Karnataka SSLC Question Paper 2025 Maths

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Papers

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[ Joár ÊÜáá©ÅñÜ ±ÜâoWÜÙÜ ÓÜíTæ : 16
A [ Total No. of Printed Pages : 16
[ Joár ±ÜÅÍæ°WÜÙÜ ÓÜíTæ : 38
CCE RF/PF
[ Total No. of Questions : 38

—⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E Code No. : 81-E
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Subject : MATHEMATICS

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General Instructions to the Candidate :
1. This question paper consists of 38 questions.
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CCE RF/PF(A)/101/1811 1 of 16

CCE RF/PF(A)/101/1811 81-E

I. Four alternatives are given for each of the following questions /

incomplete statements. Choose the correct alternative and write

the complete answer along with its letter of alphabet. 8×1=8

1. LCM of 2 and 3 is

(A) 2 (B) 3

(C) 5 (D) 6

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 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

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

2 of 16

CCE RF/PF(A)/101/1811 81-E

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

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

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

(A) x2 − y2 (B) x +y

(C) x2 + y2 (D) x −y

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CCE RF/PF(A)/101/1811 81-E

7. The common difference of the arithmetic progression

– 1, – 3, – 5 ... is

(A) –1 (B) 2

(C) –2 (D) 3

8. In the given figure ‘O’ is the centre of the circle and 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°

4 of 16

CCE RF/PF(A)/101/1811 81-E

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

9. Write the degree of a linear polynomial.

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

‘a’ units.

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

12. Write the probability of an impossible event.

13. How many solutions do the pair of linear equations

2x + 3y – 9 = 0 and 3x + 2y – 6 = 0 has ?

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CCE RF/PF(A)/101/1811 81-E

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

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

6 of 16

CCE RF/PF(A)/101/1811 81-E

16. In the given figure, write the similarity criterion used to show that

∆ ABC ~ ∆ QRP.

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

7 of 16

CCE RF/PF(A)/101/1811 81-E

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.

19. Solve the given pair of linear equations by elimination method :

2x + y = 10

x–y =2

20. Find the roots of the quadratic equation x 2 + 8x + 12 = 0.

OR

Find the discriminant of the quadratic equation x 2 + 4x + 5 = 0

and hence write the nature of the roots.

21. Find the sum of first 20 terms of the arithmetic progression

5, 9, 13, ... using formula.

8 of 16

CCE RF/PF(A)/101/1811 81-E

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.

23. According to Fundamental Theorem of Arithmetic, if 40 = x y . z ,

then find the values of x, y and z.

24. If A ( 1, y ), B ( 4, 3 ), C ( x, 6 ) and D ( 3, 5 ) are the vertices of

a parallelogram taken in an order, then find the values of x and y.

9 of 16

CCE RF/PF(A)/101/1811 81-E

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.

26. Prove that “The tangent at any point of a circle is perpendicular

to the radius through the point of contact”.

27. Prove that :
cos A 1 + sin A
+ = 2 sec A.
1 + sin A cos A

OR

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 
 

28. In the given figure ‘O’ is the centre of the circle of radius 21 cm.
If AOB = 60°, then find the area of the segment APB.

[ Take 3 = 1·73 ]

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CCE RF/PF(A)/101/1811 81-E

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 )

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

11 of 16

CCE RF/PF(A)/101/1811 81-E

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

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.

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.

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.

12 of 16

CCE RF/PF(A)/101/1811 81-E

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

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”.

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

13 of 16

CCE RF/PF(A)/101/1811 81-E

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.

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.

14 of 16

CCE RF/PF(A)/101/1811 81-E

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.

