81-E- PF+PR -2020-C- MA -Finalkseeb.kar.nic.in/docs/JUNE-2019-MODEL-ANSWERS/MATHEMATICS/… · 81-E 2 CCE PF & PR PF & PR(C)-2020 Qn. Nos. Ans. Key Value Points Marks allotted 2

PF & PR(C)-2020 [ Turn over

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KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE – 560 003

G—È.G—È.G≈È.“. Æ⁄¬fiOÊ⁄–, »⁄·¤^È% / HØ√≈È — 2020 S. S. L. C. EXAMINATION, MARCH/APRIL, 2020

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MODEL ANSWERS

¶´¤MO⁄ : 07. 04. 2020 ] —⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E

Date : 07. 04. 2020 ] CODE NO. : 81-E

…Œ⁄æ⁄fl : V⁄{}⁄ Subject : MATHEMATICS

( ‘Ê‡—⁄ Æ⁄p⁄¿O⁄√»⁄fl / New Syllabus ) ( S¤—⁄W @∫⁄¥¿£% & Æ⁄‚¥´⁄¡¤»⁄~%}⁄ S¤—⁄W @∫⁄¥¿£%/ Private Fresh & Private Repeater )

( BMW«ŒÈ ∫¤Œ¤M}⁄¡⁄ / English Version )

[ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 100

[ Max. Marks : 100

Qn. Nos.

Ans. Key

Value Points Marks

allotted

I. 1. In the pair of linear equations 0111 =++ cybxa and

0222 =++ cybxa , if 2

1

2

1bb

aa

≠ then the

(A) equations have no solution (B) equations have unique solution (C) equations have three solutions (D) equations have infinitely many solutions. Ans. :

(B) equations have unique solution 1

C CCE PF CCE PR REVISED

81-E 2 CCE PF & PR

PF & PR(C)-2020

Qn. Nos.

Ans. Key

Value Points Marks

allotted

2. In an arithmetic progression, if 12 += nan , then the common

difference of the given progression is

(A) 0 (B) 1

(C) 2 (D) 3.

Ans. :

(C) 2 1

3. The degree of a linear polynomial is

(A) 0 (B) 1

(C) 2 (D) 3.

Ans. :

(B) 1 1

4. If 13 sin θ = 12, then the value of cosec θ is

(A) 5

12 (B) 5

13

(C) 1312 (D)

1213 .

Ans. :

(D) 1213

1

5. In the figure, if Δ POQ ~ Δ SOR and PQ : RS = 1 : 2, then OP : OS

is

(A) 1 : 2 (B) 2 : 1

(C) 3 : 1 (D) 1 : 3.

Ans. :

(A) 1 : 2 1

CCE PF & PR 3 81-E


Qn. Nos.

Ans. Key

Value Points Marks

allotted

6. A straight line passing through a point on a circle is

(A) a tangent (B) a secant

(C) a radius (D) a transversal.

Ans. :

(A) a tangent 1

7. Length of an arc of a sector of a circle of radius r and angle θ is

(A) 2

360rπθ

×o

(B) 22360

rπθ×

o

(C) rπθ 2180

×o

(D) rπθ 2360

×o

.

Ans. :

(D) rπθ 2

360×

o.

1

8. If the area of the circular base of a cylinder is 22 cm2 and its

height is 10 cm, then the volume of the cylinder is

(A) 2200 cm2 (B) 2200 cm3

(C) 220 cm3 (D) 220 cm2 .

Ans. :

(C) 220 cm3 1

81-E 4 CCE PF & PR

PF & PR(C)-2020

Qn. Nos. Value Points Marks

allotted

II. Answer the following questions : 8 × 1 = 8

9. Express the denominator of 2023 in the form of mn 52 × and state

whether the given fraction is terminating or non-terminating repeating decimal.

Ans. :

52

232023

2 ×= ½

terminating decimal ½ 1

10. The following graph represents the polynomial y = p ( x ). Write the number of zeroes that p ( x ) has.

Ans. :

P ( x ) have three zeroes. 1

CCE PF & PR 5 81-E



allotted

11. Find the value of tan 45° + cot 45°.

Ans. :

tan 45° + cot 45°

= 1 + 1 ½

= 2 ½ 1

12. Find the coordinates of the mid-point of the line joining the points ),( 11 yx and ),( 22 yx .

Ans. :

P ( x, y ) = ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛ ++

2,

22121 yyxx

1

13. State “Basic proportionality theorem”.

Ans. :

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 1

14. In the figure AB and AC are the two tangents drawn from the point A to the circle with centre O. If BOC = 130° then find BAC .

