Karnataka SSLC Question Paper 2021 Maths for Urdu Medium

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Subject : MATHEMATICS

$%& '% ( !"#
!2$8 230 45 % (676 OMR # , -./ 0 + )*
40 × 1 = 40

Four choices are given for each of the following questions / incomplete
statements. Choose the correct answer among them and shade the correct
option in the OMR Answer Sheet given to you with a black / blue ball
point pen. 40 × 1 = 40

> ? @ 5A BC an = 4n + 5> ? @n'9:;<= .1

14 (B) 20 (A)
24 (D) 25 (C)

The nth term of an Arithmetic Progression is an = 4n + 5. Then its 5th
term is
(A) 20 (B) 14
(C) 25 (D) 24

> R Q @ MN OP 6 JKL F GH I 5 x 2 = 2 ( 2x + 3 ) @D E @ 2.

SN(Constant term)
6 (B) 5 (A)
–6 (D) 4 (C)

When the quadratic equation 5 x 2 = 2 ( 2x + 3 ) is expressed in the
standard form, the constant term obtained is

(A) 5 (B) 6
(C) 4 (D) –6

5 of 32 1116( )

81-U/E
U/E (RF/RR/PF/PR/NSR/NSPR)

TU ?B @D?V@ 3x + 4y – 20 = 0 @ x – 2y = 0 .3

"TUW @MNX (B) "TU`a (A)

"TUW @MN Y6Z[@ \ (D) "TUbc d (C)

x – 2y = 0 and 3x + 4y – 20 = 0 are

(A) Intersecting lines (B) Coincident lines

(C) Parallel lines (D) Perpendicular lines
f
Z] ?B @D^W @MN _, ki jheg lm .4

2x – y = 1 @ x + y = 1 (A)

x + y = 2 @ 2x + y = 2 (B)

4x – y = 4 @ 2x – y = 2 (C)

x–y=1 @ y–x=0 (D)

6 of 32 1116( )

The pair of equations of lines as shown in the graph are

(A) x+y=1 and 2x – y = 1

(B) 2x + y = 2 and x + y = 2

(C) 2x – y = 2 and 4x – y = 4

(D) y–x=0 and x – y = 1

bc dj a 2x + b2y + c 2 = 0 @ a1x + b1y + c1 = 0 Z] ?B @D^r @s@ .5
f
n o pq k?@[b 9h ?uv B?Nt$ _ITU
a1 b1 c1 a1 b1 c1
= = (B) = ≠ (A)
a2 b2 c2 a2 b2 c2

a1 b a1 b1
= 1 (D) ≠ (C)
b2 a2 a2 b2

7 of 32 1116( )

If the pair of linear equations in two variables a1x + b1y + c1 = 0 and
a 2 x + b2y + c 2 = 0 are parallel lines then the correct relation of their

coefficients is

a1 b1 c1 a1 b1 c1
(A) = ≠ (B) = =
a2 b2 c2 a2 b2 c2

a1 b1 a1 b
(C) ≠ (D) = 1
a2 b2 b2 a2

. ‘b’ @ ‘a’ B " x$ _ TU Xax + by + 14 = 0 @ 2x + 3y + 7 = 0 w @D .6

yz {

2 @ 3 (B) 3 @ 2 (A)

2 @ 1 (D) 6 @ 4 (C)

If the pair of lines 2x + 3y + 7 = 0 and ax + by + 14 = 0 are coincident
lines then the values of ‘a’ and ‘b’ are respectively equal to

(A) 2 and 3 (B) 3 and 2

(C) 4 and 6 (D) 1 and 2

C(A.P.)9:;< .7

1, – 1, – 2, .......... (A)

1, 5, 9, .............. (B)

2, – 2, 2, – 2, ............ (C)

1, 2, 4, 8, .......... (D)

8 of 32 1116( )

Which of the following is an Arithmetic Progression ?