15 of 16

CCE RF/PF(A)/101/1811 81-E

16 of 16

A [ Joár ÊÜáá©ÅñÜ ±ÜâoWÜÙÜ ÓÜíTæ : 16

CÈÉí¨Ü PÜñܤÄÔ
[ Total No. of Printed Pages : 16
CCE RF/PF [ Joár ±ÜÅÍæ°WÜÙÜ ÓÜíTæ : 38
[ Total No. of Questions : 38

ÓÜíPæàñÜ ÓÜíTæ : 81-K Code No. : 81-K
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Subject : MATHEMATICS
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6. ¯ÊÜáWæ ËñÜÄÓÜÇÝXÃÜáÊÜ ±ÜÅÍ氱ܣÅPæ¿á BÊÜ꣤ ( Version ) ÊÜáñÜᤠ¯ÊÜá¾ ±ÜÅÊæàÍÜ ±ÜñÜŨÜÈÉ
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CCE RF/PF(A)/101/1810 1 of 16

CCE RF/PF(A)/101/1810 81-K
I. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ A¥ÜÊÝ A±Üä|ì ÖæàÚPæWÜÚWæ ®ÝÆáR ±Ü¿Þì¿á EñܤÃÜWÜÙÜ®Üá°

¯àvÜÇÝX¨æ. AÊÜâWÜÙÜÈÉ ÓÜãPܤÊÝ¨Ü EñܤÃÜÊÜ®Üá° BÄÔ, A¨ÜÃÜ PÜÅÊÜÞûÜÃܨæãvÜ®æ ±Üä|ì

EñܤÃÜÊÜ®Üá° ŸÃæÀáÄ 8×1=8

1. 2 ÊÜáñÜᤠ3 ÃÜ Æ.ÓÝ.A.

(A) 2 (B) 3

(C) 5 (D) 6

2. a1x  b1y  c1  0 ÊÜáñÜᤠa 2 x  b2y  c 2  0 D ÓÜËáàPÜÃÜ|WÜÙÜ®Üá°

±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

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

2 of 16

CCE RF/PF(A)/101/1810 81-K
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

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

6. ÊÜáãƹí¨Üá˯í¨Ü P ( x, y ) ¹í¨ÜáËWæ CÃÜáÊÜ ¨ÜãÃÜÊÜâ

(A) x2  y2 (B) x y

(C) x2  y2 (D) x y

3 of 16

CCE RF/PF(A)/101/1810 81-K
7. – 1, – 3, – 5 ... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ÓÝÊÜޮܠÊÜÂñÝÂÓÜÊÜâ

(A) – 1 (B) 2

(C) – 2 (D) 3

8. PæãqrÃÜáÊÜ bñÜŨÜÈÉ ‘O’ ÊÜêñܤPæàí¨ÜÅ ÊÜáñÜᤠAPB PÜíÓÜ¨Ü E¨Üª 4  cm BX¨æ.

OB = 9 cm B¨ÜÃæ,  Pæãà®Ü¨Ü AÙÜñæ¿áá

(A) 60° (B) 80°

(C) 85° (D) 70°

4 of 16

CCE RF/PF(A)/101/1810 81-K
II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8×1=8

9. Jí¨Üá ÃæàTÝñܾPÜ ŸÖÜá±Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá [ÝñÜ wXÅ ÊÜ®Üá° ŸÃæÀáÄ.

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.

11. PæãqrÃÜáÊÜ BÊÜ꣤ ËñÜÃÜOÝ PæãàÐÜrPܨÜÈÉ ŸÖÜáÆPÜËÃÜáÊÜ ÊÜWÝìíñÜÃÜÊÜ®Üá °

ŸÃæÀáÄ

ÊÜWÝìíñÜÃÜ BÊÜ꣤

1—3 4

3—5 8

5—7 2

7—9 2

12. Jí¨Üá AÓÜí»ÜÊÜ Zo®æ¿á ÓÜí»ÜÊܯà¿áñæ¿á®Üá° ŸÃæÀáÄ.

13. 2x + 3y – 9 = 0 ÊÜáñÜᤠ3x + 2y – 6 = 0 D ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ

hæãàw¿áá GÐÜár ±ÜÄÖÝÃÜWÜÙÜ®Üá° Öæãí©¨æ

5 of 16

CCE RF/PF(A)/101/1810 81-K
14. PæãqrÃÜáÊÜ ®Üûæ¿áÈÉ, y = p ( x ) ŸÖÜá±Ü¨æãàQ¤¿á ÍÜã®ÜÂñæWÜÙÜ®Üá° ŸÃæÀáÄ.

