Karnataka SSLC Question Paper 2021 Maths for Hindi Medium

Question Booklet Serial No. : 11-

®æãàí¨Ü~ ÓÜíTæ :
Register Number :
±Ü£ÅPæ - 01 / Paper - 01
GÓ….GÓ….GÇ….Ô. ÊÜááS ±ÜÄàûæ – 2021
SSLC MAIN EXAMINATION – 2021
+ +
Subjects : MATHEMATICS + SCIENCE + SOCIAL SCIENCE
ÊÜáñÜᤠCíXÉÐ… ÊÜÞ«ÜÂÊÜá / Hindi and English Medium
CCE-RF / CCE-RR / CCE-PF / CCE-PR / NSR / NSPR
: 10-30 1-30 ] [ Time : 10-30 A.M. to 1-30 P.M.
: 40 + 40 + 40 = 120 ] [ Total No. of Questions : 40 + 40 + 40 = 120
: 40 + 40 + 40 = 120 ] [ Max. Marks : 40 + 40 + 40 = 120

TEAR HERE TO OPEN THE QUESTION BOOKLET
81-H/E — /Mathematics
Instructions to the Students :
1.

Write your eleven digit Register Number on the Question Booklet as
allotted in the admission ticket in the space provided at the top right
corner of this front page.
2.

This Question Booklet has been sealed by reverse jacket. You have to cut
on the right side to open the Question Booklet at the time of
commencement of the examination. Check whether all the pages of the
Question Booklet are intact.
3. OMR
OMR Sheet will be provided subject-wise separately.
4.

This set of Question Booklets consists of three core subjects and each
Tear here

subject has separate Question Booklet.
1112 () 1 of 32

5. 40 120

(i) – 1 40

(ii) – 41 80

(iii) – 81 120

40 questions are provided against each subject. This set of Question
Booklets contains 120 questions in all.
(i) Mathematics – Question Numbers 1 to 40
(ii) Science – Question Numbers 41 to 80
(iii) Social Science – Question Numbers 81 to 120
6.

Each question carries one mark. Answering all the questions is
compulsory and each correct answer will be awarded one mark. There will
be no negative marking for wrong answers.
7.

a)

b) OMR

OMR

20 C OMR

C

20) A B C D

c) OMR

1112 () 2 of 32

During the examination,

a) Read the questions carefully.

b) Completely darken / shade the relevant circle against Question

Number in the OMR Sheet using blue / black ball point pen. Do not

try to alter the entry and not to do any stray marks on OMR Sheet.

Example : In the question booklet, if C is the correct answer for

Question No. 20, then in the OMR Sheet shade the option C using

blue / black ball point pen as follows.

Question No. 20) A B C D ( This is an example only )

c) Do not fold, tear, wrinkle or staple on the OMR Sheet.

8.

If more than one circle is shaded for a given question, such answer is

treated as wrong and no marks will be given.

9. OMR

Student and Room Invigilator should sign in the OMR Sheet in the space

provided.

1112 () 3 of 32

10. OMR

Candidate should return the subject-wise answered OMR Sheet to the

Room Invigilator before leaving the examination hall.

11.

Rough work can be done in the space provided at the end of the Question

Booklet.

12.

Calculators, Mobiles, Smart Watches and any other electronic equipment
are not allowed inside the examination hall.

1112 () 4 of 32

Subject : MATHEMATICS

OMR
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
1. n a n = 4n + 5 5

(A) 20 (B) 14

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

term is

(A) 20 (B) 14

(C) 25 (D) 24

2. 5 x 2 = 2 ( 2x + 3 )

(A) 5 (B) 6

(C) 4 (D) –6

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

1112 () 5 of 32

3. x – 2y = 0 3x + 4y – 20 = 0

(A) (B)

(C) (D)

x – 2y = 0 and 3x + 4y – 20 = 0 are

(A) Intersecting lines (B) Coincident lines

(C) Parallel lines (D) Perpendicular lines

4.

(A) x+y=1 2x – y = 1

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

(C) 2x – y = 2 4x – y = 4

(D) y–x=0 x–y=1

1112 () 6 of 32

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

5. a1x  b1y  c1  0 a 2 x  b2y  c 2  0

a1 b1 c1 a1 b1 c1
(A)   (B)  
a2 b2 c2 a2 b2 c2

a1 b1 a1 b1
(C)  (D) 
a2 b2 b2 a2

1112 () 7 of 32

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 the

coefficients is
a1 b1 c1 a1 b1 c1
(A)   (B)  
a2 b2 c2 a2 b2 c2

a1 b1 a1 b1
(C)  (D) 
a2 b2 b2 a2

6. 2x + 3y + 7 = 0 ax + by + 14 = 0 ‘a’

‘b’

(A) 2 3 (B) 3 2

(C) 4 6 (D) 1 2

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

7.

(A) 1, – 1, – 2, ..........

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

(C) 2, – 2, 2, – 2, ............

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

1112 () 8 of 32

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, ..........