Ans. :

ooo 50130180 =−=BAC 1

15. Write, x

x 12

1=

+ in the standard form of a quadratic equation.

Ans. :

022 =−+ xx 1

81-E 6 CCE PF & PR

PF & PR(C)-2020


allotted

16. Write the formula to find the total surface area of the cone whose radius is ‘r’ units and slant height is ‘l’ units.

Ans. :

TSA of cone = π r ( r + l ) sq.units. 1

17. Solve : 2x + y = 11

x + y = 8

Ans. :

2x + y = 11

x + y = 8

Substitution method :

2x + y = 11 ... (i)

x + y = 8 ... (ii)

From equation (ii)

y = 8 – x ... (iii) ½

Substitute

y = 8 – x in equation (i)

We get,

2x + 8 – x = 11 ½

x + 8 = 11

x = 3 ½

Substitute x = 3 in equation (iii)

y = 8 – 3

y = 5 ½

∴ x = 3, y = 5. 2

Alternate method :

Elimination method :

2x + y = 11 ... (i)

x + y = 8 ... (ii)

CCE PF & PR 7 81-E



allotted

Subtract equation (ii) from equation (i) ½

2x + y = 11

x + y = 8

x = 3 ½

Substitute x = 3 is equation (ii) ½

3 + y = 8

y = 8 – 3

y = 5 ½

∴ x = 3, y = 5 2

Alternate method :

Cross multiplication method :

x y 1

1 – 11 2 1 ½

1 – 8 1 1

12

1)16(11)11(8 −=

−−−=

−−−yx ½

11

1611118=

+−=

+−yx

11

53==

yx

11

3=

x

x = 3 ½

11

5=

y

y = 5 ½

∴ x = 3, y = 5 2

18. Find the sum of 5 + 8 + 11 + ... to 10 terms using the formula.

Ans. :

5 + 8 + 11 + ............. 10 terms

81-E 8 CCE PF & PR

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allotted

a = 5

d = 3

n = 10

2nSn = [ 2a + ( n – 1 ) d ] ½

210

10 =S [ 2 ( 5 ) + 9 ( 3 ) ] ½

= 5 [ 10 + 27 ]

= 5 [ 37 ] ½

10S = 185 ½

∴ The sum of the first ten terms is 185.

2

19. Find the value of k, if the pair of linear equations 2x – 3y = 8 and

2 ( k – 4 ) x – ky = k + 3 are inconsistent.

Ans. :

2x – 3y = 8

2 ( k – 4 ) x – ky = k + 3

If these equations are inconsistent then,

2

1

2

1bb

aa

= OR 2

1

2

1bb

aa

= 21

cc

≠ ½

kk −

−=

−3

)4(22 ½

kk3

41

=−

k = 3 ( k – 4 ) ½

k = 3k – 12

12 = 3k – k

12 = 2k

k = 2

12

k = 6 ½ 2

CCE PF & PR 9 81-E



allotted

20. Find the discriminant of the equation 0352 2 =+− xx and hence

write the nature of the roots.

Ans. :

22x – 5x + 3 = 0

a = 2 b = – 5 c = 3 ½

Discriminant ( D ) = 2b – 4ac ½

= ( – 5 ) 2 – 4 ( 2 ) ( 3 )

= 25 – 24

D = 1 ½

Discriminant ( D ) > 0

∴ The roots are real and distinct. ½

Note : Roots are real, rational and distinct is also correct. 2

21. If one zero of the polynomial p ( x ) = kxx +− 62 is twice the other

then find the value of k.

OR

Find the polynomial of least degree that should be subtracted from

p ( x ) = 432 23 ++− xxx so that it is exactly divisible by

g ( x ) = 132 +− xx .

Ans. :

P ( x ) = 2x – 6x + k

Let α and β be the zeroes of P ( x )

According to the condition, β = 2α ½

Sum of the zeroes = α + β = ab− ½

α + 2α = 1

)6( −−

α + 2α = 6

3α = 6

α = 36

α = 2 ... (i) ½

81-E 10 CCE PF & PR

PF & PR(C)-2020


allotted

Product of the zeroes = αβ = ac

α ( 2α ) = 1k

2 2α = k

2 ( 2 ) 2 = k Q From (i)

8 = k ½

∴ k = 8 2

OR

p ( x ) = 23 2xx − + 3x + 4

g ( x ) = 2x – 3x + 1

2x – 3x + 1 ) 23 2xx − + 3x + 4 ( x + 1

23 3xx − + x ½ (–) (+) (–)

2x + 2x + 4 ½

2x – 3x + 1 (–) (+) (–)

5x + 3 ½

∴ 5x + 3 should be subtracted from p ( x ), so that it is completely divisible by g ( x ). ½ 2

22. Find the distance between the points ( – 5, 7 ) and ( – 1, 3 ).

OR

Find the coordinates of the point which divides the line joining the points ( 1, 6 ) and ( 4, 3 ) in the ratio 1 : 2.

