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ICSE Pariksha Mathematics-X 1

RSPL/1I ICSE Pariksha-2010 Mathematics�X

SECTION ‘A ’

Q1. (a) Salman Khan invests Rs. 25,600 for three years at the rate of 10% per annum compound interest.Find :

(i) The sum due to Salman at the end of the first year.

(ii ) The interest he earns for the second year.

(iii ) The total amount due to him at the end of the third year. [3]

(b) Using properties of proportion, solve for x : 2 2

2 2

+ −− −

x

x = 3 [3]

(c) The polynomial 2x3 + mx2 + nx – 2 when divided by (2x – 3), leaves remainder 7 and has (x + 2) as its

factor. Find m and n. [4]

Sol. (a) Given : P = Rs. 25,600, R = 10%

(i) Interest for the first year = P × R× T

100 =

25,600 ×10 ×1

100 = Rs. 2,560.

∴ Sum due to Salman at the end of the first year = Rs. 25,600 + Rs. 2,560 = Rs. 28,160.

(ii ) Principal for second year = Rs. 28,160

∴ Interest earned for the second year = 28,160 ×10 ×1

100 = Rs. 2,816.

(iii ) Principal for third year = Rs. 28,160 + Rs. 2,816 = Rs. 30,976

Interest earned for third year = 30,976 ×10 ×1

100 = Rs. 3,097.60.

∴ Total amount due to him at the end of the third year = Rs. 30,976 + Rs. 3,097.60

= Rs. 34,073.60

(b) Given : 2 2

2 2

+ −− −

x

x= 3

It can be written as,

2 2

2 2

+ −− −

x

x=

3

1

Using Componendo and Dividendo, we get

2 2 2 2

2 2 2 2

+ − + − −+ − − + −

x x

x x=

3 1

3 1

+−

⇒ 2 2

2 2− x=

4

2

⇒ 2

2 − x= 2

On squaring both sides,2

2− x= 4

2 ICSE Pariksha Mathematics-X

or 2 = 8 – 4x⇒ 4x = 8 – 2⇒ 4x = 6

⇒ x =6 3

=4 2

(c) The given polynomial is 2x3 + mx2 + nx – 2When divided by (2x – 3), it leaves remainder 7

i.e., 2x – 3 = 0 ⇒ 2x = 3 ⇒ x = 3

2

∴3 2

3 3 32 2

2 2 2 + + − m n = 7

⇒27 9 3

2 × 28 4 2

+ + −m n = 7

⇒27 9 3

24 4 2

mn+ + − = 7

⇒27 9 6 8

4

+ + −m n= 7

⇒ 9m + 6n + 19 = 28⇒ 3m + 2n = 3 ...(i)Also, (x + 2) is a factor,So, x + 2 = 0 ⇒ x = – 2∴ 2(–2)3 + m(–2)2 + n(–2) – 2 = 0or –16 + 4m – 2n – 2 = 0⇒ 4m – 2n = 18 ...(ii )(i) + (ii) ⇒ 3m + 2n = 3

+ 4m – 2n = 187m = 21

⇒ m = 3By equation (i) ⇒ 3(3) + 2n = 3or 9 + 2n = 3or 2n = – 6⇒ n = – 3∴ m = 3 and n = – 3.

Q2. (a) Solve the given inequation and graph the solution on the number line. [3]

2 12 3 , .

3 3

2− ≤ + < ∈3

y y Z

(b) Kareena purchases an article for Rs. 21,384 which includes 10% rebate on the marked price and 8%sales tax on the remaining price. Find the marked price of the article. [3]

(c) In the given figure ABCD is a rectangle. Two quadrants with centres C and D are drawn. Find the areaand perimeter of the shaded region. Given that, AB = 42 cm and BC = 28 cm [4]

D

A

C

B


Sol. (a)2 2 1

2 3 ,3 3 3

y y Z− ≤ + < ∈

⇒ 8 2 10

3 3 3− ≤ + <y ⇒

8 2 10 2

3 3 3 3y− − ≤ < −

⇒ 10 8

3 3

− ≤ ≤y ⇒ – 3.3 ≤ y ≤ 2.67 ⇒ y ∈ {– 3, – 2, – 1, 0, 1, 2}

–4 –3 –2 –1 0 1 2 3 4(b) Let the marked price of the article = Rs. x

Total amount paid =D% S.T.%

M.P. 1 1100 100

− +

⇒ Rs. 21,384 =10 8

1 1100 100

x − +

⇒ Rs. 21,384 =9 27

10 25x

⇒ x =21,384 × 25×10

27 × 9 Rs.

