Karnataka SSLC Question Paper 2025 Answer Key Mathematics

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Paper
ANSWER KEY

CCE RF/PF A
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KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003

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MARCH/APRIL 2025 EXAMINATION - 1

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MODEL ANSWERS

—⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E CODE NO. : 81-E
…Œ⁄æ⁄fl : V⁄{}⁄
Subject : MATHEMATICS
(ÍÝÇÝ A»Ü¦ì / TÝÓÜX A»Ü¦ì )
( Regular Fresh / Private Fresh )
( AMV⁄« »⁄·¤®⁄¥¿»⁄fl / English Medium )

¶´¤MO⁄ : 24. 03. 2025 ] [ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80

Date : 24. 03. 2025 ] [ Max. Marks : 80

Qn. Ans. Marks
Value Points
Nos. Key allotted

I. Multiple choice questions : 8×1=8

1. LCM of 2 and 3 is

(A) 2 (B) 3

(C) 5 (D) 6

Ans. :

(D) 6 1

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81-E 2
Qn. Ans. Marks
Value Points
Nos. Key allotted

2. If the lines represented by the equations
a1x + b1y + c1 = 0 and a 2 x + b 2y + c 2 = 0 are

coincident, then the correct relation is
a1 b c a1 b
(A) = 1 = 1 (B) ≠ 1
a2 b2 c2 a2 b2
a1 b1 c1 a1 b1 c1
(C) = ≠ (D) ≠ =
a2 b2 c2 a2 b2 c2

Ans. :
a1 b c
(A) = 1 = 1
a 2 b2 c2 1

3. The quadratic equation in the following is
(A) x 3 − 6x (B) p ( x ) = x 2 + 7x

(C) 3x = 9 (D) x 2 + 3x + 4 = 0
Ans. :

(D) x 2 + 3x + 4 = 0 1

4. In the following, the shapes which are always similar
are,
(A) any two equilateral triangles
(B) square and rectangle
(C) square and rhombus
(D) any two trapeziums
Ans. :

(A) any two equilateral triangles 1

5. The volume of a sphere of radius ‘r’ units is
2 4
(A) π r 3 cubic units (B) π r 3 cubic units
3 3
1 3
(C) π r 3 cubic units (D) π r 3 cubic units
3 2
Ans. :
4 3
(B) πr cubic units
3 1

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3 81-E
Qn. Ans. Marks
Value Points
Nos. Key allotted

The distance of a point P ( x, y ) from the origin is
6.
(A) x2 − y2 (B) x +y

(C) x2 + y2 (D) x −y

Ans. :

(C) x 2 + y2 1
The common difference of the arithmetic progression
7. – 1, – 3, – 5 ... is
(A) –1 (B) 2
(C) –2 (D) 3

Ans. :

(C) –2 1
In the given figure ‘O’ is the centre of the circle and
8. the length of the arc APB is 4 π cm. If OB = 9 cm,
then the measure of angle θ is

(A) 60° (B) 80°
(C) 85° (D) 70°

Ans. :

(B) 80° 1

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81-E 4
Qn. Marks
Value Points
Nos. allotted

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

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

9. Write the degree of a linear polynomial.

Ans. :

1 ( one ) 1

10. Write the formula to find the total surface area of a cube
of edge ‘a’ units.

Ans. :
2
6a sq.units 1

11. In the given frequency distribution table, write the modal
class :

Class-interval Frequency

1–3 4
3–5 8
5–7 2
7–9 2

Ans. :

3–5 1

12. Write the probability of an impossible event.

Ans. :

0 1

13. How many solutions do the pair of linear equations
2x + 3y – 9 = 0 and 3x + 2y – 6 = 0 has ?

Ans. :

One solution / unique solution 1

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5 81-E
Qn. Marks
Value Points
Nos. allotted

14. Write the zeroes of the polynomial y = p ( x ) in the given
graph.

Ans. :

– 1 and 4 ½+½ 1

15. Write the roots of the quadratic equation x ( x + 2 ) = 0.
Ans. :

0 and – 2 ½+½ 1

16. In the given figure, write the similarity criterion used to
show that ∆ ABC ~ ∆ QRP.

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81-E 6
Qn. Marks
Value Points
Nos. allotted

Ans. :

SSS or side – side - Side 1

Note : Q. No. from 9 to 16 give full marks for direct
answer.

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

17. In the given figure, ABC = 90°. Write the values of the

following :
i) sin α
ii) tan θ

Ans. :
3
(i) sin α = 1
5
4
(ii) tan θ = 1
3 2

18. Prove that 6 + 2 is an irrational number.

OR
The HCF and LCM of two positive integers are respectively
4 and 60. If one of the integers is 20, then find the other
integer.
Ans. :
Let us assume to the contrary that 6 + 2 is rational.
a
6+ 2 = ' a ' and ' b ' are coprimes ( b ≠ 0 ) ½
b
a
2= − 6
b
a − 6b
2= ½
b

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7 81-E
Qn. Marks
Value Points
Nos. allotted
This shows that 2 is rational.
But this contradicts the fact that 2 is irrational. ½
This contradiction has arisen because of our wrong
assumption.
∴ 6 + 2 is an irrational number. ½ 2
OR
Let 'a' and ' b ' be two positive integers
HCF ( a, b ) = 4
LCM ( a, b ) = 60
a = 20
b=?
a × b = HCF ( a, b ) × LCM ( a, b ) ½
20 × b = 4 × 60 ½
3
4× 60
b= ½
201
b = 12 ½ 2
19. Solve the given pair of linear equations by elimination
method :
2x + y = 10
x–y =2
Ans. :
2x + y = 10 ....................... ( 1 )
x – y = 2 ...........................( 2 )
Adding 3x = 12 ½
12
x=
3
x=4 ½
Substitute x = 4 in equation ( 1 )
2 ( 4 ) + y = 10 ½
8 + y = 10
y = 10 – 8
y=2 ½
Note : Marks should be given if the value of ‘x’ is
substituted in equation (2) 2
20. Find the roots of the quadratic equation
x 2 + 8x + 12 = 0.
OR

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81-E 8
Qn. Marks
Value Points
Nos. allotted
Find the discriminant of the quadratic equation
x 2 + 4x + 5 = 0 and hence write the nature of the roots.
Ans. :
x 2 + 8 x + 12 = 0

x 2 + 6x + 2x + 12 = 0 ½
x ( x + 6 ) + 2( x + 6 ) = 0
(x +6) (x +2) =0 ½

x+6=0 or x + 2 = 0 ½
x =–6 or x=–2 ½ 2

Note: If alternate method is followed to get correct
answer, then give full marks.

OR
x 2 + 4x + 5 = 0
ax 2 + bx + c = 0
a = 1, b = 4, c = 5
2
Discriminant = b − 4ac ½
2
= (4 ) − 4 (1 ) ( 5 ) ½
= 16 – 20
=–4 <0 ½ 2
Nature of roots : No real roots. ½
21. Find the sum of first 20 terms of the arithmetic
progression 5, 9, 13, ... using formula.
Ans. :
a=5
d=9–5=4
n = 20
n
Sn = [ 2a + ( n −1 )d ] ½
2
20
S 20 = [ 2 ( 5 ) + ( 20 −1 ) 4 ] ½
2
= 10 [ 10 + 76 ]
= 10 ( 86 ) ½
S20 = 860 ½
Note : If alternate method is used to get the correct
answer, then give full marks. 2

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9 81-E
Qn. Marks
Value Points
Nos. allotted

22. In the given figure, PA and PB are tangents to the circle
with centre ‘O’. If PA = 4 cm and APO = 40°, then find
the measure of AOB and length of PB.

Ans. :

In ∆OAP , ∠OAP = 90° [ Q OA ⊥ AP ] ½

∴ ∠AOP = 180° – ( 90° + 40° )

= 180° – 130°

∠AOP = 50° ½
∠AOP = ∠BOP [ Q ∆ AOP ≅ ∆ BOP ]

∴ ∠BOP = 50°

∴ ∠AOB = 50° + 50° = 100° ½

PA = PB ( By theorem ) ∴ PB = 4 cm ½

Note : If alternate method is used to get correct answer,
then give full mars. 2

23. According to Fundamental Theorem of Arithmetic, if
40 = x y . z , then find the values of x, y and z.

Ans. :

40 = 23 × 51 ½

Given 40 = x y × z

∴ x=2 y=3 z=5 ½+½+½ 2

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81-E 10
Qn. Marks
Value Points
Nos. allotted
If A ( 1, y ), B ( 4, 3 ), C ( x, 6 ) and D ( 3, 5 ) are the
24.
vertices of a parallelogram taken in an order, then find
the values of x and y.