Ans. :

A ( – 5 , 7 )

B ( – 1, 3 )

AB = 212

212 )()( yyxx −+− ½

= 22 )73())5(1( −+−−− ½

CCE PF & PR 11 81-E



allotted

= 22 )4()51( −++−

= 22 44 +

= 1616 + ½

= 2432 = cm

AB = 24 units. ½ 2

OR

A ( 1, 6 )

B ( 4, 3 )

m : n = 1 : 2

P ( x, y ) = ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+

+

+

+

nmnymy

nmnxmx 1212 , ½

= ⎥⎦

⎤⎢⎣

⎡++

++

21)6(2)3(1

,21

)1(2)4(1 ½

= ⎥⎦

⎤⎢⎣

⎡ ++3123

,3

24 = ( 2, 5 ) 1

2

23. The points A ( 1, 1 ), B ( 3, 2 ) and C ( 5, 3 ) cannot be the vertices of

the triangle ABC. Justify.

Ans. :

A ( 1, 1 ), B ( 3, 2 ), C ( 5, 3 )

Area of Δ ABC = [ ])()()(21

213132321 yyxyyxyyx −+−+− ½

ar ( Δ ABC ) = 21 [ 1 ( 2 – 3 ) + 3 ( 3 – 1 ) + 5 ( 1 – 2 ) ] ½

= 21 [ 1 ( – 1 ) + 3 ( 2 ) + 5 ( – 1 ) ]

= 21 [ – 1 + 6 – 5 ]

= 21 [ 0 ] = 0 ½

81-E 12 CCE PF & PR

PF & PR(C)-2020


allotted

Since the area of Δ ABC is 0, A ( 1, 1 ), B ( 3, 2 ) and C ( 5, 3 ) cannot be the vertices of Δ ABC. ½ 2

Alternate method :

Distance between two points P ),( 11 yx and Q ),( 22 yx is

PQ = 212

212 )()( yyxx −+− ½

AB = 51412)12()13( 2222 =+=+=−+− units

BC = 51412)23()35( 2222 =+=+=−+− units

AC = 41624)13()15( 2222 +=+=−+−

= 5220 = units ½

∴ 5 + 5 = 2 5

That is AB + BC = AC. ½

It implies that A, B, C are collinear.

∴ They cannot form a triangle. ½ 2

24. Draw a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of 60°.

Ans. :

Angle between the radii = 180° – 60° = 120°. ½

Circle — ½

Radii — ½

Tangents — ½ 2

CCE PF & PR 13 81-E



allotted

25. Show that 7 × 11 × 13 + 13 is a composite number.

Ans. :

7 × 11 × 13 + 13 = 13 ( 7 × 11 + 1 ) ½

= 13 ( 77 + 1 )

= 13 × 78 ½

= 13 × I = 13 × Integer. ½

∴ The given number has more than two factor. Therefore it is a composite number. ½ 2

26. What is Arithmetic Progression ? Write the general form of Arithmetic Progression.

Ans. :

An arithmetic progression is a sequence in which each term is

obtained by adding a fixed number to the preceding term, except the

first term. 1

The general form of A.P. is a, a + d, a + 2d, ....... . 1 2

27. Find a quadratic polynomial whose sum and product of the zeroes are

3 and 4 respectively.

Ans. :

Let α and β be the zeroes of the quadratic polynomial

Given that α + β = 3

and α β = 4 ½

The polynomial is P ( x ) = 2x – ( α + β ) x + α β ½

∴ P ( x ) = 2x – 3x + 4. 1 2

28. If tan 2A = cot ( A – 18° ), where 2A is acute angle, find the value

of A.

Ans. :

tan 2A = cot ( A – 18° )

cot ( 90° – 2A ) = cot ( A – 18° ) Q tan θ = cot ( 90° – θ ) ½

81-E 14 CCE PF & PR

PF & PR(C)-2020


allotted

90° – 2A = A – 18° ½

90° + 18° = A + 2A

∴ 3A = 108° ½

A = 3

108o

A = 36°. ½ 2

29. Find the coordinates of a point A where AB is the diameter of a circle

whose centre is ( 2, – 3 ) and B ( 1, 4 ).

