 + b2)x = 3(a2 + b2)

⇒ x =12

Substituting x = 12

in equation (i), we have

62a

+ 6by = 3a + 2b

� 6by = 2b �� y = 13

Thus, the solution is x = 12

, y = 13

.

6. a = 5, b = 1Hint: Two linear equations a1 x + b1 y + c1 = 0and a2 x + b2 y + c2 = 0 have infinite numberof solutions if

1 1 1

2 2 2= =

a b ca b c .

7. Given system of linear equations can bewritten as

(a + 2b) x + (2a – b) y – 2 = 0

(a – 2b) x + (2a + b) y – 3 = 0By cross-multiplication,

– 3 (2 – ) + 2 (2 + ) ( )

xa b a b

i

37���� ����� �� � � ��� ��

Similarly, for the equation x – y – 1 = 0

Let us draw the graph.

From the graph, the lines intersect each otherat the point A(3, 2). Therefore, the solutionis x = 3, y = 2.The lines intersect the y-axis at B(0, 8) andC(0, – 1).To find the area of the shaded portion,that is, ΔABC, draw perpendicular AMfrom A on the y-axis to intersect it in M.

Now, AM = 3 units and BC = 8 + 1 = 9 units.

... ar(ΔABC) =12

× BC × AM

=12

× 9 × 3

=272

sq. units

Hence, x = 3, y = 2; area = 13.5 sq.units.

��������– ��

1. (A) In the case of no solution,

36

=–1–2

≠–5–k

⇒ k ≠ 10.

2. (D) x = 80, y = 30

Hint: x + 2y = 140, 3x + 4y = 360.3. For unique solution,

42

≠ 2p

⇒ p ≠ 4.

4. True.According to the conditions of consistency,

either �

2 2– 5 – 5 13 3or = =

3 3– 5 – 5 32 2

Clearly, the first condition holds. Hence,the system of equations is consistent witha unique solution.

5. For infinitely many solutions,

+3

p q=

2( – )4

p q =

–(5 – 1)–12

p

⇒ 4p + 4q = 6p – 6q and– 12p – 12q = – 15p + 3

⇒ 2p – 10q = 0 and 3p – 12q = 3⇒ p = 5, q = 1.

6. x = 1, y = 1,

Hint: Take 1

3x y�= u,

13x y�

= v

∴ Given equation can be written as:

u + v = 34

⇒ 4u + 4v = 3

and12

u – 12

v = –18

⇒ 4u – 4v = – 1.

7. x =1

2�

, y =13

Hint: Put 1 x

= u and 1 y

= v.

x 4 3

y 3 2

38 �� � � � � � �� ��

8. Table for values of x and y correspondingto equation 4x – 5y – 20 = 0 is

Similarly for the equation 3x + 5y – 15 = 0

Let us draw the graphs for the two equations.

As the graphs of the two lines intersecteach other at the point A(5, 0), the requiredsolution is x = 5, y = 0.The graphs intersect the y-axis at B (0, 3)and C(0, – 4). Therefore, the coordinates ofvertices of the triangle ABC are A(5, 0),B(0, 3) and C(0, – 4).Hence the answer: x = 5, y = 0 and (5, 0),(0, 3), (0, – 4).

9. Let speeds of two cars that start from placesA and B be x km/hr and y km/hr respec-tively.Case I: When two cars travel in same direction.Let the cars meet at C

Distance travelled by the car that startsfrom A

AC = 5 × xSimilarly distance for other car

BC = 5 × y... AC – BC = 5x – 5y⇒ 5x – 5y = 100⇒ x – y = 20 ...(i)Case II: When two cars travel in oppositedirections.Let the cars meet at D

Distance travelled by the car that startsfrom A

AD = 1 × xSimilarly distance for other car

BD = 1 × y

... AD + BD = x + y

⇒ x + y = 100 ...(ii)

Solving equations (i) and (ii), we getx = 60 and y = 40

Hence, speeds of two cars that start fromplaces A and B are 60 km/h and 40 km/hrespectively.

�������� – ��

1. (B) x – y = 0 ...(i)2x – y = 2 ...(ii)– + –

– x = –2 (Subtracting)

... x = 2.Further y = x = 2.

2. The given equations represent to be parallellines if

2( – 1)3

k =

1–1

≠ –1–1

⇒ k – 1 =3

–2

⇒ k =1

–2

.

x 5 0y 0 3

x 5 0

y 0 – 4

39���� ����� �� � � ��� ��

3. m ≠ 4

Hint: �– 2

2 – 1m

.

4. For the point of intersection of any line withx-axis, put y = 0... – 3x + 7 (0) = 3 ⇒ x = – 1So the required point is (– 1, 0).

5. For inconsistency,

+ 22

k=

63

≠ 2–(3 2)

– 4k �

⇒ k + 2 = 4 and (3k + 2)2 ≠ 8

⇒ k = 2 and k ≠ 13

( )± 2 2 – 2

Hence, k = 2.

6. Given system of equations can be writtenas2x + 3y – 18 = 0 ...(i)x – 2y – 2 = 0 ...(ii)

Now, �2 31 – 2

Hence the system has unique solution.Now, by cross-multiplication on (i) and (ii),we get

– 6 – 36x

=– 4 + 18

– y=

1– 4 – 3

⇒ x = 6, y = 2Thus, the solution of given system is

x = 6, y = 2.7. x = 5, y = – 1

Hint: Take 1 1

= = + –

u, vx y x y

� � � � and solve.

8. Let Meena received x notes of � 50 and ynotes of � 100Since total number of notes is 25... x + y = 25 ...(i)Since the value of both types of notes is� 2000.... 50x + 100y = 2000⇒ x + 2y = 40 ...(ii)

Solving equations (i) and (ii), we get

x = 10, y = 15

Hence, Meena received 10 notes of � 50 and15 notes of � 100.

ORLet the length and breadth of rectangle bex units and y units respectively.Then area of rectangle = xy sq. unitsCase I. The length is increased and breadthis reduced by 2 units.

... (x + 2) (y – 2) = xy – 28

⇒ xy – 4 – 2 x + 2 y = xy – 28

⇒ x – y = 12 ...(i)

Case II. The length is reduced by 1 unitand breadth increased by 2 units.

∴ (x – 1) (y + 2) = xy + 33

⇒ xy – 2 – y + 2x = xy + 33

2x – y = 35 ...(ii)

Solving equations (i) and (ii), we get

x = 23 and y = 11

Hence, the length of the rectangle is 23 unitsand the breadth is 11 units.

9. The given linear equations are

4x – y – 8 = 0 ...(i)and 2x – 3y + 6 = 0 ...(ii)

To draw the graphs of the equations (i) and(ii), we need atleast two solutions of each ofthe equations. These solutions are givenbelow:

Plot the points A(0, – 8), B(2, 0), C(0, 2) andD(– 3, 0) on graph paper and join them toform the lines AB and CD as shown infigure.

x 0 2

y = 4x – 8 – 8 0

x 0 – 3

y =2 6

3x�

2 0

40 �� � � � � � �� ��

The graphs of these lines intersect eachother at P(3, 4). So, unique solution:

x = 3, y = 4.

Also, the graphs meet the x-axis at D(– 3, 0)and B(2, 0).

Hence, the triangle formed by the lines andthe x-axis is PBD with vertices P(3, 4), B(2,0) and D(– 3, 0).

�������� – ��

1. (C) For coincident lines,12

= 2k

= 714

⇒ k = 4.

2. (A) The given system of equations can bewritten asx + 2y = 140, 3x + 4y = 360Solving this system, we obtainx = 80, y = 30.

3. Adding the given equations, we have3x = 0 ⇒ x = 0

Substituting x = 0 in any of the givenequations, we get y = 0Hence, the required solution is x = 0, y = 0.

4. False.

As 1

2

aa =

24

, 1

2

bb

=5

10, 1

2

cc = 6

... 1

2

aa = 1

2

bb

≠ 1

2

cc

⇒ They are parallel.

5. a = – 1, b =52

Hint: 2 – (2 + 5) 5

= =2 + 1 – 9 15

ab

� � � .

6. Put 1 x

= u and 1 y

= v in given system of

equations.u + v – 7 = 0 ...(i)

2u + 3v – 17 = 0 ...(ii)By cross-multiplication,

1= =

– 17 + 21 – 17 + 14 3 – 2u – v

⇒ u = 4, v = 3

⇒ x = 14

, y = 13

Hence, x = 14

, y = 13 is the solution of the

given system of equations.

7. x = – 2, y = 5 and m = – 1

Hint: 2x + 3y = 11 ⇒ 11 – 2=

3x

y

Substitute this value of y in 2x – 4y = – 24and solve for x.

8. The given system of linear equations is2x – y – 5 = 0 ...(i)3x + y – 5 = 0 ...(ii)

To draw the graph of equations (i) and (ii),we need atleast two solutions of each of theequations, which are given below:

x 0 4

y = 2x – 5 – 5 3

x 0 3

y = – 3x + 5 5 – 4

41���� ����� �� � � ��� ��

�������� – ��

1. (A) ∠A = 70°, ∠B = 53°,∠C = 110°, ∠D = 127°.Hint: In a cyclic quadrilateral ABCD,∠A + ∠C = 180° and ∠B + ∠D = 180°.

2. (C) x = 0, y = 0Hint: Both lines are passing through theorigin.

3. For infinite number of solutions,

2p + q

=2 –

3p q

=– 21– 7

⇒ p + q = 6 and 2p – q = 9

⇒ p = 5 , q = 1.

4. False.Hint: As a + 5b = – 10.

5. False, x = 4, y = 1 does not satisfy the secondequation.

6. No solution

Hint: 2x + 3y = 7, 6x + 9y = 11

Here, 2 3 7=

6 9 11 �� Parallel lines.

7. The given system of linear equations can bewritten as

px + qy – (p – q) = 0qx – py – (p + q) = 0

To solve the system for x and y, using themethod of cross-multiplication, we have

+– ( ) – ( – )x

q p q p p q=

+ +–

– ( ) ( – )y

p p q q p q

= 2 21

– –p q

⇒ 2 2– –x

p q=

2 2

–

– –

y

p q= 2 2

1– –p q

⇒ x = 1, y = – 1.

8. The given system of equations can bewritten as

3x – 4y – 1 = 0 ...(i)6x – 8y + 15 = 0 ...(ii)

To draw the graph, we need atleast twosolutions for each of the equations (i) and(ii), which are respectively given below:

Using these points, we are drawing thegraphs of lines as shown below:

From the graph, the lines intersect each otherat the point P(2, – 1). Therefore, the solutionis x = 2, y = – 1.The lines meet the y-axis at the pointsQ(0, 5) and R(0, – 5).

9. Let the fixed charge and additional chargefor each day be � x and � y respectively.Since Saritha paid � 27 for a book kept for 7days... x + 4y = 27 ...(i)Also, Susy paid � 21 for the book kept for 5days... x + 2y = 21 ...(ii)Subtracting equation (ii) from (i), we get

2y = 6 ⇒ y = 3Again substituting y = 3 in equation (ii),we get

x = 15Hence, the fixed charge is � 15 andadditional charge for each day is � 3.

ORSon's age = 10 years, father's age = 40 years.Hint: Let the present age of father and sonbe x and y years respectively.... x + 5 = 3 (y + 5)And x – 5 = 7 (y – 5).

42 �� � � � � � �� ��

Let us draw the graph by using these points.

From the graph, it is clear that the lines areparallel. Hence, the given system ofequations is inconsistent.

9. Let the fraction be xy

On adding 1 to each of numerator and

denominator, the fraction becomes 65

... +1 6=

+1 5xy

⇒ 5x + 5 = 6y + 6

⇒ 5x – 6y = 1 ...(i)Further, on subtracting 5 from each ofnumerator and denominator, the fraction

becomes 32

... – 5 3=

– 5 2xy

⇒ 2x – 10 = 3y – 15⇒ 2x – 3y = – 5 ...(ii)Solving equations (i) and (ii), we get

x = 11, y = 9

Hence, the required fraction is 119

.

OR

� 6000, � 5250Hint: Let incomes of X and Y be 8x and 7xrespectively; and expenditures of them be19y and 16y respectively.∴ 8x – 19y = 1250 ...(i)

7x – 16y = 1250 ...(ii)

��������– ��

1. (C) The condition that the given system ofequations has unique solution is represen-ted by

1

2

aa ≠ 1

2

bb

.

2. (A) Multiplying first equation by 2 and theother one by 3 and adding, we get

21.8x = 10.9 ⇒ x = 12

Substituting x =12

in any of the given

equations, we have y = 13

.

∴ x = 12

, y = 13

.

3. k = 6

Hint: – 3 3= =

12k k

k k.

4. The condition that the given system ofequations represents parallel lines is

++

2 13 1pp

=– 23

p ≠

52

⇒ 5p = – 5 ⇒ p = – 1.

5. True.The condition for parallel lines is

26

= – 2– 6

≠ – 35

⇒13

= 13

≠ –35

The condition holds. The lines are parallel.

x 3 7

y =3 – 1

4x

2 5

x 32

112

y = 6 158

x �3 6

43���� ����� �� � � ��� ��

6. x = a2 , y = b2

Hint: Put 1x

= u and 1y

= v.

7. Given system of linear equations can bewritten as:(a – b) x + (a + b) y – (a2 – 2ab – b2) = 0

(a + b) x + (a + b) y – (a2 + b2) = 0By cross -multiplication,

2 2 2 2– ( + )( + ) + ( + ) ( – 2 – ) x

a b a b a b a ab b

2 2 2 2

– =

– ( – )( + ) + ( + ) ( – 2 – ) y

a b a b a b a ab b

1

=( – ) ( + ) – ( + ) ( + )a b a b a b a b

⇒ 2 2

– 1= =

– 2 ( + )– 2 ( + ) – 4yx

b a bb a b ab

Hence, the solution of given system ofequations is

x = a + b, y = – 2 + ab

a b.

8. To draw graph of the equation, we needatleast two solutions.Two solutions of the equation4x + 3y – 24 = 0 are mentioned in thefollowing table:

Similarly, two solutions of each of theequations 2x – y – 2 = 0 and y + 4 = 0 arerespectively

and

Using the tables obtained above, let usdraw the graph.

Observing the graph, we get the lines meeteach other pairwise in A(3, 4), B(–1, – 4)and C(9, – 4).Hence, the vertices of the triangle ABC soobtained are A(3, 4), B(–1, – 4) andC(9, – 4).

Area of ΔABC = 12

× base × height

=12

× 10 × 8 = 40 sq. units.

9. � 600, � 700Hint: Let cost price of trouser be � x andthat of shirt � y. Then

125 110 + = 1520

100 100

110 125 + = 1535

100 100

x y

x y

�������������

⇒ 25 + 22 = 3040022 + 25 = 30700

x y

x y�������

OR

6 l of 50% and 4 l of 25%.Hint: Let x litres of 50% acid and y litres of25% acid should be mixed.

x 0 6

y 8 0

x 0 1y – 2 0

x 0 4y – 4 – 4

44 �� � � � � � �� ��

8. Let the actual prices of tea-set and lemon-set be � x and � y respectively

According to the question,

Case I. Selling price – Cost price = Profit

⇒ 0.95x + 1.15y – (x + y) = 7

⇒ – 0.05x + 0.15y = 7 ...(i)

Case II. Selling price – Cost price = Profit

⇒ 1.05x + 1.10y – (x + y) = 13

⇒ 0.05x + 0.10y = 13 ...(ii)

Solving equations (i) and (ii), we get

x = 100 , y = 80

Hence, actual prices of tea-set and lemon-set are � 100 and � 80 respectively.

OR

The person invested � 500 at the rate of12% per year and � 700 at the rate of 10%per year.

Hint: Let the person invested � x at the rateof 12% per year and � y at the rate of 10 %per year

� 1012 +

100 100yx = 130

⇒ 6x + 5y = 6500 ...(i)

and12 10

+ 100 100

y x= 134

⇒ 5x + 6y = 6700 ...(ii)Adding and subtracting (i) and (ii), we get

x + y = 1200 ...(iii)x – y = – 200 ...(iv)

9. Two solutions of 6y = 5x + 10 are:

Two solutions of y = 5x – 15 are

Now, we draw the graph of the system onthe same coordinate axes.

50 25 40 + = ( + )

100 100 100+ = 10

x y x y

x y

���������

⇒ 2 = 3

+ = 10x y

x y�������

��������– �

1. (C) x = 9, y = 6

Hint: x – y = 3 and 2x + 3y = 36.

2. (A) Solving 3x – 2y = 4 and 2x + y = 5, wehave x = 2, y = 1.

Now, substituting these values of x and yin y = 2x + m, we have 1 = 2 × 2 + m∴ m = – 3.

3.3

18

p=

624

≠ 5075

⇒2

p=

32

≠ 23

∴ p = 3.

4. For inconsistency,

α12

=α3 ≠

αα– 3

⇒ α2 = 36 and 3α ≠ α2 – 3α⇒ α = ± 6 and α ≠ 0 or α ≠ 6 ⇒ α = – 6.

5. x = b, y = – a

Hint: a2x – b2y = ab(a + b), ax – by = 2abSolving the equations, we get x = b, y = – a.

6. x = 225

a, y =

265

b−

Hint: 4bx + 3ay – 2ab = 0

3bx + ay – 8ab = 0.

7. 3x + 2y = 800,12x + 8y = 3000;Not possibleHint: Let cost of 1 chair be � x and that of1 table be � y.∴ 3x + 2y = 800, 12x + 8y = 3000.

x – 2 4

y 0 5

x 3 2

y 0 – 5

45���� ����� �� � � ��� ��

3. For coincident lines

1

2

aa = 1

2

bb

= 1

2

cc

⇒ 2a b�

= 3

– 3a b� =

74a b�

⇒ a – 5b = 0.

4. False, because the given pair of equationshas infinitely many solutions at k = 40 andno solutions at k ≠ 40.

5. Given equations are

2y – x. (x + y) = 1

⇒ x + y = –1

2y x ...(i)

and (x + y)x – y = 2 ...(ii)Substituting the value of x + y from equation(i) in equation (ii), we get

⎛ ⎞⎜ ⎟⎝ ⎠

–

–1

2

x y

y x = 2

⇒ (2x – y)x – y = 21

⇒ (x – y)2 = 1

⇒ x – y = ± 1

⇒ x – y = 1 ...(iii)

or x – y = –1 ...(iv)

Substituting x – y = 1 and x – y = – 1 inequation (ii), we get respectively

x + y = 2 ...(v)

and x + y = 12

...(vi)

Solving equations (iii) and (v), we have

x = 32

; y = 12

.

Therefore, xy = 34

Solving equations (iv) and (vi), we have

x = 1–

4; y =

34

Therefore, xy = – 3

16.

Hence, xy = 34

or – 3

16.

(i) From the graph, we look that the twolines intersect each other at A(4, 5).

(ii) The vertices of the triangle: A(4, 5);B(–2, 0); C(3, 0).Height of ΔABC corresponding to thebase BC,

h = 5 unitsand base, b = BC = 5 units

Now, ar(ΔABC) = 12

× b × h

= 12

× 5 × 5

= 12.5 square units.

�������������� �

1. (C) Let us check option (C).

23

x + 52

y = 23

(– 3) + 52

(– 2) = – 2 – 5 = – 7

3x – 12

y = 3 (– 3) – 12

(– 2) = – 9 + 1 = – 8.

2. 4x – y = 42 � ⇒ x – y = 2 ...(i) x – 2y = 8 ...(ii) – + –

y = – 6

(Subtracting)∴� From (i) ⇒� x = – 4 ∴ x + y = – 10.

46 �� � � � � � �� ��

6. Given equations can be written as

xa

+ yb

– (a + b) = 0

2xa

+ 2

y

b – 2 = 0

Let us apply cross-multiplication methodto solve these equations.

+ +22 1

–

xa

b bb

=

2

–2 1

–

yb

a a a� �

=

2 2

11 1

–ab ba

⇒+

2

–b xb a

��+

2––

a ya b

���2 2

–a ba b

Taking+

2

–b xb a

=�2 2

–a ba b

and+

2––

a ya b

��2 2

–a ba b

⇒��� x = 2 2

2

( – )

( – )

a b a b

b a b and y =

2 2

2

( – )

( – )

a b a b

a a b

⇒ x = a2 and y = b2.

7. Given equations of lines are:

3x + y + 4 = 0 ...(i)

and 6x – 2y + 4 = 0 ...(ii)To draw the graphs of lines (i) and (ii), weneed atleast two solutions of each equation.For equation (i), two solutions are:

For equation (ii), two solutions are:

Let us draw the graphs of the lines (i)and (ii).

�����

From the graph it is clear that the two linesintersect each other at a point, P(– 1, – 1),therefore, the pair of equations consistent.The solution is x = – 1, y = – 1.

8. Let the cost price of the saree and the listprice of the sweater be � x and � yrespectively. Now two cases arise.Case I.

Sale price of the saree = x + x × 8

100

= 108100

x

Sale price of the sweater = y – y ×10

100

= 90

100y

∴ 108100

x + 90

100y = 1008

⇒ 108x + 90y = 100800 ...(i)

Case II.

Sale price of the saree = x + x ×10

100

= 110100

x

Sale price of the sweater = y – y ×8

100

x 0 – 3

y – 4 5

x 0 2

y 2 8

47���� ����� �� � � ��� ��

Here, a1 = 4, b1 = 5, c1 = – 2,

a2 = 2p + 2q, b2 = p + 8q,

c2 = – 2q + p – 1

For infinitely many solutions,

1

2

aa

= 1

2

bb

= 1

2

cc

⇒ 42 + 2p q

= 5

+ 8p q =

– 2– 2 + –1q p

Taking4

2 2p q+=

58p q+

⇒ 10p + 10q = 4p + 32q

⇒ 6p = 22q ⇒ 3p = 11q ...(i)

Also, taking

42 2p q+

= – 2

– 2 + –1q p

⇒ – 4p – 4q = – 8q + 4p – 4⇒ q = 2p – 1

Substitute q = 2p – 1 in equation 3p = 11q toget

p = 1119

.

Hence, p = 1119

, q = 3

19.

6. Given system of equations is43x + 67y = – 24 ...(i)67x + 43y = 24 ...(ii)

Adding (i) and (ii); and subtracting (i) from(ii), we get respectively

110x + 110y = 0and 24x – 24y = 48i.e. x + y = 0 ...(iii)and x – y = 2 ...(iv)Adding equations (iii) and (iv); andsubtracting equation (iv) from (iii), we getrespectively

x = 1 and y = – 1.7. One of the given equations is

2x + y = 4 ...(i)Here, at x = 0, y = 4and at x = 2, y = 0

= 92

100y

∴ 110100

x + 92100

y = 1028

⇒ 110x + 92y = 102800 ...(ii)Adding equations (i) and (ii), we get

218x + 182y = 203600 ...(iii)Subtracting equation (i) from (ii), we get

2x + 2y = 2000or 218x + 218y = 218000 ...(iv)

(Multiplying by 109)Solving equations (iii) and (iv), we have

x = 600 and y = 400Hence, the cost price of the saree is� 600 and the list price of the sweater is� 400.

�������������� �

1. (B) For infinitely many solutions,

1339

= 6k

= + 4k

k

Taking1339

= 6k

⇒ k = 2.

2. For no solution,

312

= 72k

≠ +4 1

kk

∴3

12=

72k

⇒ k = 14.

3. x = – a and y = k must satisfy both the givenequations. Let us substitute these values ofx and y in

bx – ay + 2ab = 0b(–a) – ak + 2ab = 0

⇒ – ak + ab = 0 ⇒ k = b.4. Yes, because

1

2

aa = 1

2

bb

= 1

2

cc , i.e.,

13

=26

= – 3– 9

.

5. Given equations are:4x + 5y = 2

(2p + 2q)x + (p + 8q)y = 2q – p + 1

48 �� � � � � � �� ��

Two solutions of equation(i) are given inthe following table:

Another given equation is2x – y = 4 ...(ii)

Two solutions of equation (ii) are given bythe following table:

Let us draw the graph of the two equations(i) and (ii) by using their correspondingtables.

From the graph, vertices of the triangle ABCare A(0, 4), B(0, – 4) and C(2, 0).

ar(ΔABC) = 12

× base × height

= 12

× AB × OC

x 0 2

y 4 0

x 0 2

y – 4 0

= 12

× (4 + 4) × 2

= 8 square units.Thus, area of the triangle is 8 square units.

8. Let Aman had a total of x oranges; and hemade lot A of p oranges and lot B ofremaining x – p oranges. There are twocases now.Case I.

Selling price of lot A = � 23

p

Selling price of lot B = � (x – p)

∴ 23

p + x – p = 400

⇒ 3x – p = 1200 ...(i)Case II.

Selling price of lot A = � p

Selling price of lot B = � 45

(x – p)

∴ p + 45

(x – p) = 460

4x + p = 2300 ...(ii)

Add equations (i) and (ii) to get

7x = 3500⇒ x = 500Hence Aman had a total number of 500oranges.

���������

1. (C) A pair of linear equations is said to beconsistent, if the lines either intersect eachother at a point or coincide.

2. (C) 6, 36

Hint: Let the son's age = x,And father's age = y∴ y = 6xand y + 4 = 4(x + 4)Solve yourself.

3. The lines are coincident

⇒ 36

=– 1– k

= 8

16 ⇒ k = 2.

49���� ����� �� � � ��� ��

4. Yes.For consistency,

either 24

aa

≠ 2bb

or 24

aa

= 2bb

= –– 2

aa

Here only the relation 24

aa

= 2bb

= –– 2

aa

,

i.e.,12

= 12

= 12

holds.

⇒ The pair is consistent.

5. 21x + 47y = 11047x + 21y = 162

68x + 68y = 272 (Adding)⇒ x + y = 4 ...(i)

Subtracting the given equations from oneanother, we get

– 26x + 26y = – 52

⇒ x – y = 2 ...(ii)Solve equations (i) and (ii) to get

x = 3, y = 1.6. We are given

+2xyx y

= 32

...(i)

and2 –

xyx y

= –310

...(ii)

Taking equation (i),

+

2xyx y

= 32

⇒ 3x + 3y = 4xy ...(iii)Now, taking equation (ii),

2 –xyx y

= – 310

⇒ – 6x + 3y = 10xy ...(iv)

Multiplying equation (iii) by 2 and addingits result to (iv), we get

9y = 18xy

∴ x = 12

Putting x = 12

in equation (iv), we get

⇒ – 3 + 3y = 5y

∴ y = – 32

Thus, x = 12

and y = – 32

.

7. The given system of equations will haveinfinite number of solutions if

1–a b

= +2

a b =

+1

– 2a b

⇒ 1–a b

= +

1– 2a b

and+2

a b=

+1

– 2a b⇒ a + b – 2 = a – b

and 2a + 2b – 4 = a + b

⇒ a + b – a + b = 2 and a + b = 4

⇒ b = 1 and a = 3

Hence, the given system of equations willhave infinite number of solutions, if

a = 3, b = 1.

8. (i) Let fixed charge = � xand charges for a distance of 1 km = � yNow, According to question,

x + 12y = 89 ...(i)x + 20y = 145 ...(ii)

We will solve equations (i) and (ii) byelimination method.Subtract equation (ii) from equation (i):

– 8y = – 56 ⇒ y = 568

= 7

Putting value of y in equation (i), we getx + 12(7) = 89

x = 89 – 84 = 5∴ x = 5; y = 7∴ For a journey of 30 km charge paid = x +

30y = 5 + 30(7) = 5 + 210 = � 215.(ii) Solution of pair of linear equations in two

variables.(iii) Love towards environment.

9. To draw the graph of a line, we are requiredatleast two solutions of its correspondingequation.At x = 0, 5x – y = 5 gives y = – 5

50 �� � � � � � � � � �

At x = 1, 5x – y = 5 gives y = 0Thus, two solutions of 5x – y = 5 are givenin the following table:

x 0 1y – 5 0 x 9 1

y – 4 0

x 2 4

y 5 – 7

Similarly, we can find the solution of eachremaining equation as given in thefollowing tables:

x + 2y = 1:

6x + y = 17:

Now, we will draw the graphs of the threelines on the same coordinate axes.From the graph, it is clear that the linesform a triangle ABC with vertices A(1, 0),B(3, – 1) and C(2, 5)

��

51���� �� �

A

D F

EB C

��������

TRIANGLES

�������� ��

1. (D) Observing the figure, we obtain∠A = ∠R, ∠B = ∠Q, ∠C = ∠P

∴ ΔABC ~ ΔRQP.

2. � ΔABC ~ ΔDEF

∴2

2ABDE

= 2

2BCEF

= 2

2ACDF

= ΔΔ

( ABC)( DEF)

arar

Taking2

2BCEF

=ΔΔ

( ABC)( DEF)

arar

⇒2

2BC

(15.4)=

64121

⇒ BC = 64 × 15.4 × 15.4121

⇒ BC = 11.2 cm.

3. Yes.Here, 262 = 242 + 102 = 676⇒ AC2 = AB2 + BC2

∴ ΔABC is a right triangle.

4. In ΔABC, LM || CB

∴ =AM ALAB AC

…(i)

(Basic Proportionality Theorem)Similarly in ΔADC,

∴ =AN ALAD AC

…(ii)

Comparing equations (i) and (ii), we have

=AM ANAB AD

. Hence proved.

5. As ∠1 = ∠2

⇒ PQ = PR …(i)

QRQS

=QTPR

(Given)

⇒ QRQS

=QTPQ

[Using (i)]

and ∠1 = ∠1 (Common)

∴ ΔPQS ~ ΔTQR (SAS criterion)Hence proved.

6. 1 : 4

Hint: ar(ΔDEF) = 14

ar(ΔABC)

⇒ΔΔ

( DEF)( ABC)

arar

= 14

.

7. 13 mHint: Distance between tops = AD

∴ AD = +2 2AE DE

= 2 2(5) (12)�

= 13 m.

8. Hint: Use Pythagoras Theorem.

9. Statement: If a line is drawn parallel toone side of a triangle to intersect the othertwo sides at distinct points, the other twosides are divided in the same ratio.Proof: ABC is a given triangle in whichDE || BC. DE intersects AB and AC at Dand E respectively.