Ans. :
Mid-point of AC = Mid-point of BD ( Diagonals of a
parallelogram bisect each other )
 x +1 6 + y   4 + 3 3 + 5 
 , = ,  ½
 2 2   2 2 
 x +1 6 + y   7 
 ,  =  , 4 ½
 2 2  2 
x +1 7 6+y
= =4
2 2 2
x+1=7 6+y=8
x=7–1 y=8–6
x=6 y=2
Finding x ½
Finding y ½
Note : If alternate method is used to get correct answer,
then give full marks. 2

IV. Answer the following questions : 9 × 3 = 27
25. Find the zeroes of the quadratic polynomial p ( x ) = x 2 +
7x + 10 and verify the relationship between the zeroes
and the coefficients.
Ans. :

p (x ) = x 2 + 7x + 10
= x 2 + 5 x + 2x + 10
= x ( x + 5 ) + 2( x + 5 )
p ( x ) = ( x + 5 ) ( x + 2) ½
(x+5) (x+2)=0
x + 5 =0 or x + 2 = 0
x=–5 or x = – 2 ½

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11 81-E
Qn. Marks
Value Points
Nos. allotted
– 5 and – 2 are the zeroes of given polynomial.
−(7 )
Sum of zeroes = – 2 + ( – 5 ) = – 7 =
1
− coefficient of x  − b 
=   1
coefficient of x 2  a 
10
Product of zeroes = ( – 2 ) × ( – 5 ) = 10 =
1
constant term  c 
2  a 
= 1
coefficien t of x 3
26. Prove that “The tangent at any point of a circle is
perpendicular to the radius through the point of contact”.
Ans. :

½
Data : 'O' is the centre of the circle. XY is the tangent at
' P '. OP is the radius. ½
To prove : OP ⊥ XY . ½
Construction : Take a point 'Q' on XY other than 'P ' and
join OQ. Let it intersect the circle at ' R '. ½
Proof : From the figure, OQ > OR.
But OR = OP ( radii of the same circle ) ½
OQ > OP.
This happens for every point on the line XY except the
point P.
∴ OP is the shortest distance from O to the points on XY.
½
∴ OP ⊥ XY
Note : If the theorem is proved as given in textbook give
full marks. 3
27. Prove that :
cos A 1 + sin A
+ = 2 sec A.
1 + sin A cos A
OR

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81-E 12
Qn. Marks
Value Points
Nos. allotted
Find the value of :

 5 cos 2 60 o + 4 sec 2 30 o − tan 2 45 o 
 
 sin2 30 o + cos 2 30 o 
 
Ans. :
cos A 1 + sin A
LHS = +
1 + sin A cos A
2 2
cos A + (1 + sin A )
= ½
cos A (1 + sin A )
2 2
cos A +1 + sin A + 2 sin A
= ½
cos A (1 + sin A )
1 + 1 + 2 sin A
= [ Qsin2 A + cos 2 A =1 ] ½
cos A (1 + sin A )
2 + 2 sin A
= ½
cos A (1 + sin A )
2 (1 + sin A )
= ½
cos A (1 + sin A )
2
=
cos A
1
= 2 sec A [Q = sec A ] ½
cos A

LHS = RHS 3
OR
1 2
cos 60° = , sec 30° = , tan 45° = 1 ½+½+½
2 3
1 3
sin 30° = , cos 30° =
2 2
2 2
1  2 
5   + 4   − (1 )2
 
2  3
= 2
½
2 3
1
  +  

2  2 
1 4
5   + 4   −1
=   3
4
½
1 3
+
4 4

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13 81-E
Qn. Marks
Value Points
Nos. allotted
5 16
+ −1
= 4 3
1
15 + 64 −12
=
12
67 3
= ½
12
Note : If directly taken as sin2 30° + cos 2 30° =1 , then also

give full marks.
In the given figure ‘O’ is the centre of the circle of radius
28.
21 cm. If AOB = 60°, then find the area of the segment
APB.
[ Take 3 = 1·73 ]

Ans. :
θ 2
Area of the sector OAPB = × πr ½
360°
60° 2211
= × × 213 × 21
360°6 7
2

= 11 × 21

= 231 cm 2 ½
∆OAB is an equilateral triangle
3 2
Area of equilateral ∆OAB = a ½
4
1 ⋅ 73
= × 21 × 21
4
762 ⋅ 93
=
4

= 190·73 cm 2 ½
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81-E 14
Qn. Marks
Value Points
Nos. allotted
Area of the segment Area of sector area of
= – ½
APB OAPB ∆OAB

= 231 – 190·73
= 40·27 cm 2 ½
3
Note : If the final answer is upto 4 decimal places
2
( 40.2675 cm ) then also give full marks.
29. Find the coordinates of a point which divides the line
segment joining the points ( – 1, 7 ) and ( 4, – 3 )
internally in the ratio 2 : 3.
OR
Find a relation between x and y such that the point (x, y)
is equidistant from the points ( 3, 6 ) and ( – 3, 4 )
Ans. :
( – 1, 7 ) ( 4, – 3 ) 2:3
x1, y1 x 2 , y2 m1 = 2, m2 = 3
 m x + m2x1 m1y2 + m2y1 
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

PA = PB
(x − 3 )2 + ( y − 6 )2 = ( x + 3)2 + ( y − 4 )2 ½
Squaring on both sides
( x − 3 )2 + ( y − 6 )2 = (x + 3 )2 + ( y − 4 )2 ½
x 2 + 9 − 6x + y 2 + 36 − 12y = x 2 + 9 + 6x + y 2 + 16 − 8y 1
− 6x − 6x −12y + 8y + 36 −16 = 0 ½

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15 81-E
Qn. Marks
Value Points
Nos. allotted
– 12x – 4y + 20 = 0
÷ –4
3x + y − 5 = 0 ½

30. Find the mean for the following data :
Class-interval Frequency
10 – 20 2
20 – 30 3
30 – 40 6
40 – 50 5
50 – 60 4
OR
Find the median for the following data :
Class-interval Frequency
15 – 20 4
20 – 25 5
25 – 30 10
30 – 35 5
35 – 40 6

Ans. :
Class interval frequency Mid-point xi f i
( fi ) xi
10 – 20 2 15 30
20 – 30 3 25 75
30 – 40 6 35 210
40 – 50 5 45 225
50 – 60 4 55 220
∑ f i = 20 ∑ f i x i = 760

2

Mean = X =
∑ f i xi ½
∑ fi
760
=
20
Mean ( X ) = 38 ½ 3
OR

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81-E 16
Qn. Marks
Value Points
Nos. allotted

Class interval frequency Cumulative
frequency

15 – 20 4 4
20 – 25 5 9
25 – 30 10 19
30 – 35 5 24
35 – 40 6 30
n = 30
1
n 30
= =15 , l = 25, c f = 9, f = 10, h = 5 ½
2 2
n 
 2 −cf 
Median = l +   ×h ½
 f 
 
 15 − 9 
= 25 +  × 5
 10 
6
= 25 + ×5 ½
10
= 25 + 3 ½
Median = 28 3

31. A box contains 20 cards numbered from 1 to 20. One
card is drawn randomly from the box. Find the
probability of getting a card bearing —
i) a perfect square number
ii) a number which is divisible by both 2 and 3.
Ans. :

n ( s ) = 20 ½
(i) A = { 1, 4, 9, 16 } ∴ n ( A ) = 4 ½
n(A)
P ( A )= ½
n (S )

4 1
P (A) = or P ( A ) = ½
20 5

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17 81-E
Qn. Marks
Value Points
Nos. allotted

(ii) B = { 6, 12, 18 } ∴n (B ) =3 ½
n (B )
P (B ) =
n (S )
3 3
P (B ) = ½
20
32. The difference between the altitude and base of a right
angled triangle is 5 cm. If the area of the triangle is
150 cm 2 , then find the base and altitude of the triangle.
OR
The sum of the squares of two consecutive even positive
integers is 164. Find the integers.
Ans. :
Let altitude = x cm, then ½
base = ( x – 5 ) cm
Area of triangle = 150 cm 2
1
. x . ( x − 5 ) =150 ½
2
x 2 − 5 x = 300
x 2 − 5 x − 300 = 0 ½
x 2 − 20 x + 15 x − 300 = 0
x ( x – 20 ) + 15 ( x – 20 ) = 0
( x – 20 ) ( x + 15 ) = 0 ½
x – 20 = 0 or x + 15 = 0
x = 20 or x = – 15 ½
Since the length can't be negative, x = 20 cm
∴ Altitude = x = 20 cm
Base = x – 5 = 20 – 5 = 15 cm ½
Note : If x and x + 5 are considered to solve the problem
to get the correct answer, then give full marks. 3
OR
Let the two consecutive even positive integers be x and
(x+2)
By data x 2 + ( x + 2 )2 =164 ½
x 2 + x 2 + 22 + 4x =164
2x 2 + 4x + 4 −164 = 0 ½
2x 2 + 4x −160 = 0

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81-E 18
Qn. Marks
Value Points
Nos. allotted
÷2
x 2 + 2x − 80 = 0 ½
x 2 + 10 x − 8 x − 80 = 0
x ( x + 10 ) − 8 ( x +10 ) = 0
( x + 10 ) ( x − 8 ) = 0
x +10 = 0 or x – 8 = 0 ½
x = – 10 or x = 8
x is positive integer ∴ x=8 ½
x + 2 = 8 + 2 = 10
3
Two consecutive even positive integers are 8 and 10. ½
33. Two line segments AB and CD intersect each other at a
point ‘O’. Join AC and BD such that AC || BD and prove
that ∆ AOC ~ ∆ BOD.
Ans.