Ans. :

Let the co-ordinates of A be ),( 11 yx

B ( 1, 4 )

O ( 2, – 3 )

Centre of a circle is mid-point of diameter ½

∴ O ( x, y ) = ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ++

2,

22121 yyxx

½

O ( 2, – 3 ) = ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ++

24

,2

1 11 yx ½

∴ 2 = 2

11 +x and – 3 =

241 +y

1x = 4 – 1 = 3 and 1y = – 6 – 4 = – 10

∴ A ( 3, – 10 ). ½ 2

30. An unbiased die whose faces are numbered from 1 to 6 is rolled once.

Find the probability of getting perfect square number on the top face

and hence find the probability of its complement event.

Ans. :

Let the event of getting a square number be A.

Then A = { 1, 4 }

∴ n ( A ) = 2

B A O

CCE PF & PR 15 81-E



allotted

Total possible outcomes S = { 1, 2, 3, 4, 5, 6 }

∴ n ( S ) = 6 ½

∴ The probability of getting a square number is

P ( A ) = )()(

SnAn

P ( A ) = 62 or

31 ½

The probability of complementary event of A is )( AP

)( AP = 1 – P ( A ) ½

= 1 – 62

= 64 or

32 . ½

2

31. Draw a line segment of length 9 cm and divide it in the ratio 1 : 2.

Ans. :

AB = 9 cm

AC : BC = 1 : 2

Drawing AB and marking 321 ,, AAA — ½

Joining 3A to B — ½

Corresponding angle — ½

Drawing AC — ½ 2

81-E 16 CCE PF & PR

PF & PR(C)-2020


allotted

Alternate method :

AC : BC = 1 : 2 Drawing AB and marking 211 ,, BBA — ½

Alternate angle — 1 Joining 21 , BA — ½ 2

32. Draw a diameter AB in a circle of radius 3 cm and construct tangents to the circle at A and B.

Ans. :

AP and PQ are required tangents

Circle — ½

Diameter — ½

Tangents — 1 2

CCE PF & PR 17 81-E



allotted

33. Find the area of a sector of a circle with radius 6 cm, if angle of sector

is 60°.

Ans. :

r = 6 cm

θ = 60°

Area of sector of circle = 2

360rπθ

×o

½

= 66722

360

60×××

o

o ½

= 722 × 6 ½

2cm7

132 . ½ 2

34. The curved surface area of a cone is 528 cm2. If the radius of its base

is 8 cm then find the height of the cone.

Ans. :

C.S.A. of cone = 528 cm2

r = 8 cm

h = ?

CSA. of cone = πrl ½

528 = 722 × 8 × l

∴ l = 8227528

××

l = 21 cm. ½

h = 22 rl − ½

= 64441−

h = 377 ½

OR h = 19·4 cm. 2

81-E 18 CCE PF & PR

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35. Prove that 5 is an irrational number.

OR

Find the HCF of 24 and 40 by using Euclid’s division algorithm. Hence find the LCM of HCF ( 24, 40 ) and 20.

Ans. :

Let us assume that 5 is rational. ½

∴ qp

=5 where p and q are co-prime, q ≠ 0.

∴ p = 5 q. ½

Squaring on both sides 22 5qp = ... (i)

⇒ 5 divides 2p ½

hence 5 divides p.

∴ p = 5 k

Squaring on both sides 22 25kp = ... (ii) ½

From (i) and (ii)

5 2q = 25 2k

2q = 5 2k

⇒ 5 divides 2q

hence 5 divides q. ½

∴ 5 is common factor of p and q. This contradicts our assumption.

∴ Our assumption is wrong.

∴ 5 is irrational number. ½

Note : Consider correct alternate method. 3

OR

a = 40, b = 24

According to Euclid’s division lemma,

a = bq + r where 0 ≤ r < b ½

i) 40 = 24 × 1 + 16

ii) 24 = 16 × 1 + 8

CCE PF & PR 19 81-E



allotted iii) 16 = 8 × 2 + 0 1

∴ H.C.F. ( 40, 24 ) = 8 ½

To find L.C.M. of 8 and 20

8 = 2 × 2 × 2 = 23

20 = 2 × 2 × 5 = 22 × 5 ½

∴ L.C.M. ( 8, 20 ) = 32 × 5

= 8 × 5

L.C.M. ( 8, 20 ) = 40 ½ 3

Alternate method to find the L.C.M. :

2 8 , 20

2 4 , 10

2 2 , 5 L.C.M. = 2 × 2 × 2 × 5 = 40

5 1 , 5

1 , 1

36. To save fuel, to avoid air pollution and for good health two persons A and B ride bicycle for a distance of 12 km to reach their office everyday. As the cycling speed of B is 2 km/h more than that of A, B takes 30 minutes less than that of A to reach the office. Find the time taken by A and B to reach the office.