(A) 1, – 1, – 2, ..........

(B) 1, 5, 9, ..............

(C) 2, – 2, 2, – 2, ............

(D) 1, 2, 4, 8, ..........

k> ? @ 11h > ? @) |' –3, –1, 1, 3, …… 9:;< .8

– 23 (B) 23 (A)

17 (D) – 17 (C)

The 11th term of the Arithmetic Progression – 3, – 1, 1, 3, ....... is

(A) 23 (B) – 23

(C) – 17 (D) 17

;<!C 126 •€ ' ' 9 } ~ @ 155 •€ ' ' 10 } ~ (A.P.) 9: ;< .9

SN> ? @10'9:
126 (B) 27 (A)

25 (D) 29 (C)

The sum of the first 10 terms of an Arithmetic Progression is 155 and the

sum of the first 9 terms of the same progression is 126 then the 10th term
of the progression is

(A) 27 (B) 126

(C) 29 (D) 25

9 of 32 1116( )

{.‘a’ BC2j ‚
• \ '2 x 2 + ax + 6 = 0 @DE @ .10

7
(B) 7 (A)
2
7
− (D) –7 (C)
2

If one root of the equation 2 x 2 + ax + 6 = 0 is 2, then the value of ‘a’

is
7
(A) 7 (B)
2
7
(C) –7 (D) −
2

SN(Discriminant)ƒ' px 2 + qx + r = 0 @DE @ .11

q 2 + 4 pr (B) q 2 − 4 pr (A)

p 2 + 4qr (D) p 2 − 4 pr (C)

The Discriminant of the quadratic equation px 2 + qx + r = 0 is

(A) q 2 − 4 pr (B) q 2 + 4 pr

(C) p 2 − 4 pr (D) p 2 + 4qr

„N{.x B?N (A.P.)9:;<4, x, 10 j .12

–6 (B) 14 (A)

7 (D) –7 (C)

If 4, x, 10 are in Arithmetic Progression the value of x is

(A) 14 (B) –6

(C) –7 (D) 7

10 of 32 1116( )

@•‚ ax 2 + bx + c = 0 @DE @ .13

−b ± b 2 − 4ac
x = (A)
2a
−b ± b 2 + 4ac
x = (B)
2a
−b − b 2 − 4c
x = (C)
2a
−b + b 2 − 4ac
x = (D)
2a

The roots of the quadratic equation ax 2 + bx + c = 0 are

−b ± b 2 − 4ac
(A) x =
2a

−b ± b 2 + 4ac
(B) x =
2a

−b − b 2 − 4c
(C) x =
2a

−b + b 2 − 4ac
(D) x =
2a

…N @•‚ (x–3)(x+2)=0 @D .14

3, –2 (B) – 3, 2 (A)

3, 2 (D) – 3, –2 (C)

The roots of the equation ( x – 3 ) ( x + 2 ) = 0 are

(A) – 3, 2 (B) 3, –2

(C) – 3, –2 (D) 3, 2

11 of 32 1116( )

f
…N 9 (!C 27 •€' 9 (g d@ .15

14 @ 13 (B) 20 @ 7 (A)

−14 @ −13 (D) 26 @ 1 (C)

If the sum of two consecutive integers is 27, then the integers are

(A) 7 and 20 (B) 13 and 14

(C) 1 and 26 (D) – 13 and – 14

{. sin θ F .16

3 1
(B) (A)
2 2
2
(D) 3 (C)
3

In the figure, the value of sin θ is

1 3
(A) (B)
2 2
2
(C) 3 (D)
3

12 of 32 1116( )

{.( sin 30° + cos 60° – tan 45° ) .17

–1 (B) 1 (A)

0 (D) 2 (C)

The value of ( sin 30° + cos 60° – tan 45° ) is
(A) 1 (B) –1

(C) 2 (D) 0

Cb@D 3 + sec 2 θ .18

4 + cot 2 θ (B) 4 + tan 2 θ (A)