⇒ x = Rs. 22,000∴ Marked price of the article = Rs. 22,000

(c) For bigger quadrant R = 28 cmFor smaller quadrant r = 42 – 28 = 14 cm

Shaded area = Area of rectangle – (Area of bigger quadrant + Area of smaller quadrant)

= 42 × 28 – 2 2

4 4

R rπ π +

= 1176 – [ ]2 2

4R r

π +

= 1176 – 2 222(28 14 )

7 × 4+

= 1176 – 22

× 9807 × 4

= 1176 – 770 = 406 cm2

Perimeter of the shaded region =R

14 422 2

π π + + + r

cm

= ( R) 56π + +2

r

=22

(14 28) 5614

× + +

=22

× 42 5614

+ = 66 + 56 = 122 cm

Q3. (a) Let A × 1 1

0 2

= [1 2], where A is a matrix :

(i) State the order of matrix A(ii ) Find the matrix A. [3]

(b) Aerika opened a recurring deposit account in a bank and deposited Rs. 150 per month for 8 months. Ifshe received Rs. 1,236 at the time of maturity, find the rate of interest per annum. [3]


(c) Use a graph paper for this question. Plot the points A(3, 2) and B(–3, –2). From A and B drawperpendiculars AM and BN on the x-axis (Take 1 cm = 1 unit on both the axes). [4]

(i) Name the image of A on reflection in the origin

(ii ) Name the figure AMBN and find its area

(iii ) Write the co-ordinates of the point to which M is mapped on reflection in x-axis and y-axis.

Sol. (a) (i)1 1

0 2

A × = [1 2 ]1 × 2 2 × 2

1 × 2

Order of A is (1 × 2)

For multiplication of two matrices number of columns of Ist matrix should be same as

the number of rows of the IInd matrix then the resulting matrix has order 1 × 2

(ii ) Let matrix A = [x y]

then [x y] 1 1

0 2

= [1 2]

⇒ [x × 1 + y × 0 x × 1 + y × 2] = [1 2]

⇒ [x x + 2y] = [1 2]

⇒ x = 1

and x + 2y = 2

or 1 + 2y = 2

⇒ 2y = 1

⇒ y = 1

2

martix A = 1

12

(b) Monthly instalment = Rs. 150

Time = 8 months

Amount deposited in 8 months = Rs. 150 × 8 = Rs. 1200

Equivalent principal for 1 month = 150 × 8(8 1)

2

+

= 150 × 8 × 9

2 = Rs. 5400

Let rate of interest be r % p.a.

Then interest on Rs. 5400 for 1 month = Rs. 5400 ×1×

12×100

r = Rs.

9

2r

Amount received on maturity = Rs. 1236

∴ 1200 + 9

2r = 1236

⇒9

2r = 1236 – 1200

⇒9

2r = 36

⇒ r = 8

Hence rate of interest = 8% p.a.


(c) (i) Image of A on reflection in origin is (– 3, – 2) i.e., the point B.

(ii ) Figure AMBN is a parallelogram.

ar (AMBN) = 2 ar (∆AMN)

= 2 × 1

(MN × AM)2

= 6 × 2 = 12 sq. units

–5 –4 –3 –2 –1 0 1 2 3 4 5XX′

1

2

3

4

Y

–1

–2

–3

–4

Y ′

A (3, 2)

B (–3, –2)

M (3,0)N (–3,0)

(iii ) The co-ordinates of the points to which M is mapped on reflection in x-axis are (3, 0) and iny-axis are (–3, 0).

Q4. (a) Without using tables, evaluate :

22 2tan 35

sin 20 sec 70 sin 50 sin 40cot 55

° − ° ° + ° + ° °[3]

(b) Kamalpreet has an account in the State Bank of India. A page from her pass book is given below :

Date Particulars Withdrawals Deposits Balance(in Rs.) (in Rs. ) (in Rs.)