Ans. :

Let the speed of A be x km/h. ∴ The time taken by A to reach the office = 1t

= speed

distance = x

12 hours. ½

The speed of B is 2 km/h more than that of A

∴ Speed of B = ( x + 2 ) km/h.

∴ Time taken by B, 2

122 +

=x

t hours. ½

Given that 21 tt − = 30 minutes

30 minutes = 21 hour

81-E 20 CCE PF & PR

PF & PR(C)-2020


allotted

21

21212

=+

−xx

½

21

)2(12)2(12

=+−+

xxxx

21

2

1224122

=+

−+

xx

xx

21

2

242

=+ xx

48 = 2x + 2x.

∴ 2x + 2x – 48 = 0

2x + 8x – 6x – 48 = 0

( x + 8 ) ( x – 6 ) = 0

x = – 8 or x = 6 ½

∴ Speed of A is 6 km/h and

Speed of B is 6 + 2 = 8 km/h.

Time taken by A = 6

12speed

distance= = 2 hours ½

Time taken by B = 8

12speed

distance= = 1·5 hours. ½

3

Alternate method :

Let the time taken by A to reach the office be t hours.

Then the speed of A will be t

S 12time

distance1 == km/h. ½

B takes 30 minutes less than that of A.

∴ Time taken by B is t – 2

1221 −

=t

hours

∴ Speed of B is

212

122 −

=t

S

12

242 −

=t

S km/h. ½

CCE PF & PR 21 81-E



allotted

Given that 12 SS − = 2

tt

1212

24−

− = 2 ½

)12(

)12(1224−

−−tt

tt = 2

tt −22

12 = 2

12 = 4 2t – 2t

i.e. 4 2t – 2t – 12 = 0 ½

OR 2 2t – t – 6 = 0

2 2t – 4t + 3t – 6 = 0

( t – 2 ) ( 2t + 3 ) = 0

t = 2 t = 23− ½

∴ t = 2 hours.

∴ The time taken by A to reach the office = 2 hours and time

taken by B to reach the office = 2 – 21 = 1·5 hours. ½

3

37. If x = p tan θ + q sec θ and y = p sec θ + q tan θ then prove that 2222 pqyx −=− .

OR

Prove that θθθθ

θ

θ

θ2222

2

2

2

cossin

1

cosecsec

cosec

1tan

)90(cot

−=

−+

−

−o

.

Ans. :

x = p tan θ + q sec θ

y = p sec θ + q tan θ 2x = ( p tan θ + q sec θ ) 2

= 2p tan2 θ + 2q sec2 θ + 2pq tan θ . sec θ ... (i) ½

2y = ( p sec θ + q tan θ ) 2

= 2p sec2 θ + 2q tan2 θ + 2pq sec θ . tan θ ... (ii) ½

81-E 22 CCE PF & PR

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allotted

Subtracting equation (ii) from (i)

22 yx − = 2p tan2 θ + 2q sec2 θ + 2pq tan θ . sec θ

– 2p sec2 θ – 2q tan2 θ – 2pq sec θ . tan θ ½

22 yx − = 2p ( tan2 θ – sec2 θ ) + 2q ( sec2 θ – tan2 θ ) ½

= 2p ( tan2 θ – sec2 θ ) – 2q ( tan2 θ – sec2 θ )

= )( 22 qp − ( tan2 θ – sec2 θ ) ½

= )( 22 qp − ( – 1 )

= 22 pq −

∴ 2222 pqyx −=− . ½ 3

OR

L.H.S. =

θθ

θ

θ

θ22

2

2

2

cosecsec

cosec

1tan

)90(cot

−+

−

−o

=

θcosecθsec

θcosec

1tan

tan22

2

2

2

−+

−θ

θ ½

=

θθ

θ

θ

θ

θ

θ

22

2

2

2

2

2

sin

1

cos

1sin

1

1cos

sin

cos

sin

−+

−

½

=

θθ

θθ

θ

θ

θθ

θ

θ

22

22

2

2

22

2

2

cos.sin

cossin

sin

1

cos

cossin

cos

sin

−+

− ½

=

θθ

θ

θθ

θ22

2

22

2

cossin

cos

cossin

sin

−+

− ½

=

θθ

θθ22

22

cossin

cossin

−

+ ½

CCE PF & PR 23 81-E



allotted

= θθ 22 cossin

1

− = RHS ½

∴

θθθθ

θ

θ

θ2222

2

2

2

cossin

1

cosecsec

cos

1tan

)90(cot

−=

−+

−

− eco

3

38. Find the median of the following data :

Class-interval Frequency

20 — 40 7

40 — 60 15

60 — 80 20

80 — 100 8

OR

Find the mode of the following data :