We have to prove

ADBD

= AECE

Let us draw EM ⊥ AB and DN ⊥ AC. JoinBE and CD.

52 �� � � � � � � � � �

2. DC2 = BC2 + BD2 = BC2 + ⎛ ⎞⎜ ⎟⎝ ⎠

2AB2

= BC2 + 14

(AC2 – BC2)

= 9 + 14

(25 – 9) = 9 + 4 = 13

⇒ DC = 13 cm.

3. x = 8cmHint: As DE || BC

∴ADDB

= AEEC

⇒−

−2 1

3x

x=

2 51

x +x �

.

4. DE || BC and DB is transversal

⇒ ∠EDA = ∠ABC(Alternate interior angles)

Similarly, ∠AED = ∠ACB

Consequently, ΔADE ~ ∠ACB

(AA similarity)

∴2

2ADAB

=ΔΔ

( ADE)( ABC)

arar

⇒2

2AD

9AD=

Δ( ADE)153

ar

⇒ ar(ΔADE) = 17cm2.

5. No.

Here,DPPE

= 5

10=

12

AndDQQF

= 6

18=

13

�DPPE

≠ DQQF

Therefore, PQ is not parallel to EF.

6. Hint: Use Basic Proportionality Theorem.

Now, ar(ΔADE) = 12

× base × height

= 12

× AD × EM ...(i)

Also, ar(ΔADE) = 12

× AE × DN ...(ii)

ar(ΔBDE) = 12

× BD × EM ...(iii)

ar(ΔCDE) = 12

× CE × DN ...(iv)

Dividing equation (i) by equation (iii) andequation (ii) by equation (iv), we have

ΔΔ

( ADE)( BDE)

arar

= ADBD

...(v)

andΔΔ

( ADE)( CDE)

arar

= AECE

...(vi)

But ar(ΔBDE) = ar(ΔCDE) ...(vii)(Triangles are on the same base DE andbetween the same parallels BC and DE)Comparing equations (v), (vi) and (vii), wehave

ADBD

=AECE

.

2nd Part:As ∠B = ∠C⇒ AB = AC⇒ AD + DB = AE + EC⇒ AD = AE (... BD = EC)

∴ ADDB

=AEEC

∴ By converse of Basic Proportionality Theorem, DE || BC.

Hence proved.

�������� ��

1. (A) ΔBAC ~ ΔADC

⇒BCAC

= ACDC

⇒ y2 = 16 × 4 ⇒ y = 8 cm.

53���� �� �

A

DB C

7. As AB = BC = AC

∴ AD ⊥ BC ⇒ BD = 12

BC

∴ Using Pythagoras Theorem

AB2 = AD2 + BD2

⇒ AD2 = AB2 − ⎛ ⎞⎜ ⎟⎝ ⎠

21BC

2

=23AB

4⇒ 4AD2 = 3AB2. Hence proved

OR

Let ABCD be a rhombusSince, diagonals of a rhombus bisect eachother at right angles,∴ AO = CO, BO = DO, ∠AOD = ∠DOC = ∠COB = ∠BOA = 90°Now, in ΔAOD

AD2 = AO2 + OD2 ...(i)Similarly, DC2 = DO2 + OC2 ...(ii)

CB2 = CO2 + BO2 ...(iii)and BA2 = BO2 + AO2 ...(iv)Adding equations (i), (ii), (iii) and (iv),we haveAD2 + DC2 + CB2 + BA2

= 2(DO2 + CO2 + BO2 + AO2)

= 2⎛ ⎞

+ + +⎜ ⎟⎝ ⎠

2 2 2 2BD AC BD CA4 4 4 4

= BD2 + CA2. Hence proved

8. Hint: BD = DE = EC = 13

BC

Use Pythagoras Theorem.

9. Statement: In a triangle, if square of thelargest side is equal to the sum of thesquares of the other two sides, then theangle opposite to the largest side is a rightangle.Proof: We are given a triangle ABC with

A′C′2 = A′B′2 + B′C′2 ...(i)We have to prove that ∠B′ = 90°

Let us construct a ΔPQR with ∠Q = 90°such that

PQ =A′B′ and QR = B′C′ ...(ii)

In ΔPQR,PR2 = PQ2 + QR2

(Pythagoras Theorem)= A′B′2 + B′C′2 ...(iii)

[From (ii)]But A′C′2 = A′B′2 + B′C′2 ...(iv)

[From (i)]From equations (iii) and (iv), we have

PR2 = A′C′2

⇒ PR = A′C′ ...(v)Now, in ΔA′B′C′ and ΔPQR,

A′B′ = PQ [From (ii)]B′C′ = QR [From (ii)]A′C′ = PR [From (v)]

Therefore, ΔA′B′C′ ≅ ΔPQR(SSS congruence rule)

⇒ ∠B′ = ∠Q (CPCT)But ∠Q = 90°... ∠B′ = 90°. Hence proved.2nd Part

In ΔADC, ∠D = 90°... AC2 = AD2 + DC2 = 62 + 82

= 36 + 64 = 100In ΔABC,

AB2 + AC2 = 242 + 100 = 676and BC2 = 262 = 676Clearly, BC2 = AB2 + AC2

Hence, by converse of Pythagoras Theorem,in ΔABC,

∠BAC = 90°⇒ ΔABC is a right triangle.

54 �� � � � � � � ��

A

X

B C

Y

�������� – �

1. (B)ΔΔ

( ADE)( ABC)

arar

= 2

2DEBC

⇒ ar(ΔADE) =

⎛ ⎞⎜ ⎟⎝ ⎠

2

2

2BC

3BC

× 81 = 36 cm2.

2. ΔOAB ~ ΔOCD

⇒ OAOC

=OBOD

⇒ OB = 4 ×32

= 6 cm.

3. In ΔABC, to make DE || AB, we have totake

⇒ ADDC

= BEEC

⇒ ++

3 193

xx

= +3 4xx

⇒ 3x2 + 19x = 3x2 + 4x + 9x + 12⇒ 6x = 12 ⇒ x = 2.

4. No,� ΔFED ~ ΔSTUCorresponding sides of the similar trianglesare in equal ratio.

∴DETU

= EFST

∴ DEST

≠ EFTU

.

5. AB || PQ ⇒ APAO

=BQBO

…(i)

AC || PR ⇒ APAO

=CRCO

…(ii)

From (i) and (ii) ,BQBO

=CRCO

⇒ BC || QR. (By converse of BPT)

6. 1 : 2.Hint: Let AB = BC = a

... AC = 2 a

...ΔΔ

ABE)ACD)

ar(ar(

= 2

2ABAC

.

7. In Δ ABC and Δ AMP,∠A = ∠A (Common)

∠ABC = ∠AMP = 90°

(i) ∴ ΔABC ~ ΔAMP, (AA criterion)

(ii) ∴ CAPA

= BCMP

.

(... Corresponding sides ofsimilar triangles are proportional.)

8. Hint:ar(Δ AXY) = ar(BXYC)

⇒ 2.ar(ΔAXY) = ar(BXYC) + ar(Δ AXY)= ar(ΔABC)

⇒ ΔΔ

( ABC) ( AXY)

arar

= 21

As ΔABC ~ Δ AXY

∴ ⎛ ⎞⎜ ⎟⎝ ⎠

2ABAX

= ΔΔ

( ABC)( AXY)

arar

=21

⇒ABAX

= 2

1 ⇒ BX

AB=

−2 12

.

9. Hint: Prove converse of PythagorasTheorem.

�������� – ��

1. (A)In triangle ABC,

ADDC

= 621

= 27

BEEC

= 18 –14

14 =

27

∴ ADDC

= BEEC

⇒ DE || AB .

2. (B) ΔABD ~ ΔBCD

⇒ABBC

= BDCD

⇒ 5.4BC

= 3.65.2

⇒ BC = 5.4 × 5.2

3.6= 7.8 cm.

A

EBC

D

55���� �� �

3.ΔΔ

( DEF)( ABC)

arar

= 2

2EFBC

⇒ ar(ΔDEF) = 54 × 169

= 96 cm2.

4. In Δ ABC and Δ ADE,∠BAC = ∠DAE (Common angle)∠ACB = ∠AED (Each 90°)

∴ Δ ABC ~ Δ ADE (AA criterion)

AB = 2 2AC + BC = 25 + 144 = 13 cm

Now,ABAD

= BCDE

= ACAE

⇒ 133

= 12DE

= 5

AE

⇒ DE = 3613

cm and AE = 1513

cm.

5. No.Ratio of areas of two similar triangles

= Square of ratio of their corresponding altitudes

= ⎛ ⎞⎜ ⎟⎝ ⎠

235

= 9

25 ≠

65

.

Hence, it is not correct to say that ratio of

areas of the triangles is 65

.

6. AE2 = AC2 + EC2 …(i)

BD2 = DC2 + BC2 …(ii)

Adding (i) and (ii), we get

AE2 + BD2 = AC2 + EC2 + DC2 + BC2

= (AC2 + BC2) + (EC2 + DC2)

⇒ AE2 + BD2 = AB2 + DE2.Hence proved.

7. In ΔAQO and ΔBPO,∠QAO = ∠PBO (Each 90°)∠AOQ = ∠BOP

(Vertical opposite angles)So, by AA rule of similarity,

ΔAQO ~ ΔBPO

⇒ AQBP

= AOBO

⇒ AQ9

= 106

⇒ AQ = 10 × 9

6⇒ AQ = 15 cm.

ORLet the height of the tower be h metres

ΔABC ~ ΔPQR.

⇒ ABPQ

=BCQR

⇒ 12h

= 840

⇒ h =12 × 40

8 = 60 metres.

8. Hint: As ΔAOB ~ ΔCOD

ΔΔ

( AOB)( COD)

arar

= 2

2ABCD

= � �

2

2

2 CD

CD =

41

.

9. Hint: Prove Pythagoras Theorem.For 2nd Part:∴ AB2 = AD2 + BD2 …(i)Also AC2 = AD2 + CD2 …(ii)From (i) and (ii),⇒ AB2 − AC2 = BD2 − CD2

⇒ AB2 + CD2 = AC2 + BD2.Hence proved.

�������� – ��

1. (A) ∠M = 180° – (∠L + ∠N) (ASP)

= 180° – (50° + 60°) = 70°

� ΔLMN ~ ΔPQR

∴ ∠M = ∠Q ⇒ ∠Q = 70°.

56 �� � � � � � � ��

2. (C) In ΔKMN, as PQ || MN,KPPM

= KQQN

⇒ KPPM

= KQ

KN – KQ

⇒ KNKQ

– 1 = PMKP

⇒ 20.4KQ

– 1 = 134

⇒ 20.4KQ

= 1 + 134

= 174

⇒ KQ = 20.4 × 4

17⇒ KQ = 4.8 cm.

3. ΔABC ~ ΔDEF.

4. � ΔABC ~ ΔPQR

∴ ΔΔ

( PRQ)( BCA)

arar

=2

2QRBC

= ⎛ ⎞⎜ ⎟⎝ ⎠

231

= 91

= 9 : 1.

5. TrueHint: Use Basic Proportionality Theorem

6. Hint:Use: ∠1 = ∠2 ∠3 = ∠4.

7. Draw EOF || AD

A D

B C

OE F

∴ OB2 = EO2 + EB2

OD2 = OF2 + DF2

∴ OB2 + OD2 = EO2 + EB2 + OF2 + DF2

= EO2 + CF2 + OF2 + AE2

[�DF = AE, EB = CF]= (EO2 + AE2) + (CF2 + OF2)

⇒ OB2 + OD2 = OA2 + OC2.

ORJoin OA, OB and OCIn right ΔAOF,

AO2 = AF2 + OF2 ...(i)

In right ΔAOE,AO2 = AE2 + OE2 ...(ii)

From equations (i) and (ii), we have

AF2 + OF2 = AE2 + OE2 ...(iii)Similarly, we can find out that

BD2 + OD2 = BF2 + OF2 ...(iv)

and CE2 + OE2 = CD2 + OD2 ...(v)

Adding equations (iii), (iv) and (v), wearrive

AF2 + BD2 + CE2 = AE2 + BF2 + CD2.Hence the result.

8. ΔABC ~ ΔPQR

⇒ABPQ

=BCQR

and ∠B = ∠Q

⇒ ABPQ

=

1BC

21

QR2

and ∠B = ∠Q

⇒ ABPQ

=BP

QM and ∠B = ∠Q

(� BD = DC and QM = MR)⇒ ΔABD ~ ΔPQM

⇒ ABPQ

= ADPM

. Hence proved.

9. Let the two given triangles be ABC andPQR such that ΔABC ~ ΔPQR

∴ABPQ

= BCQR

...(i)

57���� �� �

Let us draw perpendiculars AD and PMfrom A and P to BC and QR respectively.∴ ∠ADB = ∠PMQ = 90° ...(ii)Now, in ΔABD and ΔPQM,

∠B = ∠Q (�ΔABC~ΔPQR)∠ADB = ∠PMQ [From (ii)]

So, by AA rule of similarity, we haveΔABD ~ ΔPQM

⇒ ABPQ

= ADPM

...(iii)

From equations (i) and (iii), we get

BCQR

= ADPM

...(iv)

Now,ΔΔ

( ABC)( PQR)

arar

= × ×

× ×

1BC AD

21

QR PM2

= × ×

× ×

1BC BC

21

QR QR2

[Using (iv)]

= ⎛ ⎞⎜ ⎟⎝ ⎠

2BCQR ...(v)

Similarly, we can prove that

ΔΔ

( ABC)( PQR)

arar

= ⎛ ⎞⎜ ⎟⎝ ⎠

2ABPQ

...(vi)

and( ABC)( PQR)

arar�

�=

⎛ ⎞⎜ ⎟⎝ ⎠

2ACPR

...(vii)

From equations (v), (vi) and (vii), we obtain

( ABC)( PQR)

arar�

�=

⎛ ⎞⎜ ⎟⎝ ⎠

2ABPQ

= ⎛ ⎞⎜ ⎟⎝ ⎠

2BCQR

= ⎛ ⎞⎜ ⎟⎝ ⎠

2ACPR

.

Hence, the theorem.

Further, in the question,

( ABC)( DEF)

arar�

�=

⎛ ⎞⎜ ⎟⎝ ⎠

2BCEF

⇒ 64121

= 2BC

15.4 × 15.4

⇒ BC = × ×64 15.4 15.4

121

= 811

× 15.4 = 11.2 cm.

�������� – ��

1. (B) Ratio of areas of two similar triangles= Ratio of squares of their

corresponding sides.= 42 : 92 = 16 : 81.

2. (C) ∠M = ∠Q = 35°(Corresponding angles)

PQML

= QRMN

(Ratio of corresponding sides)

⇒ MN = 5 × 126

= 10 cm.

3. In ΔABC, DE || BC

⇒ABAC

=ADAE

⇒ AB = 21 × 57

= 15 cm.

4. Yes.

APAQ

= 5

7.5 =

23

BPBR

= 46

= 23

Here,APAQ

= BPBR

Hence, due to the converse of BasicProportionality Theorem, AB || QR.

58 �� � � � � � � ��

5. ��DB ⊥ BC and AC ⊥ BC

∴ DB || ACNow, ∠DBA = ∠BAC (Alternate angles)And, ∠DEB = ∠ACB (Each 90°)∴ ΔBDE ~ ΔABC (AA similarity)

BEAC

= DEBC

(Corresponding sides)

⇒ BEDE

= ACBC

. Hence proved.

6. AXAB

= 2 22�

Hint: See Worksheet – 35, Sol. 8.

7. Hint: AM = 12

AB;

AL = 12

AC

Use Pythagoras Theorem.

8. Hint: Join AC and use Basic ProportionalityTheorem.

A B

CD

E FX

9. As D and F are mid-points of AB and ACrespectively.

⇒ DF � BC and DF = 12

BC. ⇒ DFBC

= 12

Also, as ΔADF ~ ΔABC

⇒( ADF)( ABC)

arar

�

�=

2

2

(DF)

(BC) = 1

4...(i)

As ar(ΔADF) = ar(ΔBDE) = ar(ΔCFE) =ar(ΔDEF)

∴ (i) ⇒ ( DEF)( ABC)

arar

�

�=

14

A

B C

D F

E

(ii) Yes, as DE ��AC and DE = 12

AC

(Using mid-point theorem)

But as AB = AC = BC

DE =12

BC.

(iii) Concept of similarity of two triangles andmid-point theorem.

(iv) His ability to think rationally and takingunbiased decision.

�������� – ��

1. (A) Let the length of shadow is x metres.BE = 1.2 × 4 = 4.8 m

ΔABC ~ ΔDEC

ABDE

= BCEC

⇒ 3.60.9

= 4.8 + x

x

3.6x = 4.32 + 0.9x.

⇒ x =4.322.7

= 1.6 m.

2. (B) Here, (a)2 + ( )23a = a2 + 3a2

= 4a2 = (2a)2

According to the converse of PythagorasTheorem, the angle opposite to longest sideis of measure 90°.

3.ADDB

= 23

⇒ AB – ADAD

= 32

⇒ ABAD

– 1 = 32

⇒ ABAD

= 52

DE || BC ⇒ ΔABC ~ ΔADE

∴ BCDE

= ABAD

= 52

.

4. No.In ΔPQD and ΔRPD,

∠PDQ = ∠PDR = 90°

59���� �� �

But neither ∠PQD = ∠RPD

nor ∠PQD = ∠PRD

Therefore, ΔPQD is not similar to ΔRPD.

5. Hint: ΔBAC ~ ΔADC

⇒BAAD

= ACDC

= BCAC

⇒ CA2 = BC × CD.

6.ADDB

= 54

⇒ AD

AB – AD =

54

⇒ 5AB – 5AD = 4AD ⇒ ADAB

= 59

...(i)

As DE || BC,ΔADE ~ ΔABC

∴ DEBC

= ADAB

= 59

...(ii)

[Using (i)]��DE || BC and DC is a transversal∴ ∠EDC ~ ∠BCD

(Alternate interior angles)i.e., ∠EDF = ∠BCF ...(iii)Similarly,

∠DEF = ∠CBF ... (iv)

From equations (iii) and (iv), we have

ΔDEF ~ ΔCBF (AA similarity)

⇒ΔΔ

( DEF)( CFB)

arar

= ⎛ ⎞⎜ ⎟⎝ ⎠

2DEBC

= 2581

.

[Using equation (ii)]

7. A

B CL

D

E M F

AB = AC; DE = DF

∴ABAC

= DEDF

= 1

⇒ABDE

= ACDF

also ∠A = ∠D

⇒ ΔABC ~ ΔDEF

∴ΔΔ

( ABC)( DEF)

arar

= 2

2ALDM

⇒ALDM

= 45

∴ Ratio of corresponding heights is 4 : 5.

OR

Proof: Draw a ray DZ parallel to the rayXY.

In ΔADZ, XY || DZ

∴ AY AX 2

= =YZ XD 3

⇒ 2YZ = 3AY ... (i)

In ΔYBC, BY || DZ

∴ YZ BD 1= =

ZC DC 1(... BD = DC)

⇒ 2YZ = 2ZC ... (ii)From (i) and (ii),

2ZC = 3AY ... (iii)Now, AC = AY + YZ + ZC

= AY + 32

AY + 32

AY = 82

AY

= 4AY

Therefore, AC : AY = 4 : 1. Hence proved.

8. 2 5 cm

Hint: BD =12

BC; EB =12

AB

Use Pythagoras Theorem.

9. Statement: In a right triangle, the squareof the hypotenuse is equal to the sum ofthe squares of the other two sides.Proof: We are given a right triangle ABCright angled at B.

We need to prove that AC2 = AB2 + BC2

Let us draw BD ⊥ AC.Now, ΔADB ~ ΔABC

A

B C

2

1

D

60 �� � � � � � � ��

So,ADAB

= ABAC

(Sides are proportional)

or AD.AC = AB2 ...(i)

Also, ΔBDC ~ ΔABC

So,CDBC

= BCAC

or CD.AC = BC2 ...(ii)

Adding equations (i) and (ii), we getAD.AC + CD.AC = AB2 + BC2

⇒ AC(AD + CD) = AB2 + BC2

⇒ AC.AC = AB2 + BC2

⇒ AC2 = AB2 + BC2.

Hence proved.2nd Part:

AB2 = AD2 + BD2

= AD2 + (3CD)2

= AD2 + 9CD2

= AD2 + CD2 + 8CD2

= AC2 + 8CD2

= AC2 + 8 21

BC4

� �� �� �

1CD = BC

4� �� �� ��

∴ 2AB2 = 2AC2 + BC2. Hence proved.

�������� – ��

1. (A)ABPQ

=BCQR

= ACPR

(�� Δ ABC ~ Δ PQR)

⇒ 129

= 7x

= 10y

∴ x = 7 × 9

12 = 21

4

and y = 9 × 10

12 = 15

2.

2. Required ratio = 1625

= 45

= 4 : 5.

3. 17 m

Hint:

N

W E

S

15 m O

P

8 m

Use Pythagoras Theorem and find OP.

4. Hint:Let AB = c

AC = bBC = a

... a2 = b2 + c2

Also, ar(ΔABE) = 234

c

ar(ΔBCF) = 234

a

ar(ΔACD) = 234

b .

5. See Worksheet� 36, Sol. 6.

6. Let ABCD be a quadrilateral of whichdiagonals intersect each other at O.It is given that

AOCO

= BODO

orAOBO

= CODO

...(i)

In ΔAOB and ΔCOD,∠AOB = ∠COD

(Vertically opposite angles)

AOBO

= CODO

[From (i)]

Hence, by SAS rule of similarity, we obtainΔAOB ~ ΔCOD

⇒ ∠BAO = ∠DCO

i.e. ∠BAC = ∠DCAThese are alternate angles.Therefore, AB || CD and AC is transversal⇒ ABCD is a trapezium. Hence proved

61���� �� �

ORHint:As ∠BAC = ∠EFG ; ∠ABC = ∠FEGand ∠ACB = ∠FGE

∴12

∠ACB = 12

∠FGE

∴ ∠ACD = ∠FGHand ∠DCB = ∠HGE∴ ΔDCA ~ ΔHGFSimilarly ΔDCB ~ ΔHGE.

7. Hint:A

B C

D

E

Prove that ΔAEB ~ ΔDEC.

8. See worksheet – 33, Sol. 9 (1st part).2nd PartJoin EF and join BD to intersect EF at O.

� AB || DC, and EF || AB,∴� AB || DC || EFIn ΔABD, EO || AB,

DEAE

= DOBO

...(viii)

(Basic Proportionality Theorem)Similarly, in ΔBCD,

DOBO

= CFBF

...(ix)

Using equations (viii) and (ix), we obtainthe required result, i.e.,

AEED

= BFFC

.

�������� – ��

1. (B)DEAB

= EFBC

= DFAC

(i) (ii) (iii)

= DE + EF + DFAB + BC + CA

(iv)

⇒ 42

= ΔPerimeter of DEF

3 + 2 + 2.5

[Taking (ii) and (iv)]

⇒ Perimeter of ΔDEF = 15 cm.

2. DE || BC

⇒– 2x

x=

+ 2– 1

xx

⇒ x2 – 4 = x2 – x⇒ x = 4.

3. ΔKNP ~ ΔKML

⇒ xa

=c

b c+ ∴ x =

acb c+

.

4. Hint:

A LB

D C

P

Prove that ΔADL ~ ΔCPD.

5. Hint: 2AP = PC ⇒ AP =13

AC

Similarly, BQ = 13

BC

Use Pythagoras Theorem.

6. PQ || BC ⇒ APPB

= AQQC

= 12

∴ ΔAPQ ~ ΔABC

⇒ ΔΔ

( ABC) ( APQ)

arar

= ⎛ ⎞⎜ ⎟⎝ ⎠

2ABAP

= ( )23 = 9

� ��� �� �

�AB

3AP

⇒ ΔΔ

( ABC) ( APQ)

arar

− 1 = 8

62 �� � � � � � � ��

⇒ ( BPQC) ( APQ)

arar �

�=

81

⇒ ( APQ)

( BPQC)ar

ar�

�=

18

.

∴ Ratio of areas of ΔAPQ and trapeziumBPQC is 1 : 8.

ORLet the given square be ABCD.Let us draw an equilateral triangle APBand another equilateral triangle AQC onthe side AB and on the diagonal ACrespectively.

We need to prove

ar(ΔAPB) = 12

ar (ΔAQC)

In right ΔABC,

AC = 2 2AB + BC

= AB 2 ...(i)(��AB = BC)

Now, ar(ΔAPB) = 34

AB2 ...(ii)

And ar(ΔAQC) = 3

4 AC2

= 3

4 ( )2AB 2

= 3

2 AB2 ...(iii)

Dividing equation (ii) by equation (iii), weobtain

ΔΔ( APB)( AQC)

arar

=

2

2

3AB

43

AB2

= 12

⇒ ar(ΔAPB) = 12

ar(ΔAQC).

Hence proved.

7. Hint:

Extend AD till E such that AD = DE andsimilarly, PM = MNProve that ΔACE ~ ΔPRN

∠1 = ∠2 ...(i)But ∠3 = ∠5, (CPCT)

∠3 = ∠4 and ∠4= ∠6... ∠5 = ∠6 ...(ii)Adding (i) and (ii),

∠1 + ∠5 = ∠2 + ∠6⇒ ∠BAC = ∠QPR... ΔABC ~ ΔPQR. (By SAS)

8. Let us take two similar triangles ABC andPQR such that ΔABC ~ ΔPQR.

...ABPQ

= BCQR

= CARP

...(i)

We need to prove

ΔΔ

( ABC)( PQR)

arar =

2

2ABPQ =

2

2BCQR =

2

2CARP

Let us draw AM ⊥ BC and PN ⊥ QR.� ΔABC ~ ΔPQR∴ ∠B = ∠Q ...(ii)In ΔABM and ΔPQN,

∠B = ∠Q [From (ii)]and ∠M = ∠N (Each 90°)∴ ΔABM ~ ΔPQN (AA criterion)

∴ABPQ =

AMPN ...(iii)

63���� �� �

⇒ OBOD

= OCOA

= BCDA

= 2

⇒ BC = 2DA = 2 × 4 = 8 cm.

2. (D) Let the given areas be 2x and 3x.Required ratio = 2 : 3x x = 2 : 3 .

3. x = 11 or 8

Hint: Use ODOB

= OCOA

.

4. True.Geometrical figures which are equiangulari.e., if corresponding angles in twogeometrical figure are same, are similar.

5. In right ΔADC,AD2 = AC2 – CD2

= (2CD)2 – CD2

[� AC = BC = 2 CD]= 3 CD2.

6. Hint: BMDN is a rectangle.�BMD ~ �DMC

⇒ DNDM

=DMMC

⇒ DM2 = DN × MC

Also,��BND ~ �DNA.

⇒ DMDN

=DNAN

⇒ DN2 = DM × AN.

7. Let BE = 3x and EC = 4x.In �BCD, GE || DC∴ �BGE ~ �BDC

∴ BEBC

= GEDC

⇒ 33 + 4

xx x

= GE2AB

(∴ DC = 2 AB)

⇒ GE = 67

AB ...(i)

Similarly, �DGF ~ �DBA

⇒ FGAB

= 47

⇒ FG = 47

AB ...(ii)

Adding equations (i) and (ii), we get

GE + FG = 67

AB + 47

AB

From equations (i) and (iii), we have

AMPN =

BCQR ...(iv)

Now, ar(ΔABC) = 12

× base × height

= 12

× BC × AM

And ar(ΔPQR) = 12

× QR × PN

Therefore, ΔΔ

( ABC)( PQR)

arar

= BC × AMQR × PN

= 2

2BCQR

...(v)

[Using (iv)]From results (i) and (v), we arrive

ΔΔ

( ABC)( PQR)

arar

= 2

2ABPQ

= 2

2BCQR

= 2

2CARP

.

Hence the result.Further, consider the question in thefollowing figure.

∠ABO = ∠CDO and ∠BAO = ∠DCO(Alternate angles)

⇒ ΔAOB ~ ΔCOD (AA rule)

⇒ΔΔ

( AOB)( COD)

arar

= 2

2ABCD

⇒ ar(ΔCOD) = 84 × ⎛ ⎞⎜ ⎟⎝ ⎠

212

⎛ ⎞=⎜ ⎟⎝ ⎠�

CD 1AB 2

= 21 cm2.

�������� – ��

1. (A) ΔOBC ~ ΔODA (AA criterion)

64 �� � � � � � � � � �

⇒ EF = 107

AB

⇒ 7 EF = 10AB. Hence proved.

8. Hint: Let AB = BC = AC = a

Draw AE ⊥ BC

⇒ BE = EC = 12

a

and BD = 13

BC =13

a

∴ Using Pythagoras TheoremAD2 = AE2 + DE2

= AE2 + (BE − BD)2

⇒ AD2 = AE2 + BE2 + BD2 − 2.BE.BD

⇒ AD2 = AE2 + EC2 + ⎛ ⎞⎜ ⎟⎝ ⎠

213

a

− 2 ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 12 3

a a

= AC2 + 2 2

9 3a a

�

= a2 + 2 2

9 3a a

�=

279a

⇒ 9AD2 = 7AB2.

9. See Worksheet – 39, Sol. 9 (1st part).

Hint: 2nd Part:AC2 = AD2 + DC2

= AD2 + (3BD)2

= AD2 + 9BD2

= AD2 + BD2 + 8BD2

= AB2 + 8 21

BC4

� �� �� �

⇒ 2AC2 = 2AB2 + BC2 .

�������� – �

1 . (C) 294 cm2

Hint: Prove that ΔOBP ~ ΔOAQ.2. 6 cm

Hint: Use AA-similarity to proveΔAOB ~ ΔCOD.

3. Hint: Draw AM ⊥ BCand DN ⊥ BCAs ΔAOM ~ ΔDON

⇒ ΔΔ

( ABC)( DBC)

arar

=

1 BC AM21 BC DN2

� �

� �

= =AM AODN OD

.