To draw AB & CD → ½ + ½

Joining AC & BD → ½
In ∆ACO and ∆BDO
∠CAO = ∠ DBO [ Alternate angles AC||BD ] ½
∠ACO = ∠ BDO
∠AOC = ∠ BOD [Vertically opposite angles] ½
∴ ∆AOC ~ ∆BOD [ AAA similarity ] ½
Note : If AA similarity criterion is used to prove
∆ACO ~ ∆BDO, then also give full marks. 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 = 8
x+y = 5

CCE RF/PF(A)/101/1811 (MA)

19 81-E
Qn. Marks
Value Points
Nos. allotted

Ans. :

x + 2y = 8 For table construction 1+1

x+y=5 Drawing two lines
by marking points 1

Writing the values
of x and y 1 4

35. Prove that “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”.
Ans. :

½

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81-E 20
Qn. Marks
Value Points
Nos. allotted
Data : In ∆ABC, DE ||BC ½
AD AE
To prove : = ½
DB EC
Construction : Draw DM ⊥ AC and EN ⊥ AB .
Join BE and CD ½
1
ar ( ∆ADE ) × AD × EN AD
Proof : = 2 = ........... ( 1 ) ½
ar ( ∆BDE ) 1 DB
× DB × EN
2
1
× AE × DM
ar ( ∆ADE ) 2 AE
= = ........... ( 2 ) ½
ar ( ∆DEC ) 1 EC
× EC × DM
2
∆BDE and ∆DEC are on the same base DE and between
the same parallels BC and DE.
∴ ar ( ∆BDE ) = ar ( ∆DEC ) .............. ( 3 ) ½
From ( 1 ), ( 2 ) and ( 3 )
AD AE
= ½
DB EC 4
Note : If the theorem is proved as in the textbook, then
give full marks.
36. A solid consisting of a right circular cone of height 120
cm and radius 60 cm standing on a hemisphere of radius
60 cm is placed upright in a right circular cylinder full of
water such that it touches the bottom as shown in the
figure. If the radius of the cylinder is 60 cm and height is
180 cm, then find the volume of water left in the cylinder
in terms of π.

OR

CCE RF/PF(A)/101/1811 (MA)

21 81-E
Qn. Marks
Value Points
Nos. allotted

A solid is made of a cylinder with a hemispherical
depression having the same radius ( ‘r’ cm ) as that of
cylinder at the top end as shown in the figure. The
volume of the hemispherical depression is 18000 π cm 3 .
If the height of the cylinder is 145 cm, then find the total
surface area of the solid.

Ans. :
2
Volume of cylinder = πr h ½
= π ( 60 )2 ×180
= π ( 3600 ) × 180

= 6,48,000 π cm 3 ½
Volume of the solid = Volume of + Volume of
cone Hemisphere ½
1 2 2 3
= πr h + πr ½
3 3
1 2
= πr [ h + 2r ]
3
1 2
= π × 60 [120 + 2 ( 60 ) ] ½
3
1 2 80
= × π × 60 × 240
3
= 2,88,000 π cm 3 ½
Volume of water Volume of Volume of
= – ½
left in the cylinder Cylinder Solid
= 648000π – 288000 π
= 3,60,000π cm 3 ½ 4

OR

CCE RF/PF(A)/101/1811 (MA) [ Turn over

81-E 22
Qn. Marks
Value Points
Nos. allotted
2 3
Volume of Hemisphere = πr ½
3
2 3
18000 π = × π × r
3
3 18000 × 3
r = ½
2
r 3 = 27000
r = 30 cm ½
TSA of solid = CSA of CSA of Area of
+ + ½
Hemisphere cylinder circular
base
= 2 πr 2 + 2πrh + πr 2 ½
= πr [ 2r + 2h + r ]
22
= × 30 [ 2 × 30 + 2 × 145 + 30 ] ½
7
22
= × 30 × [ 60 + 290 + 30 ] ½
7
22
= × 30 × 380
7
250800
= cm 2 ½
7
4
≈ 35828·5 cm 2
37. An arithmetic progression consists of 16 terms. The sum
of all its terms is 768. If the last term of the progression is
93, then find the arithmetic progression. Also show that
the sum of all the terms of this progression is equal to
3 times the sum of first 16 odd natural numbers using
formula.
Ans. :
n = 16
S16 = 768
an = l = 93
n
Sn = [ a + an ] ½
2
16 8
768 = [ a + 93 ] ½
2
768
a + 93 =
8

CCE RF/PF(A)/101/1811 (MA)

23 81-E
Qn. Marks
Value Points
Nos. allotted
a + 93 = 96
a = 96 – 93
a=3 ½
an = a + ( n −1 ) d
93 = 3 + ( 16 – 1 ) d ½
93 = 3 + 15d
15d = 90
90
d=
15
d=6 ½
AP is 3, 9, 15, 21, 27 ....... ½
S16 = 3 + 9 +15 + 21+ ........ up to 16 terms
= 3 [ 1 + 3 + 5 + 7 + ......... up to 16 terms ] ½
2
= 3 × 16 2 [ Sn = n ] ½
4
↓
= 3 × 256 sum of first n odd
∴ 768 = 768 natural nos.
n
Note : If Sn = [ 2a + ( n − 1 ) d ] formula is used to get
2
the correct answer, then give full marks.
VI. Answer the following question : 1×5=5
38. A pole and a tower are standing vertically on a level
ground. The height of the pole is 6 m and the angle of
elevation to the top of the pole from the bottom of the
tower is 30°. The angle of elevation to the top of the tower
from the top of the pole is 60° as shown in the figure.
Find the height of the tower ( CD ). Also find the distance
( AC ) between the top of the pole and the top of the
tower.

CCE RF/PF(A)/101/1811 (MA) [ Turn over

81-E 24
Qn. Marks
Value Points
Nos. allotted
Ans. :
AB
In ∆ABD, tan 30° = ½
BD
1 6
=
3 BD
BD = 6 3 m ½
BD = AE = 6 3 m
CE
In ∆AEC, tan 60° = ½
AE
CE
3= ½
6 3
6 3 . 3 = CE
∴ CE = 6 ( 3 ) = 18 m ½
CE
In ∆AEC, sin 60° = ½
AC
3 18
=
2 AC
18× 2
AC = ½
3
36 3
= × ½
3 3
36 3
= ½
3
AC = 12 3 m
CD = CE + DE = 18 + 6 = 24 m ½ 5
Note : If alternate method is used to get correct answer,
then give full marks.

CCE RF/PF(A)/101/1811 (MA)

A
CCE RF/PF
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ,
ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá – 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003
ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1

ÊÜÞ¨ÜÄ EñܤÃWÜ ÜÙÜá
MODEL ANSWERS

: 81-K CODE NO. : 81-K

ËÐÜ¿á : WÜ~ñÜ
Subject : MATHEMATICS
ÍÝÇÝ A»Ü¦ì / TÝÓÜX A»Ü¦ì
Regular Fresh / Private Fresh

PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium
: 24. 03. 2025 ] [ : 80

Date : 24. 03. 2025 ] [ Max. Marks : 80

±ÜÅÍæ° EñܤÃܨÜ
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæ PÜÅÊÜÞûÜÃÜ
I. ŸÖÜá&BÁáR ±ÜÅÍæ° PÜÅÊÜÞûÜÃܨæãí©X®Ü ±ÜÅ£ ÓÜÄ¿ááñܤÃÜPæR
1 AíPÜ 8×1=8

1. 2 ÊÜáñÜᤠ3 ÃÜ Æ.ÓÝ.A.

(A) 2 (B) 3
(C) 5 (D) 6
EñܤÃÜ :
(D) 6 1

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 2
±ÜÅÍæ° EñܤÃܨÜ
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæ PÜÅÊÜÞûÜÃÜ
2. a1x  b1y  c1  0 ÊÜáñÜᤠa 2x  b2y  c 2  0 D
ÓÜËáàPÜÃÜ|WÜÙܬÜá® ¯ÜÅ£­˜ÓÜáÊÜ ÃæàTæWÜÙÜá IPÜÂWæãívÝWÜ,
ÓÜÄ¿Þ¨Ü ÓÜíŸí«ÜÊÜâ
a1 b1 c1 a1 b1
(A)   (B) 
a2 b2 c2 a2 b2
a1 b1 c1 a1 b1 c1
(C)   (D)  
a2 b2 c2 a2 b2 c2
EñܤÃÜ :
(A) a1 b c
 1  1
a2 b2 c 2 1
3. D PæÙÜX®ÜÊÜâWÜÙÜÈÉ ÊÜWÜìÓÜËáàPÜÃÜ|ÊÜâ
(A) x 3  6x (B) p ( x ) = x 2 + 7x