Class-interval Frequency

1 — 3 6

3 — 5 9

5 — 7 15

7 — 9 9

9 — 11 1

Ans. :

Class-interval f c.f.

20 — 40 7 7

40 — 60 15 22

60 — 80 20 42

80 — 100 8 50

2

502

=n = 25

Median class interval 60 – 80

∴ l = 60, c.f. = 22, f = 20, h = 20 1

½

81-E 24 CCE PF & PR

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allotted

Median = l + hf

fcn

×⎥⎦

⎤⎢⎣

⎡−

2 ½

= 60 + 2020

2225×⎥

⎦

⎤⎢⎣

⎡ − ½

= 60 + 203 × 20

= 60 + 3

= 63 ½

∴ Median = 63. 3

OR

Modal class interval is 5 – 7

∴ l = 5

0f = 9

1f = 15

2f = 9

h = 2. 1

Mode = l + hfff

ff

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

−

201

012

½

= 5 + 299)15(2

915⎥⎦

⎤⎢⎣

⎡−−

− ½

= 5 + 21830

6⎥⎦

⎤⎢⎣

⎡−

= 5 + 126 × 2 ½

= 5 + 1

= 6 ½

∴ Mode = 6 3

CCE PF & PR 25 81-E



allotted

39. The following table gives the information of daily income of 50 workers of a factory. Draw a ‘less than type ogive’ for the given data.

Daily Income Number of workers

Less than 100 0

Less than 120 8

Less than 140 20

Less than 160 34

Less than 180 44

Less than 200 50

Ans. :

x and y-axis scale — ½

Plotting points — 1½

Drawing graph — 1 3

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40. A bag contains 3 red balls, 5 white balls and 8 blue balls. One ball is taken out of the bag at random. Find the probability that the ball taken out is (a) a red ball, (b) not a white ball ?

Ans. :

Total possible outcomes = 3 + 5 + 8

n ( s ) = 16 ½

a) Let R be the event of taking out a red ball.

Then the number of favourable outcomes to the event R is 3.

∴ n ( R ) = 3. ½

Then the probability of taking out a red ball is

P ( R ) = )()(

SnRn ½

P ( R ) = 163 ½

b) Let A be the event of taking out not a white ball.

Then the number of favourable outcomes to event A is

3 + 8 = 11 ½

∴ n ( A ) = 11

∴ The probability that the ball taken out is not a white ball is

P ( A ) = 1611

)()(=

SnAn ½

3

41. Prove that the “lengths of tangents drawn from an external point to a circle are equal”.

Ans. :

½

CCE PF & PR 27 81-E



allotted

Data : PQ and PR are the tangents drawn from an external point P to the circle with centre O. ½

To prove : PQ = PR. ½

Construction : Join OP, OQ and OR. ½

Proof : In Δ POQ and Δ POR

ORPOQP = Q Radius is ⊥ lar to the tangent at

the point of contact.

OQ = OR Q Radii of the same circle

OP = OP Q Common side. ½

∴ Δ POQ ≅ Δ POR Q RHS criteria.

∴ PQ = PR Q C.P.C.T. ½

Hence proved. 3

Alternate method :

PQ and PR are the two tangents from point P to the circle with centre O. We have to prove that PQ = PR. ½

We join OP, OQ and OR. ½

OQP and ORP are right angles according to theorem 4 . 1. ½

∴ In right angled triangles OQP and ORP . ½

OQ = OR Q Radii of the same circle

OP = OP Q Common

∴ Δ OQP ≅ Δ ORP Q RHS

PQ = PR Q C.P.C.T. ½ 3

½

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42. Construct a triangle ABC with sides BC = 3 cm, AB = 6 cm and

AC = 4·5 cm. Then construct a triangle whose sides are 34 of the

corresponding sides of the triangle ABC.

Ans. :

Δ ABC ~ Δ AB l C l

Constructing given circle — 1

Drawing acute angle and dividing into 4 parts — ½

Drawing parallel lines — 1

Δ AB l C l — ½ 3

43. ABCD is a rectangle of length 20 cm and breadth 10 cm. OAPB is a sector of a circle of radius 210 cm. Calculate the area of the shaded

region. [ Take π = 3·14 ]

OR

CCE PF & PR 29 81-E



allotted A hand fan is made up of cloth fixed in between the metallic wires. It is in the shape of a sector of a circle of radius 21 cm and of angle 120° as shown in the figure. Calculate the area of the cloth used and also find the total length of the metallic wire required to make such a fan.

Ans. :

In Δ AOB

AB = 20 cm

OA = 210 cm

OB = 210 cm

2AB = 400 cm2 22 OBOA + = 100 ( 2 ) + 100 ( 2 ) = 400 cm2 ½

222 OBOAAB +=

∴ Δ AOB is a right angled triangle and AOB = 90°. ½

Area of the shaded region

= Area of rectangle – Area of segment APB ½

= l × b – [ ar ( sector OAPB ) – ar ( Δ AOB ) ]

= 20 × 10 – ⎥⎥⎦

⎤

⎢⎢⎣

⎡××−× OBOAr

21

3602π

θo

½

= 20 × 10 – ⎥⎥⎦

⎤

⎢⎢⎣

⎡××−××⋅× 210210

21210210143

360

90o

o

= 200 – [ 157 – 100 ] ½

= 200 – 57

= 143 cm2. ½ 3

OR

81-E 30 CCE PF & PR

PF & PR(C)-2020


allotted

Area of the cloth required = Area of the sector

= 2

360rπθ

×o

½

= 2121722

360

120×××

o

o ½

= 22 × 21

= 462 cm2. ½

Length of the metal wire required =

Length of the arc of the sector + 2 × radius

= o360

θ × 2πr + 2r ½

= 217222

360

120×××

o

o + 2 × 21 ½

= 44 + 42

= 86 cm. ½ 3

Area of the cloth is 462 cm2.

Length of the metal wire is 86 cm.

44. Find the solution of the pair of linear equations by graphical method. x + y = 7 3x – y = 1 Ans. :

x + y = 7

∴ y = 7 – x

x 1 2 3

y 6 5 4

3x – y = 1

y = 3x – 1

x 0 1 2

y – 1 2 5

CCE PF & PR 31 81-E



allotted

Tables — 2

Drawing two straight lines — 1

Identifying intersecting straight line

points answer — 1 4

Note : Any two points, that satisfy the equation can be considered.

45. There are five terms in an Arithmetic Progression. The sum of these terms is 55, and the fourth term is five more than the sum of the first two terms. Find the terms of the Arithmetic progression.

OR

In an Arithmetic Progression sixth term is one more than twice the third term. The sum of the fourth and fifth terms is five times the second term. Find the tenth term of the Arithmetic Progression.

Ans. :

Let the 5 terms of A.P. be a, a + d, a + 2d, a + 3d and a + 4d.

Their sum is 55

∴ a + a + d + a + 2d + a + 3d + a + 4d = 55 ½

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5a + 10d = 55 [ Note : Sum of A.P. formula can be used ] ½

a + 2d = 11 ... (i)

5214 ++= aaa

a + 3d = a + a + d + 5 ½

a – 2d = – 5 ... (ii) ½

From equations (i) and (ii)

a + 2d = 11

a – 2d = – 5

2a = 6 ½

a = 26 = 3 ½

Substitute a = 3 in equation (i)

3 + 2d = 11

2d = 11 – 3

2d = 8

d = 4 ½

∴ The 5 terms of the A.P. are 3, 7, 11, 15, 19. ½

Note : Marks should be given for any correct alternate method. 4

OR

12 36 += aa ½

a + 5d = 2 ( a + 2d ) + 1

a + 5d = 2a + 4d + 1

a – 2a + 5d – 4d = 1

– a + d = 1 ... (i) ½

254 5aaa =+ ½

a + 3d + a + 4d = 5 ( a + d )

a + 3d + a + 4d = 5a + 5d

7d – 5d = 5a – 2a

2d = 3a

CCE PF & PR 33 81-E



allotted

∴ a = 3

2d ... (ii) ½

Substitute equation (ii) in (i)

– 13

2=+ dd

– 2d + 3d = 3

d = 3. ½

Substitute the value of d in equation (ii), we get

a = 3

)3(2

a = 2 ½

∴ The 10th term of the A.P. is na = a + ( n – 1 ) d ½

10a = a + 9d

= 2 + 9 ( 3 )

= 2 + 27 10a = 29. ½

∴ The 10th term is 29. 4

46. A tower and a pole stand vertically on the same level ground. It is observed that the angles of depression of top and foot of the pole from the top of the tower of height 60 m is 30° and 60° respectively. Find the height of the pole.