4. Hint: Use concept of similarity.

5. Draw AP ⊥ BC

∴ AB2 = AP2 + BP2

= AP2 + (BD + DP)2

⇒ AB2 = AP2 + BD2 + DP2 + 2BD. DP= AD2 + BD (BD + 2DP)

⇒ AB2 − AD2 = BD × CD. [... BP = PC]Hence proved.

6. Hint:

From figure, show x =+ab

a b.

7. In ΔMDE and ΔMCB,

∠MDE = ∠MCB (Alternate angles)

MD = MC (M is mid-point of CD)

∠DME = ∠CMB(Vertically opposite angles)

∴ ΔMDE ≅ ΔMCB, (ASA criterion)⇒ DE = CB (CPCT)

A

B CD E

65���� �� �

⇒ AE – AD = BC⇒ AE = 2BC ...(i) (��BC = AD)Now, in ΔLAE and ΔLCB,⇒ ∠LAE = ∠LCB (Alternate angles)⇒ ∠ALE = ∠CLB

(Vertically opposite angles)∴ ΔLAE ~ ΔLCB (AA criterion)

⇒AEBC

= LEBL

(Corresponding sides)

⇒2BCBC =

ELBL

[Using equation (i)]

⇒ EL = 2BL. Hence proved.OR

Hint:

As AD is median

so, AB2 + AC2 = 2(AD2 + BD2)

⇒ AB2 + AC2 = 22

2 BCAD

4

� �� ��� �

� �� �

⇒ 2(AB2 + AC2) = 4AD2 + BC2 ...(i)Similarly,

2(AB2 + BC2) = 4BE2 + AC2 ...(ii)2(AC2 + BC2) = 4CF2 + AB2 ...(iii)

Add (i), (ii) and (iii),3(AB2 + AC2 + BC2)

= 4(AD2 + BE2 + CF2).8. See Worksheet – 33, Sol. 9 (1st part).

2nd Part: Draw EM || ABM is a point on CB∴ EM || AB

In ΔABC,

⇒CEAE

= CMMB ...(i)

Also in ΔBCD,

DEEB

= CMMB

...(ii)

From (i) and (ii),CEAE

= DEEB

.

�������� – ��

1. (C) BE2 = 34

a2 ⇒ a2 =43

BE2

∴ AB2 + BC2 + CA2

= a2 + a2 + a2

= 3a2

= 243 × BE

3⎛ ⎞⎜ ⎟⎝ ⎠

= 4 BE2.

2. (B)6013

cm

Hint: Use Pythagoras Theorem.

3. x = 4Hint: Use Basic Proportionality Theorem.

4. Hint: In ΔACD and ΔABC,∠A = ∠A

∠ADC = ∠ACB = 90°

⇒ ΔACD ~ ΔABC

⇒ AC2 = AB.AD …(i)

ΔBCD ~ ΔBAC

⇒ BC2 = BA.BD …(ii)

∴ Applying (ii) ÷ (i) gives the result.

5. Let the given parallelogram be ABCDWe need to prove that

AC2 + BD2 = AB2 + BC2 + CD2 + DA2

Let us draw perpendiculars DN on AB andCM on AB produced as shown in figure.

66 �� � � � � � � ��

In ΔBMC and ΔAND,BC = AD (Opposite sides of a ||gm)

∠BMC = ∠AND (Each 90°)CM = DN (Distance between

same parallels)∴ ΔBMC ≅ ΔAND (RHS criterion)⇒ BM = AN ...(i) (CPCT)In right triangle ACM,

AC2 = AM2 + CM2

= (AB + BM)2 + BC2 – BM2

= AB2 + 2AB . BM + BM2

+ BC2 – BM2

= AB2 + BC2 + 2AB. BM ...(ii)In right triangle BDN,

BD2 = BN2 + DN2

= (AB – AN)2 + (AD2 – AN2)= AB2 – 2AB .AN + AN2

+ AD2 – AN2

BD2 = AB2 + DA2 – 2AB.AN⇒ BD2 = CD2 + DA2 – 2AB.BM ...(iii)

[Using (i) and AB = CD]Adding equations (ii) and (iii), we have

AC2 + BD2 = AB2 + BC2 + CD2 + DA2.Hence proved.

6. Hint: AP || QB || RCUse Basic Proportionality Theorem.

7. (i) PQ2 = PR2 + QR2

⇒ (26)2 = (2x)2 + {2(x + 7)}2

⇒ 676 = 4x2 + 4(x2 + 49 + 14x)

⇒ 676 = 4x2 + 4x2 + 196 + 56x

⇒ 8x2 + 56x – 480 = 0

⇒ x2 + 7x – 60 = 0

⇒ x2 + 12x – 5x – 60 = 0

⇒ x(x + 12) – 5(x + 12) = 0

⇒ (x – 5)(x + 12) = 0

⇒ x = 5 or x = – 12 (reject it)

⇒ x = 5

∴ PR = 2 × 5 = 10 km

QR = 2(5 + 7) = 24 km

∴ Before construction of the highway thedistance travelled = 10 + 24 = 34 km

After construction of the highway thedistance travelled = 26 km∴ Distance saved = 34 – 26 = 8 km.

(ii) Pythagoras theorem

(iii) Yes: as it will save time and fuel. Ravi isinnovative in his thoughts, so his rationalityand social responsibility is reflected here.

8. Let us produce AD to J and PM to K so thatDJ = AD and MK = PM.Join CJ and RK.

In ΔADB and ΔJDC,AD = JD, ∠ADB = ∠JDC, BD = CD⇒ ΔADB ≅ ΔJDC

(SAS criterion of congruence)⇒ AB = JC ...(i) (CPCT)Similarly, we can prove that

PQ = KR ...(ii)According to the given conditions, we have

ABPQ

=ADPM

=ACPR

⇒ JCKR

=

AJ2

PK2

=ACPR

[Using (i) and (ii)]

⇒JCKR

=AJPK

=ACPR

⇒ ΔAJC ~ ΔPKR (SSS criterion ofsimilarity)

⇒ ∠JAC = ∠KPR (Correspondingangles)

i.e., ∠DAC = ∠MPR ...(iii)Similarly, we can prove that

∠DAB = ∠MPQ ...(iv)Adding equations (iii) and (iv), we obtain

67���� �� �

∠BAC = ∠QPR ...(v)Thus, in ΔABC and ΔPQR, we have

ABPQ

= ACPR

(Given)

and ∠BAC = ∠QPR [From (v)]Therefore, ΔABC ~ ΔPQR.

(SAS criterion of similarity)Hence proved.

ORΔABE, ΔACE and ΔADE are right angledtriangles right angle at E each.∴ AB2 = AE2 + BE2 ...(i)

AC2 = AE2 + CE2 ...(ii)and AD2 = AE2 + DE2 ...(iii)Adding equations (i) and (ii), we getAB2 + AC2 = 2AE2 + BE2 + CE2

= 2AE2 + (BD – DE)2 + (CD + DE)2

= 2AE2 + BD2 – 2BD × DE + DE2

+ CD2 + 2CD × DE + DE2

= 2AE2 + BD2 – 2BD × DE + DE2

+ BD2 + 2BD × DE + DE2

(... BD = CD)= 2AE2 + 2DE2 + 2BD2

= 2(AE2 + DE2) + 22BC

2� �� �� �

(D is a mid-point of BC)

= 2AD2 + 12

BC2 [Using (iii)]

Hence proved.

�������� – �

1. (D) ΔABC ~ ΔPQR

⇒20h

= 1050

⇒ h = ×50 2010 = 100 m.

2. (A) The ratio of similar triangles is equal tothe ratio of squares of their correspondingaltitudes.

∴10049

= 2

25h

⇒ h2 = ×25 49

100

⇒ h = ×25 49100

⇒ h =×5 710

= 3.5 cm.

3. Altitude AM divides baseBC in two equal parts. Thatis BM = MC = 7 cm.Using Pythagoras TheoremIn right ΔABM,

AM = 2 225 – 7 = +(25 7) (25 – 7)

= ×32 18 = 24 cm.

4. (i) We know that diagonal of a square

= 2 × side

In square AEFG, AF = 2 AG ...(i)

In square ABCD, AC = 2 AD ...(ii)Using equations (i) and (ii), we obtain

AFAG =

ACAD

. ...(iii)

(ii) ∠GAF = ∠DAC (Each 45°)⇒ ∠GAF – ∠GAC = ∠DAC – ∠GAC⇒ ∠CAF = ∠DAG ...(iv)From equations (iii) and (iv), we have

ΔACF ~ ΔADG.(SAS criterion)

5 . Hint: � ∠1 = ∠2

∴ PQ = PR

∴ QRQS

= QTPQ

.

6. Hint: Draw AM ⊥ BC and DN ⊥ BC.7. Hint: Fig.

A

C Ba

bc

p

68 �� � � � � � � � � �

8. Hint: For 1st part: Prove Pythagoras Theorem.For 2nd part: AC2 – AB2

= (AD2 + CD2) – (AD2 + BD2)

9. Hint: Let the DC = AB = x

Then QC = 45

x and AP = 35

x

ΔQRC ~ ΔPRA.

OR

See Worksheet – 44, Sol. 5.

��������������� �

1. (C) In ΔABC, PQ || BC

∴ APBP

= AQQC

∴2.4BP

= 23

⇒ BP = 3.6 cm

∴ AB = AP + BP = 2.4 + 3.6 = 6 cm.

2.ΔΔ

( ABC)( DEF)

arar

= 2

2BCEF

⇒94

= ⎛ ⎞⎜ ⎟⎝ ⎠

2BCEF

⇒ BCEF

= 32

.

3. Draw AM ⊥ BC and DN ⊥ BC� ∠AMO = ∠DNO = 90°and ∠AOM = ∠DON

∴ ΔAMO ~ ΔDNO(AA similarity)

∴AMDN

= AODO

...(i)

Now, ΔΔ

( ABC)( DBC)

arar =

×

×

1BC × AM

21

BC × DN2

= AODO

. [Using (i)]

Therefore � �

� �

��

�

ABC AODBC DO

arar Hence proved.

4. True, because ΔBCD ~ ΔCAD⇒ CD2 = BD .AD.

5. PQ || BC and AB is transversal∴ ∠APQ = ∠ABC ...(i)

(Corresponding angles)In ΔABC and ΔAPQ,

∠BAC = ∠PAQ (Common)∠ABC = ∠APQ [From (i)]

so, by AA criterion of similarity,

ΔABC ~ ΔAPQ

∴ΔΔ

( ABC)( APQ)

arar =

2

2ABAP

Subtracting unity from both the sides, wehave

⇒ Δ Δ

Δ( ABC) – ( APQ)

( APQ)ar ar

ar =

2

2ABAP

– 1

⇒ Δ

(trapezium BPQC)( APQ)

arar

= ⎛ ⎞⎜ ⎟⎝ ⎠

2ABAP

– 1

...(ii)It is given that

APPB

= 12

∴ PBAP

= 2

∴PBAP

+ 1 = 2 + 1

⇒PB + AP

AP= 3 ⇒

ABAP

= 3

∴⎛ ⎞⎜ ⎟⎝ ⎠

2ABAP

= 9 ...(iii)

From equations (ii) and (iii), we have

69���� �� �

Δ( APQ)(trapezium BPQC)

arar =

18

⇒ ar(ΔAPQ) : ar(trapezium BPQC)

= 1 : 8.6. See Worksheet – 36, Sol. 6.7. See Worksheet – 33, Sol. 9 (1st part).8. We are given two triangles ABC and PQR

such that ΔABC ~ ΔPQR.Draw perpendiculars AD and PM on BCand QR respectively.

We need to prove

ΔΔ

( ABC)( PQR)

arar

= 2

2ADPM

In ΔABD and ΔPQM,

∠ADB = ∠PMQ = 90°

∠ABD = ∠PQM (... ΔABC ~ ΔPQR)

∴ ΔABD ~ ΔPQM(AA criterion of similarity)

⇒ABPQ =

ADPM

...(i)

(Corresponding sides)We know that the ratio of areas of twosimilar triangles is equal to ratio of squaresof their corresponding sides

∴ ΔΔ

( ABC)( PQR)

arar

= 2

2ABPQ

...(ii)

From equations (i) and (ii), we have

ΔΔ

( ABC)( PQR)

arar

= 2

2ADPM

.Hence proved.

��������������� �

1. (A) ΔPQR ~ ΔCAB,

⇒ PQCA

= PRBC

= QRAB

orABQR

= BCPR

⇒ CAPQ

.

2.5x

= 6

6 + 2 ⇒ x =

308

, ⇒ x = 154

cm.

3. In ΔABC, LM || AB.Using Basic Proportionality Theorem, wehave

ACAL

= BCBM

⇒ 2– 3x

x=

+2 3– 2

xx

⇒ 2x2 – 4x = 2x2 + 3x – 6x – 9⇒ – x = – 9 ⇒ x = 9.

4. False, because ∠QPR ≠ 90°.

5. Let the given right angled triangle be ABCwith ∠C = 90° such that AC = b, BC = a andAB = c.

Using Pythagoras Theorem, we have

AB2 = AC2 + BC2

⇒ c2 = b2 + a2 ...(i)Area of equilateral triangle drawn on sideBC

= 34

a2 ...(ii)

Similarly, areas of equilateral trianglesdrawn on side AC and side AB arerespectively

= 34

b2 ...(iii)

and =34

c2 ...(iv)

Sum of areas of equilateral triangles drawnon the sides BC and AC

= 34

a2 + 34

b2

[Adding (ii) and (iii)]

70 �� � � � � � � ��

= 34

(a2 + b2)

= 34

c2 [Using (i)]

= Area of equilateral triangle drawn on hypotenuse AB.

Hence proved.

6.

� ΔABC ~ ΔPQR

∴ABPQ

= BCQR

=

1BC

21

QR2

⇒ ABPQ

= BDQM

...(i)

(a) In ΔABD and ΔPQM,

⇒ ABPQ

= BDQM

[From (i)]

∠ABD = ∠PQM (�ΔABC ~ ΔPQR)So, by SAS criterion of similarity, we have

ΔABD ~ ΔPQM

⇒ ABPQ

= ADPM

(b) � ΔABD ~ ΔPQM, [From part (a)]⇒ ∠ADB = ∠PMQ⇒ 180° – ∠ADC = 180° – ∠PMR

[From figure]⇒ ∠ADC = ∠PMR. Hence proved.

7. Converse of Pythagoras Theorem: In atriangle, if square of one side is equal to thesum of the squares of the other two sides,then the angle opposite the first side is aright angle.

Proof: We are given a ΔABC in which

AC2 = AB2 + BC2 ...(i)

We need to prove ∠ABC = 90°.

Let us construct a ΔPQR such that∠PQR = 90°

and PQ = AB ...(ii)QR = BC ...(iii)

Using Pythagoras Theorem in ΔPQR, wehave

PR2 = PQ2 + QR2

∴ PR2 = AB2 + BC2 ...(iv)[Using equations (ii) and (iii)]

From equations (i) and (iv), we haveAC = PR ...(v)

Now, in ΔABC and ΔPQR,AB = PQ (From (ii))BC = QR [From (iii)]AC = PR [(From (v))]

So, ΔABC � ΔPQR (SSS congruence)∴ ∠ABC = ∠PQR (CPCT)But ∠PQR = 90° (By construction)∴ ∠ABC = 90°. Hence proved

8. BD = BE ...(i) (Given)In ΔOBD, AF ⊥ OB and BD ⊥ OB∴ AF || BD∴ ΔOAF ~ ΔOBD

⇒ OAOB

= AFBD

⇒ OAOB

= AFBE

...(ii) [Using (i)]

In ΔAFC and ΔBEC,∠FAC = ∠EBC (Each 90°)∠FCA = ∠ECB

(Vertically opposite angles)

71���� �� �

So by, AA criterion of similarity,ΔAFC ~ ΔBEC

⇒ACBC

= AFBE

...(iii)

Comparing equations (ii) and (iii), we haveOAOB

= ACBC

⇒OAOB

= OC – OAOB – OC

⇒ OA × OB – OA × OC= OB × OC – OB × OA

⇒ (OA + OB) × OC= 2OA × OB

Dividing both sides by OA × OB × OC, weget

1OA

+ 1

OB=

2OC

. Hence proved.

���������

1. (B) BC = 2 25 +12 = 13 cm

ΔABD ~ ΔCBA

⇒ABBC

= ADAC

⇒ AD = 5 × 12

13 = 60

13 cm.

2. (D) ΔΔ

1

2 =

2122

PP

= 2

24050

= 1625

⇒ Δ1 : Δ2 = 16 : 25.

3.ADDB

= AEEC

⇒ 1.53

= 1

EC

⇒ EC = 31.5

= 2 cm.

4. Yes.

MQ = PQ – PM

= 15.2 – 5.7 = 9.5 cm

NR = PR – PN

= 12.8 – 4.8 = 8 cm

Now,PMMQ

= 5.79.5

= 0.6

andPNNR

= 4.88

= 0.6

Clearly,PMMQ

= PNNR

⇒ MN || QR.

5. ΔAOB ~ ΔCOD (AAA criterion ofsimilarity)

⇒AOCO

= BODO

(Corresponding sides)

⇒7 – 92 –1

xx

= 9 – 8

3x

x

⇒ 21x2 – 27x = 18x2 – 16x – 9x + 8

⇒ 3x2 – 2x – 8 = 0 ⇒ (x – 2) (3x + 4)

⇒ x = 2 or x = –43

⇒ x = 2. (Negative value rejected)

6. ∴ ΔABE ≅ ΔACD

∴ AB = AC and AE = AD (CPCT)

Consider AB = AC

⇒ AD + DB = AE + EC

⇒ DB = EC ...(i) (... AE = AD)

Also AD = AE ...(ii)

(Proved above)

Dividing equation (ii) by equation (i), wehave

ADDB

= AEEC ...(iii)

72 �� � � � � � � � � �

Hence, in ΔABC

ADDB

= AEEC

⇒ DE || BC (Converse of Basic

Proportionality Theorem)

⇒ ∠ADE = ∠ABC and ∠AED = ∠ACB

⇒ ΔADE ~ ΔABC.

7. Hint:

ΔPAC ~ ΔQBC ⇒ xz

= ACBC

ΔRCA ~ ΔQBA ⇒ yz

= ACAB

.

8. Hint:

Draw MN || AD, passing through O tointersect AB at M and DC at N.

Use Pythagoras Theorem for Δ AOM, ΔBOM,Δ CON and ΔDON.

9. (i) As DE � BC.

∴ ∠1 = ∠2 (Correspondingangles)

∠A = ∠A (Common)⇒ ΔADE � ΔABC (AA-criterion)(ii) As ΔADE � ΔABC

∴ � �

� �

�

�

ADE

ABC

ar

ar =� �

� �2

2AD

AB ⇒

12

= � �

� �2

2AD

AB

[� ar(ΔADE) = ar(DECB) ⇒ 2 ar(ΔADE) =ar(DECB) + ar(ΔADE) ⇒ 2 ar(ΔADE) =

ar(ΔABC) ⇒ � �ar�

�

ADEABC

ar( ) =

12

]

⇒1 AD

=AB2

⇒ 1

2–1 =

ADAB

–1

⇒�1 2

2=

�AD ABAB

= � �� � �� �AB – AD

AB= �

BDAB

⇒� �

�

BD 2 1 2 2 2= =

AB 22 2Hence proved.

(iii) Concept of similarity of two triangles.(iv) Honesty and rationality to divide his

land equally between his two children.��

73���� ������ ��� � � � �� � � � � �

��������

INTRODUCTION TO TRIGONOMETRY

�������� ��

1. (B) sin 3A = cos (A – 26°)⇒ sin 3A = sin {90°– (A – 26°)}⇒ 3A = 90° – A + 26°

⇒ A = 116°

4⇒ A = 29°.

2. 2 sin 2x

= 1 ⇒ sin 2x

=12

⇒ sin 2x = sin

6� ⇒

2x

= 6�

⇒ x = 3π

⇒ x = 60°.

3. sin θ = 2425

⇒ sin2 θ = 224

25⎛ ⎞⎜ ⎟⎝ ⎠

⇒ 1 – sin2 θ = 1 –2

22425

⇒ cos2 θ = 2

2725

⇒ cos θ = 725

Now, tan θ + sec θ =sin cos

θθ

+ 1cosθ

=

2425725

+ 1725

= 247

+ 257

= 497

= 7.

4. tan θ = 2

sin

1 sin

�

� �

.

5. cot 25° + tan 41°.

6. True,

LHS =cos 80sin 10

°°

+ cos 59° cosec 31°

= cos(90 – 10 )

sin10° °

°+ cos 59° cosec (90° – 59°)

= sin 10°sin 10°

+ cos 59° sec 59°

= 1 + cos 59°cos 59°

= 1 + 1 = 2

Hence, the given equation is valid.

7. 24

cot 30° + 2

1sin 60°

– cos2 45°

=( )2

4

3+ 2

1

32

⎛ ⎞⎜ ⎟⎝ ⎠

–2

12

⎛ ⎞⎜ ⎟⎝ ⎠

=43

+43

–12

= 8 8 36

� �

= 136

.

8. sin (x + y) = 1 and cos (x – y) = 3

2

⇒ sin (x + y) = sin 90° and cos (x – y) = cos 30°⇒ x + y = 90° and x – y = 30°

Adding and subtracting, we get respectively

2x = 120° and 2y = 60°

i.e., x = 60° and y = 30°.

9. cosec A = 10

sin A =1

cosec A =

110

cos A = 21 – sin A = 1

1 –10

= 310

tan A =sin Acos A

= 13

cot A =1

tan A = 3

74 �� � � � � � � �� ��

sec A =1

cos A =

103

.

10. Hint: RHS =2 2 sin A 1 – cos A

=1 – cos A 1 – cos A

= (1 – cos A) (1 + cos A)

1 – cos A

OR

Hint: LHS =2

2

sin (1 – 2 sin )cos (2cos – 1 )

θ θθ θ

�

.

�������� �

1. (A) sin (θ�+ 36°) = cos θ⇒ sin (θ�+ 36°) = sin (90°– θ)⇒ θ�+ 36° = 90° –�θ⇒ 2θ = 54°⇒ θ = 27°.

2. (C)Hint: Divide numerator and denominatorby cos θ�

3. sec θ�= 54

⇒� sec2 θ�= 2516

⇒�sec2 θ�– 1 = 2516

– 1

⇒� tan2 θ�= 9

16� ⇒ �tan θ�=

34

.

4. Hint: ∠A = 30º, ∠B = 90º, ∠C = 60º .

5. sec θ�+ tan θ�= (1 sin )1 sin

cos cos cos � ��

� �� � �

=2 2 2 2

2

11 sin

1 sin1

ab ab

a b ab

�� � �

� �

� � ��

= b a b a

b ab + a b a

� ��

��

.

6. sin A = 725

, cos A = 2425

,

sin C = 2425

and cos C = 725

.

7. cos 60° + sin 30° cot 30°tan 60° + sec 45° cosec 45°

�

�

=

1 1 + – 3

2 23 2 – 2�

= 1– 3 3

3 3�

= 3 – 33

.

8. Given expression

=cot tan (90° ) sec (90° ) cosec

sin cos (90° ) + cos sin (90° )� � � � � � �

� �� � ��

= cot cot cosec cosec

sin sin + cos cos � � � � �

� � � �

= 2 2

2 2

cot cosec

sin + cos

� � �

� �

= cosec2 θ – 1 – cosec2 θ(��sin2 ��+ cos2 � = 1)

= – 1.9. Draw Δ ABC with

AB = BC = AC = a (say)Draw AD ⊥ BC

∴ ∠BAD = ∠DAC = θ = 30°and BD = DC = a/2

∴ sin θ = BD /2AB

aa

= =12

� sin 30° = 12

.

10. LHS =

1tan tan

1 1 tan1tan

� ��

� ��

�

=� �

2tan 1tan 1 tan 1 tan

��

�� � � �

=� �

21 1tan

tan 1 tan� �

� �� �� � �� �

75���� ������ ��� � � � �� � � � � �

=31 tan 1

tan 1 tan

� �� �� �

� � �� �

=2(tan 1)(tan + 1 + tan )1

tan 1 tan

⎡ ⎤θ− θ θ⎢ ⎥θ− θ⎢ ⎥⎣ ⎦

=2tan 1

1tan

� ��

� =

2sec1

tan�

��

= 1 + sec θ cosec θ = RHS

OR

LHS =cot A – cos Acot A + cos A

=

cos Acos Asin A

cos Acos A

sin A

�

�

=

1cos A – 1

sin A1

cos A 1sin A

� �� �� �� ��� �� �

=

11

sin A1 + 1

sin A

�

=cosec A – 1cosec A + 1

= RHS

�������� ��

1. (A) Hint: tan 5º = cot 85º; tan 25º = cot 65º.2. (D) Hint: (1 + sin �) (1 – sin �)= cos2 �

= 2 1

sec �.

3. 8 tan x = 15 ⇒ tan2 x =22564

⇒ sec2 x – 1 = 22564

⇒ sec2 x =28964

⇒ sec x =178

⇒ cos x = 817

Now, sin x – cos x = 21 – cos x – cos x

=64 8

1 – –289 17

=15 – 8

17

=7

17.

4. Hint: sec � = cosec 60°⇒ cos � = sin 60° ⇒ � = 30°

... 2cos2 30°– 1 = 32× 1

4�

=3

12� =

12

5. Hint: sec 4A = cosec (90º – 4A).

6. Hint: cos (90º – �) = sin �� sin (90º – �) = cos ��

7.(1 sin )(1 – sin )(1 cos )(1 – cos )

+ θ θ+ θ θ

= 2

21 – sin1 – cos

θθ

= 2

2cossin

θθ

= cot2 θ = 27

8⎛ ⎞⎜ ⎟⎝ ⎠

= 4964

.

OR2 2

2 2cosec + cot cosec – sec

θ θθ θ

=� �

2 2

2 2

1 cot cot

1 cot 1 tan

� � � �

� � � � �

=2

2 21 2cot

cot – tan+ θ

θ θ

=1 2 3

13 –

3

� �

=783

=218

.

8. sin2 30° + sin2 45° + sin2 60° + sin2 90°

= 21

2⎛ ⎞⎜ ⎟⎝ ⎠

+2

12

⎛ ⎞⎜ ⎟⎝ ⎠

+2

32

⎛ ⎞⎜ ⎟⎝ ⎠

+ ( )21

= 1 1 3

+ +4 2 4

+ 1

= 1+ 2 + 3 + 4 10

=4 4

= 52

.

9. Hint:

LHS = 1 sin 1 sin ×

1 + sin 1 sin � � � �

� � �

= 2(sec tan )� � �

ORHint:

LHS =2cos A sin A

1 tan A sin A cos A�

� �

76 �� � � � � � � �� ��

=2 2cos A sin A

–cos A sin A cos A – sin A�

=2 2cos A sin A

cos A sin A�

�

= cos A + sin A.

10. Hint: 1 – 2sin2 � = 2cos2 � – 1 = cos2 ��– sin2 ���

�������� ��

1. (A) tan ��= 34

= Perpendicular

Base

BC = 2 23 4�

= 25 = 5

cos θ �Base

Hypotenuse�=

45

∴

41 1 cos 15= =

41 + cos 91 + 5

�

� �

��

2. (C)Hint:1 + tan θ + sec θ

1 + cot + cosec 1= 1 + + sec =

cot cot� �

�� �

3. Hint: A + B = 90º; A – B = 30º.

4. tan 2θ = cot (θ�+ 9°)

⇒ tan 2θ = tan [90° – (θ�+ 9°)]

⇒ 2θ = 90° –�θ – 9° ⇒ 3θ�= 81°

⇒ θ = 27°.

5.2

coscot

1 cos

�� �

� �

.

6. TrueHint: A6 + B6

= (A2 + B2) [(A2 + B2)2 – 3A2 B2].7. Hint:

LHS =1 2sin cos + 1 + 2sin cos

(sin + cos )(sin cos )− θ θ θ θ

θ θ θ − θ�

� �

.

8. LHS = cosA sinA 1cosA sin A 1

� �

� �

Dividing numerator and denominator bysin A, we get

= cotA 1 cosecAcotA 1 cosecA

− ++ −

= ( ) ( )2 2cot A + cosecA cosec A – cot A

cot A 1 cosec A

−

+ −

= (cosecA + cot A) [1 cosecA + cot A]

cot A cosecA + 1�

�

= cosec A + cot A = RHS.

9. Given expression

= 2sin 68cos22

�

�

– 2cot155 tan 75

°°

–3 tan 45° tan 20° tan 40° tan 50° tan 70

5°

= 2 sin (90° 22°)cos 22

�

�

– 2 cot (90° 75°)

5 tan 75°�

–

( ) ( )3×1×tan 90 – 70 tan 90 – 50

tan 50 tan 705

° ° ° °° °

=2cos 22cos 22

°°

–2 tan 755tan 75

°°

–3cot 70 cot 50 tan 50 tan 70

5° ° ° °

= 2 – 25

–

1 13 × × tan 50 tan 70

tan70 tan 505

° °° °

= 2 – 25

– 35

=10 2 3

5� �

= 55

= 1 .

10. Given expression

= 8 3 cosec2 30°. sin 60°. cos 60°. cos2 45°. sin 45°. tan 30°. cosec3 45°.

77���� ������ ��� � � � �� � � � � �

= 8 3 × 21

sin 30°. sin (90° – 30°).

cos (90° – 30°) cos2 (90° – 45°). sin 45°.

sin 30cos 30

°° 3

1sin 45°

= 8 3 × 21

sin 30°× cos 30°. sin 30°. sin2 45°.

sin 45°. 3

sin 30 1cos 30 sin 45

° ×° °

= 8 3 ×2

sin 30°. sin 30°sin 30°

⎛ ⎞⎜ ⎟⎝ ⎠

× cos 30cos 30

°°

×2

3sin 45° sin 45°

sin 45°

= 8 3 × 1 × 1 × 1 = 8 3 .