01-04-2008 BF — — 1400.00

10-05-2008 By cash — 700.00 2100.00

02-06-2008 To self 1000.00 — 1100.00

11-07-2008 To cheque 300.00 — 800.00

21-08-2008 By cash — 1700.00 2500.00

03-10-2008 By cash — 2400.00 4900.00

If the interest earned by Kamalpreet at the end of December, 2008 was Rs. 87.75, find the rate ofinterest per annum. [4]


(c) In the given figure ABCD is a quadrilateral, inscribed in a circle with centre O. [3]Side CD produced to E. If ∠ADE = 95° and ∠OBA = 30°, find :

(i) ∠OBC

(ii ) ∠OAC

Sol. (a)2

tan 35

cot 55°

° – sin 20° sec 70° + sin2 50° + sin2 40°

=

2tan 35

tan (90 55 )

° ° − °

– sin 20° cosec (90° – 70°) + cos2 (90° – 50°) + sin2 40°

=2

2 2tan 35

sin 20 cosec20 (cos 40 sin 40 )tan 35

° − ° ° + ° + ° °

= (1)2 – sin 20° × 1

1sin 20

+°

= 1 – 1 + 1 = 1

(b) Minimum balance for the month of April = Rs.1400

Minimum balance for the month of May = Rs.2100

Minimum balance for the month of June = Rs. 1100

Minimum balance for the month of July = Rs. 800

Minimum balance for the month of August = Rs. 800

Minimum balance for the month of September = Rs.2500

Minimum balance for the month of October = Rs.4900

Minimum balance for the month of November = Rs.4900

Minimum balance for the month of December = Rs.4900

Total = Rs. 23400

Now, I =P × R × T

100

⇒ 87.75 =23400 × R ×1

12 ×100

⇒ R =87.75×12

234R = 4.5%

(c) (i)

95°

30°

A O

D

CB

E

∠ADE = ∠ABC [Exterior ∠ of a cyclic quadrilateral]

⇒ 95° = ∠OBA + ∠OBC

⇒ 95° = 30° + ∠OBC

⇒ ∠OBC = 65°

(ii ) ∠ADC + ∠ADE = 180° [Linear pair ∠′s]

⇒ ∠ADC + 95° = 180°

A

B C

D

E

O


⇒ ∠ADC = 85°

Now ∠AOC = 2 ∠ADC

subtended by an arc at the centre is double the

angle subtended by it at any other point on the

remaining part of the circle

∠= 2 × 85°

∠AOC = 170°

Now OA = OC [Radii]

∴ ∠OAC = ∠OCA = x [say]

∴ In ∆AOC, x + x + ∠AOC = 180°

⇒ 2x + 170° =180°

⇒ 2x = 10°

⇒ x = 5°

∴ ∠OAC = 5°

SECTION ‘B’

Q5. (a) If x : y : z = 1 : 2 : 3, Prove that :

2 2 212 13 4 5+ + =x y z y [3]

(b) Given : sin A + sin2 A = 1,

Prove that : cos2 A + cos4 A = 1. [3](c) If a line segment with end points (3, 4) and (14, –3) meets the x-axis at P, in what ratio does P divide

the segment ? Also, find the co-ordinates of point P. [4]

Sol. (a) Let1

x= = =

2 3

y zk

Then, x = k, y = 2k, z = 3k

L.H.S. 2 2 212 13 4+ +x y z = 2 2 212( ) 13(2 ) 4(3 )+ +k k k

= 2 2 212 52 36+ +k k k

= 2100k = 10 k = 5 × 2 k = 5y

= R.H.S.

(b) Given, sin A + sin2A = 1

⇒ sin A = 1 – sin2 A

⇒ sin A = cos2 A⇒ sin2 A = cos4 A [By squaring both sides]

⇒ 1 – cos2 A = cos4 A

⇒ 1 = cos2 A + cos4 A Hence proved.

(c)

A P B(3 , 4 ) ( , )x y (1 4 , – 3 )

k 1

Let the join of A(3, 4) and B(14, – 3) be divided by the x-axis in the ratio k : 1 at P.

then, x =14 3 3 4

=1 1

+ − ++ +

k ky

k k

Since P lies on x-axis, So y = 0.