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Ans. :

Height of the tower = 60 cm

Height of the pole CD be h m and BE = CD = h

Let BD = EC = x ∴ AE = ( 60 – h ) m

and also ACE = 30°

ADB = 60° ½

In Δ le AEC

tan 30° = ECAE ½

x

h−=

6031

x = 3 ( 60 – h ) ... (i) ½

In Δ ABD

tan 60° = BDAB ½

x603 =

x = 3

60 ... (ii) ½

From equations (i) and (ii)

3 ( 60 – h ) = 3

60 ½

∴ 60 – h = 33

60×

60 – h = 360 ½

60 – h = 20

∴ h = 60 – 20

h = 40 ½

∴ The height of the pole is 40 meter. 4

CCE PF & PR 35 81-E



allotted

47. A container opened from the top is in the form of a frustum of a cone

of height 16 cm with radii of its lower and upper ends as 8 cm and

20 cm respectively. Find the cost of the milk which can completely fill

the container at the rate of Rs. 20 per litre. [ Take π = 3·14 ]

Ans. :

h = 16 cm

1r = 8 cm

2r = 20 cm. ½

Volume of frustum of cone = ⎟⎠⎞⎜

⎝⎛ ++ 21

22

213

1 rrrrhπ ½

= 31 × 3·14 × 16 { 82 + 202 + ( 8 ) ( 20 ) } ½

= 31 × 3·14 × 16 × 624

= 3·14 × 16 × 208

= 3·14 × 3328 ½

= 10449·9 cm3

~– 10450 cm3 ½

1 litre = 1000 cm3 . ½

81-E 36 CCE PF & PR

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Cost of 1000 cm3 milk is Rs. 20

Cost of 10450 cm3 milk is = 1000

1045020 × ½

= 10

2090

= Rs. 209 ½

∴ Cost of the milk which can completely fill the container is

Rs. 209. 4

48. State and prove Pythagoras theorem.

Ans. :

Statement : In a right angled triangle the square on the hypotenuse is

equal to the sum of the squares on the other two dies. 1

½

Data : In Δ ABC B = 90° ½

To prove : 222 BCABAC += ½

Construction : Draw BD ⊥ AC ½

Proof : In ABC and ADB

CCE PF & PR 37 81-E



allotted

A = A Q Common angle

ABC = ADB Q Data and construction

∴ Δ ABC ~ Δ ADB Q A.A. similarity criteria

∴ ABAC

ADAB

=

2AB = AC . AD ... (i) ½

In ABC and BDC

C = C Q Common angle

ABC = BDC Q Data and construction

∴ Δ ABC ~ Δ BDC Q A.A. similarity criteria

∴ BCAC

DCBC

=

2BC = AC . DC ... (ii) ½

Adding equations (i) and (ii)

22 BCAB + = AC . AD + AC . DC ½

= AC ( AD + DC )

= AC ( AC )

= 2AC

∴ 222 ACBCAB =+ ½

Hence proved 5

Alternative method :

81-E 38 CCE PF & PR

PF & PR(C)-2020


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Statement :

In a right triangle, the square on the hypotenuse is equal to the sum

of the squares of the other two sides. 1

½

Proof : We are given a right triangle ABC, right angled at B ½

We need to prove 222 BCABAC += ½

Draw BD ⊥ AC ½

Now Δ ADB ~ Δ ABC

So ACAB

ABAD

=

∴ AD . AC = 2AB ... (i) ½

Also, Δ BDC ~ Δ ABC

ACBC

BCCD

=

CD . AC = 2BC ... (ii) ½

Adding equations (i) and (ii)

AD . AC + CD . AC = 22 BCAB + ½

AC ( AD + CD ) = 22 BCAB + 5

CCE PF & PR 39 81-E



allotted

AC ( AC ) = 22 BCAB +

222 BCABAC += ½

Hence proved.

Documents

81-E- PF+PR -2020-C- MA -Finalkseeb.kar.nic.in/docs/JUNE-2019-MODEL-ANSWERS/MATHEMATICS/… · 81-E 2 CCE PF & PR PF & PR(C)-2020 Qn. Nos. Ans. Key Value Points Marks allotted 2