3 + cot 2 θ (D) 2 + cot 2 θ (C)

3 + sec 2 θ is equal to

(A) 4 + tan 2 θ (B) 4 + cot 2 θ

(C) 2 + cot 2 θ (D) 3 + cot 2 θ

ˆ ˆ ˆ
•• ‘†@ c'’“. @ ,‰C6•Ž kŒh 30 metres ,†‡ @ ‰Š \ 6‹c .19

ˆ
}”@ . @ BC30°

30 m (B) 10 m (A)

30 3 m (D) 10 3 m (C)

The angle of elevation of the top of a tower from a point on the ground,
which is 30 metres away from the foot of the tower, is 30°. Then the height
of the tower is
(A) 10 m (B) 30 m

(C) 10 3 m (D) 30 3 m

13 of 32 1116( )

„N •. ( sin θ × cosec θ ) .20

1 (B) 2 (A)

3 1
(D) − (C)
2 2

The value of ( sin θ × cosec θ ) is

(A) 2 (B) 1

1 3
(C) – (D)
2 2

˜™'Šš@ —`aW @M] IB ( x 2 , y2 ) @ A ( x1 , y1 )T– .21

 x 2 − x1 y 2 − y1   x 2 + x1 y 2 + y1 
 ,  (B)  ,  (A)
 2 2   2 2 
   

 x 2 + x1 y 2 + y1   x 2 + y2 x + y1 
 ,  (D)  , 1  (C)
 3 3   3 3 
   

The formula to find the mid-point of the line segment joining the points
A ( x1 , y1 ) and B ( x 2 , y2 ) is

 x 2 + x1 y 2 + y1   x 2 − x1 y 2 − y1 
(A)  ,  (B)  , 
 2 2   2 2 
   

 x 2 + y2 x + y1   x 2 + x1 y 2 + y1 
(C)  , 1  (D)  , 
 3 3   3 3 
   

14 of 32 1116( )

•Ž›œ ' ( x 2 , y 2 ) @ ( x1 , y1 ) T– .22

( x 2 − x1 )2 − ( y 2 − y1 )2 (B) ( x1 − x 2 )2 + ( y1 − y 2 )2 (A)

( x 2 + x1 )2 + ( y 2 + y1 )2 (D) ( x1 + x 2 )2 − ( y1 + y 2 )2 (C)

The distance between the points ( x1 , y1 ) and ( x 2 , y 2 ) is

(A) ( x1 − x 2 )2 + ( y1 − y 2 )2 (B) ( x 2 − x1 )2 − ( y 2 − y1 )2

(C) ( x1 + x 2 )2 − ( y1 + y 2 )2 (D) ( x 2 + x1 )2 + ( y 2 + y1 )2

•l Q @Mž @ Ÿ ) c,o ¡¢ .23

L£ (B) 4œ (A)

( Range )¦ (D) ¤¥@ (C)

The value among the observations of most repeated scores of the data is
(A) the mean (B) the mode
(C) the median (D) the range

4œ '§ • .24

¨q 1 3 5 7

5 (B) 16 (A)
4 (D) 1.6 (C)

The Mean of the following scores is

Marks 1 3 5 7

(A) 16 (B) 5

(C) 1.6 (D) 4

15 of 32 1116( )

C˜™ @Dr @M$ _I©›œ ¤¥@ @ L£‰4œ ¡¢ª .25

¤¥@ 3 =4œ 2 + L£ (A)

4œ 3 =¤¥@2 +L£ (B)

4œ =¤¥@3 +L£ (C)

L£ =4œ 3 +¤¥@ 2 (D)

The relation among the Mean, Mode and Median is

(A) 3 Median = 2 Mean + Mode (B) 3 Mean = 2 Median + Mode

(C) Mean = 3 Median + Mode (D) Mode = 3 Mean + 2 Median

°')±$)²OPB ž³ KL F.)±$6 ¯ ,« $¬j IM - =®,G£ .26

SN´@ '° M - (A)