8. – 3, – 1, 1, 3, ....... 11

(A) 23 (B) – 23

(C) – 17 (D) 17

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

(A) 23 (B) – 23

(C) – 17 (D) 17
9. 10 155 9

126 10

(A) 27 (B) 126

(C) 29 (D) 25

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

1112 () 9 of 32

10. 2 x 2 + ax + 6 = 0 2 ‘a’

7
(A) 7 (B)
2
7
(C) –7 (D) 
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

11. px 2  qx  r  0

(A) q 2  4 pr (B) q 2  4 pr

(C) p 2  4 pr (D) p 2  4qr

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

12. 4, x, 10 x

(A) 14 (B) –6

(C) –7 (D) 7

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

(A) 14 (B) –6

(C) –7 (D) 7

1112 () 10 of 32

13. ax 2 + bx + c = 0

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

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

14. (x–3)(x+2)=0

(A) – 3, 2 (B) 3, – 2

(C) – 3, – 2 (D) 3, 2

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

(A) – 3, 2 (B) 3, – 2

(C) – 3, – 2 (D) 3, 2

1112 () 11 of 32

15. 27

(A) 7 20 (B) 13 14

(C) 1 26 (D) – 13 – 14

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

16. sin 

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

In the figure, the value of sin  is

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

1112 () 12 of 32

17. ( sin 30° + cos 60° – tan 45° )

(A) 1 (B) –1

(C) 2 (D) 0

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

(A) 1 (B) –1

(C) 2 (D) 0

18. 3 + sec 2 

(A) 4 + tan 2  (B) 4 + cot 2 

(C) 2 + cot 2  (D) 3 + cot 2 

3 + sec 2  is equal to

(A) 4 + tan 2  (B) 4 + cot 2 

(C) 2 + cot 2  (D) 3 + cot 2 

19. 30
30°

(A) 10 (B) 30

(C) 10 3 (D) 30 3

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

1112 () 13 of 32

20. ( sin   cosec  )

(A) 2 (B) 1

1 3
(C) – (D)
2 2

The value of ( sin   cosec  ) is

(A) 2 (B) 1

1 3
(C) – (D)
2 2

21. A ( x1 , y1 ) B ( x 2 , y2 )

 x 2  x1 y 2  y1   x 2  x1 y 2  y1 
(A)  ,  (B)  , 
 2 2   2 2 
   

 x 2  y2 x1  y1   x 2  x1 y 2  y1 
(C)  ,  (D)  , 
 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 , y 2 ) is

 x 2  x1 y 2  y1   x 2  x1 y 2  y1 
(A)  ,  (B)  , 
 2 2   2 2 
   

 x 2  y2 x1  y1   x 2  x1 y 2  y1 
(C)  ,  (D)  , 
 3 3   3 3 
   

1112 () 14 of 32

22. ( x1 , y1 ) ( x 2 , y2 )

(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

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

23.

(A) (B)

(C) (D)

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

24.

1 3 5 7

(A) 16 (B) 5

(C) 1.6 (D) 4

The Mean of the following scores is

Marks 1 3 5 7

(A) 16 (B) 5

(C) 1.6 (D) 4

1112 () 15 of 32

25.

(A) 3 =2 +

(B) 3 =2 +

(C) =3 +

(D) =3 +2

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

26.

(A)

(B)

(C)

(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

1112 () 16 of 32

27.

–
(A) 2

×
(B) 3

+
(C) 2

+
(D) 3

The formula to find the mid-point of the class interval is
Upper limit  low er limit
(A)
2
Upper limit  low er limit
(B)
3
Upper limit  low er limit
(C)
2
Upper limit  low er limit
(D)
3

28.  ABC XY || BC

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

1112 () 17 of 32

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

29. DF

(A) 6 2 (B) 3 2

(C) 4.2 (D) 8.4

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

1112 () 18 of 32

30.  ABC ~  PQR  ABC = 64  PQR

= 100 AB = 8 PQ

(A) 12 (B) 15

(C) 10 (D) 8

 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

31.  ABC B = 90° BD  AC AB = 6 BC = 8 CD

10

6

8
(A) 10 (B) 6.4

(C) 4.8 (D) 3.6

1112 () 19 of 32

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

32. A O AT

OT = 4  OTA = 30° AT

(A) 4 (B) 2

(C) 2 3 (D) 4 3

1112 () 20 of 32

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

33. O PA, PBC CD PC = 8

AP = 5 CD

(A) 5 (B) 3

(C) 8 (D) 13

1112 () 21 of 32

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

34.

(A)

(B)

(C)

(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
1112 () 22 of 32

35. 'T'

(A)

(B)

(C)

(D)

1112 () 23 of 32

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)

1112 () 24 of 32

36. 616

(A) 49 (B) 14

(C) 21 (D) 7

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

37.

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

The volume of a cone as shown in the figure is

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

1112 () 25 of 32

38. r h

(A) (  r 2 + 2rh ) (B) 2rh

1
(C)  r 2h (D) ( r 2 + h )
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) (  r 2 + 2rh ) cm 2 (B) 2rh cm 2

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

39.

1 1
(A)  l ( r1  r2 ) (B)  h ( r12  r22  r1 r2 )
3 3

(C) l ( r1  r2 ) (D) l ( r1  r2 )

1112 () 26 of 32

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

figure the formula used is

1 1
(A)  l ( r1  r2 ) (B)  h ( r12  r22  r1 r2 )
3 3

(C) l ( r1  r2 ) (D) l ( r1  r2 )

40. 462

308

(A) 308 (B) 231

(C) 154 (D) 1078

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

1112 () 27 of 32

( SPACE FOR ROUGH WORK )

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( SPACE FOR ROUGH WORK )

1112 () 29 of 32

( SPACE FOR ROUGH WORK )

1112 () 30 of 32

( SPACE FOR ROUGH WORK )

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1112 () 32 of 32