OR

Hint: sec2 θ = 22

1 1216

xx

� �

∴ sec2 θ – 1 = 22

1 1216

xx

� �

∴ tan2 θ = 21

4x

x� ��� �� �

⇒ tan θ = ± 14

xx

� ��� �� �

.

�������� ��

1. (B) As sin A = 34

,

let BC = 3x and CA = 4x

∴ AB = ( )2 24 – (3 )x x = 7 x

Now, tan A = BCAB

= 37xx

= 37

.

2. (C) 22 tan 30

1 tan 30°

+ ° = 2

12

3

11

3

×

⎛ ⎞+ ⎜ ⎟⎝ ⎠

=

23

43

= 3

2.

3. Hint: tan x = 158

⇒ sin x = 1517

, cos x = 817

∴ sin2 x – cos2 x = 225289

– 64289

= 161289

.

4. sin 30 tan 45 – cosec 60sec 30°+ cos 60° + cot 45°

° + ° °

=

1 21 –

2 32 1

123

+

+ + = 3 2 3 – 4

4 3 2 3+

+ +

=3 3 – 43 3 4+

× 3 3 – 43 3 – 4

= 27 16 – 24 327 – 16

+ = 43 – 24 311

.

5. � A + B + C = 180º

∴ LHS = C + A 180º – B cot = cot

2 2

= cot (90º – B2

) = tan B2

= RHS.6. Yes.

Hint: Both sides =725

.

7. LHS = (cosec A – sin A) (sec A – cos A)

=1 1

– sin A – cos Asin A cos A

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=21– sin A

sin A ×

21 – cos Acos A

=2 2cos A sin A

sin A cos A×

= sin A cos A ... (i)

78 �� � � � � � � �� ��

RHS =1

tan A cot A+ =

1sin A cos Acos A sin A

+

= 2 2sin Acos A

sin A cos A+= sin A cos A ...(ii)

(��sin2A + cos2A = 1)From equations (i) and (ii), we obtainLHS = RHS.

8. 7 sin2 θ + 3(1 – sin2 θ) = 4Let sin θ = x∴ 7x2 + 3 – 3x2 = 4

⇒ 4x2 = 1 ⇒ x2 = 14

⇒ x = ± 12

∴ sin θ = 12

or sin θ = 1

2�

sin θ = –12

is not possible as θ is acute.

⇒ cosec θ = 2 ∴ cos θ = 3

2

∴ sec θ + cosec θ = 2

23� . Hence proved

9. 1Hint: cos (40º + θ) = sin {90° – (40° + θ)}

= sin (50° – θ)and cos 40° = sin 50°.

10. LHS = m2 – n2 = (tan θ + sin θ)2

– (tan θ – sin θ)2 = 4sin θ tan θ ...(i)

RHS = 42

22

sin= 4 – sin

cosmn

θ θθ

�

�

= 24 sin sec –1θ θ

= 4 sin θ tan θ .... (ii)From (i) and (ii), LHS = RHS.

�������� ��

1. (A) Required value = 25 64 36 8

2 –100 100 6

⎛ ⎞+ ×⎜ ⎟⎝ ⎠

= 25 × 1

300 (192 + 216 – 400)

=1

12 × 8 =

23

.

2. (C) sin θ = 21 – cos θ = 1 0.36�

= 0.8

And tan θ = sincos

θθ

= 0.80.6

= 43

Now, 5 sin θ – 3 tan θ = 5 × 0.8 – 3 ×43

= 0

3. Hint: Divide numerator and denominatorby sin A.

31 +1+ cot A 4=

31 – cot A 1 –4

.

4. sec A =23

⇒ sec A = sec 30° ⇒ A = 30°

⇒ A + B = 90° ⇒ B = 90°– 30° = 60°

Now, cosec B = cosec 60° = 23

5. Given expression

= ( )

( )

222

2

1 23 + 4 3 5 0

2 3

2 2 – 3

⎛ ⎞⎛ ⎞ + + ×⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

+

= 3 2 4

4 – 3+ +

= 9.

6. FalseHint: �A = 30º, �B = 60º.

7. sec – cosec sec + cosec

θ θθ θ

=

1 1 –

cos sin 1 1

+ cos sin

θ θ

θ θ

= sin cos sin + cos

� � �

� �

=� �

� �

1sin cos

sin1

sin + cos sin

� � � ��

� � ��

=1 – cot 1 + cot

θθ

=

31 –

43

1 + 4

= 17

.

Hence proved.

79���� ������ ��� � � � �� � � � � �

8. cos sin cos + sin

� � �

� �=

cot – 1cot + 1

θθ

.

(Dividing numerator and denominator bysin θ)

=– 1

1

pqpq +

=–p q

p q+.

9. Given expression

= 2 2sin 35° cos 55°

cos 55° sin 35°⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

– 2cos 60°

= ( )2sin 35°

cos 90° 35°⎧ ⎫⎨ ⎬−⎩ ⎭

+ ( )

2cos 55°sin 90° 55°

⎧ ⎫⎨ ⎬−⎩ ⎭

– 2cos 60°

= 2sin 35°

sin 35°⎛ ⎞⎜ ⎟⎝ ⎠

+ 2cos 55°

cos 55°⎛ ⎞⎜ ⎟⎝ ⎠

– 2cos 60°

= 1 + 1 – 2 cos 60°

= 1

2 22

� � = 2 – 1 = 1

ORGiven expression

= cos 58°sin 32°

+sin 22°cos 68°

–cos 38°cosec 52°

tan 18° tan 35° tan 60° tan 72° tan 55°

= ( )cos 58°

sin 90° – 58° +

( )sin 22°

cos 90° – 22°

– ( )

( )( )

cos 38°cosec 90° – 38°tan 18° tan 35° tan 60° tan 90° – 18°

tan 90° – 35°

= cos 58°cos 58°

+ sin 22°sin 22°

– cos 38° sec 38°

tan 18° tan 35° tan 60° cot 18° cot 35°

= 2 –

1cos 38° × × tan18 ×tan 35

cos 38°tan 18° tan 35° tan 60°

° °

= 2 – 1

tan 60°

= 2 – 13

= 2 3 – 1

3 ×

33

= 6 – 3

3.

10. tan A = n tan B

⇒ cot B = tanA

n and sin A = m sin B

⇒ sin B = 1sin A

m

⇒ cosec B = sin A

m

... cosec2 B – cot2 B = 1

⇒2 2

2 2sin A tan Am n− = 1

⇒2 2 2

2cos A

sin Am n� = 1

⇒ m2 – 1 = (n2 – 1) cos2 A

⇒2

21

– 1mn

� = cos2 A.

Hence proved

OR

Consider an equilateral tri-angle PQR in which PS ⊥ QR.Since PS ⊥ QR so PS bisects∠P as well as base QR.We observe that ΔPQS is aright triangle, right-angled atS with ∠QPS = 30° and ∠PQS = 60°.For finding the trigonometric ratios, we needto know the length of the sides of thetriangle. So, let us suppose PQ = x

Then , QS = 12

QR = 2x

and (PS)2 = (PQ)2 – (QS)2

= x2 – 2

4x

= 23

4x

∴ PS = 32

x

80 �� � � � � � � �� ��

(i) cos 60° =QSPQ

= 2x

x =

12

(ii) sin 60° =PSPQ

=

32

x

x =

32

(iii) tan 30° =QSPS

= 232

x

x =

13

.

�������� ��

1. (B) b2x2 + a2y2 = b2a2 cos2θ + a2b2 sin2 θ= a2b2.

2. (A) A = 90° – 60° = 30°∴ cosec A = cosec 30° = 2.

3. (C) tan θ = 125

⇒ 1 + tan2 θ = 1 +2

2125

⇒ sec θ = 135

Now, 1 + sin 1 – sin

θθ

=

1 + sin cos

1 sin cos

�

�

� �

�

= sec tansec – tan

θ + θθ θ

=

13 125 5

13 12–

5 5

�

=

25515

= 25.

4.sin 29°cos 61°

⎛ ⎞⎜ ⎟⎝ ⎠

+ 2cos 27°

sin 63°⎛ ⎞⎜ ⎟⎝ ⎠

– 4cos2 45°

= � �

sin 29°cos 90° 29°�

+ ( )

2cos 27°

sin 90° 27°

⎧ ⎫⎪ ⎪⎨ ⎬−⎪ ⎪⎩ ⎭

– 4 × 2

12

⎛ ⎞⎜ ⎟⎝ ⎠

= 1 + 12 – 42

= 0.

5. Given expression

= 44 21 1

2 2

⎧ ⎫⎪ ⎪⎛ ⎞ ⎛ ⎞+⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭ – 3 � �

221

12

� �� �� ��� � � �� � �

–2

32

⎛ ⎞⎜ ⎟⎝ ⎠

= 41 1

+16 4

⎛ ⎞⎜ ⎟⎝ ⎠

– 31

– 12

⎛ ⎞⎜ ⎟⎝ ⎠

–34

= 14

+ 1 – 32

+ 3 – 34

= 174

– 94

= 84

= 2.

6.1

tan A +

sin A1 + cos A

= cos Asin A

+ sin A

1 + cos A

=� �

2 2cos A + cos A + sin Asin A 1 + cos A

= ( )1 + cos A

sin A 1 + cos A = cosec A

= 2.

7. � sin θ = 34

∴ cosec θ = 43

� cos θ = 21 – sin θ = 91–

16 =

74

∴ sec θ = 47

and cot θ = 73

Now, LHS

= 2 2

2cosec – cot

sec – 1θ θ

θ=

16 7–

9 916

– 17

=

9997

= 79

= 73

= RHS. Hence proved.

81���� ������ ��� � � � �� � � � � �

8. Hint: LHS = 11 cos A

–sin A sin A

– 1sin A

=sin A

1 – cos A –

1sin A

=( )

2sin A – 1 + cos Asin A 1 – cos A

=( )

21 – cos A – 1 + cos Asin A 1 – cos A

=( )( )

cos A 1 – cos A

sin A 1 – cos A = cot A.

OR

Using a3 + b3 = (a2 + b2 – ab) (a + b), we get3 3sin cos .sin cos

sin cosθ + θ

+ θ θθ + θ�

�

=2 2 .(sin cos )(sin cos sin cos )

sin cosθ + θ θ + θ − θ θ

θ + θ+ sin � ��cos �

= . .1 sin cos sin cos 1− θ θ + θ θ = .

9. ( )( )( )( )

2 + 2 sin 1 – sin 1 + cos 2 – 2 cos

θ θθ θ

= ( )( )( )( )

2 1 + sin 1 – sin 2 1 + cos 1 – cos

θ θθ θ

= 2

21 – sin1 – cos

θθ

= 2

2cossin

θθ

= cot2 θ

= 215

8� �� �� �

= 22564

.

10. Hint: p2 – 1 = sec2 θ + tan2 θ + 2 sec θ tan θ – 1= 1 + tan2 θ + tan2 θ + 2 sec θ tan θ – 1= 2 tan θ (tan θ + sec θ)Similarly p2 + 1 = 2 sec θ (tan θ + sec θ).

�������� ��

1. (B)Hint: x + y = 2 cot A

x – y = 2 cos A

∴2

x yx y

⎛ ⎞−⎜ ⎟+⎝ ⎠

= sin2 A

and2

2x y−⎛ ⎞

⎜ ⎟⎝ ⎠= cos2 A

∴ sin2 A + cos2 A = 1.

2. (A) 5

Hint: (x + 1)2 = x2 + 52

3. (A) tan A = 13

= tan 30°

⇒ ∴ ∠A = 30°

∴ ∠C = 180° – ∠A – ∠B = 180° – 120°

= 60°Now, sin A cos C + cos A sin C= sin 30° cos 60° + cos 30° sin 60°

= 12

× 12

+ 3

2 ×

32

= 1.

4. cos α= 12

⇒ cos α = cos 60°

⇒ α = 60°

tan β =13

⇒ tan β = tan 30°

⇒ β = 30°.

Now, sin (α + β) = sin (60° + 30°) = sin 90° = 1.

5. tan 1° tan 2°.... tan 43° tan 44° tan 45°tan 46° tan 47°..... tan 88° tan 89°

= (tan 1° tan 89°)(tan 2° tan 88°)....(tan 43° tan 47°)(tan 44° tan 46°) tan 45°

= (tan 1° cot 1°)(tan 2° cot 2°)....(tan 43°

cot 43°)(tan 44° cot 44°) tan 45°

= (1) × (1) × .... × (1) × (1) × tan 45°

= (1 × 1 ×.... × 1 × 1) × tan 45°

= 1 × 1 = 1.

6. Given expression

= tan 50° + sec 50°

cot 40° + cosec 40° + cos 40° cosec 50°

82 �� � � � � � � �� ��

= tan 50° + sec 50°

cot (90° 50°) + cosec (90° 50°)� �

+

cos 40°cosec (90° – 40°)

= tan 50° + sec 50°tan 50° + sec 50°

+ cos 40°. 1

cos 40°

= 1 + 1 = 2.

7. LHS = tan (A – B) = tan (60° – 30°) = tan 30°

= 13

.

RHS =tan A tan B

1 + tan A tan B�

�

=tan 60° tan 30°

1 tan 60° tan 30°�

� �

=

13 –

31

1 + 3 . 3

=

3 – 13

1 1+ =

23

2

= 13

= LHS. Hence verified.

8. RHS = 6 2

6 2 2 sin 3 sin 1.+ + 1cos cos cos

θ θθ θ� �

� �

= 6 6 2 2

6

sin + cos + 3 sin coscos

� � �� �

�

= � �32 2

6

sin cos

cos

�� �

�

6 2 2= sec = LHS. [ sin +cos = 1]� � ��

ORHint: Numerator ofLHS = tan θ + sec θ – (sec2 θ – tan2 θ)= (tan θ + sec θ) – (tan θ + sec θ) (sec θ – tan θ)= (tan θ + sec θ) (1 – sec θ + tan θ).

9. cos θ + sin θ = 2 cos θSquaring both sides, we getcos2 θ + 2 cos θ sin θ + sin2 θ = 2 cos2 θ⇒ 2 cos2 θ – cos2 θ – 2 cos θ sin θ = sin2 θ⇒ cos2 θ – 2 cos θ sin θ = sin2 θAdding sin2 θ to both sides, we havesin2 θ + cos2 θ – 2 cos θ sin θ = sin2 θ + sin2 θ

⇒ (cos θ – sin θ)2 = 2 sin2 θ

⇒ cos θ – sin θ = 2 sin θ Hence proved.

10. Hint: l tan θ + m sec θ = n ...(i) × l′l′ tan θ – m′ sec θ = n′ ...(ii) × l

⇒ ll′ tan θ + ml′ sec θ = nl′l′ l tan θ – m′ l sec θ = n′ l

– + – (m′ l + ml′) sec θ = nl′ – n′ l

⇒ sec θ = nl n lm l ml

’ ’’ ’�

�

Similarly, tan θ = nm mnlm ml

’ ’’ ’�

�

.

�������� ��

1. (D) Given expression

= 2 2

2 2

cos (90 70 ) cos 70

sec (90 40 ) cot 40

° − ° + °° − ° − °

+ 2 {cosec2 58° – cot 58° tan (90° – 58°)}

= 2 2

2 2sin 70° + cos 70°

cosec 40° – cot 40°+ 2 (cosec2 58° – cot2 58°)

= 11

+ 2(1) = 1 + 2 = 3.

2. (A) sec 5A = cosec (A – 36°)⇒ sec 5A = sec {90° – (A – 36°)}⇒ 5A = – A + 126° ⇒ A = 21°.

3. Given expression

= sin2 5° + sin2 10° ... + sin2 40° + sin2 45°+ sin2 50° + ... + sin2 80° + sin2 85° + sin2 90°

= cos2 85° + cos2 80° + .... + cos2 50° +2

12

⎛ ⎞⎜ ⎟⎝ ⎠

+ sin2 50° + .... + sin2 80° + sin2 85° + (1)2

= (cos2 85° + sin2 85°) + (cos2 80° + sin2 80°)

+ .... + (cos2 50° + sin2 50°) + 12

+ 1

= (1 + 1 + .... 8 terms) + 12

+ 1

= 8 +12

+ 1 = 19

2.

83���� ������ ��� � � � �� � � � � �

4. tan 3x = 12

.12

+12

= 12

+12

= 1

⇒ tan 3x = tan 45° ⇒ x = 453�

= 15°.

5. cosec A = 2 ⇒ sin A = 12

cos A = 21 – sin A = 1

1–2

= 12

tan A = 1, cot A = 1

Now, ( )

2 2

2 2

2 sin A + 3 cot A

4 tan A – cos A =

12 × + 3 × 12

14 1

2� ��� �� �

= 42

= 2.

6. TrueHint:

a cos θ + b sin θ = 4 ...(i) × sin θa sin θ – b cos θ = 3 ...(ii) × cos θ

⇒ a cos θ sin θ + b sin2 θ = 4 sin θa sin θ cos θ – b cos2 θ = 3 cos θ

– + –

b = 4 sin θ – 3 cos θSimilarly, a = 4 cos θ + 3 sin θ

∴ a2 + b2 = 16 sin2 θ + 9 cos2 θ – 12 sin θcos θ + 16 cos2 θ + 9 sin2 θ

+ 12 sin θ cos θ = 16 + 9 = 25.

7. (a2 − b2) sin θ + 2ab . cos θ = a2 + b2

Divide by cos θ

(a2 − b2) tan θ + 2ab = 2 2

cosa b�

�

⇒ (a2 − b2) tan θ + 2ab = (a2 + b2) . sec θ

= (a2 + b2). 21 tan� �

Squaring both sides:

(a2 − b2)2 tan2 θ + 4 a2b2 + 4 ab (a2 − b2) tan θ= (a2 + b2)2 (1 + tan2 θ)

= (a2 + b2)2 + (a2 + b2)2 tan2 θ

[(a2 − b2)2 − (a2 + b2)2] tan2 θ + 4 a2b2 + 4 ab(a2 − b2) tan θ − (a2 + b2)2 = 0

⇒ – 4a2b2 tan2 θ + 4ab (a2 – b2) tan θ − a4 − b4

+ 2a2b2 = 0⇒ − 4a2b2 tan2 θ + 4ab (a2 − b2) tan θ

− (a2 − b2)2 = 0⇒ 4a2b2 tan2 θ − 4ab (a2 − b2)tan θ

+ (a2 − b2)2 = 0⇒ [2ab tan θ − (a2 − b2)]2 = 0⇒ 2ab tan θ = a2 − b2

⇒ tan θ = 2 2

2a b

ab�

.

8. Hint: Use (a2 + b2)3 = a6 + b6 + 3a2b2(a2 + b2).

9. LHS

=( )( )

3 3

1+ cot A + tan A sin A – cos A

sec A – cosec A

=( )

3 3

cos A sin A1 + + sin A – cos A

sin A cos A1 1

–cos A sin A

⎛ ⎞⎜ ⎟⎝ ⎠

=

2 2

2 2

3 3

(sin A cos A + cos A + sin A)(sin A cos A)sin A cos A

(sin A – cos A)(sin A + cos A + sin Acos A)sin A cos A

−

= sin2 A . cos2 A = RHS. Hence proved.

10. m = cosec θ – sin θ = 1sin θ

– sin θ

=21– sin

sinθ

θ =

2cossin

θθ

n = sec θ – cos θ = 1cos�

– cos θ

=21– cos

cos�

� =

2sincos

�

�

Now, LHS = ( ) ( )2 2

2 23 3m n mn+

=

24 2 3

2cos sin

×cossin

⎛ ⎞θ θ⎜ ⎟⎜ ⎟θθ⎝ ⎠

+

22 4 3

2cos sin

×sin cos

⎛ ⎞θ θ⎜ ⎟⎜ ⎟θ θ⎝ ⎠

84 �� � � � � � � �� ��

= ( )2

3 3cos θ + ( )2

3 3sin θ

= cos2 θ + sin2 θ = 1 = RHS.OR

LHS= (1 + cot A – cosec A)(1 + tan A + sec A)

= cos A 1

1sin A sin A

� �� �� �� �sin A 1

1cos A cos A

� �� �� �� �

= sin A + cos A –1sin A

× cos A + sin A + 1cos A

= ( )2 2sin A + cos A – 1

sin A cos A

= 2 2sin A + 2 sin A cos A + cos A – 1

sin A cos A

= 2 sin A cos Asin A cos A

= 2

= RHS. Hence proved.

�������������� �

1. (A) cos 30° = BCAC

⇒ 32

= 10x

⇒ x = 20

3 ⇒ x =

20 33

cm

sin 30° =ABAC

⇒ 12

=3

20 3y ×

⇒ y = 10 33

cm.

2. 25°Hint: tan 2θ = cot (θ + 15°) ⇒ tan 2θ = tan [90°– (θ + 15°)].

3. We know: sin θ = cos (90° – θ) so, givenexpression

= � �2cos(90 35 )

cos 55� � �

�

+ 2cos 55

cos(90 35 )�� �

� �� � �� �

– 2 cos 60°

= 2cos 55

cos 55°⎛ ⎞

⎜ ⎟°⎝ ⎠+

2cos 55cos 55

°⎛ ⎞⎜ ⎟°⎝ ⎠

– 2 × 12

= 1 + 1 – 1 = 1.

4. True, because LHS = tan 60° = 3 and

RHS = 22 tan 30

1 – tan 30°

° =

231

1 –3

= 3 .

5. 2 (sin6 A + cos6 A) – 3 (cos4 A + sin4 A) + 1= 2{(sin2 A + cos2 A)3 – 3sin2 A cos2A

(sin2 A + cos2 A)} – 3{(sin2 A + cos2 A)2 – 2sin2 A cos2 A} + 1

[� (a + b)3 = a3 + b3 + 3ab(a + b) and (a + b)2 = a2 + b2 + 2ab]

= 2(1 – 3sin2 A cos2 A) – 3(1 – 2sin2 Acos2 A) + 1

[� sin2 A + cos2 A = 1]= 2 – 6 sin2 A cos2 A – 3 + 6 sin2 A cos2 A

+ 1 = 0

6. 2sin x + cos y = 1 (Given)⇒ 2sin x + cos y = 20

⇒ sin x + cos y = 0 ... (i)2 2sin cos16 x y� = 4 (Given)

⇒ � �2 2sin cos24

x y� = 4

⇒ � �2 22 sin cos4

x y�

= 41

⇒ 2(sin2 x + cos2 y) = 1 ... (ii)Substituting cos y = – sin x from (i) in (ii),we get

2(sin2 x + sin2 x) = 1 ⇒ sin2 x = 14

⇒ sin x = ± 12

When sin x = – 12

, cos y = 12

When sin x = 12

, cos y = – 12

Hence, sin x = – 12

, cos y = 12

or sin x = 12

,

cos y = – 12

.

7. We know thatsin (90°– θ) = cos θ, tan (90°– θ) = cot θ,

sec(90°– θ) = cosec θ

85���� ������ ��� � � � �� � � � � �

Now, 2 2

2 2sec (90 – ) – cot2(sin 25 sin 65 )

° θ θ° + °

+ 2 2 2

2 22cos 60 tan 28 tan 62

3(sec 43 – cot 47 )° ° °

° °

= ( ){ }

2 2

2 2

sec (90 ) cot

2 sin 25 sin 90 25

° − θ − θ° + ° − °

+( ){ }

2 2 2

2 2

2cos 60 tan 28 tan (90 28 )

3 sec 43 cot 90 43

° ° ° − °

° − ° − °

= � �

2 2

2 2

cosec cot

2 sin 25 cos 25

� � �

� � �

+ 2 2 2

2 2

2cos 60° tan 28 cot 28

3(sec 43 tan 43 )

° °° − °

=

12× ×11 4+

2×1 3×1

= 1 1

+2 6

= 23

.

8. Given equations are: sin θ + cos θ = p ... (i)sec θ + cosec θ = q ... (ii)Squaring both the sides of equation (i),we getsin2 θ + cos2 θ + 2sin θ cos θ = p2

Subtract unity from both the sides to getp2 – 1 = 2sin θ cos θ ... (iii)Equation (ii) can be written as

q =1

cos θ +

1sin θ

⇒ q =sin cossin cos

� � �

� �... (iv)

From equations (iii) and (iv), we get

q (p2 – 1) =sin cossin cos

� � �

� �× 2sin θ cos θ

⇒ q (p2 – 1) = 2(sin θ + cos θ)⇒ q (p2 – 1) = 2p. Hence proved.

�������������� �

1. (A) Given expression= sin 25° cos (90° – 25°) + cos 25°

sin (90°– 25°)= sin2 25° + cos2 25° = 1.

2.2sin – cos2sin cos

θ θθ + θ

=

2sin cos–cos cos

cos2sincos cos

� �

� �

���

� �

= 2 tan – 12tan 1

θθ +

=

42× – 1

34

2× 13�

= 511

.

3.cos 45

sec 30 cosec 30°�

� � =

12

22

3+

=� �

12

2 1 3

3

�

= 12

× ( )

3

2 1 3+

=( )

( )( )3 3 – 1

2 2 3 1 3 – 1

×

+ =

3 – 34 2

×22

= 3 2 – 6

8.

4. False, because cos2 23° – sin2 67° = 0, 0 isnot a positive value.

5. LHS =cos A

1+ sin A +

1 sin Acos A+

=� �

� �

22cos A + 1 + sinA

1 + sin A cosA

=� �

2 2cos A + 1 + sin A + 2sin A1 + sin A cos A

=� �

2 2 sin A1 sin A cos A

�

�

= � �

� �

2 1 sin A

1 sin A cos A

�

�

=2

cos A = 2 sec A

= RHS. Hence proved.

86 �� � � � � � � �� ��

6. Let us construct a triangle ABC in whichAB = BC = AC = a (say). Draw AD ⊥ BC.AD bisects BC

⇒ BD = DC = 2a

AD bisects ∠BAC⇒ θ = 30°In right angled ΔABD.

AD2 = AB2 – BD2 = a2 – 2

2a⎛ ⎞

⎜ ⎟⎝ ⎠

= a2 – 2

4a =

234a

⇒ AD = 3

2a

Now, in Δ ABD,

tan θ = BDAD

⇒ tan 30° = 2

32

a

a

⇒ tan 30° = 13

.

7. ( a2 – b2) sin θ + 2ab cos θ = a2 + b2 (Given)Divide both sides by cos θ to get( a2 – b2) tan θ + 2ab = (a2 + b2) sec θSquaring both sides, we get(a2 – b2)2 tan2 θ + 4a2 b2 + 4ab(a2 – b2) tan θ

= (a2 + b2)2 sec2 θ

⇒ (a2 – b2)2 tan2 θ – (a2 + b2)2 tan2 θ + 4ab

(a2 – b2) tan θ – (a2 + b2)2 + 4a2b2 = 0

(��sec2 θ = 1 + tan2 θ)⇒ – 4a2b2 tan2 θ + 4ab (a2 – b2) tan θ

– (a2 – b2)2 = 0⇒ – 4a2b2 x2 + 4ab (a2 – b2) x – (a2 – b2)2 = 0

where x = tan θThis is a quadratic equation in x.Here, discriminant,

D = � � � �2 22 2 2 2 2 2 2 216 4 4a b a b a b a b� � � �

= 0

∴ x = 2 2 2 2

2 2

4 ( ) 022 ( 4 )

ab a b a baba b

� � ��

�

� �

⇒ tan θ = 2 2

2a b

ab� . Hence proved.

8. Since ABC is a acute angled triangleso, ∠A < 90°, ∠B < 90° and ∠C < 90°.Also ∠A + ∠B + ∠C = 180° ... (i )

sin (A + B – C) = 12

(Given)

⇒ sin (A + B – C) = sin 30°⇒ ∠A + ∠B – ∠C = 30° ... (ii )Similarly, ∠B + ∠C – ∠A = 45° ... (iii )Add equations (ii ) and (iii ) to get

2∠B = 75° ⇒ ∠B = 137

2�

Subtract equation (ii) from equation (i) to get2∠C = 150° ⇒ ∠C = 75°

Subtract equation (iii) from equation (i) to get

2∠A = 135° ⇒ ∠A = 1

672�

Thus, ∠A = 1

672�

, ∠B = 1

372�

and ∠C = 75°.

���������

1. (A) x = sec

2θ

and 1x

= tan

2θ

∴ 22

12 –x

x� �� �� �

= 2 2 2sec tan

–4 4

� �� �� �� �� �

= 2 2sec – tan

24

� �� �� �� �� �

= 12

.

2.� �

2 2

2 2

cos 20° + cos 70

2 sin 59 sin 31

�

� �=

2k

⇒� �

2 2

2 2

sin 70 cos 70

2 sin 59 cos 59

�� �

�� �

= 2k

⇒1 22 k� ⇒ k = 4.

3. sin4 θ + cos4 θ = 1 + 4k sin2 θ cos2 θ⇒ (sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ

= 1 + 4k sin2 θ cos2 θ⇒ 2 sin2 θ cos2 θ (–1 – 2k) = 0

⇒ – 1 – 2k = 0 ⇒ k = – 12

.

87���� ������ ��� � � � �� � � � � �

4. tan θ = 4⇒ tan2 θ + 1 = 42 + 1⇒ sec2 θ = 17

∴ 110

(tan2 θ + 2 sec2 θ) =1

10(16 + 2 × 17)

= 5.5. False.

Suppose A = 30° and B = 60°Then, LHS = tan (A + B) = tan (30° + 60°)

= tan 90°⇒ LHS = undefined .... (i)and RHS = tan A + tan B = tan 30°

+ tan 60°

=13

+ 3 = 1 3

3+

= 43

⇒ RHS = a real number .... (ii)From results (i) and (ii), it is clear that thegiven identity is false.

6. –17

Hint: cos 55° = cos (90° – 35°) = sin 35°cos 70° = sin 20°

and tan 5° = cot 85°.

7. 134

.

Hint: sin 30° = 12

= cos 60°, sin 60° = 3

2,

cos 45° = 12

= sin 45°, sin 90° = 1.

