Karnataka SSLC Model Question Paper 2024 Maths

ACADEMIC YEAR

2024

Karnataka
Board
Model Paper

PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ
ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
Malleshwaram, Bengaluru – 560 003

2023-24 ÃÜ GÓ….GÓ….GÇ….Ô. ÊÜÞ¨ÜÄ ±ÜÅÍ氱ܣÅPæ
S.S.L.C. MODEL QUESTION PAPER – 2023-24

…Œ⁄æ⁄fl : V⁄{}⁄
Subject : MATHEMATICS
( AMV⁄« »⁄·¤®⁄¥¿»⁄fl / English Medium )

ÓÜÊÜá¿á : 3 WÜípæ 15 ¯ËáÐÜWÜÙÜá ] ËÐÜ¿á ÓÜíPæàñÜ : 81-E
WÜÄÐÜu AíPÜWÜÙÜá : 80 ] Subject Code : 81-E

CCE-RF : ÍÝÇÝ Ë¨Ý¦ìWÜÙÜá / Regular Fresh

General Instructions to the Candidate :
1. This question paper consists of 38 questions.
2. Question paper has been sealed by reverse jacket. You have to cut on
the right side to open the paper at the time of commencement of the
examination. Check whether all the pages of the question paper are
intact.
3. Follow the instructions given against the questions.
4. Figures in the right hand margin indicate maximum marks for the
questions.
5. The maximum time to answer the paper is given at the top of the
question paper. It includes 15 minutes for reading the question paper.

[ Turn over

81-E 2 CCE RF

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. Every positive odd integer is of the form ( where q is a positive

integer )

(A) 2q + 1

(B) 2q + 2

(C) 2q + 4

(D) 2q

2. The lines represented by the pair of linear equations x + 2y = 8

and 2x + 4y = 10 are

(A) intersecting each other

(B) perpendicular to each other

(C) coincident

(D) parallel to each other

81-E 3 CCE RF

3. The n th term ( a n ) of the Arithmetic progression whose first

term is ‘a’ and common difference ‘d’ is

(A) an = a + ( n + 1 ) d

n
(B) an = [a + (n −1)d ]
2

(C) an = a + ( n – 1 ) d

(D) an = a ( n – 1 ) d

4. Sum of the zeroes of the polynomial p ( x ) = x 2 – 2x – 8 is

(A) –8 (B) 2

(C) –2 (D) 8

5. If tan θ = 1, then the value of sec θ is

1
(A) (B) 3
3

1
(C) 2 (D)
2

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81-E 4 CCE RF

6. The correct relation related to the ∆ PQR given in the figure is

(A) PQ 2 = PR 2 + QR 2 (B) PR 2 = PQ 2 + QR 2

(C) QR 2 = PR 2 + PQ 2 (D) PQ 2 = PR 2 − QR 2

7. The volume of a cone having radius ‘r’ and height ‘h’ is

(A) πr 2 h (B) 2 πrh

2 1
(C) πr 2 h (D) πr 2 h
3 3

8. In ∆ ABC, if AB = 3 units, BC = 1 unit, AC = 2 units and

ACB = θ , then the value of ‘θ’ is

(A) 0° (B) 60°

(C) 45° (D) 90°

81-E 5 CCE RF

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

9. The HCF and LCM of two numbers are 4 and 60 respectively. If

one of the numbers is 12, then find the other number.

10. Write the degree of the polynomial

g ( p ) = 7 p 4 − 2p 3 + 3p 2 + p − 3

11. Find the 5 th term of the Arithmetic progression 3, 1, – 1, .... .

12. Express the quadratic equation 2x = 3 x 2 – 5 in the standard

form.

1 3
13. If sin A = , cos A = , then find the value of tan A.
2 2

14. A fair coin is tossed once. Find the probability of getting Head.

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81-E 6 CCE RF

15. In the given figure, if AOB = 2 APB , then find the value of

APB .

16. Write the formula to find the curved surface area of the frustum

of a cone given in the figure.

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

17. Prove that 2 + 3 is irrational.

OR

Find the HCF of 64 and 332 by using Euclid’s division algorithm.

81-E 7 CCE RF

18. Solve by elimination method :

2x + 3y = 14

2x + y = 10

19. Find the sum of first 30 terms of the Arithmetic progression

3, 7, 11, .... using formula.

20. Find the roots of the equation x 2 – 7x + 12 = 0 using quadratic

formula.

21. Prove that sin 30° + cos 60° + tan 45° = sec 60°.

OR

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

22. Find the coordinates of the point which divides the line segment

joining the points ( 2, 1 ) and ( 7, 6 ) in the ratio 3 : 2.

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81-E 8 CCE RF

23. A box contains tokens which are numbered from 1 to 15. A token

is drawn at random from the box. Find the probability that the

token does not bear a prime number.

24. Draw a pair of tangents to the circle of radius 4 cm which are

inclined to each other at an angle of 60°.

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

25. Divide p ( x ) = x 4 – 3 x 2 + 4x + 5 by g ( x ) = x 2 – 1 and find the

quotient [ q ( x ) ] and remainder [ r ( x ) ].

OR

On dividing x 3 − 3x 2 + x + 2 by a polynomial g ( x ) the quotient

and remainder are ( x – 2 ) and ( – 2x + 4 ) respectively. Find

g ( x ).

26. The diagonal of a rectangular field is 20 m more than the shorter

side of it. If the shorter side is 10 m less than the longer side,

then find the sides of the rectangular field.

81-E 9 CCE RF

27. Find the area of a triangle PQR whose vertices are P ( 1, 6 ),

Q ( 3, 2 ) and R ( 10, 8 ).

OR

ABC is a triangle whose vertices are A ( 1, 4 ), B ( – 2, – 2 ),

C ( 4, – 2 ). If AD is median to BC, then find the length of AD.

28. Find the mean for the distribution given below :

Class-interval Frequency

0 – 10 4

10 – 20 6

20 – 30 17

30 – 40 13

40 – 50 7

50 – 60 3

OR

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81-E 10 CCE RF

Find the mode for the following data :

Class-interval Frequency

1–5 1

5 – 10 2

10 – 15 13

15 – 20 15

20 – 25 7

25 – 30 2

29. The following table gives production yield per hectare of paddy of

100 farms of a village. Draw a ‘more than type ogive’ for the given

data :

Production yield Number of farms

( In kg/hectare ) ( cumulative frequency )

50 or more than 50 100

55 or more than 55 98

60 or more than 60 90

65 or more than 65 77

70 or more than 70 49

75 or more than 75 15

81-E 11 CCE RF

30. In the figure BAC = ADB , BC = 8 cm and AB = 6 cm.

Area of ∆ ABC 16
Prove that = .
Area of ∆ ABD 9

31. Prove that “The lengths of tangents drawn from an external point

to a circle are equal”.

32. Construct a triangle with sides 5 cm, 6 cm and 9 cm and then

2
construct another triangle whose sides are of the
3

corresponding sides of the first triangle.

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81-E 12 CCE RF

33. In the figure, ‘O’ is the centre of the circle of radius 5 cm and

APB is an equilateral triangle of side 8 cm. AP and BP are

tangents. Find the area of the shaded region.

OR

ABCD is a rectangle and APB is a semicircle as shown in the

figure. The length ( BC ) of the rectangle is 3 times the radius of

the semicircle and the total area of the figure APBCDA is

371 cm 2 , then find the length of the semicircular arc.

81-E 13 CCE RF

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

34. Find the solutions of the given pair of linear equations by

graphical method :

x+y=4

2x + y = 7

35. There are 20 terms in an Arithmetic progression. The sum of the

first term and 6th term of the progression is zero. The 4 th and

5 th terms of the progression are 2 and 6 respectively. Find the

Arithmetic progression and also find which term of the

progression is 62.

36. A man standing at the point ‘A’ on the building (AD) observes a

car at point ‘C’ on a straight road from the foot of the building.

The car moves 500 m towards the building and reaches the point

‘B’, now he observes the car from point ‘A’. The angles of

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81-E 14 CCE RF

depression in these cases are complementary to each other. If the

car takes 9 minutes to reach from point ‘C’ to point ‘D’ at the

speed of 100 m/minute, then find the height of the building.

OR

In ∆ ABC, AD ⊥ BC. If ABC = 60°, ACB = 30° and

BC = 36 cm, then find measures of AB, AC and AD.

37. Prove “Basic proportionality theorem.” ( Thale's theorem ).

81-E 15 CCE RF

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

38. A test tube is made up of a cylinder and a hemisphere as shown

in the figure. If the diameter of the hemisphere is 3·5 cm and the

total height of the test tube is 17·5 cm, then find the curved

surface area of the test tube and the quantity of the solution that

could be completely filled in the hemispherical part.

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81-E 16 CCE RF

PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ
ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
Malleshwaram, Bengaluru – 560 003

2023-24 ÃÜ GÓ….GÓ….GÇ….Ô. ÊÜÞ¨ÜÄ ±ÜÅÍ氱ܣÅPæ
S.S.L.C. MODEL QUESTION PAPER – 2023-24
ËÐÜ¿á : WÜ~ñÜ
Subject : MATHEMATICS
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium

ÓÜÊÜá¿á 3 WÜípæ 15 ¯ËáÐÜWÜÙÜá ] ËÐÜ¿á ÓÜíPæàñÜ : 81-K
WÜÄÐÜu AíPÜWÜÙÜá 80 ] Subject Code : 81-K
CCE-RF : ÍÝÇÝ Ë¨Ý¦ìWÜÙÜá / Regular Fresh

±ÜÄàûݦìWÝX ÓÝÊÜޮܠÓÜãaÜ®æWÜÙÜá
1. D ±ÜÅÍ氱ܣÅPæ¿áá Joár 38 ±ÜÅÍæ°WÜÙÜ®Üá° Öæãí©¨æ.
2. ±ÜÅÍ氱ܣÅPæ¿á®Üá° ×ÊÜáá¾S hÝPæp… ÊÜáãÆPÜ ÊæãÖÜÃÜá ÔàÇ… ÊÜÞvÜÇÝX¨æ. ±ÜÄàûæ
±ÝÅÃÜí»ÜÊÝWÜáÊÜ ÓÜÊÜá¿áPæR ¯ÊÜá¾ ±ÜÅÍ氱ܣÅPæ¿á ŸÆŸ© ±ÝÍÜÌìÊÜ®Üá° PÜñܤÄÔ,
±ÜÅÍ氱ܣÅPæ¿áÈÉ GÇÝÉ ±ÜâoWÜÙÜá CÊæÁáà Gí¨Üá ±ÜÄàüÔPæãÚÛ.
3. ±ÜÅÍæ°WÜÚWæ PæãqrÃÜáÊÜ ÓÜãaÜ®æWÜÙÜ®Üá° ±ÝÈÔ.
4. ŸÆ »ÝWܨÜÈÉ PæãqrÃÜáÊÜ AíQWÜÙÜá ±ÜÅÍæ°WÜÚXÃÜáÊÜ ±Üä|ì AíPÜWÜÙÜ®Üá° ñæãàÄÓÜáñܤÊæ.
5. ±ÜÅÍ氱ܣÅPæ¿á®Üá° K©PæãÙÜÛÆá 15 ¯ËáÐÜWÜÙÜ PÝÇÝÊÜPÝÍÜÊÜâ ÓæàĨÜíñæ, EñܤÄÓÜÆá
¯WÜ©±ÜwÓÜÇÝ¨Ü ÓÜÊÜá¿áÊÜ®Üá° ±ÜÅÍ氱ܣÅPæ¿á ÊæáàÇݽWܨÜÈÉ ¯àvÜÇÝX¨æ.

[ Turn over

81-K 2 CCE RF
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. ‘q’ Jí¨Üá «Ü®Ü ±ÜäOÝìíPÜÊݨÝWÜ ¿ÞÊÜâ¨æà «Ü®Ü ¸æÓÜ ±ÜäOÝìíPÜ¨Ü ÓÝÊÜÞ®ÜÂ

ÃÜã±ÜÊÜâ

(A) 2q + 1

(B) 2q + 2

(C) 2q + 4

(D) 2q

2. x + 2y = 8 ÊÜáñÜᤠ2x + 4y = 10 D hæãàw ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ®Üá°

±ÜÅ£¯˜ÓÜáÊÜ ÃæàTæWÜÙÜá

(A) ±ÜÃÜÓܳÃÜ dæà©ÓÜáñܤÊæ

(B) ±ÜÃÜÓܳÃÜ ÆíŸÊÝXÃÜáñܤÊæ

(C) IPÜÂWæãÙÜáÛñܤÊæ

(D) ±ÜÃÜÓܳÃÜ ÓÜÊÜÞíñÜÃÜÊÝXÃÜáñܤÊæ

81-K 3 CCE RF
3. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ ±Ü¨Ü ‘a’ ÊÜáñÜᤠÓÝÊÜޮܠÊÜÂñÝÂÓÜ ‘d’

B¨ÝWÜ, A¨ÜÃÜ ‘n’ ®æà ±Ü¨ÜÊÜâ ( a n )

(A) an = a + ( n + 1 ) d

(B) an  n [ a  ( n  1 ) d ]
2

(C) an = a + ( n – 1 ) d

(D) an = a ( n – 1 ) d

4. p ( x ) = x 2 – 2x – 8 D ŸÖÜá±Ü¨æãàQ¤¿á ÍÜã®ÜÂñæWÜÙÜ ÊæãñܤÊÜâ

(A) –8 (B) 2

(C) –2 (D) 8

5. tan  = 1 B¨ÝWÜ, sec  ¨Ü ¸æÇæ¿áá

(A) 1 (B) 3
3

(C) 2 (D) 1
2

[ Turn over

81-K 4 CCE RF
6. bñÜŨÜÈÉ PæãqrÃÜáÊÜ £Å»Üág PQR Wæ ÓÜÄÖæãí¨ÜáÊÜ ÓÜíŸí«ÜÊÜâ

(A) PQ2  PR2  QR 2 (B) PR2  PQ2  QR 2

(C) QR 2  PR2  PQ2 (D) PQ2  PR2  QR 2

7. £Åg ‘r’ ÊÜáñÜᤠGñܤÃÜ ‘h’ BXÃÜáÊÜ ÍÜíPÜá訆 Z®Ü¶ÜÆÊÜâ

(A) r 2 h (B) 2 rh

(C) 2 r 2 h (D) 1 r 2 h
3 3

8.  ABC ¿áÈÉ AB = 3 ÊÜÞ®ÜWÜÙÜá, BC = 1 ÊÜÞ®Ü, AC = 2 ÊÜÞ®ÜWÜÙÜá ÊÜáñÜá¤

ACB   B¨ÝWÜ, ‘’ ¨Ü ¸æÇæ¿áá

(A) 0° (B) 60°

(C) 45° (D) 90°

81-K 5 CCE RF
II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8×1=8

9. GÃÜvÜá ÓÜíTæÂWÜÙÜ ÊÜá.ÓÝ.A. ÊÜáñÜᤠÆ.ÓÝ.A.WÜÙÜá PÜÅÊÜáÊÝX 4 ÊÜáñÜᤠ60 BXÊæ.

Jí¨Üá ÓÜíTæ 12 B¨ÜÃæ C®æã°í¨Üá ÓÜíTæ¿á®Üá° PÜívÜá×wÀáÄ.

10. g ( p ) = 7 p 4  2 p 3  3 p 2  p  3 ŸÖÜá±Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá [ÝñÜÊÜ®Üá°

wXÅ ŸÃæÀáÄ.

11. 3, 1, – 1, .... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á 5 ®æà ±Ü¨ÜÊÜ®Üá° PÜívÜá×wÀáÄ.

12. 2x = 3 x 2 – 5 D ÊÜWÜìÓÜËáàPÜÃÜ|ÊÜ®Üá° B¨ÜÍÜìÃÜã±Ü¨ÜÈÉ ÊÜÂPܤ±ÜwÔ.

13. sin A = 1 , cos A = 3 B¨ÝWÜ, tan A ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.
2 2

14. Jí¨Üá PÜáí©ÆÉ¨Ü ®Ý|ÂÊÜ®Üá° Jí¨Üá ¸ÝÄ bËá¾Ô¨ÝWÜ ÎÃÜ ÊÜ®Üá° ±Üvæ¿ááÊÜ

ÓÜí»ÜÊܯà¿áñæ¿á®Üá° PÜívÜá×wÀáÄ.

[ Turn over

81-K 6 CCE RF
15. PæãqrÃÜáÊÜ bñÜŨÜÈÉ AOB  2 APB B¨ÜÃæ, APB ¿á ¸æÇæ¿á®Üá°
PÜívÜá×wÀáÄ.

16. bñÜŨÜÈÉ PæãqrÃÜáÊÜ ÍÜíPÜá訆 ¼®Ü°PÜ¨Ü ±ÝÍÜÌì ÊæáàÇ澄 ËÔ¤à|ìÊÜ®Üá°
PÜívÜá×w¿ááÊÜ ÓÜãñÜÅÊÜ®Üá° ŸÃæÀáÄ.

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

17. 2 + 3 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ÓݘÔ.

A¥ÜÊÝ

64 ÊÜáñÜᤠ332 ÃÜ ÊÜá.ÓÝ.A.ÊÜ®Üá° ¿ááQÉv…®Ü »ÝWÝPