Karnataka SSLC Question Paper 2022 Answer Key Mathematics

CCE RF
CCE RR A
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KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESHWARAM,
BANGALORE – 560 003

G—È.G—È.G≈È.“. Æ⁄¬fiOÊ⁄–, »⁄·¤^È% / HØ√≈È — 2022
S. S. L. C. EXAMINATION, MARCH/APRIL, 2022

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MODEL ANSWERS

¶´¤MO⁄ : 04. 04. 2022 ] —⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E
Date : 04. 04. 2022 ] CODE NO. : 81-E
…Œ⁄æ⁄fl : V⁄{}⁄
Subject : MATHEMATICS
( À¤≈¤ @∫⁄¥¿£% & Æ⁄‚¥´⁄¡¤»⁄~%}⁄ À¤≈¤ @∫⁄¥¿£% / Regular Fresh & Regular Repeater )
( BMW«ŒÈ »⁄·¤®⁄¥¿»⁄fl / English Medium )

[ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80

[ Max. Marks : 80

Qn. Ans. Marks
Nos. Key allotted

I. Multiple choice : 8×1=8

1. The graphical representation of the pair of lines x + 2y – 4 = 0

and 2x + 4y – 12 = 0 is

(A) intersecting lines (B) parallel lines

(C) coincident lines (D) perpendicular lines.

(B) parallel lines 1

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81-E 2 CCE RF & RR

Qn. Ans. Marks
Nos. Key allotted

2. The common difference of the Arithmetic progression

8, 5, 2, – 1, ... is

(A) – 3 (B) –2

(C) 3 (D) 8.

(A) –3 1

3. The standard form of 2x 2 = x − 7 is

(A) 2x 2 − x = − 7 (B) 2x 2 + x − 7 = 0

(C) 2x 2 − x + 7 = 0 (D) 2x 2 + x + 7 = 0 .

(C) 2x 2 − x + 7 = 0 1

4. The value of cos ( 90° – 30° ) is
1
(A) – 1 (B)
2

(C) 0 (D) 1.

1
(B)
2 1

5. The distance of the point P ( x, y ) from the origin is

(A) x2 + y2 (B) x2 + y2

(C) x 2 − y 2 (D) x2 − y2 .

(A) x2 + y2 1

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 3 81-E

Qn. Ans. Marks
Nos. Key allotted

6. In a circle, the angle between the tangent and the radius at the
point of contact is
(A) 30° (B) 60°
(C) 90° (D) 180°.

(C) 90° 1

In the given figure, the volume of the frustum of a cone is
7.

(A) π ( r 1 + r 2 ) l (B) π ( r1 − r 2 )l

1 1
(C) π h ( r12 − r2 2 − r 1 r 2 ) (D) π h ( r12 + r2 2 + r 1 r 2 )
3 3

1 2 2
πh ( r1 + r2 + r1 r2 )
(D) 3 1

Surface area of a sphere of radius ‘r’ unit is
8.
(A) π r 2 sq.units (B) 2 π r 2 sq.units
(C) 3 π r 2 sq.units (D) 4 π r 2 sq.units.

2
(D) 4πr sq.units 1

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81-E 4 CCE RF & RR

Qn. Marks
Nos. allotted

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

( Direct answers from Q. Nos. 9 to 16 full marks should be given )

9. If the pair of linear equations in two variables are inconsistent, then
how many solutions do they have ?

No solution 1

10. In an Arithmetic progression if ‘a’ is the first term and ‘d’ is the
common difference, then write its n th term.

an = a + ( n − 1 ) d 1

11. Write the standard form of quadratic equation.

ax 2 + bx + c = 0 1

sin 18 o
12. Write the value of .
cos 72o

1 1

13. Write the distance of the point ( 4, 3 ) from x-axis.

3 1

14. Find the median of the scores 6, 4, 2, 10 and 7.

6 1

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 5 81-E

Qn. Marks
Nos. allotted

15. Write the statement of “Basic Proportionality” theorem ( Thales

theorem ).