(C) 3x = 9 (D) x 2 + 3x + 4 = 0
EñܤÃÜ :
(D) x 2  3x  4  0 1
4. CÊÜâWÜÙÜÈÉ ¿ÞÊÝWÜÆã ÓÜÊÜáÃÜã±ÜÊÝXÃÜáÊÜ BPÜê£WÜÙÜá
(A) ¿ÞÊÜâ¨æà GÃÜvÜá ÓÜÊÜá¸ÝÖÜá £Å»ÜágWÜÙÜá
(B) ÊÜWÜì ÊÜáñÜᤠB¿áñÜ
(C) ÊÜWÜì ÊÜáñÜᤠÊÜhÝÅPÜê£
(D) ¿ÞÊÜâ¨æà GÃÜvÜá ñÝŲgÂWÜÙÜá
EñܤÃÜ :
(A) ¿ÞÊÜâ¨æà GÃÜvÜá ÓÜÊÜá¸ÝÖÜá £Å»ÜágWÜÙÜá 1

5. £Åg ‘r’ ÊÜÞ®ÜWÜÚÃÜáÊÜ Jí¨Üá WæãàÙÜ¨Ü Z®Ü¶ÜÆÊÜâ
2 4
(A)  r 3 Z®ÜÊÜÞ®Ü (B)  r 3 Z®ÜÊÜÞ®Ü
3 3
1 3
(C)  r 3 Z®ÜÊÜÞ®Ü (D)  r 3 Z®ÜÊÜÞ®Ü
3 2
EñܤÃÜ :
4
(B)  r 3 Z®ÜÊÜÞ®Ü
3 1

CCE RF/PF(A)/101/1810 (MA)

3 81-K
±ÜÅÍæ° EñܤÃܨÜ
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæ PÜÅÊÜÞûÜÃÜ
6. ÊÜáãƹí¨Üá˯í¨Ü P ( x, y ) ¹í¨ÜáËWæ CÃÜáÊÜ ¨ÜãÃÜÊÜâ

(A) x2  y2 (B) x y

(C) x2  y2 (D) x y

EñܤÃÜ :

(C) 2 2
x y 1

7. – 1, – 3, – 5 ... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ÓÝÊÜޮܠÊÜÂñÝÂÓÜÊÜâ

(A) –1 (B) 2

(C) –2 (D) 3
EñܤÃÜ :

(C) –2 1

8. PæãqrÃÜáÊÜ bñÜŨÜÈÉ ‘O’ ÊÜêñܤPæàí¨ÜÅ ÊÜáñÜᤠAPB PÜíÓܨÜ
E¨Üª 4  cm BX¨æ. OB = 9 cm B¨ÜÃæ,  Pæãà®Ü¨Ü
AÙÜñæ¿áá

(A) 60° (B) 80°

(C) 85° (D) 70°

EñܤÃÜ :

(B) 80° 1

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 4

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
II. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ 8×1=8

±ÜÅÍæ° ÓÜíTæ 9 Äí¨Ü 16 ÃÜÊÜÃæWæ ®æàÃÜ EñܤÃÜPæR ±Üä|ì AíPÜWÜÙÜ®Üá°
¯àvÜáÊÜâ¨Üá
9. Jí¨Üá ÃæàTÝñܾPÜ ŸÖÜá¯Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá [ÝñÜ wXÅ ÊÜ®Üá°
ŸÃæÀáÄ.
EñܤÃÜ :
1 (Jí¨Üá)
1
10. Aíb®Ü E¨Üª ‘a’ ÊÜÞ¬ÜWÜÚÃÜáÊÜ Jí¨Üá ÊÜWÜì Z¬Ü¨Ü ¯Üä|ìÊæáàÇ澄
ËÔ¤à|ìÊÜ®Üá° PÜívÜá×w¿ááÊÜ ÓÜãñÜÅ ŸÃæÀáÄ.
EñܤÃÜ :
6a 2 aܨÜÃÜ ÊÜÞ®ÜWÜÙÜá
1
11. PæãqrÃÜáÊÜ BÊÜ꣤ ËñÜÃÜOÝ PæãàÐÜrPܨÜÈÉ ŸÖÜáÆPÜËÃÜáÊÜ
ÊÜWÝìíñÜÃÜÊܬÜá® ŸÃæÀáÄ
ÊÜWÝìíñÜÃÜ BÊÜ꣤
1—3 4
3—5 8
5—7 2
7—9 2
EñܤÃÜ :
3–5 1
12. Jí¨Üá AÓÜí»ÜÊÜ Zo®æ¿á ÓÜí»ÜÊÜ­à¿áñæ¿á¬Üá® ŸÃæÀáÄ.
EñܤÃÜ :
0 1
13. 2x + 3y – 9 = 0 ÊÜáñÜᤠ3x + 2y – 6 = 0 D ÃæàTÝñܾPÜ
ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿áá GÐÜár ±ÜÄÖÝÃÜWÜÙÜ®Üá° Öæãí©¨æ
EñܤÃÜ :
Jí¨Üá ±ÜÄÖÝÃÜ / B®Ü®Ü ±ÜÄÖÝÃÜ 1

CCE RF/PF(A)/101/1810 (MA)

5 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
14. PæãqrÃÜáÊÜ ®Üûæ¿áÈÉ, y = p ( x ) ŸÖÜá¯Ü¨æãàQ¤¿á ÍÜã¬ÜÂñæWÜÙܬÜá®
ŸÃæÀáÄ.

EñܤÃÜ :
– 1 ÊÜáñÜᤠ4 ½+½ 1
15. x ( x + 2 ) = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙܬÜá® ŸÃæÀáÄ.
EñܤÃÜ :
0 ÊÜáñÜᤠ– 2 ½+½ 1
16. PæãqrÃÜáÊÜ bñÜŨÜÈÉ  ABC ~  QRP Gí¨Üá ñæãàÄÓÜÆá ŸÙÜÔÃÜáÊÜ

ÓÜÊÜáÃÜã¯Üñæ¿á ­«ÝìÃÜPÜ WÜá|ÊܬÜá® ŸÃæÀáÄ.

EñܤÃÜ :
¸Ý.¸Ý.¸Ý. A¥ÜÊÝ ¸ÝÖÜá&¸ÝÖÜá&¸ÝÖÜá 1

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 6
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
III. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ 8 × 2 = 16
17. PæãqrÃÜáÊÜ bñÜŨÜÈÉ ABC = 90° BX¨æ. PæÙX
Ü ®ÜÊÜâWÜÙÜ ¸æÇæ¿á®Üá°
ŸÃæÀáÄ

i) sin 

ii) tan 

EñܤÃÜ :
3
(i) sin   1
5
4
(ii) tan   1
3 2
18. 6 + 2 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ÓݘÔ.

A¥ÜÊÝ
GÃÜvÜá «Ü®Ü ±ÜäOÝìíPÜWÜÙÜ ÊÜá.ÓÝ.A. ÊÜáñÜᤠÆ.ÓÝ.A.WÜÙÜá PÜÅÊÜáÊÝX
4 ÊÜáñÜᤠ60 BX¨æ. Jí¨Üá ±ÜäOÝìíPÜÊÜâ 20 B¨ÜÃæ, ÊÜáñæã¤í¨Üá
±ÜäOÝìíPÜÊÜ®Üá° PÜívÜá×wÀáÄ.
EñܤÃÜ :

6  2 Jí¨Üá »ÝWÜÆŸœ ÓÜíTæ Gí¨Üá F×Óæãà|.

a
6 2  ' a ' ÊÜáñÜᤠ' b ' ÓÜÖÜ AË»ÝgÂWÜÙÜá ( b  0 ) ½
b
a
2 6
b
a  6b
2 ½
b

CCE RF/PF(A)/101/1810 (MA)

7 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
C¨Üá 2 Jí¨Üá »ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ñæãàÄÓÜáñܤ¨æ. B¨ÜÃæ C¨Üá
2 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ GíŸ ÓÜñÜ ÓÜíWÜ£Wæ ËÃÜá¨ÜœÊÝX¨æ. ½
®ÜÊÜá¾ ñܱÜâ³ FÖæÀáí¨Ü, D ËÃæãà«Ý»ÝÓÜ EípÝX¨æ.
2
 6  2 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ ½
A¥ÜÊÝ
'a' ÊÜáñÜᤠ' b ' GÃÜvÜá «Ü®Ü ±ÜäOÝìíPÜWÜÙÝXÃÜÈ.
ÊÜá.ÓÝ.A. ( a, b ) = 4
Æ.ÓÝ.A. ( a, b ) = 60
a = 20
b=?
a  b = ÊÜá.ÓÝ.A. ( a, b )  Æ.ÓÝ.A. ( a, b ) ½
20  b = 4  60 ½
3
4 60
b ½
201
b = 12 ½ 2
19. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á®Üá° ÊÜiìÓÜáÊÜ Ë«Ý®Ü©í¨Ü
¹wÔ
2x + y = 10
x–y =2
EñܤÃÜ :
. 2x + y = 10 ....................... ( 1 )
x – y = 2 ...........................( 2 )
PÜãw¨ÝWÜ 3x = 12 ½
12
x=
3
x=4 ½
x = 4 ®Üá° ÓÜËáàPÜÃÜ| ( 1 ) ÃÜÈÉ B¨æàÎÔ¨ÝWÜ
2 ( 4 ) + y = 10 ½
8 + y = 10
y = 10 – 8
y=2 ½ 2
ÓÜãaÜ®æ ‘x’ ¸æÇæ¿á®Üá° ÓÜËáàPÜÃÜ| (2) ÃÜÈÉ B¨æàÎԨܪÃÜã,
AíPÜWÜÙÜ®Üá° ¯àvÜáÊÜâ¨Üá.
CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 8
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ

20. x2 + 8x + 12 = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá°
PÜívÜá×wÀáÄ.
A¥ÜÊÝ
x 2 + 4x + 5 = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü Íæãà«ÜPÜÊÜ®Üá° PÜívÜá×wÀáÄ
ÊÜáñÜᤠÊÜáãÆWÜÙÜ ÓÜÌ»ÝÊÜÊܬÜá® ŸÃæÀáÄ.
EñܤÃÜ :

x 2  8x  12  0

x 2  6x  2x 12  0 ½
x ( x  6 )  2( x  6 )  0

( x 6) ( x 2) 0 ½
x+6=0 A¥ÜÊÝ x + 2 = 0 ½
2
x =–6 A¥ÜÊÝ x=–2 ½

ÓÜãaÜ®æ ¯Ü¿Þì¿á ˫ݬÜÊ¬Ü Üá® ŸÙÜÔ, ÓÜÄ EñܤÃÜÊܬÜá® ¯Üv©
æ ¨ÜªÈÉ
±Üä|ì AíPÜÊÜ®Üá° ¯àvÜáÊÜâ¨Üá.

A¥ÜÊÝ
2
x  4x  5  0

ax 2  bx  c  0

a = 1, b = 4, c = 5
2
Íæãà«ÜPÜ = b  4ac ½
2
= (4 )  4 (1 ) ( 5 ) ½
= 16 – 20

=–4 <0 ½
ÊÜáãÆWÜÙÜ ÓÜÌ»ÝÊÜ ÊÝÓܤÊÜ ÊÜáãÆWÜÙÜ®Üá° Öæãí©ÃÜáÊÜâ©ÆÉ. ½ 2

CCE RF/PF(A)/101/1810 (MA)

9 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
21. 5, 9, 13, ... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ 20 ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá°
ÓÜãñÜÅ E±ÜÁãàXÔ PÜívÜá×wÀáÄ.
EñܤÃÜ :
a=5
d=9–5=4
n = 20
n
Sn  [ 2a  ( n 1 )d ] ½
2
20
S 20  [ 2 ( 5 )  ( 20 1 ) 4 ] ½
2
= 10 [ 10 + 76 ]
= 10 ( 86 ) ½
S20  860 ½ 2
ÓÜãaÜ®æ ¯Ü¿Þì¿á ˫ݬÜÊ¬Ü Üá® ŸÙÜÔ, ÓÜÄ EñܤÃÜÊܬÜá® ¯Üv© æ ¨ÜªÈÉ
±Üä|ì AíPÜÊÜ®Üá° ¯àvÜáÊÜâ¨Üá.
22. PæãqrÃÜáÊÜ bñÜŨÜÈÉ ‘O’ Pæàí¨ÜÅËÃÜáÊÜ ÊÜêñܤPæR PA ÊÜáñÜᤠPB WÜÙÜá
ÓܳÍÜìPÜWÜÙÝXÊæ. PA = 4 cm ÊÜáñÜᤠAPO = 40° B¨ÜÃæ, AOB
¿á AÙÜñæ ÊÜáñÜᤠPB ¿á E¨ÜªÊÜ®Üá° PÜívÜá×wÀáÄ.

EñܤÃÜ :
∆OAP ¿áÈÉ OAP = 90° [  OA  AP ] ½
 AOP = 180° – ( 90° + 40° )
= 180° – 130°
AOP = 50° ½
AOP = BOP [  AOP   BOP ]
 BOP = 50°
 AOB = 50° + 50° = 100° ½
AP = PB (±ÜÅÊæáà¿á©í¨Ü)  PB = 4 cm ½ 2
ÓÜãaÜ®æ ¯Ü¿Þì¿á ˫ݬÜÊ¬Ü Üá® ŸÙÜÔ, ÓÜÄ EñܤÃÜÊܬÜá® ¯Üv©
æ ¨ÜªÈÉ
±Üä|ì AíPÜÊÜ®Üá° ¯àvÜáÊÜâ¨Üá.
CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 10
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
23. AíPÜWÜ~ñÜ¨Ü ÊÜáãÆ ±ÜÅÊæáà¿á¨Ü ±ÜÅPÝÃÜ, 40 = x y . z B¨ÜÃæ, x, y
ÊÜáñÜᤠz ¸æÇæWÜÙÜ®Üá° PÜívÜá×wÀáÄ.
EñܤÃÜ :

40  23  51 ½
¨Üñܤ 40  x y  z

 x=2 y=3 z=5 ½+½+½ 2
24. A ( 1, y ), B ( 4, 3 ), C ( x, 6 ) ÊÜáñÜᤠD ( 3, 5 ) CÊÜâ Jí¨Üá
ÓÜÊÜÞíñÜÃÜ aÜñÜá»Üáìg¨Ü A®ÜáPÜÅÊÜá ÍÜêíWÜWÜÙݨÜÃæ, x ÊÜáñÜᤠy
¸æÇæWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

EñܤÃÜ :
ÓÜÊÜÞíñÜÃÜ aÜñÜá»Üáìg¨Ü PÜ|ìWÜÙÜá ¯ÜÃÜÓܳÃÜ A˜ìÓÜáñܤÊæ
AC ¿á ÊÜá«Ü¹í¨Üá = BD ¿á ÊÜá«Ü¹í¨Üá

 x 1 6  y   4  3 3  5 
 ,  ,  ½
 2 2   2 2 
 x 1 6  y   7 
 ,    , 4 ½
 2 2   2 
x 1 7 6y
 4
2 2 2
x+1=7 6+y=8
x=7–1 y=8–6
x=6 y=2
x PÜívÜá×w¿ááÊÜâ¨Üá ½
y PÜívÜá×w¿ááÊÜâ¨Üá ½ 2
ÓÜãaÜ®æ ¯Ü¿Þì¿á ˫ݬÜÊܬÜá® ŸÙÜÔ, EñܤÃÜÊܬÜá® ¯Üv©
æ ¨ÜªÈÉ
±Üä|ì AíPÜÊÜ®Üá° ¯àvÜáÊÜâ¨Üá.

CCE RF/PF(A)/101/1810 (MA)

11 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ

IV. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ 9 × 3 = 27

25. p ( x ) = x + 7x + 10 D ÊÜWÜìŸÖÜá¯Ü¨æãàQ¤¿á ÍÜã¬ÜÂñæWÜÙ¬
Ü Üá®
2

PÜívÜá×wÀáÄ ÖÝWÜã ÍÜã®ÜÂñæWÜÙÜá ÊÜáñÜᤠAÊÜâWÜÙÜ ÓÜÖÜWÜá|PÜWÜÙÜ
¬ÜvÜáË¬Ü ÓÜíŸí«ÜÊܬÜá® ñÝÙæ ¬æãàw.

EñܤÃÜ :

p (x )  x 2  7x 10

 x 2  5x  2x 10

 x ( x  5 )  2( x  5 )

p ( x )  ( x  5 ) ( x  2) ½

(x+5) (x+2)=0

x + 5 =0 A¥ÜÊÝ x + 2 = 0

x=–5 A¥ÜÊÝ x = – 2 ½

– 5 ÊÜáñÜᤠ– 2 ÍÜã®ÜÂñæWÜÙÝXÊæ.

ÍÜã®ÜÂñæWÜÙÜ Êæãñܤ = – 2 + ( – 5 ) = – 7 =  ( 7 )
1

b
=  
 a  1
 

ÍÜã¬ÜÂñæWÜÙÜ WÜá|ÆŸœ = ( – 2 )  ( – 5 ) = 10 = 10
1

c 
=   1
a  3

26. ÊÜêñܤ¨Ü Êæáà騆 ¿ÞÊÜâ¨æà ¹í¨ÜáË®ÜÈÉ GÙæ¨Ü ÓܳÍÜìPÜÊÜâ, ÓܳÍÜì
¹í¨ÜáˬÜÈÉ GÙæ¨Ü £ÅgÂPæR ÆíŸÊÝXÃÜáñܤ¨æ Gí¨Üá ÓݘÔ.
CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 12
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
EñܤÃÜ :