SNK µ'° M - (B)

SN´3'° M - (C)

SNg g ° M - (D)

A cylinder made of wax is melted and recast completely into a sphere.
Then the volume of the sphere is

(A) two times the volume of the cylinder

(B) half the volume of the cylinder

(C) 3 times the volume of the cylinder

(D) equal to the volume of the cylinder

16 of 32 1116( )

˜™'M$¶ Šš@'·@§¸ .27

(A)

(B)

(C)

(D)

The formula to find the mid-point of the class interval is
Upper limit − lower limit
(A)
2
Upper limit × lower limit
(B)
3
Upper limit + lower limit
(C)
2
Upper limit + lower limit
(D)
3

( BN XY || BC ∆ ABC .28

AX AY AX AC
= (B) = (A)
BX CY AB AY
AB AC AX XY
= (D) = (C)
BX AY BX AY

17 of 32 1116( )

In the ∆ ABC, XY || BC then

AX AC AX AY
(A) = (B) =
AB AY BX CY
AX XY AB AC
(C) = (D) =
BX AY BX AY
f » »
!¿À.}Á.DF, m Nt$ ¹6 º¼½¼¾@ # .29

3 2 cm (B) 6 2 cm (A)
8.4 cm (D) 4.2 cm (C)

Observe the given two triangles and then identify the length of DF in the
following :

(A) 6 2 cm (B) 3 2 cm

(C) 4.2 cm (D) 8.4 cm

18 of 32 1116( )

j C 100 cm 2 Ã ' ∆ PQR @ ‰ 64 cm 2 Ã ' ∆ ABC !C ∆ ABC ~ ∆ PQR .30

}Á.PQBNAB = 8 cm

15 cm (B) 12 cm (A)

8 cm (D) 10 cm (C)

∆ ABC ~ ∆ PQR. Area of ∆ ABC = 64 cm 2 and the area of ∆ PQR = 100 cm 2
. If AB = 8 cm then the length of PQ is

(A) 12 cm (B) 15 cm

(C) 10 cm (D) 8 cm

. CD B N BC = 8 cm @ AB = 6 cmj C BD ⊥ AC @ B = 90° ∆ ABC Ä .31

}Á

6.4 cm (B) 10 cm (A)

3.6 cm (D) 4.8 cm (C)

19 of 32 1116( )

In the ∆ ABC, B = 90° and BD ⊥ AC. If AB = 6 cm, BC = 8 cm

then the length of CD is

(A) 10 cm (B) 6.4 cm

(C) 4.8 cm (D) 3.6 cm

Å
C|Ê,ÈÉ AT§Çl— !C A §ÇŠ' AT §Ç—'Zg W @ Æ O F .32

{.ATBN ∠OTA = 30°j !COT = 4 cmË

2 cm (B) 4 cm (A)

4 3 cm (D) 2 3 cm (C)

20 of 32 1116( )

In the given figure AT is a tangent drawn at the point A to the circle with

centre O such that OT = 4 cm. If ∠OTA = 30° then AT is

(A) 4 cm (B) 2 cm

(C) 2 3 cm (D) 4 3 cm

Å
@ PC = 8 cm j !CO Æ Í " §Ç TU Zg CD @ PBC ‰PA, FÌb .33

}Á.CD§Çl—BNAP = 5 cm

3 cm (B) 5 cm (A)

13 cm (D) 8 cm (C)

21 of 32 1116( )

In the given figure PA, PBC and CD are the tangents to a circle with
centre O. If PC = 8 cm and AP = 5 cm, the length of the tangent CD is