8. sin θ + cos θ = aSquaring both sides.

sin2 θ + cos2 θ + 2 sin θ cos θ = a2

⇒ 2 sin θ cos θ = a2 – 1

⇒ sin θ cos θ = 2 – 12

a... (i)

Now, sin6 θ + cos6 θ = ( )32 2sin cosθ + θ

– 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)

= 13 – 322 – 1

2a⎛ ⎞

⎜ ⎟⎜ ⎟⎝ ⎠(1)

[Using equation (i)]

= 1 – ( )223– 1

4a =

2 24 – 3 ( – 1)4a

.

Hence proved.

9. LHS =( )( )

2

2

sec tan – 1

sec tan 1

θ + θθ + θ +

=2 2

2 2sec + tan + 2 sec tan – 1sec + tan + 2 sec tan + 1

θ θ θ θθ θ θ θ

=( )

( )2 2

2 2

sec 1 + tan + 2 sec tan

sec + tan + 1 2 sec tan

θ − θ θ θ

θ θ + θ θ

=2 2

2 2

tan + tan + 2 sec tan

sec + sec + 2 sec tan

� � � �

� � � �

= 2 tan (tan + sec )2 sec (sec + tan )

θ θ θθ θ θ

=tan sec

θθ

= tan θ cos θ

=sin cos

θθ

. cos θ = sin θ = RHS.

Hence proved.OR

sin A – sin B cos A – cos Bcos A + cos B sin A + sin B

+

=

� �� �� �� �

� �� �

sin A – sin B sin A + sin Bcos A – cos B cos A cos B

cos A + cos B sin A + sin B� �

=� �� �

2 2 2 2sin A – sin B cos A – cos Bcos A + cos B sin A + sin B

�

=� �

� �� �

2 2 2 2sin A cos A) (sin B cos B

cos A + cos B sin A + sin B

� � �

= ( )( )1 – 1

cos A + cos B sin A + sin B

= 0 which is an integer.

��

88 �� � � � � � � � � �

��������

STATISTICS

�������� ��

1. (B) 21.1Hint: 3 Median = Mode + 2 Mean.

2. Since the mode is 7∴ 2k – 1 = 7 ⇒ k = 4.

3. In such case, mean will increase by 3.∴ New mean = 18 + 3 = 21.

4. x = 26

Hint: Mean = i i

i

f xf

�

�.

5. Class Frequency ( f ) Cumulativeinterval Frequency (cf )

0-8 8 88-16 10 18

16-24 16 3424-32 24 5832-40 15 7340-48 7 80

N = 80

For median class, N 80=

2 2= 40

In the cumulative frequency column, 58 isjust greater than 40.

So, 24-32 is the median class.

Here, l = 24, cf = 34, f = 24, N2

= 40, h = 8

Using formula:

Median = l +

N–

2 ×cf

hf

� ��� �� �� �� �� �� ��� �� ��� �

= 24 +40 – 34

× 8 = 2624

� ��� �� ��� �

Hence, median of the given distribution is 26.6. 36.25

Hint: Here maximum class frequency is 32.So, the modal class is 30-40.

Now, l = 30, f1 = 32, f0 = 12, f2 = 20, h = 10Use the formula:

Mode = l + 1 0

1 0 2

–2 – –

f ff f f

� ��� �� ���� �× h.

7. Hint:Production No. of Production yield

yield farms (in kg/ha) cf(in kg/ha) more than or

equal to

50-55 2 50 10055-60 8 55 9860-65 12 60 9065-70 24 65 7870-75 38 70 5475-80 16 75 16

100 80 0∴ For more than ogive, plot following points:(50, 100), (55, 98), (60, 90), (65, 78), (70, 54),(75, 16), (80, 0).

�������� �

1. (B) median 2. 43. The given distribution can be represented as:

Marks obtained No. of students 0-10 510-20 320-30 430-40 340-50 6

More than 50 42

Clearly, the frequency of the class 30-40 is 3.4. Let us rewrite the given table with cumu-

lative frequencies.Class interval f cf

0-5 10 105-10 15 25

10-15 12 3715-20 20 5720-25 9 66

N = 66

89���� ��� ��

� N = 66

∴ N2

= 33

∴ Median class = 10-15Modal class = 15-20

Required sum = 10 + 15 = 25.

5. In the given distribution, maximum classfrequency is 20, so the modal class is 40-50.Here, lower limit of modal class: l = 40Frequency of the modal class: f1 = 20Frequency of the class preceding the modalclass: f0 = 12Frequency of the class succeeding the modalclass: f2 = 11Size of class: h = 10Using the formula:

Mode = l + 1 0

1 0 2

–2 – –

f ff f f

� ��� �� ��� �× h

= 40 +20 – 12

× 102 × 20 – 12 – 11

� ��� �� ��� �

= 40 + 4.70 = 44.70.

Hence, mode of the given data is about 45cars.

6. Let us rewritten the table with class intervals.

Class interval f cf

36-38 0 038-40 3 340-42 2 542-44 4 944-46 5 1446-48 14 2848-50 4 3250-52 3 35

N = 35

We mark the upper class limits on x-axisand cumulative frequencies on y-axis with asuitable scale.

We plot the points (38, 0); (40, 3); (42, 5);(44, 9); (46, 14); (48, 28); (50, 32) and (52, 35).

These points are joined by a free handsmooth curve to obtain a less than type ogiveas shown in the figure.

Figure: Less than type ogive

To obtain median from the graph:

We first locate the point corresponding to

N2

= 352

= 17.5 students on the y-axis. From

this point, draw a line parallel to the x-axis

to cut the curve at P. From the point P, drawa perpendicular PQ on the x-axis to meet itat Q. The x-coordinate of Q is 46.5. Hence,

the median is 46.5 kg.

Let us verify this median using the formula.

Median = l +N2

cf

f

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

× h

= 46 + 17.5 14

14�� �

� �� � × 2

= 46 + 714

= 46 + 0.5

= 46.5 kg.

Thus, the median is the same in bothmethods.

90 �� � � � � � � � � �

�������� – ��

1. (C) mid-points of the classes.

2. (D) We have

Mode = 3 Median – 2 Mean

⇒ 45 = 3 Median – 2 × 27⇒ Median = 33.

3. 25-30

Hint: 5 + 8+ 3+ 2N 9= =

2 2.

4. Required number of athletes= 2 + 4 + 5 + 71 = 82.

5.

Let us assumed mean be a = 52Here, h = 20

Using the formula:

Mean = a + ×i i

i

f uh

f��

⇒ 50 = 52 + 1 2– 2.8 – 1.1 + 0.920

120f f

�

⇒ 1.1 f1 – 0.9 f2 = 9.2 ...(i)

But68 + f1 + f2 = 120

⇒ f1 + f2 = 52 ...(ii)

Solving (i) and (ii), we obtain

f1 = 28and f2 = 24.

6. Let us convert the given data into less thantype distribution.

Class f Lifetimes cfinterval (in hrs.)

0-20 10 less than 20 10

20-40 35 less than 40 45

40-60 52 less than 60 97

60-80 61 less than 80 158

80-100 38 less than 100 196

100-120 29 less than 120 225

7. (i) By making the given data continuous, we get: a = 57, h = 3.

No. of mangoes No. of boxes (fi) Mid-points (xi)i

i

x au

h�

� fiui

49.5-52.5 15 51 – 2 – 3052.5-55.5 110 54 – 1 – 11055.5-58.5 135 a = 57 0 058.5-61.5 115 60 1 11561.5-64.5 25 63 2 50

400if� � 25i if u� �

� Mean = a + h i i

i

f uf

� ��� ��� � = 57 + 3 ×

25400

� �� �� � = 57 +

75400

� 57.19.

(ii) Step devitation method(iii) Vikram Singh believes in quality serving, fruits will remian fresh and free from germs and

flies.

91���� ��� ��

We mark the upper class limits along thex-axis with a suitable scale and the cumu-lative frequencies along the y-axis with asuitable scale. For this, we plot the pointsA(20, 10), B(40, 45), C(60, 97), D(80, 158),E(100, 196) and F(120, 225) on a graph paper.These points are joined by a free handsmooth curve to obtain a less than type ogiveas shown in the given figure.

Figure: Less than type ogive

7. The given distribution can be againrepresented with the cumulative frequenciesas given below:

Class interval fi xi cf fi xi

100-120 12 110 12 1320120-140 14 130 26 1820140-160 8 150 34 1200160-180 6 170 40 1020180-200 10 190 50 1900

50 7260

Mean: Mean = i i

i

f x

f

ΣΣ

� Σ fi = 50 and Σfixi = 7260

∴ Mean =7260

50= 145.20.

Hence, the mean is � 145.20Median:

Median = l + N2

cf

f

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

× h

��N = 50, N2

= 25, f = 14, cf = 12,

l = 120 and h = 20

∴ Median = 120 + 25 1214�� �

� �� �× 20

= 120 + 18.57 = 138.57Hence, the median is � 138.57.Mode:

Mode = l + 1 0

1 0 22f f

f f f�� �

� �� �� � × h

��l = 120, f1 = 14, f0 = 12

f2 = 8 and h = 20

∴ Mode = 120 + 14 12

2 14 12 8�� �

� �� �� � �× 20

= 120 + 408

= 125

Hence, the mode is � 125.

8. C.I. No. of consumers (fi) (c.f.)

65-85 4 4

85-105 5 9

105-125 13 22 = c.f.

125-145 20 = f 42

145-165 14 56

165-185 8 64

185-205 4 68

N = 68

∴N2

= 682

= 34∴ c.f. just greater than 34 is 42.

∴ Median class is 125-145.

92 �� � � � � � � � � �

∴ Median = l +

N– . .

2c f

f× h

= 125 + 34 22

20�

× 20

= 125 + 12 = 137.(ii) 20 + 14 + 8 + 4 = 46 families.

(iii) Since, Mr. Sharma is saving electricity so hisconsumption is less, which means hismonthly bill will also be less. So, he believesin saving and hence is responsible also.

�������� – ��

1. (B) 30-40Hint:

Class interval Frequency Cumulative(C.I.) ( f ) Frequency

0-10 4 410-20 4 820-30 8 1630-40 10 2640-50 12 3850-60 8 4660-70 4 50

2. 45Hint:Draw a line parallel to the x-axis at the point

y =402

= 20. This line cuts the curve at a

point. From this point, draw a perpendi-cular to the x-axis. The abscissa of the pointof intersection of this perpendicular with thex-axis determines the median of the data.

3. The given distribution can also be representedas follows:

Class interval Frequency

0-10 310-20 920-30 1530-40 3040-50 1850-60 5

As the maximum frequency is 30, the modalclass is 30-40.

4. C.I. fi xi fixi

1-3 9 2 183-5 22 4 885-7 27 6 162

7-10 17 8.5 144.5

Σfi = 75 Σfixi = 412.5

Mean = 412.55.5

75i i

i

f x

f

Σ= =

Σ.

5. In the given distribution, the classes are inthe inclusive form. Let us convert them into

exclusive form by subtracting 163 162,

2� i.e.,

0.5 from lower limit and adding the same toupper limit of each class.

Class interval f

159.5-162.5 15

162.5-165.5 118

165.5-168.5 142

168.5-171.5 127

171.5-174.5 18

Here, the maximum frequency is 142.

∴ l = 165.5, fl = 142, f0 = 118, f2 = 127, h = 3Now,

mode = l + 1 0

1 0 22f f

f f f−⎛ ⎞

⎜ ⎟− −⎝ ⎠× h

= 165.5 + 142 118

284 118 127�� �

� �� �� � × 3

= 165.5 + 1.85 = 167.35

Hence, the modal height of the students is167.35 cm.

6. The given data may be re-tabulated by thefollowing manner with correspondingcumulative frequencies.

93���� ��� ��

Heights (in cm.) No. of girls CumulativeC.I. ( f ) frequency

(cf )

Below 140 4 4140-145 7 11145-150 18 29150-155 11 40155-160 6 46160-165 5 51

N = 51

Now, N = 51. So, N2

= 25.5.

This observation lies in the class 145-150.Then l = 145, cf = 11, f = 18, h = 5

Now, median = l +

N –

2 ×

cfh

f

� ��� �� �� �� �� �� ��� �� ��� �

= 145 + 25.5 – 11

× 518

� ��� �� ��� �= 149.03.

Hence, the median height of the girls is149.03 cm.

7. C.I. fi xi fixi

10-12 7 11 77

12-14 12 13 156

14-16 18 15 270

16-18 13 17 221

Σ fi = 50 Σ fixi = 724

∴ Mean mileage = i i

i

f xf

�

� =

72450

= 14.48 km/l.

(ii) No, the manufacturer is claiming mileage1.52 km/l more than average mileage.

(iii) The manufacturer should be honest withhis customer.

8. 69.5.Hint: Change the given distribution intoless than type and more than typedistributions. For drawing the ‘less than type’

ogive, take upper class limits andcorresponding cumulative frequencies; andfor drawing the ‘more than type’ ogive takelower class limits and correspondingcumulative frequencies.

�������������� ��

1. (A) Here, a = 25, h = 10.

∴ x = a + h i i

i

f u

f

Σ⎛ ⎞⎜ ⎟Σ⎝ ⎠

= 25 + 10 20

100� �� �� � = 27.

2. Sum of 11 numbers = 11 × 35 = 385

Sum of first 6 numbers = 6 × 32 = 192

Sum of last 6 numbers = 6 × 37 = 222

∴ 6th number = 192 + 222 – 385 = 29.

3. The modal class is 30-40.

h = 10, f1 = 32, f0 = 12, f2 = 20, l = 30.

Mode = l + 1 0

1 0 22f f

f f f−⎛ ⎞

⎜ ⎟− −⎝ ⎠ × h

= 30 + 32 12

64 12 20�� �

� �� �� � × 10

= 30 + 6.25

= 36.25.4. False,

Hint: N = 5 + 15 + 30 + 8 + 2 = 60

∴N2

= 30

5. xi fi fixi

3 5 156 2 127 3 214 2 8

p + 3 4 4p + 128 6 48

Σfi = 22 Σfixi = 4p + 116

94 �� � � � � � � � � �

Mean = i i

i

f x

f

ΣΣ

⇒ 6 =4 116

22p �

⇒ 132 = 4p + 116⇒ 4p = 16 ⇒ p = 4.

6. Class interval Frequency( f )

0-20 420-40 640-60 1860-80 8

80-100 14

The class corresponding to the maximumfrequency is 40-60. So, 40-60 is the modalclass.

Mode = l + 1 0

1 0 22f f

f f f−⎛ ⎞

⎜ ⎟− −⎝ ⎠× h

Here, l = 40, f1 = 18, f0 = 6, f2 = 8 and h = 20

∴ Mode = 40 + 18 6

2×18 6 8−⎛ ⎞

⎜ ⎟− −⎝ ⎠ × 20

= 40 + 12 2022� = 50.91.

7. We notice classes are continuous. We formcumulative frequency table by less thanmethod.

MarksNumber Marks

(C.I.)of stu- less cf Pointdents than

0-10 5 10 5 (10, 5)10-20 8 20 13 (20, 13)20-30 10 30 23 (30, 23)30-40 9 40 32 (40, 32)40-50 6 50 38 (50, 38)50-60 7 60 45 (60, 45)

On plotting these points on a graph paperand joining them by a free hand smoothcurve, we get a curve called less than ogive.

8. The cumulative frequency table for the givendata is given below:

Marks No. of Cumulative

(C.I.) students frequency( f ) (cf )

0-10 10 1010-20 f1 10 + f120-30 25 35 + f130-40 30 65 + f140-50 f2 65 + f1 + f250-60 10 72 + f1 + f2

N = 75 + f1 + f2

Clearly, N = 75 + f1 + f2But N = 100∴ f1 + f2 = 25 ... (i)

∴N2

= 50.

The median is 32 which lies in the class30-40.So, l = 30, f = 30, cf = 35 + f1, h = 10.Using the formula:

Median = l +

N2

cf

f

� ��� �� �� �

× h

⇒ 32 = 30 + 150 3530

f� �� �� �� �

× 10

95���� ��� ��

⇒2

10= 115

30f�

⇒ 75 – 5f1 = 30

⇒ f1 =75 30

5�

⇒ f1 = 9

Substituting f1 = 9 in equation (i), we get

9 + f2 = 25 ⇒ f2 = 16

Hence, f1 = 9 and f2 = 16.

�������������� ��

1. (D) x1 + x2 + ................ + xn = n × x

⇒ 1 2x xk k

� + .......... + nx nx

k k�

(Dividing throughout by k)

⇒ 1 2 .........

x x xnk k k

n

� � �

= xk

(Dividing throughout by n)

⇒ Required mean =xk

.

2. The first ten prime numbers are:2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Median =11 13

2�

= 242

= 12.

3. Mean = i i

i

f xf

�

�

⇒ 15 =

5 × 6 + 10 × + 15 × 6 + 20× 10 + 25 × 5

6 + + 6 + 10 + 5

k

k

⇒445 10

27k

k�

�

= 15

⇒ k = 8.

4. False, because the values of these threemeasures depend upon the type of data, soit can be the same.

5. Let us use the assumed mean method to findthe mean of the given data.

Marks No. of Class di = fidi(C.I.) students mark xi – 35( fi) (xi)

0-10 4 5 – 30 – 12010-20 6 15 – 20 – 12020-30 8 25 – 10 – 8030-40 10 35 0 040-50 12 45 10 12050-60 30 55 20 600

Σfi = 70 Σfidi = 400

Here, assumed mean, a = 35

Now, required mean = a + i i

i

f df

ΣΣ

= 35 + 40070

= 35 + 5.71 = 40.71.

6. Since mode = 36, which lies in the classinterval 30-40, so the modal class is 30-40.∴ f1 = 16, f0 = f, f2 = 12, l = 30 and h = 10.

Now, mode = l + 1 0

1 0 22f f

f f f�� �

� �� �� �× h

⇒ 36 = 30 + 16

32 12f

f�� �

� �� �� �× 10

⇒ 610

= 1620

ff

�

�

⇒ 120 – 6f = 160 – 10 f⇒ 4 f = 40 ⇒ f = 10.

7. 31.5 marks.Hint:

Classes No. of Cumulativestudents frequency

0-10 5 510-20 8 1320-30 6 1930-40 10 2940-50 6 3550-60 6 41

Draw the ogive by plotting the points:

(10, 5), (20, 13), (30, 19), (40, 29), (50, 35)

96 �� � � � � � � � � �

and (60, 41). Here N2

= 20.5. Locate the

point on the ogive whose ordinate is 20.5.The x-coordinate of this point will be themedian..

8. We prepare the cumulative frequency tableby less than method as given below:

Cumu-Fre- Score lative

Scores quency less fre- Point ( f ) than quency

( f )

200-250 30 250 30 (250, 30)

250-300 15 300 45 (300, 45)

300-350 45 350 90 (350, 90)

350-400 20 400 110 (400, 110)

400-450 25 450 135 (450, 135)

450-500 40 500 175 (500, 175)

500-550 10 550 185 (550, 185)

550-600 15 600 200 (600, 200)

We plot the points given in above table on agraph paper and then join them by freehand smooth curve to draw the cumulativefrequency curve by less than method.

Similarly for the cumulative frequencycurve by more than method, we preparethe corresponding frequency table.

Cumu-Fre- Score lative

Scores quency more fre- Point ( f ) than quency

( cf )

200-250 30 200 200 (200, 200)

250-300 15 250 170 (250, 170)

300-350 45 300 155 (300, 155)

350-400 20 350 110 (350, 110)

400-450 25 400 90 (400, 90)

450-500 40 450 65 (450, 65)

500-550 10 500 25 (500, 25)

550-600 15 550 15 (550, 15)

We plot the points given in this last tableon the same graph and join them by freehand smooth curve to draw the cumulativefrequency curve by more than method (seefigure).

Median: The two curves intersect eachother at a point. From this point, we drawa perpendicular on the x-axis. The foot ofthis perpendicular is P(375, 0). The abscissaof the point P, i.e., 375 is the requiredmedian.

Hence, the median is 375.

Figure: Less than and more than typecumulative frequency curves

97���� ��� ��

���������

1. (A) Let Σfi = N

Σ(fixi – x ) = Σfixi – N x

= N N

i if xx

�� ��� �� �

= N ( x – x ) = 0.2. Let us rewrite the given distribution in the

other manner.

Marks No. of students

0-10 310-20 920-30 1530-40 3040-50 1850-60 5

Clearly, the modal class is 30-40.

3. 17.5Hint: First, transform the given class-intervalsinto exclusive form and then find thecumulative frequency table.

Here, N = 13 + 10 + 15 + 8 + 11 = 57

∴ N2

= 28.5.

4. Monthly income No. of(in �) families

10000-13000 1513000-16000 1616000-19000 1919000-22000 1722000-25000 18

25000 or more 15

Hence, required number of families is 19.

5. No, because an ogive is a graphical repre-sentation of a cumulative frequency distri-bution.

6. Yes; as we knowmode = 3 median – 2 mean⇒ 3 median = mode + 2 mean

⇒ Median =13

mode + 23

mean

= mode –23

mode +23

mean

= mode +23

(mean – mode).

7. ui =C.I. xi fi Aix

h− fiui

800-820 810 740

220

� � � – 14

820-840 830 1420

120

� � � – 14

840-860 850 190

020

� 0

860-880 870 1520

120

� 15

880-900 890 940

220

� 18

Σfi = 64 Σfiui = 5

Let assumed mean beA = 850h = 20

Mean = A + i i

i

u ff

Σ⎛ ⎞⎜ ⎟Σ⎝ ⎠

× h

= 850 + 564

� �� �� �

× 20

= 850 + 1.5625 = 851.5625.

Hence, the required mean is 851.5625.

8. Mode = l + 1 0

1 0 22f f

f f f−⎛ ⎞

⎜ ⎟− −⎝ ⎠ × h

Here, l = 30, f1 = 45, f0 = 30, f2 = 12, h = 10

∴ Mode = 30 + 45 30

90 30 12−⎛ ⎞

⎜ ⎟− −⎝ ⎠ × 10

= 30 + 3.125 = 33.125 marks.

98 �� � � � � � � � � �

9. (i) Class intervals Frequency Mid-points fixi(in daily pocket (No. of children) of C.I.

allowances) (in �) ( fi ) (xi )

11-13 7 12 8413-15 6 14 8415-17 9 16 14417-19 13 18 23419-21 x 20 2021-23 5 22 11023-25 4 24 96

Σfi = 44 + x Σfixi = 752 + 20x

∴ Mean = i i

i

f xf

�

� =

752 2044

xx

�

�

As Mean = � 18 (given) ∴ 18 = 752 20

44x

x�

�

⇒ 792 + 18x = 752 + 20x ⇒ 40 = 2x ⇒ x = 20.

(ii) Arithmetic mean of grouped data.

(iii) One shouldn't be spend thrift, but should save his money for future use.

��

����������������

Practice Paper-1

SECTION–A

1. (D) 1

2

aa =

62

= 3, 1

2

bb

= – 3–1

= 3, 1

2

cc =

109

��i.e., 1

2

aa = 1

2

bb

� 1

2

cc

��The given lines are parallel.

2. (B) �D = �Q and �E = �R

� �DEF ~ �QRP (AA rule of similarity)

� DEPQ

� EFRP

.

3. ... LCM (x, y) = HCF( , )

x yx y

�

= 180012

= 150.4. 9��< 90° � ��< 10°�����is an acute angle.

cos 9� = sin �

� cos 9� = cos –2�� ��� �� �

� 9� = –2�

� ���� = π20

� tan 5� = tan π4

= 1.

SECTION– B5. True, because out of any two consecutive

positive integers, one is even and the otherone is odd; and the product of an even andan odd is even.

6. (i) We know that the factors of a primeare 1 and the prime itself only.Therefore, the common factor of p and qwill be 1 only. Hence, HCF (p, q) = 1.(ii) LCM (p, q) = pq

7. No, if two zeroes are � and � of polynomialx2 + kx + k, then

� + � = – k and � . � = k

�������� �����

� 2� = – k and �2 = k (when � = �)

� � = –2k

and �2 = k.

�2

4k

= k (Comparing both)

� k2 = 4k �� k2 – 4k = 0

k(– k – 4) = 0 �� k = 4, 0.

8. For infinitely many solutions,

1339

= 6k

= + 4k

k ��

1339

= 6k

and1339

= + 4k

k ��k = 2

and 3k = k + 4 ��k = 2and k = 2, i.e., k = 2.

9. Yes.

Here, 262 = 242 + 102 = 676

� AC2 = AB2 + BC2

���ABC is a right triangle.

OR

No,� �FED ~ �STUCorresponding sides of the similar trianglesare in equal ratio.

�DETU

= EFST

� DEST

� EFTU

.

10. xi fi fixi

3 6 185 8 407 15 1059 p 9p

11 8 8813 4 52

fi = p + 41 fixi = 9p + 303

� �����������

Mean = i i

i

f xf

�

� � 7.5 =

++

9 30341

pp

� 7.5p + 307.5 = 9p + 303 �� 1.5p = 4.5� p = 3.

SECTION – C

11. On the contrary let us assume that 2 3 –

3 2 is a rational number. Then, we cantake coprime a and b such that

ab

= 2 3 – 3 2

�2

2ab

= 12 + 18 – 12 6

(Squaring both sides)

� 12 6 = 30 – 2

2ab

��� 6 = 2 2

230 –

12b a

b.

Since, a and b are integers, therefore, RHSof this last equation is rational and so LHSmust be rational.

But this contradicts the fact that 6 isirrational.This contradiction has arisen due to

incorrect assumption that 2 3 – 3 2 is arational number.

So, we conclude that 2 3 – 3 2 is anirrational number.

12. Let a be any odd positive integer. Then itis of the form 6m + 1, 6m + 3 or 6m + 5,where m is an integer.Here, 3 cases arise.Case I. When a = 6m + 1,

a2 = (6m + 1)2 = 36m2 + 12m + 1= 12m (3m + 1) + 1= 6q + 1, where q = 2m (3m + 1).

Case II. When a = 6m + 3,a2 = (6m + 3)2 = 36m2 + 36m + 9

= 36m2 + 36m + 6 + 3= 6(6m2 + 6m + 1) + 3 = 6q + 3,

where q = 6m2 + 6m + 1.Case III. When a = 6m + 5,

a2 = (6m + 5)2 = 36m2 + 60m + 25= 36m2 + 60m + 24 + 1= 12(3m2 + 5m + 2) + 1 = 6q + 1,

where q = 2(3m2 + 5m + 2).Hence, a2 is of the form 6q + 1 or 6q + 3.

ORAs: 1032 = 408 × 2 + 216 ...(i)

408 = 216 × 1 + 192 ...(ii)216 = 192 × 1 + 24 ...(iii)192 = 24 × 8 + 0 ...(iv)

� HCF = 24 � From (iii)� 24 = 216 – 192

= 216 – [408 – 216] [Using (ii)]= 2 × 216 – 408= 2[1032 – 2 × 408] – 408

[Using (i)]� 24 = 1032 × 2 – 5 × 408 � m = 2.

13. Since – 5x is a factor of

f (x) = x3 – 3 5 x2 + 13x – 3 5 ,

so as f (x) may be rewritten.

f (x) = x3 – 3 5 x2 + 13x – 3 5= x3 – 5 x2 – 2 5 x2 + 10x

+ 3x – 3 5

= x2 � �– 5x – ( )2 5 – 5x x

+ ( )3 5x –

= ( )– 5x ( )2 – 2 5 + 3x x

To find zeroes of f (x), put f (x) = 0.

( )– 5x ( )+2 – 2 5 3x x = 0

� – 5x = 0 or x2 – 2 5 + 3 = 0

� x = 5

or x = ± × ×2 5 20 – 4 1 3

2

� x = 5 or x = ±2 5 2 22

� x = 5 or x = 5 + 2

or 5 – 2

� ���������������

Hence all the zeroes of f (x) are 5 , 5 +

2 and 5 – 2 .

14. The given pair of equations may berewritten as

+

2x y

xy=

23

; 2 –x y

xy =

10–

3

i.e, 1x

+ 1y

= 43

; – 1x

+ 2y

= 10

–3

Adding this last pair, we get

3y

= – 2 �� y = 3

–2

Substituting y = 32

– in the first equation

of last pair, we get

1x +

1– 3

2

= 43

��� 1x –

23

= 43

�� 1x = 2 � x =

12

Hence, x = 12

, y = 3

–2

is the required

solution.15. Let the length of each side of the given

equilateral triangle be a, thenAB = BC = CA = a ...(i)

� BD = 3a ...(ii)

Draw AP BC to meet BC at P. P will bethe mid-point of BC, that is

BP = 2a

...(iii)

� DP = BP – BD =2a

–3a

=6a

...(iv)

[Using (ii) and (iii)]Now, in right-angled triangle APB,

AP2 = AB2 – BP2 �� AP2 = a2 – 2

4a

� AP2 = 23

4a ...(v)

Also, in right-angled triangle APD,

AD2 = AP2 + DP2

� AD2 = 23

4a

+ 2

36a =

2 22736

a a�

[From (iv) and (v)]

� 36AD2 = 28a2 � 9AD2 = 7a2

� 9AD2 = 7AB2. Hence proved.

16. �BAD = 90° – �CAD(���BAC = 90°)

= 90° – (90° – �ACD) (���ADC = 90°)

��BAD = �ACD ...(i)�BDA = �ADC = 90° ...(ii)

Using equations (i) and (ii) in �ABD and�CAD, we have

�ABD ~ �CAD (AA rule of similarity)

�BDAD

= ADCD

(Corresponding parts)

� BD . CD = AD2.

OR

Given: AB ��CD; AB = 2CD

To find: �

�

( AOB )( COD )

ar ar

As �1 = �2[����Alternate interior angles]

�3 =��4� �AOB ~ �COD (AA – Criteria)

��

�

( AOB )( COD )

ar ar

= 2

2AB

CD =

2ABCD

� �� �� �

= 22CD

CD� �� �� �

(� AB = 2CD)

= 41

.

� � �����������

17. θ θθ + θ

cos – sincos sin

= +

1 – 3

1 3

��

cos sin–

cos cos 1– 3sincos 1 3

cos cos

� ��

� ��

�� �� �

� �

� 1 – tan1 tan

�

� �=

+

1 – 31 3

� tan � �� 3 ������= 60°.