⇒3 4

1

− ++

k

k= 0 ⇒ k =

4

3

So, ratio = 4 : 3


Hence the co-ordinates of point P

=

414 × 3

3 , 04

13

+

+

=65

,07

Q6. (a) Solve for x : 2x2 + x – 4 = 0 (By using quadratic formula) [3]

(b) In the given figure, XY is diameter of the circle, PQ is a tangent to the circle at Y. If ∠AXB = 50° and∠ABX = 70°, calculate ∠BAY and ∠APY. [3]

P Y

O

B

Q

C7 0 °

50°

A

X

(c) Mr. Negi sold a certain number of shares of Rs. 20 paying 8% dividend at Rs. 18 and invested theproceeds in Rs. 10 shares, paying 12% dividend at 50% premium. If the change in his annual incomeis Rs. 120. Find the number of shares sold by Mr. Negi. [4]

Sol. (a) 2x2 + x – 4 = 0

a = 2, b = 1, c = – 4

By using formula,

x =2 4

2− ± −b b ac

a

=21 (1) 4 2 4

2 2

− ± − × × −×

= 1 1 32

4

− ± +

=1 33

4

− ± =

1 5.74

4

− ±

=1 5.74 1 5.74

,4 4

− + − −

=4.74 6.74

,4 4

−

=2.37 3.37

,2 2

− = 1.18, – 1.68

(b) In ∆AXB,

∠XAB + ∠AXB + ∠ABX = 180° [Triangle property]

⇒ ∠XAB + 50° + 70° =180°

⇒ ∠XAB = 180° – 120° = 60°

⇒ ∠XAY = 90° [Angle of semi-circle]

∴ ∠BAY = ∠XAY – ∠XAB

= 90° – 60° = 30°


and ∠BXY = ∠BAY = 30° [Angle of same segment]

∴ ∠ACX = ∠BXY + ∠ABX

= 30° + 70° = 100°

also, ∠XYP = 90° [∴ Diameter ⊥ tangent]∠APY = ∠ACX – ∠CYP

∠APY = 100° – 90°

∠APY = 10°.

(c) Let the number of shares sold be x

Annual income = Rs.8

× × 20100

x = Rs.8

5

x

S.P. of x shares = Rs. 18x

Number of new shares bought = 18

15

x

∴ Annual income = Rs.12 18

10100 15

× × x

= Rs. 36

25

x

So,8 36

5 25−x x

= 120

⇒ 40 36

25

−x x= 120

⇒ 4x = 120 × 25

⇒ x =120 × 25

4

⇒ x = 750

Hence, number of shares sold by Mr. Negi = 750.

Q7. (a) The following table shows the age distribution of cases of certain disease admitted during a year inparticular hospital. [6]

Age (in years) No. of cases

5 – 14 6

15 – 24 11

25 – 34 21

35 – 44 23

45 – 54 14

55 – 64 5

Total 80

Draw a histogram, frequency polygon and hence find the mode from the graph.

(b) A man from the top of a vertical tower observe a car moving at a uniform speed coming directlytowards him. If it takes 12 minutes to change the angle of depression from 30° to 45°, how soon afterit, will the car reach the tower. [4]


Sol. (a) Age (in years) No. of cases

4 .5 1 4.5 2 4.5 3 4.5 4 4.5 5 4.5 6 4.5C .I (ag e in y ea rs)

2

4

6

8

1 0

2 4

1 2

1 4

1 6

1 8

2 0

2 2

No.

of

case

s

H is to g ra m

4.5–14.5 6

14.5 – 24.5 11

24.5 – 34.5 21

34.5 – 44.5 23

44.5 – 54.5 14

54.5 – 64.5 5

Hence, mode from the graph = 36.3

(b) Let AB be the tower of height h metres,

let the speed of the car be x m/min.

Distance = Speed × Time

So, CD = 12x

Let, car takes t minutes to reach the tower AB, then

AD = tx

In ∆ABD, tan θ =AB

AD

⇒ tan 45° =h

tx3 0 ° 4 5 °

C D A1 2x t x

B45° 30°

h

⇒ 1 =h

tx⇒ h = tx

In ∆ABC, tan θ =AB

AC

⇒ tan 30° =12 +

h

x tx

⇒1

3=

(12 )+tx

x t

⇒1

3=

12+t

t

⇒ 3 t = 12 + t

⇒ 3 t – t = 12

⇒ ( )3 1t − = 12

⇒ t =12 12 12 12000

= = =1.732 1 0.732 7323 1 −−

= 16.39 min

Thus, the car will reach the tower in = 16.39 min.