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.
Note : If correct alternate statement is written, give full marks. 1

16. In the given figure, write the formula used to find the curved surface
area of the cone.

Curved surface area of cone = πrl sq units 1

III. Answer the following questions : 8 × 2 = 16

17. Solve the given pair of linear equations by Elimination method :
2x + y = 8
x – y = 1

2x + y = 8 ................ (1 )
Adding x – y = 1 ................ (2) ½
3x = 9

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81-E 6 CCE RF & RR

Qn. Marks
Nos. allotted
9
x=
3
x=3 ½
Substituting x = 3 in (1)
2(3) + y = 8 ½
6+y=8
y=8–6
2
y=2 ½
18. Find the 30th term of the arithmetic progression 5, 8, 11, ..... by
using formula.

5, 8, 11 .............
Here a = 5, d = 8 – 5 = 3, n = 30 ½
nth term of arithmetic progression
an = a + (n − 1 ) d ½
a 30 = 5 + (30 − 1) 3 ½

= 5 + 29 × 3
= 5 + 87
a 30 = 92 ½
2

19. Find the sum of first 20 terms of the Arithmetic progression
10, 15, 20, .......... by using formula.

OR

Find the sum of first 20 positive integers using formula.

a = 10, d = 15 – 10 = 5, n = 20, S 20 = ?

n
Sn = [ 2a + ( n − 1) d ] ½
2

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 7 81-E

Qn. Marks
Nos. allotted
20
S 20 = [ 2 (10 ) + ( 20 − 1 ) 5 ] ½
2

= 10 [ 20 + 19 × 5 ]
= 10 [ 20 + 95 ]
= 10 × 115 ½
S20 =1150 ½

Note : Any other suitable method is followed to get the correct answer, 2
full marks should be given.

OR
n ( n +1)
Sn = ½
2
n = 20
20 ( 20 + 1 )
S 20 = ½
2
20 × 21
=
2
= 10 × 21 ½
S 20 = 210 ½ 2

20. Find the roots of x 2 + 5x + 2 = 0 by using quadratic formula.

x 2 + 5x + 2 = 0

ax 2 + bx + c = 0
a = 1, b = 5, c = 2

− b ± b 2 − 4ac
x= ½
2a

− 5 ± 5 2 − 4(1 ) (2)
= ½
2 (1)
− 5 ± 25 − 8
= ½
2
− 5 ± 17
= ½
2 2

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Qn. Marks
Nos. allotted
21. Find the value of the discriminant and hence write the nature of roots
of the quadratic equation x 2 + 4x + 4 = 0.

x 2 + 4x + 4 = 0

ax 2 + bx + c + 0
a = 1, b = 4, c = 4
2
Discriminant = b − 4ac ½
2
= 4 − 4 (1 ) ( 4 ) ½
= 16 – 16
=0 ½
Nature of roots : Two equal real roots. ½ 2
22. Find the distance between the points A ( 2, 6 ) and B ( 5, 10 ) by using
distance formula.
OR
Find the coordinates of the mid-point of the line segment joining the
points P ( 3, 4 ) and Q ( 5, 6 ) by using ‘mid-point’ formula.
A (2, 6 ) B ( 5, 10 )
x1,y1 x 2 , y2

Distance formula d = ( x 2 − x1 )2 + ( y 2 − y1 )2 ½

= ( 5 − 2)2 + ( 10 − 6 )2 ½
= 32 + 42

= 9 + 16

= 25 ½
d = 5 units ½ 2

OR

P (3, 4 ) Q ( 5, 6 )
x1,y1 x 2 , y2
 x + x 2 y1 + y 2 
Mid-point formula P ( x , y ) =  1 ,  ½
 2 2 
 

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 9 81-E

Qn. Marks
Nos. allotted
3+5 4+6
=  , 
 ½
 2 2 
 8 10 
=  , 
 ½
 2 2 
P ( x , y ) = ( 4, 5 ) ½
2

23. Draw a line segment of length 10 cm and divide it in the ratio 2 : 3 by
geometric construction.

AC : CB = 2 : 3
Drawing line segment (10 cm ) ½
Constructing acute angle at A ½
Marking 5 arcs ½
Constructing A2C || A5B ½
Note : If correct alternate method is followed, give full marks. 2

24. In the given figure find the values of
i) sin θ
ii) tan α.

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81-E 10 CCE RF & RR

Qn. Marks
Nos. allotted
12
(i) sin θ = 1
13
5
(ii) tan α = 1
12 2

IV. Answer the following questions : 9 × 3 = 27

25. The sum of first 9 terms of an Arithmetic progression is 144 and its
9th term is 28 then find the first term and common difference of the
Arithmetic progression.

n
Sn = [a+l) ½
2
9
S9 = [ a + 28 ]
2
9
144 = [ a + 28 ] ½
2
16
144 × 2
= a + 28
9

32 = a + 28

a = 32 – 28 ½

a=4
an = a + ( n – 1 ) d ½

a9 = 4 + ( 9 – 1 ) d

28 = 4 + 8d ½

24 = 8d
24
d =
8

d= 3 ½ 3

Any other correct alternate, method may be given full marks.

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 11 81-E

Qn. Marks
Nos. allotted

26. The diagonal of a rectangular field is 60 m more than its shorter side.
If the longer side is 30 m more than the shorter side, then find the
sides of the field.
OR
In a right angled triangle, the length of the hypotenuse is 13 cm.
Among the remaining two sides, the length of one side is 7 cm more
than the other side. Find the sides of the triangle.

ABCD → rectangular field
Let AB = x m then BC = ( x + 30 ) m, AC = ( x + 60 ) m

AC 2 = AB 2 + BC 2 ½
( x + 60 )2 = x 2 + ( x + 30 )2 ½
x 2 + 602 + 2 × x × 60 = x 2 + x 2 + 302 + 2 × x × 30

3600 + 120x = x 2 + 900 + 60x

x 2 + 900 + 60x − 3600 −120x = 0

x 2 − 60x − 2700 = 0 ½
x 2 − 90x + 30x − 2700 = 0
x ( x − 90 ) + 30 (x − 90 ) = 0
(x − 90 ) ( x + 30 ) = 0 ½
x – 90 = 0 or x + 30 = 0
x = 90 or x = – 30 ( not considered ) ½
∴ x = 90
AB = x = 90 m
BC = ( x + 30 ) = 90 + 30 = 120 m ½ 3

OR

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Qn. Marks
Nos. allotted

Let ABC be a right angled triangle.

Let AC = 13 cm, AB = x cm and BC = ( x + 7 ) cm

AC 2 = AB 2 + BC 2 ½

132 = x 2 + ( x + 7)2 ½
2 2
⇒ 169 = x + x + 49 + 14x

2
⇒ 169 = 2x + 49 + 14x

2
⇒ 2x + 49 + 14x − 169 = 0

2
⇒ 2x + 14x − 120 = 0 ½

÷2, x 2 + 7x − 60 = 0

2
⇒ x + 12x − 5x − 60 = 0

⇒ x ( x + 12 ) − 5 ( x + 12 ) = 0

⇒ ( x + 12 ) ( x − 5 ) = 0 ½

x + 12 = 0 or x – 5 = 0

x = – 12 or x = 5

( not considered ) ∴ x = 5 ½

AB = x = 5 cm

BC = ( x + 7 ) = 5 + 7 = 12 cm ½ 3

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 13 81-E

Qn. Marks
Nos. allotted

27. Prove that
( sin A + cosec A ) 2 + ( cos A + sec A ) 2 = 7 + tan 2 A + cot 2 A.
OR
Prove that : sec θ ( 1 – sin θ ) ( sec θ + tan θ ) = 1.
2 2
LHS = ( sin A + cosec A ) + ( cos A + sec A )
2 2 2 2
= sin A + cos ec A + 2 sin A.cosec A + cos A + sec A
+ 2 cos A. sec A 1
1
= sin2 A + cos 2 A + cosec 2 A + 2 sin A . 2
+ sec A
sin A
1
+ 2 cos A . 1
cos A
2 2
= 1 + (1 + cot A ) + 2 + (1 + tan A ) + 2