15. x ( x + 2 ) = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá° ŸÃæÀáÄ.

6 of 16

CCE RF/PF(A)/101/1810 81-K

16. PæãqrÃÜáÊÜ bñÜŨÜÈÉ  ABC ~  QRP Gí¨Üá ñæãàÄÓÜÆá ŸÙÜÔÃÜáÊÜ

ÓÜÊÜáÃÜã±Üñæ¿á ¯«ÝìÃÜPÜ WÜá|ÊÜ®Üá° ŸÃæÀáÄ.

III. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8 × 2 = 16

17. PæãqrÃÜáÊÜ bñÜŨÜÈÉ ABC = 90° BX¨æ. PæÙÜX®ÜÊÜâWÜÙÜ ¸æÇæ¿á®Üá° ŸÃæÀáÄ

i) sin 

ii) tan 

7 of 16

CCE RF/PF(A)/101/1810 81-K
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 ÊÜáñÜá¤

60 BX¨æ. Jí¨Üá ±ÜäOÝìíPÜÊÜâ 20 B¨ÜÃæ, ÊÜáñæã¤í¨Üá ±ÜäOÝìíPÜÊÜ®Üá°

PÜívÜá×wÀáÄ.

19. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á®Üá° ÊÜiìÓÜáÊÜ Ë«Ý®Ü©í¨Ü

¹wÔ

2x + y = 10

x–y =2

20. x 2 + 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 ÊÜáñÜá¤

ÊÜáãÆWÜÙÜ ÓÜÌ»ÝÊÜÊÜ®Üá° ŸÃæÀáÄ.

8 of 16

CCE RF/PF(A)/101/1810 81-K
21. 5, 9, 13, ... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ 20 ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá° ÓÜãñÜÅ

E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

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.

23. AíPÜWÜ~ñÜ¨Ü ÊÜáãÆ ±ÜÅÊæáà¿á¨Ü ±ÜÅPÝÃÜ, 40 = x y . z B¨ÜÃæ, x, y ÊÜáñÜᤠz
¸æÇæWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

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.

9 of 16

CCE RF/PF(A)/101/1810 81-K
IV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 9 × 3 = 27

25. p ( x ) = x 2 + 7x + 10 D ÊÜWÜìŸÖÜá±Ü¨æãàQ¤¿á ÍÜã®ÜÂñæWÜÙÜ®Üá°
PÜívÜá×wÀáÄ ÖÝWÜã ÍÜã®ÜÂñæWÜÙÜá ÊÜáñÜᤠAÊÜâWÜÙÜ ÓÜÖÜWÜá|PÜWÜÙÜ ®ÜvÜáË®Ü
ÓÜíŸí«ÜÊÜ®Üá° ñÝÙæ ®æãàw.

26. ÊÜêñܤ¨Ü Êæáà騆 ¿ÞÊÜâ¨æà ¹í¨ÜáË®ÜÈÉ GÙæ¨Ü ÓܳÍÜìPÜÊÜâ, ÓܳÍÜì ¹í¨ÜáË®ÜÈÉ
GÙæ¨Ü £ÅgÂPæR ÆíŸÊÝXÃÜáñܤ¨æ Gí¨Üá ÓݘÔ.

cos A 1  sin A
27.  = 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.

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 ]

10 of 16

CCE RF/PF(A)/101/1810 81-K
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«ÜÊÜ®Üá° PÜívÜá×wÀáÄ.

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

11 of 16

CCE RF/PF(A)/101/1810 81-K
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

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.

32. Jí¨Üá ÆíŸPæãà®Ü £Å»Üág¨Ü GñܤÃÜ ÊÜáñÜᤠ±Ý¨Ü¨Ü ®ÜvÜá訆 ÊÜÂñÝÂÓÜÊÜâ 5 cm
BX¨æ. £Å»Üág¨Ü ËÔ¤à|ìÊÜâ 150 cm 2 B¨ÜÃæ, £Å»Üág¨Ü ±Ý¨Ü ÊÜáñÜá¤
GñܤÃÜÊÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

GÃÜvÜá A®ÜáPÜÅÊÜá «Ü®Ü ÓÜÊÜá±ÜäOÝìíPÜWÜÙÜ ÊÜWÜìWÜÙÜ ÊæãñܤÊÜâ 164 B¨ÜÃæ, B
±ÜäOÝìíPÜWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

33. AB ÊÜáñÜᤠCD GÃÜvÜá ÃæàTÝSívÜWÜÙÜá ‘O’ ¹í¨ÜáË®ÜÈÉ ±ÜÃÜÓܳÃÜ dæà©ÓÜáñܤÊæ.
AC || BD BWÜáÊÜíñæ AC ÊÜáñÜᤠBD WÜÙÜ®Üá° ÓæàÄÔ ÊÜáñÜá¤
 AOC ~  BOD Gí¨Üá ÓݘÔ.

12 of 16

CCE RF/PF(A)/101/1810 81-K
V. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 4 × 4 = 16

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

x + 2y = 8

x+y = 5

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.
36. 60 cm £ÅgÂËÃÜáÊÜ A«ÜìWæãàÙÜ¨Ü ±Ý¨Ü¨Ü ÊæáàÇæ 120 cm GñܤÃÜ ÊÜáñÜá¤
60 cm £ÅgÂÊÜ®Üá° Öæãí©ÃÜáÊÜ Jí¨Üá ®æàÃÜ ÊÜêñܤ±Ý¨Ü ÍÜíPÜáÊÜ®Üá° hæãàwÔ¨Ü
Z®ÝPÜꣿá®Üá° ÓÜí±Üä|ìÊÝX ¯àįí¨Ü ñÜáí¹¨Ü ®æàÃÜ ÊÜêñܤ±Ý¨Ü ÔÈívÜÃ…®ÜÈÉ
ñÜÙÜÊÜ®Üá° ÊÜááoárÊÜíñæ ®æàÃÜÊÝX bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ ÊÜááÙÜáXÔ¨æ.
ÔÈívÜÃ…®Ü £ÅgÂÊÜâ 60 cm ÊÜáñÜᤠGñܤÃÜÊÜâ 180 cm B¨ÜÃæ, ÔÈívÜÃ…®ÜÈÉ
EÚ©ÃÜáÊÜ ¯àÄ®Ü ±ÜÅÊÜÞ|ÊÜ®Üá°  ¿áÈÉ ÊÜÂPܤ±ÜwÔ.

A¥ÜÊÝ

13 of 16

CCE RF/PF(A)/101/1810 81-K
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|ìÊÜ®Üá° PÜívÜá×wÀáÄ.

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.

14 of 16

CCE RF/PF(A)/101/1810 81-K
VI. 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. PÜíŸ¨Ü ÊæáàÆᤩÀáí¨Ü Wæãà±ÜâÃܨÜ
ÊæáàÆᤩWæ CÃÜáÊÜ E®Ü°ñÜ Pæãà®ÜÊÜâ bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ 60° BX¨æ.
Wæãà±ÜâÃÜ¨Ü GñܤÃÜÊÜ®Üá° ( CD ) PÜívÜá×wÀáÄ. ÖÝWÜã PÜíŸ¨Ü ÊæáàÆᤩ ÊÜáñÜá¤
Wæãà±ÜâÃÜ¨Ü ÊæáàÆᤩXÃÜáÊÜ ¨ÜãÃÜÊÜ®Üá° ( AC ) PÜívÜá×wÀáÄ.

15 of 16

CCE RF/PF(A)/101/1810 81-K

16 of 16

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