(A) 5 cm (B) 3 cm

(C) 8 cm (D) 13 cm

Î, m .34
f Å
C $Ï6Š\ iÐIZg §Çl— (A)
f Å Å
C Ñ$N,Š Zg ÒÓ— l |ÊIZg (B)
f Å
C ԧǑŠŠÕÖ'§Çl— @ Zg (C)
f Å
C N Y6§Ç‘Š×Ø ÙÚ‰§Çl—|Ê6Š=IZg (D)
The wrong statement in the following is

(A) a tangent to a circle touches the circle exactly at one point

(B) when a straight line is drawn to a circle it always passes through a
point on the circle

(C) the point common to the circle and its tangent is called the point of
contact

(D) the tangent drawn at any point to a circle is perpendicular to the
radius drawn at the point of contact

22 of 32 1116( )

f Å
_I¿Ø.Z] §ÇTl UW @MžÜIZg \ , ‘T’ Š›@Û FÌb .35
f
!CF.G•Ý W @Mž Þ« ¿Ø?ß!Cx$

(A)

(B)

(C)

(D)

23 of 32 1116( )

Which is the next step of construction while constructing a pair of

tangents to a circle from an external point ‘T’ given in the figure ?

(A)

(B)

(C)

(D)

24 of 32 1116( )

ÙÚ')±$§ BN616 sq.cm.Ã à')±$= .36

14 cm (B) 49 cm (A)
7 cm (D) 21 cm (C)
The surface area of a sphere is 616 sq.cm. Then the radius of the same
sphere is
(A) 49 cm (B) 14 cm
(C) 21 cm (D) 7 cm

SN°'T@á# â F .37

πr ( r + l ) (B) πr 2 h (A)
1
πr l (D) πr 2 h (C)
3

The volume of a cone as shown in the figure is

(A) πr 2 h (B) πr ( r + l )

1
(C) πr 2 h (D) πr l
3

25 of 32 1116( )

Å
å'æg F› - Z9ab@g äa W @ã }”@ h cm @ )9a ÙÚ r cm .38

SN‰N \ '瘙'à à

2πrh cm 2 (B) ( π r 2 + 2πrh ) cm 2 (A)

1
( π r 2 + h ) cm 2 (D) π r 2h cm 3 (C)
3

The formula to find the total surface area of a right circular based

cylindrical vessel of base radius r cm and height h cm opened at one end is

(A) ( π r 2 + 2πrh ) cm 2 (B) 2πrh cm 2

1
(C) π r 2h cm 3 (D) ( π r 2 + h ) cm 2
3

ˆ
SN˜™'M$Gë (CSA) à à)è'éê T@á, -. qëÌb F .39

1 1
π h ( r12 + r2 2 + r1 r 2 ) (B) π l ( r1 + r 2 ) (A)
3 3

π l ( r1 − r 2 ) (D) πl ( r1 + r 2 ) (C)

26 of 32 1116( )

To find the curved surface area of a frustum of a cone as shown in the

figure the formula used is

1 1
(A) π l ( r1 + r 2 ) (B) π h ( r12 + r2 2 + r1 r 2 )
3 3

(C) πl ( r1 + r 2 ) (D) π l ( r1 − r 2 )

Z9a )$Ú!C 308 cm 2 à à)è @ 462 cm 2 à àå')$Ú§ì .40

SNÃ '
231 cm 2 (B) 308 cm 2 (A)

1078 cm 2 (D) 154 cm 2 (C)

The total surface area of solid hemisphere is 462 cm 2 . If the curved

surface area of it is 308 cm 2 , then the area of the base of the hemisphere

is

(A) 308 cm 2 (B) 231 cm 2

(C) 154 cm 2 (D) 1078 cm 2

27 of 32 1116( )

[ í G'î ]
[Space for Rough Work]

28 of 32 1116( )

[ í G'î ]
[Space for Rough Work]

29 of 32 1116( )

[ í G'î ]
[Space for Rough Work]

30 of 32 1116( )

[ í G'î ]
[Space for Rough Work]

31 of 32 1116( )

32 of 32 1116( )