[ Q cosec2 A =1 + cot2 A
sec2 A = 1 + tan2 A
sin2 A + cos2 A =1 ] ½
2 2
= 7 + tan A + cot A ½
LHS = RHS 3

OR
LHS = sec θ (1 − sin θ ) ( sec θ + tan θ )
1  1 sin θ 
= (1 − sin θ )  + 
 1
cos θ  cos θ cos θ 
( 1 − sin θ ) ( 1 + sin θ )
= × ½
cos θ cos θ
1 − sin2 θ
= ½
cos 2 θ
cos2 θ
= 2
[ Q 1 − sin2 θ = cos 2 θ ] ½
cos θ
=1 ½
∴ L.H.S. = R.H.S 3

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81-E 14 CCE RF & RR

Qn. Marks
Nos. allotted

28. Find the coordinates of the point on the line segment joining the points
A ( – 1, 7 ) and B ( 4, – 3 ) which divides AB internally in the ratio
2 : 3.

OR

Find the area of triangle PQR with vertices P ( 0, 4 ), Q ( 3, 0 ) and
R ( 3, 5 ).

A ( – 1, 7 ), B ( 4, – 3 ) 2:3
x1,y1 x 2 , y2 m1 m2

 m x + m 2 x1 m1y 2 + m 2y1 
P ( x, y ) =  1 2 ,  1
 m +m m + m 
 1 2 1 2 
 2( 4 ) + 3 ( − 1) 2( − 3 ) + 3 ( 7 ) 
=  , 
 ½
 2+3 2+3 
 8 − 3 − 6 + 21 
=  , 
 ½
 5 5 
 5 15 
=  , 
 ½
5 5 

P ( x , y ) = ( 1, 3 ) ½
3

OR

P ( 0, 4 ), Q ( 3, 0 ) R ( 3, 5 )
x1,y1 x 2 , y2 x3 , y3

1
A= [ x1 ( y 2 − y 3 ) + x 2 ( y 3 − y1 ) + x 3 ( y1 − y 2 ) ] 1
2
1
= [ 0 ( 0 − 5 )+ 3( 5 − 4 )+ 3( 4 − 0 )] ½
2

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 15 81-E

Qn. Marks
Nos. allotted
1
= [ 0 ( − 5 ) + 3 (1 ) + 3 ( 4 ) ] ½
2
1
= [ 0 + 3 + 12 ] ½
2
1 3
= × 15
2
15
A= or 7·5 sq. units ½
2

29. Find the mean for the following grouped data by Direct method :

Class-interval Frequency

10 — 20 2

20 — 30 3

30 — 40 5

40 — 50 7

50 — 60 3

OR

Find the mode for the following grouped data :

Class-interval Frequency

5 — 15 3

15 — 25 4

25 — 35 8

35 — 45 7

45 — 55 3

RF/RR (A)-(200)-9020 (MA) [ Turn over

81-E 16 CCE RF & RR

Qn. Marks
Nos. allotted

C-I fi xi f i xi

10-20 2 15 30

20-30 3 25 75

30-40 5 35 175

40-50 7 45 315

50-60 3 55 165

N = 20 ∑ f i x i = 760
Table 2
[ Mid points – 01
finding f i x i - 01 ]