Q8. (a) In the given figure, ABCD is a parallelogram and E is a pointon BC. The diagonal BD intersect AE at F.

Prove that : DF × FE = FB × FA. [3]

D

A

C

EF

B


(b) Construct a ∆ABC, in which BC = 7.5 cm, ∠B = 60°, altitude AD = 3 cm. Construct a circle to touchBC at its mid-point and to pass through A. [4]

(c) P, Q and R have co-ordinates (–2, 1), (2, 2) and (6, –2) respectively, write down : [3](i) The gradient of QR.

(ii ) The equation of the line passing through P and perpendicular to QR.

Sol. (a) Draw GF || AD || BC cutting AB in G.

In ∆ABE, GF || BED

A

C

EF

G B

∴AG

GB=

AF

FE...(i)

In ∆BAD, GF || AD

∴ AG

GB=

DF

FB...(ii )

From equation (i) and (ii ), we get

AF

FE=

DF

FB⇒ AF × FB = FE × DF

(b) Draw BC = 7.5 cm. Construct ∠ZBC = 60°. Draw LN. parallelto BC at a distance of 3 cm, cutting BZ at A. Draw ADperpendicular to BC. Then AD = 3 cm. Join A to C. Then ABC isthe required triangle. Bisect BC at M by drawing the right bisectorXY of BC. Draw the right bisector PQ of AM to meet XY at O.With O as centre and OA or OM as radius draw the requiredcircle which will touch BC at its mid-point M and also passthrough A.

(c) (i) P(–2, 1), Q(2, 2) and R(6, – 2)Gradient of QR

m1

= 2 1

2 1

−−

y y

x x

m1

=2 2

6 2

− −−

= 4

4

− = –1

(ii ) Slope of the line ⊥ to QR (m2) =

1

1 1

1m

− −=−

= 1

So, equation of line ⊥ to QR and passing through P.y – y

1= m

2 (x – x

1)

⇒ y – 1 = 1(x + 2)⇒ x – y + 3 = 0

Q9. (a) If X = 4 1

1 2

−

, show that 6X – X2 = 9I, where I is the unit matrix. [3]

(b) Prove that : 2 2

2 2

1 sin .sec

1 cos .cosec

+ θ θ+ θ θ

= tan θ [3]

(c) Find the number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circularcylinder of height 10 cm and diameter 4.5 cm. [4]

Sol. (a) X =4 1

,1 2

−

I = 1 0

0 1

P

D M

Y

X

C

NQ

O

Z

L

B


L.H.S., 6X – X2 = 64 1 4 1 4 1

1 2 1 2 1 2

− − − −

=24 6 15 6

6 12 6 3

− − −

=9 0

0 9

= 1 0

90 1

= 9I, R.H.S.

(b) L.H.S.2 2

2 2

1 sin sec

1 cos cosec+ θ⋅ θ

+ θ⋅ θ=

22

22

11 sin .

cos1

1 cos .sin

+ θθ

+ θθ

=2 2 2 2

2 2

cos sin sin cos

cos sin/θ + θ θ + θ

θ θ

=2

2

sin

cosθθ

= sin

cos

θθ

= tan θ = R.H.S.

(c) Since each coin will be in cylindrical form.

So, r = 0.75 cm and h = 0.2 cm

∴ Volume of each coin =πr2h

= π × (0.75)2 × 0.2 cm3

For the required cylinder, R = 2.25 cm, and H = 10 cm

∴ Volume of new cylinder =πR2H

= π × (2.25)2 × 10 cm3

∴ Number of coins =Volume of new cylinder

Volume of each coin

=2

2

× (2.25) ×10

× (0.75) × 0.2ππ

= 450.