18. In �ABC, tan A =1

2 2=

BCAB

� AC = 2 2(1) (2 2)� = 1 8� = 3

sin A = 13

, cos A = 2 23

,

sin C = 2 23

, cos C = 13

Now, sin A. cos C + cos A. sin C

=13

×13

+ 2 23

× 2 23

=19

+89

= 1.

19. True,

LHS = cos 80sin 10

°°

+ cos 59° cosec 31°

= cos(90 – 10 )

sin10° °

° + cos 59° cosec (90° – 59°)

= sin 10°sin 10°

+ cos 59° sec 59°

= 1 + cos 59°cos 59°

= 1 + 1 = 2

Hence, the given equation is valid.

20. The given data is

C.I. 0-5 5-10 10-15 15-20 20-25 25-30 30-35

f 10 15 30 80 40 20 5

From the table, maximum occuringfrequency is 80. So, modal class is 15-20.

� Mode = l + 1 0

1 0 22f f

f f f

� ��� �

� �� � × h

Here, l = 15, f1 = 80, f0 = 30, f2 = 40, h = 5

�Mode = 15 + ⎛ ⎞−⎜ ⎟− −⎝ ⎠

80 30160 30 40

× 5

= 15 + 25090

= 17.78

Hence, modal size is 17.78 hectares.

OR

Class Frequency Cumulativeinterval ( f ) Frequency (cf )

0-8 8 88-16 10 18

16-24 16 3424-32 24 5832-40 15 7340-48 7 80

N = 80

For median class, N 80=

2 2= 40

In the cumulative frequency column, 58 isjust greater than 40.

So, 24-32 is the median class.

Here, l = 24, cf = 34, f = 24, N2

= 40, h = 8

Using formula:

Median = l +

N–

2 ×cf

hf

� ��� �� �� �� �� �� ��� �� ��� �

= 24 +40 – 34

× 8 = 2624

� ��� �� ��� �

Hence, median of the given distribution is 26.

� ���������������

SECTION – D

21. Let f (x) = 3 2 x2 + 13x + 6 2

= 3 2 x2 + 9x + 4x + 6 2(Spliting middle term)

= 3x( 2 x + 3) + 2 2 ( 2 x + 3)

= (3x + 2 2 ) ( 2 x + 3)To find the zeroes of f (x), we have

3x + 2 2 = 0 or 2 x + 3 = 0

� x = – 2 2

3 or

– 3

2� Zeroes of the given polynomial are

– 2 23

and – 3

2.

Now,sum of zeroes =�– 2 2

3+

– 3

2 =

– 13

3 2

= – 2

Coefficient ofCoefficient of

xx

Product of zeroes

= – 2 2

3 ×

– 3

2 =

6 2

3 2= 2

Constant termCoefficient of x

.

Hence proved.

22. (i) Let speed of the train be x km/hr andthat of the bus be y km/hr.

DistanceSpeed

= Time

Case I. According to question, we get

�60x

+ 300 – 60

y= 4 �

60x

+ 240

y = 4

� 15x

+ 60y

= 1 � 1x

+ 4y

= 1

15 ...(i)

Case II. According to the given conditions,we get

300 – 100100x y

� ��� �� � = 4 +

1060

�100

x +200y =

256

� 4x +

8y =

16

Dividing by 4, we get

1x +

2y =

124

... (ii)

Use equation (i) – equation (ii),

4y –

2y =

115

–124

2y =

8 – 5120

= 3

120

� y = 120×2

3= 80 km/hr

Put y = 80 in equation (i), we get

1x +

480

= 115

�1x =

115

–120

= 4 – 360

= 160

� x = 60 km/hr.Hence, speed of the train = 60 km/hrand speed of the bus = 80 km/hr.

(ii) Solution of system of linear equations intwo variables.

(iii) By opting for public transport it depictsthat she is a responsibile citizen, so herresponsibility and rationality have beendepicted here.

OR(i) Let l = length of the rectangle

b = breadth of the rectangleAccording to question,

(l + 7)(b – 3) = lb …(i)(l – 7)(b + 5) = lb …(ii)

From equation (i),lb + 7b – 3l – 21 = lb� 7b – 3l = 21 …(iii)

From equation (ii),lb – 7b + 5l – 35 = lb

� – 7b + 5l = 35 …(iv)Adding equations (iii) and (iv), we get

2l = 56 �� l = 28 mPutting the value of l in equation (iii), weget

b = 15 m.� l = 28 m, b = 15 m.

(ii) Solution of system of linear equationsin two variables.

(iii) Love for environment and human beings.

� � �����������

23. Table for values of x and y as regardingequation 3x + y – 5 = 0 is

x 0 1y 5 2

Similarly table for equation 2x – y – 5 = 0 is

x 0 1y – 5 – 3

Let us draw the graph of lines using thetables obtained above.

The lines intersect y-axis at (0, 5) and (0, – 5).

24. We are given a triangle ABC in which aline PQ parallel to BC is drawn to intersectthe sides AB and AC at P and Q respectively.

We need to prove ,APPB

= AQQC

Join PC and QB. Draw QM AB and PN AC.

Now, area of a triangle

= 12

× base × height

� ar(�APQ) = 12

× AP × MQ ...(i)

Also ar(�APQ)= 12

× AQ × NP ...(ii)

Comparing equations (i) and (ii), we get12

× AP × MQ = 12

× AQ × NP

� APAQ

= NPMQ

...(iii)

Further, ar(�BPQ) = 12

× PB × MQ ...(iv)

And ar(�CQP) = 12

× QC × NP ...(v)

But triangles BPQ and CQP are on thesame base PQ and between the sameparallels PQ and BC, so their areas mustbe equal.i.e., ar(�BPQ) = ar(�CQP)

�12

× PB × MQ = 12

× QC × NP

[Using equations (iv) and (v)]

�PBQC

= NPMQ

...(vi)

From, equations (iii) and (vi), we get

APAQ

= PBQC

� APPB

= AQQC

.

Hence proved.25. We are given two triangles ABC and PQR

such that �ABC ~ �PQR.Draw perpendiculars AD and PM on BCand QR respectively.

We need to prove

ΔΔ

( ABC)( PQR)

arar

= 2

2ADPM

In �ABD and �PQM,�ADB = �PMQ = 90°�ABD = �PQM ( ��ABC ~ �PQR)

� �ABD ~ �PQM (AA criterion of similarity)

�ABPQ =

ADPM

...(i) (Corresponding sides)

� ���������������

We know that the ratio of areas of twosimilar triangles is equal to ratio of squaresof their corresponding sides

�ΔΔ

( ABC)( PQR)

arar

=�2

2ABPQ

...(ii)

From equations (i) and (ii), we have

ΔΔ

( ABC)( PQR)

arar

=�2

2ADPM

Hence proved.

26. sin � + cos � = 3 (Given)

� sin2 � + cos2 � + 2 sin � cos � = 3 (Squaring both the sides)

� 1 + 2 sin ��cos � = 3

� sin ��cos � = 1 ...(i)

Now, tan ��+ cot � =sin coscos sin

� ��

� �

= 2 2sin cos

sin cos� � �

�� �=

11

[Using (i)]

= 1.Hence proved.

27. Let us take LHS of the given identity.2 2

2 2 2tan cosec

tan –1 sec – cosec� �

�� � �

=

2

2

2

2

sincos

sin– 1

cos

�

�

�

�

+ 2

2 2

1sin

1 1–

cos sin

�

� �

=

2

2 2

2 2 2 2

2 2 2

sin 1cos sin

sin – cos sin – coscos sin .cos

�

� ��

� � � �

� � �

= 2 2

2 2 2 2

sin cos

sin – cos sin – cos

� ��

� � � �

= 2 2

2 2

sin cos

sin – cos

� � �

� �= 2 2

1sin – cos� �

[ �sin2 � + cos2 � = 1]= RHS. Hence proved.

ORtan A = n tan B

� cot B = tanA

n and sin A = m sin B

� sin B = 1sin A

m���cosec B =

sin Am

... cosec2 B – cot2 B = 1

�2 2

2 2sin A tan Am n− = 1 ��

2 2 2

2cos A

sin Am n� = 1

� m2 – 1 = (n2 – 1) cos2 A

�2

21

– 1mn

� = cos2 A.Hence proved

28. m2 – n2= (tan � + sin �)2 – (tan � – sin �)2

= (tan � + sin � + tan ��– sin �)(tan ��+ sin ��– tan � + sin �)[�� A2 – B2 = (A + B) (A – B)]

= 2 tan �� . 2 sin �= 4sin �� tan � ...(i)

4 mn = 4 (tan sin )(tan sin )�� � � � �

= 2 24 tan sin�� �

[� (A + B) (A – B) = A2 – B2]

=2

22

sin4 sin

cos�

���

= 4 sin � ��

2

11

cos

= 24 sin sec –1θ θ= 4 sin � tan � .... (ii)

From (i) and (ii), m2 – n2 = 4 mn .

29. Let us prepare the cumulative frequencytable by more than method as givenbelow:

Production Productionyeild fi yeild (in kg/ha) cf Point

(in kg/ha) more thanor equal to

50-55 2 50 50 (50, 50)55-60 6 55 48 (55, 48)60-65 8 60 42 (60, 42)65-70 14 65 34 (65, 34)70-75 15 70 20 (70, 20)75-80 5 75 5 (75, 5)

Total 50

� � �����������

We plot the points mentioning in the tablesuch that lower class limits are on thex-axis and the cumulative frequencies areon the y-axis. By joining these points byfree hand smooth curve, We obtain morethan type ogive as shown in the adjoininggraph.

To obtain median from graph:

Draw a line parallel to x-axis passing

through y = 502

= 25. This line meets the

ogive at (68.2, 25).

�� Median = 68.2.

30. Given that median is 28.5. It lies in the classinterval 20-30, so 20-30 is the median class.Let us prepare frequency distribution table.

C. I. f cf

0-10 5 510-20 x 5 + x20-30 20 25 + x30-40 15 40 + x40-50 y 40 + x + y50-60 5 45 + x + y

x + y + 45

N = 60

������ x + y + 45 = 60

� x + y = 15 ... (i)

Median = l + N

–2

cfh

f

� ��� �� �� ���� �� ��

� 28.5 = � �30 – 5 +

20 + ×1020

x

� �� �� ��

On simplifying, x = 8 ...(ii)

From equations (i) and (ii), we have

x = 8 and y = 7.

31. We will use the step-deviation method.

Marks No. of Class- di ui fiui

(C.I.) students mark = xi – 45 = 10

id

(fi) xi

0-10 5 5 – 40 – 4 – 20

10-20 4 15 – 30 – 3 – 12

20-30 8 25 – 20 – 2 – 16

30-40 12 35 – 10 – 1 – 12

40-50 16 45 0 0 0

50-60 15 55 10 1 15

60-70 10 65 20 2 20

70-80 8 75 30 3 24

80-90 5 85 40 4 20

90-100 2 95 50 5 10

�fi = 85 ��fiui= 29

Let a = 45. Here h = 10

Mean = a + i i

i

f uh

f

� ���� �

�� �

= 45 + 2985

× 10

= 45 + 3.41

= 48.41.

��������������� �

Practice Paper-2

SECTION–A

1. (C) ... 3 3 3

343

2 × 5 ×7=

3 3

343

2 × 5 × 343 =

11000

= 0.001.

2. (D) The condition for the line parallel is:

1

2

aa

= 1

2

bb

≠ 1

2

cc

⇒ 32

= 25c ≠ 2

1

⇒ c = 154

.

3. As( ABC)( DEF)ΔΔ

arar

= 2

2

BC

EF(Result on areas of similar triangles)

A

B C

D

E F

⇒54

( DEF)Δar=

2

2

3

4 ⇒ ar(ΔDEF) = 96 cm2.

4.o o

o

cos (90 – 70 )

sin 70 +

2 coscos

θθ

=2k

⇒ o

o

sin 70

sin 70+ 2 =

2k

⇒ 3 =2k

⇒ k = 6.

SECTION–B

5. False, because 3 or 2 3 is an irrationaland sum or difference of a rational numberand an irrational number is an irrationalnumber.

6. No, because prime factorisation of anynumber of type 2n can not have 5 as one ofits prime factor.

7. False, since the discriminant is zero for k =

±12

.

8. The condition for infinitely many solutions

is 1

2

aa = 1

2

bb

= 1

2

cc

. ⇒ 35

= – ( + 1)1 – 2

aa

=2 – 1

3b

b

⇒ 35

=– ( + 1)1 – 2

aa

and 35

=2 – 1

3b

b

⇒ 3 – 6a = – 5a – 5 and 9b = 10b – 5⇒ a = 8 and b = 5.

9. Yes, because ΔPBC and ΔPDE are similar

by SAS rule. AsBPDP

=PCPE

=12

and ∠BPC =

∠DPE.

ORIn ΔAQO and ΔBPO,

∠QAO = ∠PBO (Each 90°)

∠AOQ = ∠BOP(Vertically opposite angles)

So, by AA rule of similarity,

ΔAQO ~ ΔBPO

⇒AQBP

= AOBO

⇒ AQ9

= 106

⇒ AQ = 10 × 9

6⇒ AQ = 15 cm.

10. The empirical relationship among the threemeasures of central tendency is:

Mode = 3 Median – 2 Mean= 3 × 55 – 2 × 50= 165 – 100 = 65.

SECTION–C

11. To prove that the pair of numbers (847,2160) is coprime by using Euclid’salgorithm, we have to prove that the highestcommon factor of the pair is 1.

Since 2160 > 847

∴ 2160 = 847 × 2 + 466

Since the remainder 466 ≠ 0.

∴ 847 = 466 × 1 + 381

Since the new remainder 381 ≠ 0.

��� ��������� �

∴ 466 = 381 × 1 + 85Since the new remainder 85 ≠ 0.∴ 381 = 85 × 4 + 41Since the new remainder 41 ≠ 0.∴ 85 = 41 × 2 + 3Since the new remainder 3 ≠ 0.∴ 41 = 3 × 13 + 2Since the new remainder 2 ≠ 0.∴ 3 = 2 × 1 + 1Since the new remainder 1 ≠ 0.∴ 2 = 1 × 2 + 0Since, the remainder has now become zero,the divisor at this stage is 1, the HCF of847 and 2160 is 1.

ORLet us assume, to the contrary, that 3 +

2 5 is rational.So we can find coprimes a and b such that

3 + 2 5 = ab

Rearranging, 5 = 3

2a b

b�

a and b are integers ⇒ a – 3b is an integer

⇒ 3

2a b

b�

is rational number

∴ 5 should be rational. But we

know that 5 is irrational. So our

assumption that 3 + 2 5 is rational iswrong.

Hence 3 + 2 5 is irrational.12. Let x be any positive integer.

Then it is either of the form 3q or 3q + 1 or 3q+ 2.Case I. x = 3qCubing both sides, we get

x3 = (3q)3 = 27q3 = 9m …(i)m = 3q3

Case II. x = 3q + 1Cubing both sides, we get

x3 = (3q + 1)3

= 27q3 + 1 + 3 (3q + 1) × 3q= 27q3 + 27q2 + 9q + 1= 9q (3q2 + 3q + 1) + 1

= 9m + 1 …(ii)

m = q(3q2 + 3q + 1).

Case III. x = 3q + 2

Cubing both sides, we get

x3 = (3q + 2)3

= 27q3 + 8 + 3 (3q + 2) 6q

= 27q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9m + 8; …(iii)

m = q (3q2 + 6q + 4).Thus, from equations (i), (ii) and (iii), it isclear that cube of any positive integer iseither of the form 9m, or 9m + 1 or 9m + 8.

13. We have time =distance

speedTime taken by Abhay to cover one complete

round = 36012

= 30 hours

Time taken by Ravi to cover one complete

round = 36015

= 24 hours

Abhay and Ravi reach the starting pointrespectively after 30 hours and 24 hours, andtheir respective multiples. Therefore, they willmeet again at the starting point after the timegiven by least common multiple of 30 hoursand 24 hours Let us determine the LCM of 30hours and 24 hours.

30 = 2 × 3 × 5,24 = 23 × 3 ⇒ LCM = 23 × 3 × 5 = 120

Hence, the required time is 120 hours.

14. Let α and β be the zeroes of 6x2 + x + k.

∴ α2 + β2 = (α + β)2 – 2αβ

= 21

– – 26 6

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

k =

1–

36 3k

But it is given that α2 + β2 = 2536

∴ 2536

= 136

–3k ⇒

3k

= – 2436

⇒ k = – 2.

��������������� �

OR

Since a = 2 is a zero of a3 – 3a2 – 10a + 24,therefore a3 – 3a2 – 10a + 24 is divisible bya – 2. Further the obtained quotient willprovide the other two zeroes.

a2 – a – 12 = (a – 4) (a + 3)For other zeroes, put a – 4 = 0 and a + 3 = 0

a = – 3, 4Thus, the other two zeroes are – 3 and 4.

15. In right-angled triangle PQS,

PS2 = PQ2 + QS2 ⇒ QS2 = PS2 – PQ2

⇒2QR

4= PS2 – PQ2

⇒ QR2 = 4PS2 – 4PQ2 …(i)

In right-angled triangle PQR,

PR2 = PQ2 + QR2

= PQ2 + 4PS2 – 4PQ2 [Using (i)]

⇒ PR2 = 4PS2 – 3PQ2. Hence proved.

16. Draw RM ⊥ PQ.In ΔPRS, ∠PSR > 90°

⇒ PR2 = PS2 + RS2 + 2PS.SM ...(i)[Using result on obtuse-angled triangle]

In ΔQRS, ∠QSR < 90°⇒ QR2 = RS2 + SQ2 – 2SQ.SM ...(ii)

[Using result on acute-angled triangle]

Add equations (i) and (ii) to get

PR2 + QR2 = 4PS2 – 2SM(PS – SQ)

= 42PQ

2⎛ ⎞⎜ ⎟⎝ ⎠

(� PS = SQ)

⇒ PR2 + QR2 = PQ2. Hence proved.

OR

Given:ABPQ

= ACPR

= ADPM

and

AD, PM are medians.

To show: ΔABC ~ ΔPQR.

Construction: Extend AD up to D' such thatAD = DD' and extend PM up to M' suchthat PM = MM'Join BD', D'C; QM', M' R.Proof: As AD = DD' and BD = DC and PM= MM' and QM = MR∴ ABD' C and PQM' R are �gm∴ AB = D' C and PQ = RM' {����Opposite sides of a �gm are equal}

As it is given that: ABPQ =

ACPR

= ADPM

⇒ D C AC 2AD ADRM PR 2PM PM

’ ’’ ’� � �

⇒ ΔACD' ~ ΔPRM' (SSS criteria)∴ ∠1 = ∠2 (CPCT)

��� ����������

Similarly ∠3 = ∠4 ∴ ∠1 + ∠3 = ∠2 + ∠4⇒ ∠BAC = ∠QPRNow : In ΔABC and ΔPQR

ABPQ

= ACPR

…(i)

and ∠BAC = ∠QPR …(ii)∴ ΔABC ~ ΔPQR (SAS)

Hence proved.

17. sin θ = 34

... (i) Given

� sin2 θ + cos2 θ = 1∴ cos2 θ = 1 – sin2 θ

⇒ cos2 θ = 1 –23

4⎛ ⎞⎜ ⎟⎝ ⎠

[Using (i)]

⇒ cos2 θ = 716

⇒ sec2 θ = 167

... (ii)

Let us take LHS of the given equation.

LHS =2 2

2

cosec – cot

sec – 1

� �

�

= 2

1

sec – 1θ

[... cosec2 θ – cot2 θ = 1]

=1

16– 1

7

=79

= 73

[Using (ii)]

= RHS. Hence proved.

18. We have sin 60° = 32

, cos 60° = 12

,

sec 30° =2

3, cosec 30° = 2

∴ 2 2

2 2

3 + sin 60° + cosec 30°

5 + cos 60° + sec 30°

=

( )2

2

22

33 + + 2

2

1 25 + +

2 3

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

=

33 + + 4

41 4

5 + +4 3

Multiplying Num. and Deno. by 12,

=36 + 9 + 4860 + 3 + 16

=9379

.

19. Consider, left hand side of the givenequation,

tan cot+

1 – cot 1 – tanθ θ

θ θ=

1tan tan

+1 1 – tan1 –

tan

θ θθ

θ=

( )2tan 1

+tan – 1 tan 1 – tan

θθ θ θ

= � �

3tan 1tan tan 1

� �

� � �

= ( ) ( )

( )

2tan – 1 tan + 1 + tan

tan tan – 1

θ θ θ

θ θ

[� a3 – b3 = (a – b) (a2 + b2 + ab]

= 1

tan + + 1tan

θθ

= sin cos

+ +1cos sin

θ θθ θ

= 2 2sin + cos

+ 1sin cos

θ θθ θ

= sec θ. cosec θ + 1

= RHS. Hence proved.

20. Let us convert the more than typedistribution to the normal distribution.

Marks No. of students

0-20 3

20-40 7

40-60 20

60-80 15

80-100 5

We observe from the table that the value 20is the maximum frequency. So, the modalclass is 40-60.

Now, mode = l + 1 0

0 2×

2f – f

hf f f

� �� �� �1 – –

Here, l = 40, f1 = 20, f0 = 7, f2 = 15,

h = 20

∴ Mode = 40 +20 – 7

× 2040 – 7 – 15

⎛ ⎞⎜ ⎟⎝ ⎠

��������������� �

= 40 + 26018

= 40 + 14.44

= 54.44 marks.

SECTION – D

21. To solve a system of equations graphicallywe need atleast two solutions of eachequation.Two solutions of the equation 2x – y = 2 aregiven in the following table:

x 0 3

y – 2 4

Two solutions of the equation 4x – y = 8 aregiven in the following table:

x 2 1

y 0 – 4

Let us draw the graph of the two givenequations.

X'

Y'

From the graph, it is clear that the two linesintersect each other at the point (3, 4).

Hence, the solution is x = 3, y = 4.

22. (i) First we divide x4 + x3 + 8x2 + ax + b byx2 + 1 as follows:

2

2 4 3 2

4 2

3 2

3

2

2

7

1 8

7

7 ( 1)7 7

( 1) ( 7)

x x

x x x x ax b

x x

x x ax bx x

x a x bx

a x b

� �

� � � � �

�

� �

� � �

�

� �

� � �

�

� �

� � �

Since, x4 + x3 + 8x2 + ax + b is divisible byx2 + 1, therefore remainder = 0

i.e., (a – 1)x + (b – 7) = 0

or (a – 1)x + (b – 7) = 0.x + 0

Equating the corresponding terms, we have

a – 1 = 0 and b – 7 = 0

i.e., a = 1 and b = 7.

(ii) Common good, Social responsibility.

23. Let the speed of rowing in still water andthe speed of the current be u km/hr and vkm/hr respectively.The speed of rowing in downstream

= (u + v) km/hr.The speed of rowing upstream

= (u – v) km/hr.

Using the formula: Distance

Speed= Time

According to first condition of the question,

18+u v

+12–u v

= 3 ... (i)

According to second condition of thequestion,

36+u v

+40–u v

= 8 ... (ii)

Let us put, 1+u v

= x and 1–u v

= y such

that equations (i) and (ii) reduce to18x + 12y = 3 ... (iii)

And 36x + 40y = 8 ... (iv)

��� �������� �

Equations (iii) and (iv) form a pair of linearequations.Multiply equation (iii) by 2 and subtractthe result from equation (iv) to get

16y = 2 ⇒ y = 18

Substitute y = 18

in equation (iv) to get

36x = 8 – 5 ⇒ x = 1

12

� x = 1+u v

∴ u + v = 12 � y = 1–u v

∴ u – v = 8This last system gives u = 10 and v = 2Hence, the speed of the rowing in still water= 10 km/hr and the speed of the current= 2 km/hr.

24. ar(ΔAXY) = ar(BXYC)

⇒ 2.ar(ΔAXY) = ar(BXYC) + ar(ΔAXY)= ar(ΔABC)

A

X

B C

Y

⇒ ΔΔ

( ABC) ( AXY)

arar

= 21

As ΔABC ~ ΔAXY

∴ ⎛ ⎞⎜ ⎟⎝ ⎠

2ABAX

= ΔΔ

( ABC)( AXY)

arar

=21

⇒ABAX

= 2

1 ⇒

AXAB

= 12

⇒AB BX

AB�

= 12

⇒ BX

1AB

� = 12

⇒ BXAB

=−2 12

.

25. Hint:

ΔABC ~ ΔPQR

⇒ ABPQ

= BCQR

and ∠B = ∠Q

⇒ ABPQ

= 2BD2QM

and ∠B = ∠Q

⇒ ABPQ

= BDQM

and ∠B = ∠Q

⇒ ΔABD ~ ΔPQM = AB ADPQ PM

�

26. m = cosec θ – sin θ (Given)

⇒ m = 1

sin θ– sin θ =

21 – sinsin

θθ

⇒ m = 2cos

sinθ

θ... (i) (� 1 – sin2 θ = cos2 θ)

⇒ m2 = 4

2

cos

sin

θθ

... (ii) (Squaring)

Further, n = sec θ – cos θ (Given)

⇒ n = 1

– coscos

θθ

⇒ n =21 – cos

cosθ

θ

⇒ n = 2sin

cosθθ

... (iii) ⇒ n2= 4

2

sin

cos

θθ

... (iv)

Multiplying equations (ii) and (iii), we get

m2n = 4 2

2

cos sin×

cossin

θ θθθ

= cos3 θ

∴ (m2n)2/3 = cos2 θ ... (v)Multiplying equations (i) and (iv), we get

n2m = 4

2

sin

cos

θθ

×2cos

sinθ

θ = sin3 θ

∴ (n2m)2/3 = sin2 θ ... (vi)

Adding equations (v) and (vi), we get

(m2n)2/3 + (n2m)2/3 = cos2 θ + sin2 θi.e., (m2n)2/3 + (n2m)2/3 = 1. Hence proved.

27. Given equations are: sin θ + cos θ = p ... (i)and sec θ + cosec θ = q ... (ii)

Squaring both the sides of equation (i),we get sin2 θ + cos2 θ + 2sin θ cos θ = p2

Subtract unity from both the sides to get

p2 – 1 = 2sin θ cos θ ... (iii)

�����������������

Equation (ii) can be written as

q = 1

cos θ +

1sin θ

⇒ q = sin cossin cos

� � �

� �... (iv)

From equations (iii) and (iv), we get

q (p2 – 1) =sin cossin cos

� � �

� �× 2sin θ cos θ

⇒ q (p2 – 1) = 2(sin θ + cos θ)⇒ q (p2 – 1) = 2p. Hence proved.

OR

As x = a sin θ ⇒ ax

= 1

sin θ

and y = b tan θ ⇒ by

= 1

tan θ

∴ LHS = 2

2

a

x–

2

2

b

y = 2

1sin θ

– 2

1tan θ

= 2

1sin θ

– 2

2

cossin

θθ

= 2

2

1 cossin− θ

θ =

2

2

sinsin

θθ

= 1 = RHS Hence proved.28. As ∠A + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° – ∠A

⇒B C

2∠ + ∠

= 180 – A

2° ∠

= 90° – 12

∠A

∴ sinB C

2+⎛ ⎞

⎜ ⎟⎝ ⎠= sin

A90

2⎛ ⎞° −⎜ ⎟⎝ ⎠

= cosA2

⇒ sin2B C

2+⎛ ⎞

⎜ ⎟⎝ ⎠= cos2

A2

…(i)

Now, sin2 A2

+ sin2 B C

2+

= sin2 A2

+ cos2 A2

[Using (i)]= 1. Hence Proved.

29. The cumulative frequency table for thegiven data is given below:

Marks No. of students Cumulative(C.I.) ( f ) frequency (cf )

0-10 10 1010-20 f1 10 + f120-30 25 35 + f130-40 30 65 + f140-50 f2 65 + f1 + f250-60 10 72 + f1 + f2

N = 75 + f1 + f2

Clearly, N = 75 + f1 + f2But N = 100∴ f1 + f2 = 25 ... (i)

∴N2

= 50.

The median is 32 which lies in the class30-40.

So, l = 30, f = 30, cf = 35 + f1, h = 10.

Using the formula:

Median = l +

N2

cf

f

� ��� �� �� �

× h

⇒ 32 = 30 + 150 3530

f� �� �� �� �

× 10

⇒2

10= 115

30f� ⇒ 75 – 5f1 = 30

⇒ f1 = 75 30

5�

⇒ f1 = 9

Substituting f1 = 9 in equation (i), we get

9 + f2 = 25 ⇒ f2 = 16

Hence, f1 = 9 and f2 = 16.

30. Class-interval Frequency (C.I.) ( f )

0-10 310-20 920-30 1530-40 3040-50 1850-60 5

��� ����������

The class corresponding to the maximumfrequency is 30-40. So, 30-40 is the modalclass.

Mode = l +1 0

1 0 22f f

f f f�� �

� �� �� � × h

Here, l = 30, f1 = 30, f0 = 15, f2 = 18 andh = 10

� Mode = 30 + 30 15

2 30 15 18�� �

� �� �� � �×10

= 30 + 15

60 15 18� �� �� �� �

× 10

= 30 + 15

60 33� �� �� ��

× 10

= 30 + 1527

× 10 = 30 + 15027

= 30 + 5.5 = 35.5

OR

Class Mid- Freq- ui = fiui

interval values uencyAix

h

−

(xi) (fi)

10-30 20 540

220

� � � – 10

30-50 40 820

120

� � � – 8

50-70 60 = A 120

020

� 0

70-90 80 2020

120

� 20

90-110 100 340

220

� 6

110-130 120 260

320

� 6

Σfi = 50 Σfiui = 14

Let assumed mean be A = 60Here, h = 20

Mean = A + h × i i

i

f u

f

Σ⎛ ⎞⎜ ⎟Σ⎝ ⎠

= 60 + 20 × 1450

⎛ ⎞⎜ ⎟⎝ ⎠

= 60 + 5.6 = 65.6Hence, the required arithmetic mean is 65.6.