Mean, X =
∑ f i xi OR
∑ FX ½
N N
760
=
20
X = 38 ½ 3

OR

From the frequency distribution table we find that
f 0 = 4, f1 = 8, f 2 = 7, h =10 and l = 25
f1 − f 0
Mode = l + [ ]×h 1
2 f1 − f 0 − f 2
8−4
= 25 + [ ] × 10 ½
2( 8 ) − 4 − 7
4
= 25 + [ ] × 10 ½
16 − 11
4 2
= 25 + × 10 ½
51

= 25 + 8
Mode = 33 ½ 3

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 17 81-E

Qn. Marks
Nos. allotted

30. During a medical check-up
check up of 50 students of a class, their heights were
recorded as follows :
Draw “less than type” ogive for the given data :

Height in cm Number of students
( Cumulative frequency )

Less than 140 5

Less than 145 10

Less than 150 15

Less than 155 25

Less than 160 40

Less than 165 50

RF/RR (A)-(200)-9020 (MA) [ Turn over

81-E 18 CCE RF & RR

Qn. Marks
Nos. allotted
Drawing axes and writing scale ½ + ½ =1

Marking points 1

Drawing Ogive 1 3

31. Prove that “the lengths of tangents drawn from an external point to a

circle are equal”.

½

Data : O is the centre of the circle. PQ and PR are tangents drawn from

external point 'P'. ½

To Prove : PQ = PR ½

Construction : Join OP, OQ and OR. ½

Proof : In the figure

∠OQP = ∠ORP = 90° OQ ⊥ PQ

OR ⊥ PR

OQ = OR [ radii of same circle ] ½

OP = OP [ common side ]

∆ OQP ≅ ∆ORP [ RHS ] ½

PQ = PR [ CPCT ]

Note : If the theorem is proved as given in the text-book,
text book, give full

marks. 3

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 19 81-E

Qn. Marks
Nos. allotted

32. Construct two tangents to a circle of radius 3 cm from a point 8 cm

away from its centre.

Drawing a circle C1 of radius 3 cm ½

Drawing OP = 8 cm ½
Constructing perpendicular bisector of OP 1
Drawing C2 circle ½

Joining PA and PB ½ 3

33. The volume of a solid right circular cylinder is 2156 cm 3 . If the height
of the cylinder is 14 cm, then find its curved surface area.
22
[ Take π = ]
7

RF/RR (A)-(200)-9020 (MA) [ Turn over

81-E 20 CCE RF & RR

Qn. Marks
Nos. allotted

V = 2156 cm 3

h = 14 cm

r=?

CSA = ?
2
Volume of cylinder = πr h ½
22 2 2
2156 = ×r × 14 ½
71

2156 = 44 r 2
2156
r2 =
44

r 2 = 49

r = 49

r = 7 cm ½

Curved surface area of = 2πrh ½
22
cylinder = 2× × 7 × 14 ½
7

= 2 × 22 × 14

= 616 cm 2 ½ 3

V. Answer the following questions : 4 × 4 = 16

34. Find the solution of the given pair of linear equations by graphical
method :

x + 2y = 6

x+y= 5

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 21 81-E

Qn. Marks
Nos. allotted

x + 2y = 6

x+y=5 For table construction ( 1 + 1 ) 2

Drawing two lines by marking points 1
Marking point of intersection & writing
values of x and y 1
Note : Any other points also may be considered to get straight lines. 4

35. The angle of elevation of the top of a building from the foot of a tower is
30° and the angle of elevation of the top of the tower from the foot of
the building is 60°. Both the tower and building are on the same level

RF/RR (A)-(200)-9020 (MA) [ Turn over

81-E 22 CCE RF & RR

Qn. Marks
Nos. allotted
ground. If the height of the tower is 50 m, then find the height of the
building.

OR
As observed from the top of a 75 m high light house from the sea-
level, the angles of depression of two ships are 30° and 45°. If one ship
is exactly behind the other on the same side of the light house, then
find the distance between the two ships.

CD
In ∆BDC, tan 60° = ½
BD
50
3= ½
BD
50
∴ BD = .............. (1) ½
3
AB
In ∆ABD, tan 30° = ½
BD

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 23 81-E

Qn. Marks
Nos. allotted
1 AB
=
3 BD

BD = 3 . AB .............. (2) ½

From (1) and (2)
50
3 . AB = ½
3
50
AB = ½
3. 3
50 2
AB = or 16 m ½
3 3 4

OR

Distance between the two ships is PQ
AB
In ∆ABP, tan 45° = ½
BP
75
1= ½
BP
∴ BP = 75 ½
AB
In ∆ABQ, tan 30° = ½
BQ
1 75
= ½
3 BP + PQ
1 75
= ½
3 75 + PQ

75 + PQ = 75 3

PQ = 75 3 − 75 ½
PQ = 75 ( 3 − 1 ) m ½ 4

36. Construct a triangle with sides 4·5 cm, 6 cm and 8 cm. Then construct
3
another triangle whose sides are of the corresponding sides of the
4

first triangle.

RF/RR (A)-(200)-9020 (MA) [ Turn over

81-E 24 CCE RF & RR

Qn. Marks
Nos. allotted

Construction of given triangle 1

Construction of acute angle with division 1

Drawing parallel lines 1

Obtaining required triangle 1 4

In the figure AXB and CYD are the arcs of two concentric circles with
37.
centre O. The length of the arc AXB is 11 cm. If OC = 7 cm and
∠ AOB = 30°, then find the area of the shaded region.
22
[ Take π = ]
7

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 25 81-E

Qn. Marks
Nos. allotted

θ
Length of the arc = × 2πr ½
360°
30° 2211
11 = × 2× ×r ½
360 o126 7
3
11r
11 =
21
11× 21
r=
11
r = 21 cm ½
θ 2
Area of the sector OAXB = A1 = × πr ½
360°
30° 22 2
= × × 21
360 ° 7
11
1 22 3
1
= × × 21 × 21
126 71
2
231
= cm 2 ½
2
θ 2
Area of the sector OCYD = A2 = × πr
360°
30° 22 2
= × ×7
360° 7
1 2211
= × ×7×7
126 7
77
A2 = cm 2 ½
6
Area of the shaded region = A1 − A2
231 77
= −
2 6
693 − 77
= ½
6
616
=
6

RF/RR (A)-(200)-9020 (MA) [ Turn over

81-E 26 CCE RF & RR

Qn. Marks
Nos. allotted
308
= cm 2
3
or ½
= 102·66 cm 2
4
OR 102·7 cm 2

VI. Answer the following question : 1×5=5

38. Prove that “the ratio of the areas of two similar triangles is equal to the
square of the ratio of their corresponding sides”.

½
Data : ∆ABC ~ ∆PQR
AB BC AC
∴ = = ½
PQ QR PR

Area of ∆ABC BC 2
To prove : = ½
Area of ∆PQR QR 2

Construction : Draw AM ⊥ BC and PN ⊥ QR ½
1
Area of ∆ABC × BC × AM
Proof : = 2 ½
Area of ∆PQR 1
× QR × PN
2
Area of ∆ABC BC AM
= × ................... (1) ½
Area of ∆PQR QR PN

In ∆ABM and ∆PQN

∠B = ∠Q

RF/RR (A)-(200)-9020 (MA)

CCE RF & RR 27 81-E

Qn. Marks
Nos. allotted

∠M = ∠N = 90° [ By construction ]

∆ABM ~ ∆PQN [ AA similarity criterion ] ½
AM AB
= ................... (2) ½
PN PQ
BC AB
But = .............. (3) (data )
QR PQ

From (2) and (3)
AM BC
= .................... (4) ½
PN QR

Substituting (4) in (1)
Area of ∆ABC BC BC
= ×
Area of ∆PQR QR QR

Area of ∆ABC BC 2
= ½ 5
Area of ∆PQR QR 2

Note : Proving the theorem as it is in the textbook give full marks.

RF/RR (A)-(200)-9020 (MA) [ Turn over