½
¨Üñܤ : ‘O’ ÊÜêñܤ Pæàí¨ÜÅ ‘P’ ¹í¨ÜáË®ÜÈÉ XY ¿áá ÓܳÍÜìPÜÊÝX¨æ.
OP £ÅgÂÊÝX¨æ. ½
Óݫܯà¿á : OP  XY . ½
ÃÜaÜ®æ : XY ÊæáàÇæ ‘P’ ¹í¨ÜáÊÜ®Üá° ÖæãÃÜñÜá±ÜwÔ ‘Q’ ¹í¨ÜáÊÜ®Üá°
ñæWæ¨ÜáPæãÚÛ. OQ ÊÜ®Üá° ÓæàÄÔ. C¨Üá ÊÜêñܤÊÜ®Üá° ‘R’ ¹í¨ÜáË®ÜÈÉ
dæà©ÓÜÈ. ½
ÓÝ«Ü®æ : bñÜÅ©í¨Ü, OQ > OR.
B¨ÜÃæ, OR = OP (£ÅgÂWÜÙÜá) ½
OQ > OP.
P ¹í¨ÜáÊÜ®Üá° ÖæãÃÜñÜá±ÜwÔ, XY Êæáà騆 GÇÝÉ ¹í¨ÜáWÜÚWÜã C¨Üá
A®ÜÌÀáÓÜáÊÜâ¨ÜÄí¨Ü O ¹í¨Üá˯í¨Ü XY Êæáà騆 CñÜÃæ ¹í¨ÜáWÜÚí¨Ü
¨ÜãÃÜQRíñÜ OP ¿á PܯÐÜu E¨ÜªÊÜ®Üá° Öæãí©¨æ. ½
 OP  XY
ÓÜãaÜ®æ ¯ÜsܯÜâÓܤPܨÜÈÉÃÜáÊÜíñæ ¯ÜÅÊæáà¿áÊܬÜá® ÓݘԨܪÃÜã ¯Üä|ì
AíPÜWÜÙÜ®Üá° ¯àvÜáÊÜâ¨Üá. 3
27. cos A 1  sin A
 = 2 sec A Gí¨Üá ÓݘÔ.
1  sin A cos A

A¥ÜÊÝ
 5 cos 2 60   4 sec2 30   tan 2 45  
  C¨ÜÃÜ ¸æÇæ¿á®Üá°
 sin2 30   cos 2 30  
 
PÜívÜá×wÀáÄ.

CCE RF/PF(A)/101/1810 (MA)

13 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
EñܤÃÜ :
cos A 1  sin A
GvÜ»ÝWÜ = 
1  sin A cos A

cos2 A  (1  sin A )2
= ½
cos A (1  sin A )

cos2 A 1  sin2 A  2 sin A
= ½
cos A (1  sin A )
1 1  2 sin A 2 2
= [ sin A  cos A 1 ] ½
cos A (1  sin A )
2  2 sin A
= ½
cos A (1  sin A )
2 (1  sin A )
= ½
cos A (1  sin A )
2
=
cos A
1
= 2 sec A [  sec A ] ½
cos A
GvÜ»ÝWÜ = ŸÆ»ÝWÜ 3
A¥ÜÊÝ
1 2
cos 60  , sec 30  , tan 45° = 1 ½+½+½
2 3
1 3
sin 30  , cos 30 
2 2
2 2
1  2 
5    4    (1 )2
2  3
= 2
½
2 
1 3 
    
2  2 
1 4
5    4   1
4 3
=   ½
1 3

4 4
5 16
 1
= 4 3
1

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 14
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
15  64 12
=
12
67
= ½ 3
12
ÓÜãaÜ®æ sin2 30  cos2 30 1 Gí¨Üá ®æàÃÜÊÝX ñæWæ¨ÜáPæãíw¨ÜªÃÜã
±Üä|ì AíPÜWÜÙÜ®Üá° ¯àvÜáÊÜâ¨Üá.
28. PæãqrÃÜáÊÜ bñÜŨÜÈÉ ‘O’ Pæàí¨ÜÅËÃÜáÊÜ ÊÜêñܤ¨Ü £Åg 21 cm BX¨æ.
AOB = 60° B¨ÜÃæ, APB ÊÜêñܤSívÜ¨Ü ËÔ¤à|ìÊÜ®Üá°
PÜívÜá×wÀáÄ.
[ 3 = 1·73 Gí¨Üá ñæWæ¨ÜáPæãÚÛ ]

EñܤÃÜ :
OAPB £ÅhÝÂíñÜÃÜ SívÜ¨Ü ËÔ¤à|ì =   r 2 ½
360
11
60 22 3
=   21  21
3606 7
2

= 11  21

= 231 cm 2 ½
∆OAB ÓÜÊÜá¸ÝÖÜá £Å»ÜágÊÝX¨æ.

3 2
ÓÜÊÜá¸ÝÖÜá £Å»Üág, ∆OAB ËÔ¤à|ì = a ½
4
1  73
=  21  21
4
762  93
=
4

= 190∙73 cm 2 ½

CCE RF/PF(A)/101/1810 (MA)

15 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
APB ÊÜêñܤSívÜ¨Ü ËÔ¤à|ì = OAPB £ÅhÝÂíñÜÃÜ ∆OAB
–
SívÜ¨Ü ËÔ¤à|ì ËÔ¤à|ì ½
= 231 – 190∙73
= 40∙27 cm 2 ½ 3
2
ÓÜãaÜ®æ Aí£ÊÜá EñܤÃÜÊÜâ 4 ¨ÜÍÜÊÜÞíÍÜ Óݧ®Ü¨ÜÈɨܪÃæ (40∙2675 cm )
±Üä|ì AíPÜÊÜ®Üá° ¯àvÜáÊÜâ¨Üá.
29. ( – 1, 7 ) ÊÜáñÜᤠ( 4, – 3 ) ¹í¨ÜáWÜÙÜ®Üá° ÓæàÄÓÜáÊÜ ÃæàTÝSívÜÊÜ®Üá°
BíñÜÄPÜÊÝX 2 : 3 A®Üá±ÝñܨÜÈÉ Ë»ÝXÓÜáÊÜ ¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜ®Üá°
PÜívÜá×wÀáÄ.
A¥ÜÊÝ
( x, y ) ¹í¨ÜáÊÜâ ( 3, 6 ) ÊÜáñÜᤠ( – 3, 4 ) ¹í¨ÜáWÜÚí¨Ü ÓÜÊÜÞ®Ü
¨ÜãÃܨÜÈɨܪÃæ, x ÊÜáñÜᤠy WÜÙÜ ¬ÜvÜáÊæ Jí¨Üá ÓÜíŸí«ÜÊܬÜá® PÜívÜá×wÀáÄ.
EñܤÃÜ :
( – 1, 7 ) ( 4, – 3 ) 2:3
x1, y1 x 2 , y2 m1  2, m2  3
 m x  m2x1 m1y2  m2y1 
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
A¥ÜÊÝ

PA = PB
2 2 2 2
(x  3 )  ( y  6 )  ( x  3)  ( y  4 ) ½

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 16
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
GÃÜvÜã PÜvæ ÊÜWÜì ÊÜÞw¨ÝWÜ
( x  3 )2  ( y  6 )2  (x  3 )2  ( y  4 )2 ½
x 2  9  6x  y 2  36  12y  x 2  9  6x  y 2 16  8y 1
 6x  6x 12y  8y  36 16  0 ½
12x  4y  20  0

 –4
3x y  5  0 ½
3

30. D PæÙÜX®Ü ¨ÜñݤíÍÜWÜÚWæ ÓÜÃÝÓÜÄ¿á®Üá° PÜívÜá×wÀáÄ

ÊÜWÝìíñÜÃÜ BÊÜ꣤

10 — 20 2

20 — 30 3

30 — 40 6

40 — 50 5

50 — 60 4

A¥ÜÊÝ
D PæÙÜX®Ü ¨ÜñݤíÍÜWÜÚWæ ÊÜá«ÝÂíPÜÊÜ®Üá° PÜívÜá×wÀáÄ
ÊÜWÝìíñÜÃÜ BÊÜ꣤

15 — 20 4

20 — 25 5

25 — 30 10

30 — 35 5

35 — 40 6

CCE RF/PF(A)/101/1810 (MA)

17 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
EñܤÃÜ :
ÊÜWÝìíñÜÃÜ BÊÜ꣤ ÓÜíbñÜ BÊÜ꣤ xi f i
( fi ) xi
10 – 20 2 15 30

20 – 30 3 25 75

30 – 40 6 35 210

40 – 50 5 45 225

50 – 60 4 55 220

 f i  20  f i xi  760

2

ÓÜÃÝÓÜÄ = X   i i
f x
½
 fi
760
=
20
ÓÜÃÝÓÜÄ ( X )  38 ½ 3
A¥ÜÊÝ
ÊÜWÝìíñÜÃÜ BÊÜ꣤ ÓÜíbñÜ BÊÜ꣤
15 – 20 4 4
20 – 25 5 9
25 – 30 10 19
30 – 35 5 24
35 – 40 6 30
n = 30
1
n 30
 15 , l = 25, c f  9, f = 10, h = 5 ½
2 2
n 
 2 cf 
ÊÜá«ÝÂíPÜ = l    h ½
 f 
 