31. Let us convert the given data into less thantype distribution.

Class f Lifetimes cfinterval (in hrs.)

0-20 10 less than 20 10

20-40 35 less than 40 45

40-60 52 less than 60 97

60-80 61 less than 80 158

80-100 38 less than 100 196

100-120 29 less than 120 225

�����������������

3. �ABC ~ �DEF

�B = �E =180° – (40° + 65°) = 75°.

4. AC2 = BC2 – AB2

� AC2 = 2 – 1 = 1AC = 1

cosec C = BCAB

=2

1= 2 .

SECTION–B5. 8n can be rewritten as 23n. Clearly, the prime

factor of 8n is only 2. To end with the digit 0,one of the prime factors of 8n must be 5.Hence, 8n cannot end with the digit zero forany n �N.

6. LCM=First number × Second number

HCF

=96 × 404

4= 24 × 404 = 9696.

7. True, because we find the remainder zerowhen 3x4 + 5x3 – 7x2 + 2x + 2 is divided byx2 + 3x + 1.

8. Infinite number of solutions because thesystem obeys the following condition:

1

2

aa

= 1

2

bb

= 1

2

cc

, i.e., 13

=– 3– 9

=– 3– 9

.

9. In �AOD and �COB,

AOOC

= DOOB

= 12

and ��AOD =��COB� �AOD ~ �COB

�ADBC

=�12��� �

4BC

��12

� BC = 8 cm.OR

Let the height of the tower be h metres�ABC ~ �PQR

�ABPQ

=BCQR

���12h

= 840

� h = 12 × 40

8 = 60 metres.

10. Since the maximum frequency is 8, somodal class is 4-8.� l = 4, f1 = 8, f0 = 4, f2 = 5, h = 4

Now, mode = l + 1 0

1 0 2

–×

2 – –⎛ ⎞⎜ ⎟⎝ ⎠

f fh

f f f

= 4 +8 – 4

× 416 – 4 – 5

⎛ ⎞⎜ ⎟⎝ ⎠

= 4 + 2.29

= 6.29.

We mark the upper class limits along thex-axis with a suitable scale and thecumulative frequencies along the y-axiswith a suitable scale. For this, we plot thepoints A(20, 10), B(40, 45), C(60, 97), D(80,158), E(100, 196) and F(120, 225) on a graphpaper. These points are joined by a freehand smooth curve to obtain a less thantype ogive as shown in the given figure.

Practice Paper-3

SECTION–A

1. (D) There are infinitely many real numbersof both types rational and irrational

between 3 and 5 .

2. (B) For infinite number of solutions:

3k

= 5

10, i.e., k = 6.

��� ����������

SECTION–C

11. Let us assume on the contrary that 2 isa rational number. Then 2 can be written

as 2 = ab

, where a and b are coprime

and b � 0.

2 = 2

2ab

(Squaring)

� a2 = 2b2 ... (i)�� a2 is divisible by 2 ...(ii)� a is divisible by 2 ...(iii)[If a prime (here 2) divides d2, then thesame prime divides d, where d is a positiveinteger.]� a = 2c � a2 = 4c2 ... (iv)

From (i) and (iv), we get4c2 = 2b2 � b2 = 2c2

� b2 is divisible by 2� b is divisible by 2 ....(v)From results (ii) and (v), we have a and bboth are divisible by 2.But this contradict the fact that a and b arecoprime. This contradiction has arisenbecause of our incorrect assumption that

2 is a rational number. Thus, we concludethat 2 is an irrational number.

OR(i) The given fraction can be written as

4 3 443 43 × 5

= = 0.02152 ·5 10

Hence, the given number terminates afterfour places of decimal.(ii) The given fraction can be written as

4

5 5 5359 2 × 359

=2 × 5 2 × 5

= 5744

100000= 0.05744

Hence, the given number terminates afterfive places of decimal.

12. (i) LCM of 2, 3, 4, 5, 6 = 60� Required number = 60p + 1; p is

positive integer= (7 × 8 + 4)p + 1= (7 × 8p) + (4p + 1)

Now, this number is to be divisible by 7.Since 7 × 8p is always divisible by 7; so,we must choose the least value of pwhich will make 4p + 1 divisible by 7.Putting p = 1, 2, 3, 4, 5 etc. in successionwe find that p = 5.� Required number 60p + 1

= 60 × 5 + 1 = 301(ii) LCM of two real numbers.

(iii) Kindness and love towards society.

13. p(t) = t2 – 15To obtain zeroes of p(t), put p(t) = 0i.e., t2 – 15 = 0

t2 – ( )215 = 0 � ( ) ( )15 – 15+t t = 0

� t = – 15 , 15

So, zeroes of p(t) are – 15 and 15

Sum of zeroes = – 15 + 15 = 0

= – 01

= 2

– Coefficient of

Coefficient of

t

t

Product of zeroes = – 15 15× = – 15

= –151

= Constant termCoefficient of t

.

Hence verified.14. The given system of equations can be

rewritten as 4x + 3y – 48 = 0

40x – 6y – 192 = 0Applying the method of cross multiplica-tion to solve the system.

��– 576 – 288

x =

–– 768 + 1920

y=

1– 24 – 120

�– 864

x=

– 1152y

= 1

– 144

� x = 864144

and y = 1152144

� x = 6 and y = 8.

15. We are given a square of side length a.Then length of its diagonal will be a 2 .We know that area of an equilateral

triangle of side length x is 234

x .

�����������������

� Area of the equilateral triangle describedof the side of the square.

Aside = 234

a ... (i)�����a2 =43

Aside

And area of the equilateral triangledescribed on the diagonal of the square

Adiagonal = 3

4(a 2 )2 = 23

2a

�Adiagonal = 3

2× ⎛ ⎞

⎜ ⎟⎝ ⎠side

4A

3[Using (i)]

� Aside = diagonal1

× A2

. Hence proved.

16. In the figure drawn,AB DC and �AED ~�BEC.�ADC and �BDCboth are on the same base DC and liebetween same parallels AB and DC.So, ar(�ADC) = ar(�BDC)�� ar(�AED) + ar(�DEC)

= ar(�BEC) + ar(�DEC)� ar(�AED) = ar(�BEC) …(i)

Now,2

2( AED) (AD)

=( BEC) (BC)

arar�

�

(� �AED ~ �BEC)

� 1 = ( )( )

2

2

AD

BC[From (i)]

� AD = BC. Hence proved.17. Given expression

=4 41 1

22 2

� �� �� � � ��� � � � � �� � �– ( ) ( ){ }2 23 1+

+ 3⎛ ⎞⎜ ⎟⎝ ⎠

223

= 2 1 1

16 16⎛ ⎞+⎜ ⎟⎝ ⎠

– ( )3 1+ + 3 × 43

= 14

– 4 + 4 = 14

.

18. Consider left hand side of the givenequation.LHS = (cosec A – sin A) (sec A – cos A)

= 1

– sin Asin A

⎛ ⎞⎜ ⎟⎝ ⎠

1– cos A

cos A⎛ ⎞⎜ ⎟⎝ ⎠

= 2 21 – sin A 1 – cos A

.sin A cos A

= 2 2cos A sin A

.sin A cos A

= sin A cos A

Also, taking right hand side,

RHS = 1

tan A cot A+ =1

sin A cosAcosA sin A

+

= 2 2

sin A cos A

sin A cos A+ = sin A cos A

Hence, LHS = RHS.OR

Draw ��ABC withAB = BC = AC = a (say)

Draw AD BC� �BAD = �DAC = ��= 30°and BD = DC = a/2

� sin ����BD /2AB

aa

= �12�� sin 30����

12

.

19. Given equations are: sin �� + cos ��= p ... (i)and sec �� + cosec ��= q ... (ii)Squaring both the sides of equation (i),we get sin2 ��+ cos2 �� �2sin ��cos ��= p2

Subtract unity from both the sides to getp2 – 1 = 2sin ��cos � ... (iii)

Equation (ii) can be written as

q = 1

cos θ +

1sin θ

� q = sin cossin cos

� � �

� �... (iv)

From equations (iii) and (iv), we get

q �p2 – 1� =sin cossin cos

� � �

� �× 2sin � cos �

� q �p2 – 1� = 2(sin � + cos �)� q �p2 – 1� = 2p. Hence proved.

��� ����������

20. Let us use step-deviation method to obtainthe mean.

C.I. fi xi di ui=id

hdiui

= xi–a

0-20 17 10 – 40 – 2 – 3420-40 p 30 – 20 – 1 – p40-60 32 50 0 0 060-80 24 70 20 1 24

80-100 19 90 40 2 38

�fi � � � � ��fiui= 92+p = 28 – p

Here a = 50, h = 20

Mean = a +� ×⎛ ⎞∑⎜ ⎟

∑⎝ ⎠i i

i

f uh

f

� 50 = 50 + 28 –92 +

pp

� 28 – p = 0 � p = 28.

ORSince mode = 36, which lies in the classinterval 30-40, so the modal class is 30-40.

��f1 = 16, f0 = f, f2 = 12, l = 30 and h = 10.

Now, mode = l + 1 0

1 0 22f f

f f f�� �

� �� �� �× h

� 36 = 30 + 16

32 12f

f�� �

� �� �� �× 10

�6

10=

1620

ff

�

�

� 120 – 6f = 160 – 10 f

� 4f = 40

������f = 10.

SECTION – D21. We have

x3 = sec A – cos A

= 1

– cos Acos A

= 21 – cos A

cos A

� x =

12 3sin A

cos A

⎛ ⎞⎜ ⎟⎝ ⎠

Similarly, y3 = cosec A – sin A

= 21 – sin A1

– sin Asin A sin A

=

y =

12 3cos A

sin A

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

LHS = x2 y2 (x2 + y2) = x4y2 + x2y4

=

4 21 12 23 3sin A cos A

cos A sin A

� � � �� � � �� � � �� � � � � � � �� � � �� �

+

2 41 12 23 3sin A cos A

cos A sin A

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

=

4 22 23 3sin A cos A

cos A sin A

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+

2 42 23 3sin A cos A

cos A sin A

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

18 4 3

4 2

sin A cos A

cos A sin A

⎛ ⎞×⎜ ⎟⎜ ⎟⎝ ⎠

+

14 8 3

2 4

sin A cos A

cos A sin A

⎛ ⎞×⎜ ⎟⎜ ⎟⎝ ⎠

= ( ) ( )1 1

6 63 3sin A cos A+ = sin2A + cos2 A

= 1 = RHS. Hence proved.

22. Let the original fraction be xy .

On adding 1 to both the numerator and

the denominator of xy , it becomes 4

5

i.e. ,+ 1+ 1

xy

= 45

, i.e., 5x + 5 = 4y + 4

i.e., 5x – 4y = –1 ...(i)

On subtracting 5 from both the numerator

and the denominator of xy , it becomes

12

�����������������

i.e., – 5– 5

xy

= 12

, i.e., 2x – 10 = y – 5

i.e., 8x – 4y = 20 ...(ii)Subtracting equation (i) from equation (ii),we get 3x = 21 � x = 7Substituting x = 7 in equation (i), we get 5 × 7 – 4y = –1 � y = 9

Hence, the required fraction is79

.OR

Let incomes of X and Y be 8x and 7xrespectively; and their expenditures be 19yand 16y respectively.We know that:Income – Expenditure = Savings... 8x – 19y = 1250 …(i)and 7x – 16y = 1250 …(ii)Comparing equations (i) and (ii), we have

8x – 19y = 7x – 16y or x = 3y …(iii)Substituting this value of x in equation (i),we get 24y – 19y = 1250� 5y = 1250 � y = 250Substituting this value of y in equation (iii),we get... x = 3 × 250 = 750.Now, 8x = 8 × 750 = 6000and 7x = 7 × 750 = 5250hence X’s income is 6000 and Y’s income is5250.

23. Let us make the table for the values of x andcorresponding values of y to the equation

2x + y – 8 = 0

x 2 4 y 4 0

Similarly, for the equation x – y – 1 = 0

x 4 3 y 3 2

Let us draw the graph.From the graph, the lines intersect eachother at the point A(3, 2). Therefore, thesolution is x = 3, y = 2.The lines intersect the y-axis at B(0, 8) andC(0, – 1).

To find the area of the shaded portion, thatis, �ABC, draw perpendicular AM from Aon the y-axis to intersect it in M.

Now, AM = 3 units and BC = 8 + 1 = 9 units.

... ar(�ABC) =12

× BC × AM

=12

× 9 × 3 =272

sq. units

Hence, x = 3, y = 2; area = 13.5 sq.units.24. Pythagoras Theorem:

In a right triangle, the square of thehypotenuse is equal to the sum of thesquares of the other two sides.Proof: We are given, a �ABC in which �A= 90°

We need to prove BC2 = AB2 + AC2.Draw AD BC (see figure)In �ABC and �DBA,

�ABC = �DBA (Common)

��� ����������

�BAC = �BDA (Each 90°)So, �ABC ~ �DBA (A A criterion of similarity)

Therefore,ABBD

= BCAB

or AB2 = BD . BC ...(i)Similarly, �ABC ~ �DAC

Therefore,ACDC

=BCAC

AB2 = DC . BC ...(ii)Adding equations (i) and (ii), we getBD . BC + DC . BC = AB2 + AC2

or (BD + DC) . BC = AB2 + AC2

or BC . BC = AB2 + AC2

(��BC = BD + DC)or BC2 = AB2 + AC2. Hence proved.

ORConverse of Pythagoras Theorem: In atriangle, if square of one side is equal tothe sum of the squares of the other twosides, then the angle opposite the first sideis a right angle.

Proof: We are given a �ABC in whichAC2 = AB2 + BC2 ...(i)

We need to prove �ABC = 90°.Let us construct a �PQR such that �PQR =90° and PQ = AB ...(ii)

QR = BC ...(iii)Using Pythagoras Theorem in �PQR, wehave PR2 = PQ2 + QR2

� PR2 = AB2 + BC2 ...(iv)[Using equations (ii) and (iii)]

From equations (i) and (iv), we haveAC = PR ...(v)

Now, in �ABC and �PQR,AB = PQ (From (ii)]BC = QR [From (iii)]

AC = PR [(From (v)]So, �ABC � �PQR (SSS congruence)� �ABC = �PQR (CPCT)But �PQR = 90° (By construction)� �ABC = 90°. Hence proved.

25. Let the given parallelogram be ABCDWe need to prove thatAC2 + BD2 = AB2 + BC2 + CD2 + DA2

Let us draw perpendiculars DN on AB andCM on AB produced as shown in figure.

In �BMC and �AND,BC = AD (Opposite sides of a gm)

�BMC = �AND (Each 90°)CM = DN

(Distance between same parallels)� �BMC � �AND (RHS criterion)� BM = AN ...(i) (CPCT)In right triangle ACM, AC2 = AM2 + CM2

= (AB + BM)2 + BC2 – BM2

= AB2 + 2AB . BM + BM2 + BC2 – BM2

= AB2 + BC2 + 2AB. BM ...(ii)In right triangle BDN,BD2 = BN2 + DN2

= (AB – AN)2 + (AD2 – AN2)= AB2 – 2AB .AN + AN2 + AD2 – AN2

BD2 = AB2 + DA2 – 2AB.AN� BD2 = CD2 + DA2 – 2AB .BM ...(iii)

[Using (i) and AB = CD]Adding equations (ii) and (iii), we haveAC2 + BD2 = AB2 + BC2 + CD2 + DA2.

Hence proved.26. As��� � are zeros of p(x) = x2 – px + q.

��� + � = p;�� = q

�����������������

� Consider 2 2

2 2� �

�� �

= 4 4

2 2� ��

� �

= 2 2 2 2 2

2 2

( ) 2� � � � � �

� � =

2 2 2 2

2

( ) 2q

q

� � � �

=

22 2

2

( ) 2 2q

q

� �� � � � �� �� � =2 2 2

2

( 2 ) 2p q q

q

� �

= 4 2 2 2 4 2 2

2 2

4 4 2 2 4p q p q q p q p q

q q

� � � � �

�

= 4 2

2

42

p pqq

� � = RHS. Hence proved.

27. sec � = x +1

4x ... (i) (Given)

or sec2 � = 21

4⎛ ⎞+⎜ ⎟⎝ ⎠

xx

(Squaring)

or 1 + tan2 � = x2 + 21 1

216+

x

or tan2 � = x2 + 21 1

–216x

or tan2 � = 21

–4

⎛ ⎞⎜ ⎟⎝ ⎠

xx

or tan � = + 1–

4⎛ ⎞⎜ ⎟⎝ ⎠

xx

... (ii)

Add equations (i) and (ii) to get

sec ��+ tan � = x + 1

4x + x –

14x

or x + 1

4x – x +

14x

= 2x or 1

2x.

Hence proved.

28. LHS = (1 + cot A – cosec A)

(1 + tan A + sec A)

= cos A 1

1sin A sin A

� �� �� �� �sin A 1

1cos A cos A

� �� �� �� �

= sin A + cos A –1sin A

× cos A + sin A + 1cos A

= ( )2 2sin A + cos A – 1

sin A cos A

= 2 2sin A + 2 sin A cos A + cos A – 1

sin A cos A

= 2 sin A cos Asin A cos A

= 2 = RHS. Hence proved.

29. We prepare the cumulative frequency tableby less than method as given below:

Marks Freq. Marks cf Pointless than

0-10 4 10 04 (10, 4)10-20 10 20 14 (20, 14)20-30 16 30 30 (30, 30)30-40 22 40 52 (40, 52)40-50 20 50 72 (50, 72)50-60 18 60 90 (60, 90)60-70 8 70 98 (70, 98)70-80 2 80 100 (80, 100)

We take marks on the x-axis andcumulative frequency on the y-axis andthen plot the points mentioned in the table.On joining these points by free handsmooth curve, we get less than ogive.Further, we prepare the cumulativefrequency table by more than method asgiven below:

Marks Freq. Marks cf Pointmore thanor equal to

0-10 4 0 100 (0, 100)10-20 10 10 96 (10, 96)20-30 16 20 86 (20, 86)30-40 22 30 70 (30, 70)40-50 20 40 48 (40, 48)50-60 18 50 28 (50, 28)60-70 8 60 10 (60, 10)70-80 2 70 2 (70, 2)

We will plot the points mentioned in thistable on the same graph. On joining thesepoints by free hand smooth curve, we getmore than ogive.

��� ����������

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

30

25

20

15

10

5

010 20 30 40 50 60 70 80 90 100

(0, 100)

(10, 96)

(20, 86)

(30, 70)

(40, 48)

(50, 28)

(60, 10)

(70, 2)

more than ogive

less than ogive

(80, 100)(70, 98)

(60, 90)

(50, 72)

(40, 52)

(30, 30)

(20, 14)

(10, 4)(39, 0)

Y

X

Median: The abscissa of the point ofintersection of both the ogivesdetermines the median of the givendistribution. To find such abscissa, wedraw a perpendicular from the point ofintersection of both the ogives to thex-axis, which meet the axis at (39, 0).Hence the required median is 39 marks.

30. The given distribution can be again repre-sented with the cumulative frequencies asgiven below:

Class interval fi xi cf fi xi

100-120 12 110 12 1320120-140 14 130 26 1820140-160 8 150 34 1200160-180 6 170 40 1020180-200 10 190 50 1900

50 7260

Mean: Mean = i i

i

f x

f

ΣΣ

� � fi = 50 and �fixi = 7260

� Mean = 7260

50= � 145.20.

Median: Median = l +

N2

cf

f

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

× h

� N = 50, N2

= 25, f = 14, cf = 12,

l = 120 and h = 20

� Median = 120 + 25 1214�� �

� �� � × 20 = � 138.57

Mode: Mode = l + 1 0

1 0 22f f

f f f�� �

� �� �� � × h

� l = 120, f1 = 14, f0 = 12, f2 = 8 and h = 20

� Mode =120 + 14 12

2 14 12 8�� �

� �� �� � �× 20 = �125

31. We prepare cumulative frequency tablefrom the given data.

C.I. Frequency Cumulative( f ) Frequency (cf )

0-8 8 88-16 10 18

16-24 16 3424-32 24 5832-40 15 7340-48 7 80

N = 80

Here, N = 80 � N2

= 40

Cumulative frequency just more than 40 is58. So 24-32 is the median class.� l= 24, cf = 34, f = 24, h = 8

� Median = l +

N–

2 ×

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

cfh

f

= 24 +40 – 34

24⎛ ⎞⎜ ⎟⎝ ⎠

× 8 = 24 + 4824

= 26.

��������������� �

4. If x = 30°,3 cos 30° – 4 cos3 30°

= 3 ×3

2 – 4 ×

3 38

= 0.

SECTION–B

5. As p2 and p3 are common factors.

��HCF (x, y) is 2 3n lp p .

6. The required highest number will be theHCF of 120, 224 and 256.120 = 23 × 3 × 5; 224 = 25 × 7; 256 = 28

Therefore, HCF = 23 = 8.7. Let the required polynomial be f(x).

Then f (x) = k[x2 – (sum of zeroes) x + product of zeroes]

= k[x2 – (–5 + 2)x + (–5)(2)]= k(x2 + 3x – 10), k being a real number.

f (x) is not a unique polynomial as k is anyreal number.

8. Condition for infinite number of solutions:

1

2

aa �� 1

2

bb��� 1

2

cc i.e.,

2 34

k ���

2 13

k ��

4( 1)3

k ��

i.e., k = 52

.

9. No. In �AOB and �DOC, �AOB = �DOC (Vertically opposite angles)

AO OBDO OC

� as 5 63 10� , i.e., 5 3

3 5�

Therefore, �AOB is not similar to�DOC.

10. Yes. Let us take left hand side of the givenequation.

LHS = 2

2

1 + sin

cos

�

�= 2

1

cos �+

2

2

sin

cos

�

�

= sec2 � + tan2 ��= (1 + tan �)2 + tan2 � = 1 + 2 tan2 ��= RHS.

OR�� A + B + C = 180º

�� LHS = C + A 180º – B cot = cot

2 2

= cot (90º – B2

) = tan B2

= RHS.

SECTION–C

11. Let us assume to the contrary that 3 4 5�

is rational. Then we can take integers a andb � 0 such that

ab

= 3 – 4 5 ,

i.e., 4 5 = 3 – ab

i.e., 5 = 3

4b a

b�

Practice Paper-4

SECTION–A

1. (D) 2 32 3�

�

= 2 3 2 32 3 2 3� �

�

� �

= 2 3 2 6

2 3� �

�

= 2 6 5�

As 2 6 5� is irrational, it has non-terminating, non-repeating decimal form.

2. (C) 2 + k = 1 and p + 3 = 2, i.e., k = – 1and p = – 1. So, p + k = – 2.

3. ADBD

= AECE

= 13

� DE BC

���ADE���ABC � ��� DE ADBC AB

�

����7.5x

= 3.514

� � �� x = 158

cm

��� �������� �

RHS of this last equation is rational as aand b are integers, but LHS of it isirrational. This is an incorrect statementdue to our wrong assumption that 3 4 5�

is rational.So, we conclude that 3 4 5�

is irrati-onal.

12. Any positive integer is either of the form3q, 3q + 1 or 3q + 2.There are 3 cases now:Case I. When n = 3q, n + 1 = 3q + 1and n + 2 = 3q + 2.Here, only n is divisible by 3.Case II. When n = 3q + 1, n + 1 = 3q + 2and n + 2 = 3q + 3 = 3(q + 1)Here, only n + 2 is divisible by 3.Case III. When n = 3q + 2,n + 1 = 3q + 3 = 3(q + 1)and n + 2 = 3q + 5 = 3(q + 1) + 2Hence only n + 1 is divisible by 3.

ORLet x be any positive integer. Then x can be ofthe form: 5m, 5m + 1, 5m + 2, 5m + 3, 5m + 4� If x = 5m x2 = 25m2 = 5(5m2) = 5q. If x = 5m + 1 ���x2 = 25m2 + 1 + 10m

= 5(5m2 + 2m) + 1 = 5q + 1 If x = 5m + 2 � x2 = 25m2 + 4 + 20m

= 5(5m2 + 4m) + 4 = 5q + 4 If x = 5m + 3 � x2 = 25m2 + 9 + 30m

= 5(5m2 + 6m + 1) + 4 = 5q + 4 If x = 5m + 4 x2 = 25m2 + 16 + 40m

= 5(5m2 + 8m + 3) + 1 = 5q + 1� Square of x can be of the form: 5q, 5q +1 or 5q + 4.

13. By the division algorithm,Dividend = Divisor × Quotient

+ Remainder� 6x3 + 8x2 – 3x + 8

= g(x) × (3x + 4) + 6x + 20

g(x) = 3 26 8 9 12

3 4x x x

x� � �

�

= 22 (3 4) 3(3 4)

3 4x x x

x� � �

�

g(x) = 2(3 4)(2 3)

3 4x x

x� �

�

g(x) = 2x2 – 3.

14. As��� are zeros of p(x) = x2 – px + q.

�� + = p;

� = q

� Consider 2 2

2 2

� ��

� �=

4 4

2 2

� ��

� �

= 2 2 2 2 2

2 2

( ) 2� � � � � �

� � =

2 2 2 2

2

( ) 2q

q

� � � �

=

22 2

2

( ) 2 2q

q

� �� � � � �� �� � =2 2 2

2

( 2 ) 2p q q

q

� �

= 4 2 2 2 4 2 2

2 2

4 4 2 2 4p q p q q p q p q

q q

� � � � �

�

= 4 2

2

42

p pqq

� � = RHS. Hence proved.

ORLet the present ages of father and his sonare x years and y years respectively.According to the given conditions:

x + y = 65After 5 years, the father’s age

= (x + 5) yearsAfter 5 years, the son’s age = (y + 5) yearsTherefore, x + 5 = 2 (y + 5)i.e., x – 2y = 5Thus, the required pair of linear equations isx + y = 65.... (i) and x – 2y = 5.... (ii)Subtracting equation (ii) from equation (i),we get 3y = 60 � y = 20Substituting y = 20 in equation (i), we get

x + 20 = 65 �x = 45Thus, present age of father = 45 years andpresent age of his son = 20 years.

15. Let us draw MN parallel to AB, whichpasses through P. So, AM = BN andDM = CN.

��������������� �

From right-angled triangles APM, BPN,CPN, DPM; we have respectively

PA2 = PM2 + AM2 .... (i)

PB2 = PN2 + BN2 .... (ii)

PC2 = PN2 + CN2 .... (iii)

PD2 = PM2 + DM2 .... (iv)

From equations (i) and (ii),

PA2 – PB2 = PM2 – PN2 .... (v)

From equations (iii) and (iv),

PC2 – PD2 = PN2 – PM2 .... (vi)

Add equations (v) and (vi) to get

PA2 + PC2 = PB2 + PD2. Hence proved.

16. Let median AD passes through the pointO on PQ.

To prove: PO = QO

Proof: In �APO and �ABD,

�PAO = �BAD

(Common angle)

� � �

�� � �

APO = ABDand AOP = ADB

(Corresponding angles)���APO ~ �ABD (By AAA similarity)

POBD

= AOAD

...(i)

Similarly, in �AQO and �ACD,

QOCD

= AOAD

...(ii)

From equations (i) and (ii), we have

POBD

= QOCD

PO = QO (... BD = CD)

Hence, median AD also bisects PQ.Hence proved.

ORLet the given parallelogram be ABCDWe need to prove thatAC2 + BD2 = AB2 + BC2 + CD2 + DA2

Let us draw perpendiculars DN on AB andCM on AB produced as shown in figure.

In �BMC and �AND,BC = AD (Opposite sides of a gm)

�BMC = �AND (Each 90°)CM = DN

(Distance between same parallels)��BMC � �AND (RHS criterion) BM = AN ...(i) (CPCT)In right triangle ACM,AC2 = AM2 + CM2

= (AB + BM)2 + BC2 – BM2

= AB2 + 2AB . BM + BM2 + BC2 – BM2

= AB2 + BC2 + 2AB. BM ...(ii)In right triangle BDN,BD2 = BN2 + DN2

= (AB – AN)2 + (AD2 – AN2)= AB2 – 2AB .AN + AN2 + AD2 – AN2

BD2 = AB2 + DA2 – 2AB.AN BD2 = CD2 + DA2 – 2AB .BM ...(iii)

[Using (i) and AB = CD]Adding equations (ii) and (iii), we haveAC2 + BD2 = AB2 + BC2 + CD2 + DA2.

Hence proved.

17. cos sin 1 3cos sin 1 3

� � � ��

�� � �

Using componendo and dividendo, we get

cos sin cos sincos sin cos sin

� � � � � � �

� � � � � � �=

1 3 1 31 3 1 3� � �

� � �

�2 cos2 sin

�

� �=

22 3�

�cot ��= 13

� cot ��= cot 60° ���= 60°.

18. LHS = 2 4 2 4

2 1 2 1

cos cos sin sin� � �

� � � �

= 2 sec2 ��– sec4 ��– 2 cosec2 ��+ cosec4 �

��� �������� �

= 2(1 + tan2 �) – (1 + tan2 �)2

– 2(1 + cot2 �) + (1 + cot2 �)2

= (1 + tan2 �) × (2 – 1 – tan2 �)

– (1 + cot2 �) × (2 – 1 – cot2 �)

= 1 – tan4 ��– (1 – cot4 �)

= cot4 ��– tan4 ��= RHS. Hence proved.

19. We know that cosec (90° – �) = sec �

cot (90° – �) = tan � tan (90° – �) = cot �

Now,2 2

sec .cosec (90° – ) tan .cot(90 )sin 55 sin 35

tan 10 .tan 20 . tan 60 .tan 70 .tan 80

θ θ − θ ° − θ+ ° + °

° ° ° ° °

=

2

2sec . sec tan .tan sin 55 sin (90 55 )

tan10 tan 20 . 3 .tan(90 20 ) tan(90 10 )

θ θ − θ θ + °+ ° − °

°× ° ° − °° − °

= 2 2 2 2sec tan sin 55 cos 55

tan 10 .tan 20 . 3 cot 20 cot10� � � � � � �

� � � � �

= 1 1

1 1tan 10 .tan 20 . 3

tan 10 tan 20

�

� � � �� �

= 23

20. We prepare the cumulative frequency tablefor the given data.

Lifetimes Frequency Cumulative(in hrs.) ( f ) frequency (cf )

1500-2000 24 242000-2500 86 1102500-3000 90 2003000-3500 115 3153500-4000 95 4104000-4500 72 4824500-5000 18 500

N = 500

Here, h = 500

� N = 500, ��N2

= 250.

So, cf = 200, f = 115, l = 3000.

Median = l +

N2

cf

f

� ��� �� �� �

× h

= 3000 +250 200

115�� �

� �� � × 500

= 3000 + 217.39 = 3217.39 hours.

SECTION – D

21. To draw a line, we need atleast twosolutions of its corresponding equation.

x 0 2 x 1 0

y – 4 0 y 0 – 1

� � � � � Two solutions of Two solutions of 2x – y = 4 x – y = 1From the graph, the two lines intersecteach other at the point A(3, 2).� x = 3 and y = 2Shaded region is �ABC.��Height of �ABC = 3 unitsAnd its base = BC = 3 units.

� ar(�ABC) = 12

× 3 × 3 = 92

square units.

��������������� �

22. Given system of linear equations may bewritten as

bx + ay – ab (a + b) = 0b2x + a2y – 2 a2b2 = 0

Applying cross-multiplication, we get

3 2 4 3 22x

a b a b a b� � �

= 2 3 2 3 42

y

a b a b ab

�

� � �

= 2 2

1

a b ab�

4 3 2

x

a b a b�

= 4 2 3

y

ab a b

�

�

= 2 2

1

a b ab�

(i) (ii) (iii)Taking (i) and (iii),

x = 34 3 2

22 2

( )( )

a b a ba b a ba

ab a ba b ab

��

� �

��

Taking (ii) and (iii),

y = �

�

� �

��

34 2 32

2 2

( )– –

( )ab b aab a b

bab a ba b ab

Thus, the required solution is x = a2, y = b2.23. (i) Let p(x)= Total Relief Fund

g(x) = Number of families whoreceived Relief Fund

q(x) = Amount each family received r(x) = Amount left after distributionWhen the polynomial p(x) is divided bya polynomial g(x) such that q(x) and r(x)are respectively the quotient and theremainder, the division algorithm is

p(x) = g(x) . q(x) + r(x) …(i)According to the question,

p(x) = 3x3 + x2 + 2x + 5q(x) = 3x – 5

and r(x) = 9x + 10Substituting these values of p(x), q(x) andr(x) in the equation (i), we get

3x3 + x2 + 2x + 5 = g(x) (3x – 5) + 9x + 10

(3x – 5) g(x) = 3x3 + x2 + 2x + 5 – 9x – 10

= 3x3 + x2 – 7x – 5

g(x) = 3 23 7 5

3 5x x x

x� � �

�

To find g(x), we proceed as following:

2

3 2

3 2

2

2

2 13 5 3 7 5

3 5

6 7 56 10

3 53 5

0

x xx x x x

x x

x xx x

xx

� �

� � � �

�

� �

� �

�

� �

�

�

� �

Thus, g(x) = x2 + 2x + 1.

(ii) Common good, Accountability, socialresponsibility.

24. We are given a �ABC in which a linePQ BC intersects the sides AB at P andAC at Q.

We need to prove AP AQPB QC

� .

Draw QM ��AB and PN ��AC.

Join PC and BQ.

Proof: Area of a triangle

= 12

× Base × Height

� ar(�APQ) = 12

× AP × QM

= 12

× AQ × PN

AP × QM = AQ × PN

QMPN

= AQAP

... (i)

Since �BPQ and �CQP are on the samebase PQ and between the same parallelsPQ and BC, therefore, their areas shouldbe equal.i.e., ar(�BPQ) = ar(�CQP)

12

× PB × QM = 12

× QC × PN

QMPN

=�QCPB

... (ii)

��� �������� �

Form equations (i) and (ii), it is clear that

AQAP

=�QCPB

��i.e., APPB

=�

AQQC

�

Hence proved.25. Let us produce AD to J and PM to K so

that DJ = AD and MK = PM.Join CJ and RK.

In �ADB and �JDC,

AD = JD, �ADB = �JDC, BD = CD

�ADB � �JDC

(SAS criterion of congruence)

AB = JC ...(i) (CPCT)Similarly, we can prove that

PQ = KR ...(ii)According to the given conditions, we have

ABPQ

= ADPM

= ACPR

JCKR

=

AJ2

PK2

= ACPR

[Using (i) and (ii)]

JCKR

= AJPK

= ACPR

�AJC ~ �PKR

(SSS criterion of similarity)

�JAC = �KPR (Corresponding angles)

i.e., �DAC = �MPR ...(iii)

Similarly, we can prove that

�DAB = �MPQ ...(iv)

Adding equations (iii) and (iv), we obtain

�BAC = �QPR ...(v)

Thus, in �ABC and �PQR, we have

ABPQ

= ACPR

(Given)

and�BAC = �QPR [From (v)]

Therefore, �ABC ~ �PQR. (SAS criterion of similarity)

Hence proved.

26. m = cosec � – sin ��=1

sin θ – sin �

= 21– sin

sinθ

θ =

2cossin

θθ

n = sec � – cos ��= 1

cos� – cos �

= 21– cos

cos�

� =

2sincos

�

�

Now, LHS = ( ) ( )2 2

2 23 3m n mn+

=

24 2 3

2cos sin

×cossin

⎛ ⎞θ θ⎜ ⎟⎜ ⎟θθ⎝ ⎠

+

22 4 3

2cos sin

×sin cos

⎛ ⎞θ θ⎜ ⎟⎜ ⎟θ θ⎝ ⎠

= ( )2

3 3cos θ + ( )2

3 3sin θ

= cos2 ��+ sin2 ��= 1 = RHS.

27.cot tan

+1 – tan 1 – cot

x x

x x= 1 + sec x . cosec x

LHS = cot tan

+1 – tan 1 – cot

x x

x x

=

cos sinsin cos

+sin cos

1 – 1 –cos sin

x x

x xx x

x x

=cos cos

×sin cos – sin

x xx x x

+ sin sin

×cos sin – cos

x xx x x

=2cos

sin (cos – sin )x

x x x+

2sincos (sin – cos )

xx x x

= 3 3cos – sin

.sin cos (cos – sin )x x

x x x x

��������������� �

=2 2.(cos – sin )(cos cos sin sin )

.sin cos (cos – sin )x x x x x x

x x x x� �

=.1 cos sin

.sin cosx x

x x�

= 1.sin cosx x

+ 1

= 1 + sec x . cosec x = RHS.Hence proved.

28. It is given that sin � + cos � = aSquaring both sides, we getsin2 ��+ cos2 ��+ 2 sin ��cos ��= a2

But sin2 ��+ cos2 � = 1 .... (i)� 1 + 2 sin ��cos � = a2

2 sin ��cos � = a2 – 1

sin ��cos � = 2 12

a �

.... (ii)

Cubing both sides of equation (i), we getsin6 ��+cos6 � + 3 sin2 � cos2 �� (sin2 �

+ cos2 �) = 1

sin6 ��+ cos6 ��+ 3 (sin � cos �)2 = 1[Using (i)]

sin6 ��+ cos6 � + 3 22 1

2a� ��

� �� �= 1

sin6 ��+ cos6 � = 1 – 3 × 2 2( 1)

4a �

� sin6 ��+ cos6 � = 2 24 3( 1)

4a� �

.

Hence proved.OR

Considera sin3 � + b cos3 � = sin � . cos � a sin � . sin2 � + b cos � . cos2 �

= sin � . cos � b cos � . sin2 � + b cos � . cos2 �= sin � . cos � (... a sin � = b cos �) b sin2 � + b cos2 � = sin �

b (sin2 � + cos2 �) = sin �� b = sin � ...(i)Again, a sin � = b cos � a . b = b cos � [From (i)]� a = cos � ...(ii)Now, squaring and adding equations (i)and (ii), we get

b2 + a2 = sin2 � + cos2 �

� a2 + b2 = 1.Hence proved.

29. We prepare a table for less than series withcorresponding cumulative frequencies andpoints.

Marks Freq. Marks Cumul- Pointless ativethan Freq.

0-10 5 10 5 (10, 5)10-20 8 20 13 (20, 13)20-30 6 30 19 (30, 19)30-40 10 40 29 (40, 29)40-50 6 50 35 (50, 35)50-60 5 60 40 (60, 40)

We take upper limits on the x-axis andcumulative frequencies on the y-axis. Thenwe plot the points on the graph paper. Byjoining these points by free hand smoothcurve, we obtain less than ogive as shownin the above figure.

ORA more than type distribution from thegiven distribution is given below:

��� �������� �

Production yield (in kg/ha) Cumulativefreq.

More than or equal to 50 100More than or equal to 55 98More than or equal to 60 90More than or equal to 65 78More than or equal to 70 54More than or equal to 75 16

To draw more than type ogive, we marklower limits from the given table onx-axis and cumulative frequency from theabove table on y-axis with suitablescale(s). We plot the points (50, 100), (55,98), (60, 90), (65, 78), (70, 54) and (75, 16).Now, we join these points with free handsmooth curve as shown. in the adjoiningfigure.

Fig. : More than type ogive.

50-70 60 = A 120

020

� 0

70-90 80 2020

120

� 20

90-110 100 340

220

� 6

110-130 120 260

320

� 6

�fi = 50 �fiui = 14

30.

Class Mid- Freq. ( fi ) ui = fiui

interval valuesAix

h

−

(xi )

10-30 20 540

220

� � � – 10

30-50 40 820

120

� � � – 8

�����������������

Let assumed mean be A = 60Here, h = 20

Mean = A + h × i i

i

f u

f

Σ⎛ ⎞⎜ ⎟Σ⎝ ⎠

= 60 + 20 × 1450

⎛ ⎞⎜ ⎟⎝ ⎠

= 60 + 5.6 = 65.6Hence, the required arithmetic mean is65.6.

31. Class-interval Frequency

0-10 510-20 1120-30 1930-40 3040-50 15

The class corresponding to the maximumfrequency is 30-40. So, 30-40 is the modalclass.

Mode = l + 1 0

1 0 22f f

f f f�� �

� �� �� � × h

Here, l = 30, f1 = 30, f0 = 19, f2 = 15, h = 10

Mode = 30 + 30 19

2 30 19 15�� �

� �� �� � � × 10

= 30 + 11

60 34� �� �� �� × 10

= 30 + 1126

× 10

= 30 + 11026

= 30 + 4.23 = 34.23.

Practice Paper-5

SECTION–A

1. (B) Let us assume that x + y is rationalnumber and let x + y = z; when z isrational.� x + y + 2 xy = z2 � 2 xy = z2 – x – y

xy = 2 – –

2z x y

which given a contradiction as LHS isirrational but RHS is rational.

� x + y is an irrational number.

2. (C) Coincident lines is given by

1

2

aa

= 1

2

bb

= 1

2

cc

Here, a1 = p, b1 = q, c1 = – 4a2 = 4, b2 = 3, c2 = – 5

Now,4p

= 3q

= – 4– 5

���4p

= 3q

= 0.8

� p = 3.2, q = 2.4

Therefore, 3p + q = 12.

3. Consider 3A = 90°� � �� A = 30°So, sin 30° – cos 2 × 30° = sin 30° – cos 60°

= 12

–12

= 0.

4.DE EF FDAB BC CA

+ ++ +

=EFBC

� Perimeter of �DEF = 42

× (3 + 2 + 2.5)

� Perimeter of �DEF = 15 cm.

SECTION–B

5. True, because product of an even numberand an odd number is an even number.

6. Going in opposite direction to the factortree, we obtain(i) 2 × 330 = 660 and (ii) 2 × 165 = 330

7. For infinite number of solutions,

1

2

aa

= 1

2

bb

= 1

2

cc

i.e., 22a

= 5+a b

= 189

,

i.e., a = 2 and a + b = 10 i.e., a = 2, b = 8.

��� ����������

8.ABDF

= BCEF

= CADE

= 12

� �ABC ~ �DFE

����B�=��F. But �B = 60°, so, �F = 60°.

9. Let p (x) = ax2 + bx + cSo, y = ax2 + bx + c should be satisfied by(–1, 0), (0, –3) and (4, 0)Therefore, 0 = a – b + c; – 3 = c;

0 = 16a + 4b + c

� c = – 3, a = 34

, b = –94

.

Hence, p(x) =34

x2 –94

x – 3

� p(x) = 34

(x2 – 3x – 4)

This is the required expression.10.

xi 15 17 19 20 + p 23

fi 2 3 4 5p 6 �fi=15 + 5p

fixi 30 51 76 100p + 5p2 138 �fixi = 295

+100p + 5p2

Mean = ∑∑

i i

i

f xf

����20 = 2295 100 5

15 5+ +

+p pp

� 300 + 100p = 295 + 100p + 5p2

� 5p2 = 5 � p = � 1.But frequency cannot take negative value.So, p ��–1. Hence, p = 1.

ORRelation among mean, median and modeis given by the following impirical formula:Mode = 3 Median – 2 Mean

SECTION–C

11. Let a be any positive integer. We knowthat any positive integer is either of theform 2q or 2q + 1 for some integers q.� a = 2q or 2q + 1Case I. When a = 2q, a + 1 = 2q + 1.

� a (a + 1) = 2q × 2q + 12q (2q +1) = 2r, where r = q (2q + 1)

So, a (a + 1) is divisible by 2.Case II. When a = 2q + 1,

a + 1 = 2q + 2 = 2 (q + 1)� a (a + 1) = 2 (2q + 1) (q + 1)

where r = (2q + 1) (q + 1)So, a (a + 1) is divisible by 2.Hence, multiplication of any twoconsecutive positive integers is divisibleby 2.

12. Ram, Ravi and Nitin will meet next afterthe time given by the LCM of 5 days, 24days and 9 days.Now, we find out the LCM of 5, 24 and 9.5 = 5; 24 = 23 × 3; 9 = 3 × 3� LCM = 23 × 3 × 3 × 5 = 360They met last on Sunday. So, it will beSunday after 7n days, where n is a naturalnumber.So, it will be Sunday after 357 days.Therefore, it will be Wednesday after 360days. Hence, they will meet on nextWednesday.

ORWe represent 6, 72 and 120 in their primefactors.6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5Now, HCF = 2 × 3 = 6And LCM = 23 × 32 × 5 = 360.

13. Let zeroes are �, , Let � = 8 ... (i)Also we know � + + = 9 ... (ii)� + � + = 26

� = 24 � 8 ()�= 24 � �= 3From (ii), ���� + = 6 ... (iii)

From (i) , �� = 8�

� Use if in (iii) we get � + 8�

= 6

� �2 – 6� + 8 = 0(� – 4) (� – 2) = 0 ��� = 4 ���� = 2.

If � = 4 ���� = 2 and if � = 2� = 4 ��Zeroes are 2, 4 and 3.

14. Put 1+x y

= u and 1–x y

= v in the given

system of equation, we get

�����������������

10u + 2v = 415u – 5v = – 2

i.e., 5u + v = 2 ... (i)15u – 5v = – 2 ... (ii)

Multiply equation (i) by 5 and add theresult to (ii)

40u = 8 or u = 15

Substitute u = 15

in equation (i)

v = 1

u = 15

and v = 1 give x + y = 5

and x – y = 1On solving, we get x = 3 and y = 2Hence, x = 3, y = 2 is the required solution.

15. In �ABC, DE �� BC� �ABC ~ �ADE

D E

A

B C

�( )( )

ABCADE

ΔΔ

arar

=2

2ABAD

...(i)

Again, DE �� BC� ar (�ADE) = ar ( �BCED)� ar (�ABC) = 2 ar (�ADE)

�( )( )

ABCADE

ΔΔ

arar

= 2 ... (ii)

From equations (i) and (ii), we get

ABAD

= 2

1

Let AB = 2 x and AD = x,

then from the figure,

BD = 2 x – x = ( )2 – 1 x

Now,BDAB

= ( )2 – 1

2

x

x

= 2 –1

2×

22

= 2 – 2

2.

16. In right-angled �ABC,AB2 + BC2 = AC2 ...(i)

(By Pythagoras Theorem)

A

M

B N C

In �ABN,

AN2 = AB2 + BN2

= AB2 + 2BC

2� �� �� �

(... N is the mid-point of BC)

= AB2 +14

BC2

� 4 AN2= 4AB2 + BC2 ...(ii)

Similarly, in �CBM,

4 CM2 = AB2 + 4 BC2 ...(iii)

Adding equations (ii) and (iii), we get

4AN2 + 4 CM2 = (4AB2 + AB2) + (BC2 + 4BC2) = 5AB2 + 5BC2

� 4(AN2 + CM2) = 5 (AB2 + BC2) = 5 AC2

[From (i)]

Hence proved.

17. cos 582

sin 32⎛ ⎞°⎜ ⎟°⎝ ⎠

– 3cos 18 cosec 52

tan 15 tan 60 tan 75� �� �� �

� � �� �

=( )cos 90 – 32

2sin 32

⎧ ⎫° °⎨ ⎬°⎩ ⎭

– 3� �

� �

cos 90 – 52 cosec 52

tan 90 – 75 3 tan 75

� �� � �� �� �

� � � � �� ��

��� ����������

= 2sin 32sin 32

⎛ ⎞°⎜ ⎟°⎝ ⎠

– 2

1sin 52

sin 521

cot 75 3cot 75

� �� �� ��� �� �� � �� ��� �

= 2 × 1 –1

3 ×3

= 2 – 1 = 1.

OR

LHS = cos 2��= cos (2 × 30°) = cos 60° = 12

RHS = 2

21 – tan1 tan

θ+ θ

= 2

21 – tan 301 tan 30

°+ °

=

2

2

11 –

3

11

3

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞+ ⎜ ⎟⎝ ⎠

=

11–

31

13

+=

2343

= 24

= 12

Thus, LHS = RHS. Hence proved.

18. LHS = tan sec – 1tan – sec 1

θ + θθ θ+

= ( )2 2tan sec – sec – tan

tan – sec 1

θ + θ θ θ

θ θ +

= ( ) ( ) ( )tan sec – sec tan sec – tan

tan – sec 1

θ + θ θ + θ θ θθ θ +

= ( )( )tan sec 1 – sec tan

tan – sec 1

θ + θ θ + θθ θ +

= tan � + sec � = sincos

θθ

+ 1

cos θ

= 1 sin

cos+ θ

θ= RHS. Hence proved.

19. p2 – q2 = (a cot ��+ b cosec �)2

– (b cot ��+ a cosec �)2

= a2 cot2 � + 2ab cot � cosec � + b2 cosec2 �

– b2 cot2 ��– 2ab cot ��cosec ��– a2 cosec2 �

= – a2 (cosec2 ��– cot2 �)+ b2 (cosec2 ��– cot2 �)

= – a2 (1) + b2 (1) = b2 – a2.20. Since, the maximum frequency is 41, so

the modal class is 10000-15000.

� l = 10000, f1= 41, f0 = 26, f2 = 16, h = 5000

Now, mode = l + 1 0

1 0 22f f

hf f f

� ���� �� ��

= 10000 +41– 26

500082 – 26 – 16

� � �� �� �

= 10000 +75000

40= 10000 + 1875 = 11875.

Thus, the monthly modal income is � 11875.OR

Let a = 50; h = 20

C.I. fi xi ui =–ix ah

fiui

0-20 4 10 – 2 – 820-40 10 30 – 1 – 1040-60 28 50 0 060-80 36 70 1 36

80-100 50 90 2 100

128 �fiui=118

x = a + h i i

i

f uf

� ��� �� � �

= 50 + 20 ×118128

� �� �� �

= 50 + 29516

= 50 + 18.4 = 68.4.

SECTION – D

21. Given that p(x) = a(x2 + 1) – x(a2 + 1)

i.e., p (x) = ax2 – (a2 + 1)x + a

To find zeroes of p (x), put p (x) = 0.

i.e., ax2 + a – a2x – x = 0

i.e., (ax2 – a2x) – (x – a) = 0

i.e., ax (x – a) – 1 (x – a) = 0

i.e., (x –a) (ax –1) = 0

i.e., x = a, 1a

Thus, a and 1a

are the zeroes of p (x).

Sum of zeroes = 1a a� =

2 1aa�

�� ��������������

= –� �2– 1a

a�

= 2

Coefficient of–

Coefficient of

x

x

Product of zeroes = a ×1a = a

a

= 2Constant term

Coefficient of x. Hence proved.

ORLet p(x) = 3x3 + x2 + 2x + 5

q(x) = 3x – 5r(x) = 9x + 10g(x) = ?

We know p(x) = q(x) . g(x) + r(x)� 3x3 + x2 + 2x + 5 = (3x – 5) g(x) + 9x + 10� 3x3 + x2 + 2x + 5 – 9x – 10 = g(x) . (3x – 5)

�3 23 7 5

3 5x x x

x� � �

�

= g(x)

Consider2

3 2

3 2

2

2

2 13 5 3 7 5

3 5

6 7 56 10

3 53 5

x xx x x x

x x

x xx x

xx

� �

� � � �

�

� �

� �

�

� �

�

�

�

g(x) = x2 + 2x + 1.22. Let the actual length be x and breadth be y.

Then according to question,xy – 9 = (x – 5) (y + 3) ...(i)

and xy + 67 = (x + 3) (y + 2) ...(ii)Simplifying equations (i) and (ii), we have3x – 5y = 6 ...(iii) and 2x + 3y = 61 ...(iv)On solving (iii) and (iv), x = 17, y = 9Hence, length of rectangle is 17 units andthat of breadth is 9 units.

23. Table for values of x and y correspondingto equation 4x – 5y – 20 = 0 is

Similarly for the equation 3x + 5y – 15 = 0

Let us draw the graphs for the two equations.

As the graphs of the two lines intersecteach other at the point A(5, 0), the requiredsolution is x = 5, y = 0.The graphs intersect the y-axis at B (0, 3)and C(0, – 4). Therefore, the coordinates ofvertices of the triangle ABC are A(5, 0),B(0, 3) and C(0, – 4).Hence the answer: x = 5, y = 0 and (5, 0),(0, 3), (0, – 4).

24. Statement: In a triangle, if square of thelargest side is equal to the sum of the squaresof the other two sides, then the angleopposite to the largest side is a right angle.Proof: We are given a triangle ABC with

A�C�2 = A�B�2 + B�C�2 ...(i)We have to prove that �B� = 90°Let us construct a �PQR with �Q = 90°such that, PQ= A�B� and QR = B�C� ...(ii)

x 5 0y 0 – 4

x 5 0y 0 3

��� ����������

In �PQR, PR2 = PQ2 + QR2

(Pythagoras Theorem)= A�B�2 + B�C�2[From (ii)] ...(iii)

But A�C�2 = A�B�2 + B�C�2 [From (i)] ...(iv)From equations (iii) and (iv), we have

PR2 = A�C�2

� PR = A�C� ...(v)Now, in �A�B�C� and �PQR,

A�B� = PQ [From (ii)]B�C� = QR [From (ii)]A�C� = PR [From (v)]

Therefore, �A�B�C� � �PQR(SSS congruence rule)

� �B� = �Q (CPCT)But �Q = 90°... �B� = 90°. Hence proved.

ORWe are given two triangles ABC andPQR such that �ABC ~ �PQR.Draw perpendiculars AD and PM on BCand QR respectively.

We need to prove, ΔΔ

( ABC)( PQR)

arar

= 2

2ADPM

In �ABD and �PQM,�ADB = �PMQ = 90°�ABD = �PQM (�����ABC ~ �PQR)

� �ABD ~ �PQM(AA criterion of similarity)

��ABPQ =

ADPM

(Corresponding sides) ...(i)

We know that the ratio of areas of twosimilar triangles is equal to ratio of squaresof their corresponding sides

�ΔΔ

( ABC)( PQR)

arar

=�2

2ABPQ

...(ii)

From equations (i) and (ii), we have

ΔΔ

( ABC)( PQR)

arar

=�2

2ADPM

� Hence proved.

25. Given: �ABC and �DBC are on the samebase BC and O is the point of intersectionAD and BC

To prove:�

�

( ABC) ( DBC)

arar =

AODO

Construction: Draw AM � BC and DN � BCNow in �AMO and �DNO,

�AMO = �DNO [Each 90°]

�AOM = �DON

[Vertically opposite angles]

� �MOA ~ �NOD [AA Similarity]

�AMDN

= AODO

[Ratio of corresponding sides ofsimilar triangles]

Now,�

�

( ABC) ( DBC)

arar =

1BC AM

21

BC DN2

� �

� �

or�

�

( ABC) ( DBC)

arar =

AMDN

or�

�

( ABC) ( DBC)

arar =

AODO

Proved.

AM AODN DO

� ��� �� �

�

26. We haveq sin � = p and p cos � = q

� sin � = pq

and cos � = qp

(i)6

pq

� �� �� �

+6

qp

� �� �� �

= sin6 � + cos6 �

= (sin2 �)3 + (cos2 �)3

= (sin2 ��+ cos2 �)3 – 3 sin2 � . cos2 �

(sin2 � + cos2 �)[... a3 + b3 = (a + b)3 – 3ab (a + b)]

��������������� �

= (1)3 – 3 .2

2

p

q.

2

2

q

p. 1= 1 – 3 = – 2.

Hence proved.

(ii) 6

6

p

q +

6

6

q

p= – 2 ��

12 12

6 6

p q

p q

�

= – 2

� �� p12 + q12 = – 2p6 q6

� p12 + q12 + 2p6 q6 = 0 � (p6 + q6)2 = 0

�� p6 + q6 = 0. Hence proved.

27. We have to prove

sec – 1sec 1

θθ +

+sec 1sec – 1

θ +θ

= 2 cosec �

LHS

=sec 1 sec 1sec 1 sec 1

� � � ��

� � � �+

sec 1 sec 1sec 1 sec 1

� � � ��

� � � �

= ( )2

2

sec – 1

sec – 1

θθ

+ ( )2

2

sec 1

sec – 1

θ +θ

= ( )2

2

sec – 1

tan

θθ

+ ( )2

2

sec 1

tan

θ +θ

= sec – 1 sec 1tan tan� � �

�� �

= sec – 1 sec 1

tanθ + θ +

θ=

2 sectan

θθ

=

2cossincos

θθθ

= 2

sin θ= 2 cosec � = RHS.

Hence proved.28. Consider an equilateral triangle PQR in

which PS � QR. Since PS � QR so PSbisects �P as well as base QR.We observe that �PQS is a right triangle,right-angled at S with �QPS = 30° and�PQS = 60°.For finding the trigonometric ratios, weneed to know the length of the sides ofthe triangle. So, let us suppose PQ = x

Then , QS = 12

QR = 2x

and (PS)2 ��(PQ)2 – (QS)2 = x2 – 2

4x

= 23

4x

� PS = 32

x

(i) sin 60° = PSPQ

=

32

x

x =

32

(ii) tan 30° = QSPS

= 232

x

x =

13

.

29. To draw the less than type and more thantype ogives, we prepare the cumulativefrequency table by less than and more thanmethods as given below:Less than type cumulative frequency table:

Marks No. of Marks c f pointstudents less than

0-10 7 10 7 (10, 7)10-20 10 20 17 (20, 17)20-30 23 30 40 (30, 40)30-40 51 40 91 (40, 91)40-50 6 50 97 (50, 97)50-60 3 60 100 (60, 100)

More than type cumulative frequency table:

Marks No. of Marks c f pointstudents more than

or equal to0-10 7 0 100 (0, 100)

10-20 10 10 97 (10, 97)20-30 23 20 83 (20, 83)30-40 51 30 60 (30, 60)40-50 6 40 9 (40, 9)50-60 3 50 3 (50, 3)

��� ��������� �

100

95

90

85

80

75

70

65

60

55

50

45

40

35

30

25

20

15

10

5

010 20 30 40 50

(0, 100)

(10, 97)

(20, 83)

(40, 9)

(50, 97)

(40, 91)

(30, 40)

(20, 17)

(10, 7)

Y

X60

(60, 100)

(50, 3)

(30, 60)

(32, 0)

More than ogive

evigo naht sseL

We plot the points as given in both of thetables on a graph, taking marks on thex-axis and the cumulative frequencies on they-axis. On joining these points by free handsmooth curve, we obtain the less than andmore than type ogives as shown in the figure.Median: The abscissa of the point ofintersection of the two ogives determines

the median of the given data. To obtainthe coordinates of this point of intersection,we draw a perpendicular from this pointon the x-axis. The abscissa of the foot ofthis perpendicular is the required median.Here the coordinates of the foot of theperpendicular are (32,0), where 32 is theapproximate value.Hence, the required median is nearly 32marks.

30. (i) By making the given data continuous, we get: a = 57, h = 3

No. of mangoes No. of boxes (fi) Mid-points (xi) ui = ix ah�

fiui

49.5-52.5 15 51 – 2 – 3052.5-55.5 110 54 – 1 – 11055.5-58.5 135 a = 57 0 0

��������������� �

58.5-61.5 115 60 1 11561.5-64.5 25 63 2 50

� fi = 400 � fiui = 25

� Mean = a + hi i

i

f uf

� ��� ��� �

= 57 + 3 × 25400

� �� �� � = 57 +

75400

� 57.19.(ii) Step deviation method

(iii) Vikram Singh believes in qualityserving, fruits will remain fresh and freefrom germs and flies.

31. First, we prepare the cumulative frequencytable as given below:

Class Frequency Cumulativeinterval ( f ) frequency (cf )

85-100 11 11100-115 9 20115-130 8 28130-145 5 33

N = 33

� N = 33

�N2

= 16.5

Cumulative frequency just greater than16.5 is 20. So, median class is 100-115.

� cf = 11, f = 9, l = 100, h = 15

Now, median = l +

N–

2cf

hf

� �� � �� �� �

= 100 +16.5 – 11

159

� � �� �� �

= 100 + 9.17

= 109.17

Hence, the median speed is 109.17 km/hr.

��

��� ����������

NOTE

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________