Q10. (a) A two digit number is such that the product of the digits is fourteen. If 45 is added to the number thenthe digits interchange their places. Find the number. [3]

(b) The bisectors of ∠B and ∠C of a quadrilateral ABCD, intersect at P. Show that P is equidistant fromthe opposite sides AB and CD. [4]

(c) What point (or points) on the x-axis are at a distance of 5 units from the point (5, – 4). [3]

Sol. (a) Let the unit's digit be = x

Then, ten's digit according the question be = 14

x

the original number = 14

10 xx

× + = 140

xx

+

The number obtained by interchanging the places of digit = 14

10 xx

× + = 14

10xx

+


Now, according the question

14045x

x+ + =

1410x

x+

⇒2140 45x x

x

+ +=

210 14x

x

+

⇒ 140 + x2 + 45x = 10x2 + 14

⇒ 9x2 – 45x – 126 = 0

⇒ x2 – 5x – 14 = 0

⇒ x2 – 7x + 2x – 14 = 0

x(x – 7) + 2(x – 7) = 0

⇒ (x – 7) (x + 2) = 0

⇒ So, x – 7 = 0 or x + 2 = 0

x = 7 or x = – 2 (neglected)

Hence, original number = 140

xx

+ = 140

77

+ = 20 + 7 = 27

(b) Draw PS ⊥ BCP lies on the bisector of ∠B

PM ⊥ AB [By construction]and PS⊥ BCSo, PM = PS ...(i)

B S C

D

A

NP

MNow, P lies on the bisector of ∠C

PS ⊥ BCand PN ⊥ CDSo, PN = PS ...(ii )From equation (i) and (ii ), we get

PM = PN(c) Let (k, 0) be any point on x-axis, then according the question,

2 2(5 ) ( 4 0)− + − −k = 5

⇒ 25 + k2 – 10k + 16 = 25⇒ k2 – 10k + 16 = 0⇒ (k – 8) (k – 2) = 0⇒ k – 8 = 0 or k – 2 = 0⇒ k = 8 or k = 2∴ The required points are (8, 0) and (2, 0).

Q11. (a) Compute mean for the following data. (By Step Deviation Method).

Marks No. of students

less than 10 12

less than 20 19

less than 30 35

less than 40 47

less than 50 58

less than 60 65

less than 70 84

less than 80 100 [4]


(b) Draw an ogive from the following frequency table.

Class interval Frequency

8 – 12 10

12 – 16 16

16 – 20 22

20 – 24 18

24 – 28 12

28 – 32 4

Total 82

from this ogive, calculate :

(i) The median.

(ii ) The upper and lower quartile.

(iii ) The inter quartile range. [6]

Sol. (a) Marks c.f. f x d = x – A u = d/h fu

0 – 10 12 12 5 – 30 – 3 – 36

10 – 20 19 7 15 – 20 – 2 – 14

20 – 30 35 16 25 – 10 – 1 – 16

30 – 40 47 12 35 0 0 0

40 – 50 58 11 45 10 1 11

50 – 60 65 7 55 20 2 14

60 – 70 84 19 65 30 3 57

70 – 80 100 16 75 40 4 64

Σf = 100 Σfu = 80

Here, A = 35, h = 10

Mean = A + ×ΣΣfu

hf

=80

35 ×10100

+

= 35 + 8 = 43


(b) Class interval frequency (f) c.f.

8 – 12 10 10

12 – 16 16 26

16 – 20 22 48

20 – 24 18 66

24 – 28 12 78

28 – 32 4 82

N = 82

(i) N = 82 (even)

The position of median is given by = N 82

412 2

= =

08 1 2 1 5 1 6 2 0 2 4 2 8 3 2 3 6

1 0

2 0

3 0

4 0

5 0

6 0

7 0

8 0

9 0

Y

X

(1 2, 1 0)

(1 6, 2 6)

(2 0, 4 8)

2 3

1 9

(2 4, 6 6)

(3 2, 8 2)

(2 8, 7 8)

Cum

ulat

ive

Fre

quen

cy

C la ss-In te rv a l

So, from the graph, median = 19

(ii ) The position of Upper Quartile is given by = 3N 3× 82 246

= =4 4 4

= 61.5

So, from the graph, Upper Quartile = 23

The position of lower Quartile is given by = N 82

20.54 4

= =

So, from the graph, lower Quartile = 15

(iii ) Inter quartile range = Q3 – Q

1

= 23 – 15 = 8

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