 15  9 
= 25   5
10 
 
CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 18
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
6
= 25  5 ½
10
= 25 + 3 ½ 3
ÊÜá«ÝÂíPÜ = 28
31. Jí¨Üá ±æqrWæ¿áÈÉ 1 Äí¨Ü 20 ÃÜÊÜÃæWæ ®ÜÊÜáã¨ÝXÃÜáÊÜ 20 PÝv…ìWÜÚÊæ.
±æqrWæÀáí¨Ü Jí¨Üá PÝvÜì®Üá° ¿Þ¨ÜêbfPÜÊÝX ñæWæ¨ÝWÜ —
i) Jí¨Üá ±Üä|ìÊÜWÜì ÓÜíTæ¿á®Üá° ±Üvæ¿ááÊÜ ÓÜí»ÜÊܯà¿áñæ
ii) 2 ÊÜáñÜᤠ3 Äí¨Ü »ÝWÜÊÝWÜáÊÜ ÓÜíTæ¿á®Üá° ±Üvæ¿ááÊÜ
ÓÜí»ÜÊܯà¿áñæ CÊÜâWÜÙÜ®Üá° PÜívÜá×wÀáÄ.
EñܤÃÜ :
n ( s ) = 20 ½
(i) A = { 1, 4, 9, 16 }  n ( A )  4 ½
n(A)
P ( A ) ½
n (S )
4
P (A)  A¥ÜÊÝ P ( A )  1 ½
20 5
(ii) B = { 6, 12, 18 } n (B ) 3 ½
n (B )
P (B ) 
n (S )
3
P (B )  ½
20 3

32. Jí¨Üá ÆíŸPæãà¬Ü £Å»Üág¨Ü GñܤÃÜ ÊÜáñÜᤠ¯Ý¨Ü¨Ü ¬ÜvÜáË¬Ü ÊÜÂñÝÂÓÜÊÜâ
5 cm BX¨æ. £Å»Üág¨Ü ËÔ¤à|ìÊÜâ 150 cm 2 B¨ÜÃæ, £Å»Üág¨Ü ±Ý¨Ü
ÊÜáñÜᤠGñܤÃÜÊÜ®Üá° PÜívÜá×wÀáÄ.
A¥ÜÊÝ
GÃÜvÜá A®ÜáPÜÅÊÜá «Ü®Ü ÓÜÊÜá±ÜäOÝìíPÜWÜÙÜ ÊÜWÜìWÜÙÜ ÊæãñܤÊÜâ 164
B¨ÜÃæ, B ±ÜäOÝìíPÜWÜÙÜ®Üá° PÜívÜá×wÀáÄ.
EñܤÃÜ :
±Ý¨Ü = x cm ÊÜáñÜá¤
GñܤÃÜ = ( x – 5 ) cm BXÃÜÈ ½
£Å»Üág¨Ü ËÔ¤à|ì = 150 cm 2
CCE RF/PF(A)/101/1810 (MA)

19 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
1
. x . ( x  5 ) 150 ½
2
x 2  5x  300

x 2  5x  300  0 ½
x 2  20x 15x  300  0
x ( x – 20 ) + 15 ( x – 20 ) = 0
( x – 20 ) ( x + 15 ) = 0 ½
x – 20 = 0 A¥ÜÊÝ x + 15 = 0
x = 20 A¥ÜÊÝ x = – 15 ½
E¨ÜªÊÜâ MáOÝñܾPÜÊÝXÃÜÆá ÓÝ«ÜÂËÆÉ x = 20 cm
 ±Ý¨Ü = x = 20 cm 3
GñܤÃÜ = x – 5 = 20 – 5 = 15 cm ½
WÜÊÜá¯Ô x ÊÜáñÜᤠx + 5 Gí¨Üá ±ÜÄWÜ~Ô, ÓÜÄ¿Þ¨Ü EñܤÃÜ ±Üv橨ܪÃæ,
±Üä|ì AíPÜWÜÙÜ®Üá° ¯àvÜáÊÜâ¨Üá.
A¥ÜÊÝ
GÃÜvÜá «Ü®Ü ÓÜÊÜá ±ÜäOÝìíPÜWÜÙÜá x ÊÜáñÜᤠ( x + 2 ) BXÃÜÈ.
¨Üñܤ, x 2  ( x  2 )2 164 ½
x 2  x 2  22  4x 164
2x 2  4x  4 164  0 ½
2x 2  4x 160  0
2

x 2  2x  80  0 ½
x 2 10x  8x  80  0
x ( x 10 )  8 ( x 10 )  0
( x  10 ) ( x  8 )  0
x 10  0 A¥ÜÊÝ x – 8 = 0 ½
x = – 10 A¥ÜÊÝ x = 8
x C¨Üá «Ü®Ü ±ÜäOÝìíPÜÊÝXÃÜáÊÜâ¨ÜÄí¨Ü, x = 8 ½
x + 2 = 8 + 2 = 10
GÃÜvÜá «Ü®Ü ÓÜÊÜá ±ÜäOÝìíPÜWÜÙÜá 8 ÊÜáñÜᤠ10 BXÊæ. ½ 3

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 20
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
33. AB ÊÜáñÜᤠCD GÃÜvÜá ÃæàTÝSívÜWÜÙÜá ‘O’ ¹í¨ÜáË®ÜÈÉ ±ÜÃÜÓܳÃÜ
dæà©ÓÜáñܤÊæ. AC || BD BWÜáÊÜíñæ AC ÊÜáñÜᤠBD WÜÙ®Ü Üá° ÓæàÄÔ ÊÜáñÜá¤
 AOC ~  BOD Gí¨Üá ÓݘÔ.

EñܤÃÜ :

AB ÊÜáñÜᤠCD WÜÙÜ®Üá° GÙæ¿áÆá ½+½
AC ÊÜáñÜᤠBD WÜÙÜ®Üá° ÓæàÄÓÜÆá ½

∆ACO ÊÜáñÜᤠ∆BDO WÜÙÈ
Ü É

CAO = DBO [ ±Ü¿Þì¿á Pæãà®ÜWÜÙÜá AC ||BD ] ½

ACO = BDO

AOC = BOD [ ÍÜêíWݼÊÜááS Pæãà®ÜWÜÙÜá ] ½

 ∆AOC ~ ∆BOD [ Pæãà.Pæãà.Pæãà. ¯«ÝìÃÜPÜ WÜá| ] ½

WÜÊÜá¯Ô ∆ACO ~ ∆BDO Gí¨Üá ÓݘÓÜÆá Pæãà.Pæãà. ÓÜÊÜáÃÜã±Üñæ¿á

­«ÝìÃÜPÜ WÜá| ŸÙÜԨܪÃÜã ¯Üä|ì AíPÜWÜÙܬÜá® ­àvÜáÊÜâ¨Üá.
3

V. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ 4 × 4 = 16

34. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á ±ÜÄÖÝÃÜÊÜ®Üá° ®Üûæ¿á
˫ݮܩí¨Ü PÜívÜá×wÀáÄ

x + 2y = 8

x+y = 5

CCE RF/PF(A)/101/1810 (MA)

21 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
EñܤÃÜ :

x + 2y = 8 PæãàÐÜrPÜWÜÙÜ®Üá° ÃÜbÓÜáÊÜâ¨Üá 1+1

x+y=5 ¹í¨ÜáWÜÙÜ®Üá° WÜáÃÜá£Ô GÃÜvÜá
ÃæàTæWÜÙÜ®Üá° GÙæ¿ááÊÜâ¨Üá 1

x ÊÜáñÜᤠy ¸æÇæWÜÙÜ®Üá°

ŸÃæ¿ááÊÜâ¨Üá 1
4
35. £Å»Üág¨Ü GÃÜvÜá ¸ÝÖÜáWÜÙÜ®Üá° GÃÜvÜá ˼®Ü° ¹í¨ÜáWÜÙÜÈÉ
dæà©ÓÜáÊÜíñæ Jí¨Üá ¸ÝÖÜáËWæ ÓÜÊÜÞ®ÝíñÜÃÜÊÝX GÙæ¨Ü ÓÜÃÜÙÜÃæàTæ¿áá
EÚ¨æÃÜvÜá ¸ÝÖÜáWÜÙÜ®Üá° ÓÜÊÜÞ®Üá±ÝñܨÜÈÉ Ë»ÝXÓÜáñܤ¨æ. Gí¨Üá
ÓݘÔ.
EñܤÃÜ :

CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 22
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ

½
¨Üñܤ : ∆ABC ¿áÈÉ DE ||BC ½
AD AE
Óݫܯà¿á :  ½
DB EC
ÃÜaÜ®æ : DM  AC ÊÜáñÜᤠEN  AB . ½
BE ÊÜáñÜᤠCD WÜÙÜ®Üá° ÓæàÄÔ.

1
Ë ( ADE )  2  AD  EN  AD ........... ( 1 )
ÓÝ«Ü®æ : Ë ½
( BDE ) 1  DB  EN DB
2
1
Ë ( ADE )  2  AE  DM  AE ........... ( 2 ) ½
Ë ( DEC ) 1  EC  DM EC
2
∆BDE ÊÜáñÜᤠ∆DEC WÜÙáÜ Jí¨æà ±Ý¨Ü DE ÊÜáñÜᤠJí¨æà ÓÜÊÜÞíñÜÃÜ
ÃæàTæWÜÙÝ¨Ü BC ÊÜáñÜᤠDE WÜÙÜ ®ÜvÜáÊæ C¨æ.
 Ë ( BDE )  Ë ( DEC ) .............. ( 3 ) ½
( 1 ), ( 2 ) ÊÜáñÜᤠ( 3 ) Äí¨Ü

AD AE
 ½
DB EC
ÓÜãaÜ®æ ¯ÜsܯÜâÓܤPܨÜÈÉÃÜáÊÜíñæ ¯ÜÅÊæáà¿áÊܬÜá® ÓݘԨܪÃÜã ¯Üä|ì
AíPÜWÜÙÜ®Üá° ¯àvÜáÊÜâ¨Üá. 4
36. 60 cm £ÅgÂËÃÜáÊÜ A«ÜìWæãàÙÜ¨Ü ±Ý¨Ü¨Ü ÊæáàÇæ 120 cm GñܤÃÜ
ÊÜáñÜᤠ60 cm £ÅgÂÊÜ®Üá° Öæãí©ÃÜáÊÜ Jí¨Üá ®æàÃÜ ÊÜêñܤ±Ý¨Ü ÍÜíPÜáÊÜ®Üá°
hæãàwst Z®ÝPÜꣿá®Üá° ÓÜí±Üä|ìÊÝX ¯àįí¨Ü ñÜáí¹¨Ü ®æàÃÜ
ÊÜêñܤ¯Ý¨Ü ÔÈívÜÃ…¬ÜÈÉ ñÜÙÜÊܬÜá® ÊÜááoárÊÜíñæ ¬æàÃÜÊÝX bñÜŨÜÈÉ
ñæãàÄÔÃÜáÊÜíñæ ÊÜááÙÜáXÔ¨æ. ÔÈívÜÃ…¬Ü £ÅgÂÊÜâ 60 cm ÊÜáñÜá¤
CCE RF/PF(A)/101/1810 (MA)

23 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
GñܤÃÜÊÜâ 180 cm B¨ÜÃæ, ÔÈívÜÃ…¬ÜÈÉ EÚ©ÃÜáÊÜ ­àĬÜ
±ÜÅÊÜÞ|ÊÜ®Üá°  ¿áÈÉ ÊÜÂPܤ±ÜwÔ.

A¥ÜÊÝ
ÔÈívÜÃ…¬Ü ÊæáàÇݽWܨÜÈÉ ÔÈívÜÃ…¬ÜÐærà £ÅgÂËÃÜáÊÜ ( ‘r’ cm )
A«ÜìWæãàÙÝPÜꣿá®Üá°, PæãÃæ¨Üá bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ Jí¨Üá
Z®ÝPÜꣿá®Üá° ñÜ¿ÞÄÔ¨æ. PæãÃæ¿áÇÝ¨Ü A«ÜìWæãàÙÝPÜꣿá
Z®Ü¶ÜÆÊÜâ 18000  cm 3 BX¨æ. ÔÈívÜÃ…¬Ü GñܤÃÜ 145 cm B¨ÜÃæ,
Z¬ÝPÜꣿá Joár ÊæáàÇ澄 ËÔ¤à|ìÊܬÜá® PÜívÜá×wÀáÄ.

EñܤÃÜ :

ÔÈívÜÃ…¬Ü Z¬Ü¶ÜÆ = r 2h ½
2
=  ( 60 ) 180
=  ( 3600 )  180

= 6,48,000  cm 3 ½
CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 24
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
½
1 2 2 3
= r h  r ½
3 3
1 2
= r [ h  2r ]
3
1
=   602 [120  2 ( 60 ) ] ½
3
1
=    602  24080
3
= 2,88,000  cm 3 ½
½
= 648000 – 288000 
= 3,60,000 cm 3 ½
A¥ÜÊÝ
A«ÜìWæãàÙÜ¨Ü Z®Ü¶ÜÆ = 2 r 3 ½
3
2 3
18000  =   r
3
18000  3
r3  ½
2
r 3  27000
r = 30 cm ½
½
2 2
= 2 r  2rh  r ½
= r [ 2r  2h  r ]
22
=  30 [ 2  30  2  145  30 ] ½
7
22
=  30  [ 60  290  30 ] ½
7
22
=  30  380
7
250800
= cm 2 ½ 4
7
A¥ÜÊÝ
 35828∙5 cm 2
CCE RF/PF(A)/101/1810 (MA)

25 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
37. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿áÈÉ 16 ±Ü¨ÜWÜÚÊæ. A¨ÜÃÜ GÇÝÉ ±Ü¨ÜWÜÙÜ
ÊæãñܤÊÜâ 768 BX¨æ. ÍæÅà{¿á Pæã®æ¿á ±Ü¨ÜÊÜâ 93 B¨ÜÃæ, B
ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á®Üá° PÜívÜá×wÀáÄ ÖÝWÜã D ÍæÅà{¿á GÇÝÉ
±Ü¨ÜWÜÙÜ ÊæãñܤÊÜâ, Êæã¨ÜÆ 16 ¸æÓÜ ÓÝÌ»ÝËPÜ ÓÜíTæÂWÜÙÜ Êæãñܤ¨Ü
ÊÜáãÃÜÃÜÐÜrPæR ÓÜÊÜáÊÝXÃÜáñܤ¨æ Gí¨Üá ÓÜãñÜÅ E±ÜÁãàXÔ ñæãàÄÔ.
EñܤÃÜ :
n = 16
S16  768
an l  93
n
Sn  [ a  an ] ½
2
8
16
768  [ a  93 ] ½
2
768
a  93 
8
a + 93 = 96
a = 96 – 93
a=3 ½
an  a  ( n 1) d
93 = 3 + ( 16 – 1 ) d ½
93 = 3 + 15d
15d = 90
90
d
15
d=6 ½
ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿áá 3, 9, 15, 21, 27 ....... BX¨æ. ½
S16  3  9 15  21 ........ 16 ±Ü¨ÜWÜÙÜÊÃ
Ü æWæ
= 3 [ 1 + 3 + 5 + 7 + ......... 16 ±Ü¨ÜWÜÙÜÊÜÃæWæ ] ½
2
= 3  16 2 [ Sn  n ] ½

= 3  216 Êæã¨ÜÆ n ÓÝÌ»ÝËPÜ
 768 = 768 ¸æÓÜ ÓÜíTæÂWÜÙÜ Êæãñܤ
ÓÜãaÜ®æ Sn  n [ 2a  ( n 1 ) d ] ÓÜãñÜÅ E±ÜÁãàXÔ ÓÜÄ EñܤÃÜ
2
±Üv橨ܪÈÉ ±Üä|ì AíPÜWÜÙÜ®Üá° ¯àvÜáÊÜâ¨Üá. 4
CCE RF/PF(A)/101/1810 (MA) [ Turn over

81-K 26
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
VI. D PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×5=5

38. Jí¨Üá PÜíŸ ÊÜáñÜᤠJí¨Üá Wæãà¯ÜâÃÜÊÜâ ÓÜÊÜáñÜpÝr¨Ü ¬æÆ¨Ü ÊæáàÇæ
¬æàÃÜÊÝX ­í£Êæ. PÜíŸ¨Ü GñܤÃÜ 6 m ÊÜáñÜᤠWæãà±ÜâÃÜ¨Ü ±Ý¨Ü©í¨Ü
PÜíŸ¨Ü ÊæáàÆᤩWæ CÃÜáÊÜ E¬Ü®ñÜ Pæãà¬ÜÊÜâ 30° BX¨æ. PÜퟨÜ
ÊæáàÆᤩÀáí¨Ü Wæãà±ÜâÃÜ¨Ü ÊæáàÆᤩWæ CÃÜáÊÜ E®Ü°ñÜ Pæãà®ÜÊÜâ
bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ 60° BX¨æ. Wæãà±ÜâÃÜ¨Ü GñܤÃÜÊÜ®Üá° ( CD )
PÜívÜá×wÀáÄ. ÖÝWÜã PÜíŸ¨Ü ÊæáàÆᤩ ÊÜáñÜᤠWæãà¯ÜâÃܨÜ
ÊæáàÆᤩXÃÜáÊÜ ¨ÜãÃÜÊÜ®Üá° ( AC ) PÜívÜá×wÀáÄ.

EñܤÃÜ :

AB
∆ABD ¿áÈÉ tan 30  ½
BD
1 6

3 BD
BD  6 3 m ½
BD  AE  6 3 m
CE
∆AEC ¿áÈÉ tan 60  ½
AE
CE
3 ½
6 3
6 3 . 3  CE
 CE = 6 ( 3 ) = 18 m ½

CCE RF/PF(A)/101/1810 (MA)

27 81-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
CE
∆AEC ¿áÈÉ sin60  ½
AC
3 18

2 AC
18 2
AC  ½
3
36 3
=  ½
3 3
36 3
= ½
3
AC = 12 3 m
5
CD = CE + DE = 18 + 6 = 24 m ½
ÓÜãaÜ®æ ¯Ü¿Þì¿á ˫ݬÜÊܬÜá® ŸÙÜÔ, EñܤÃÜÊܬÜá® ¯Üv©
æ ¨ÜªÈÉ
±Üä|ì AíPÜÊÜ®Üá° ¯àvÜáÊÜâ¨Üá.

CCE RF/PF(A)/101/1810 (MA) [ Turn over

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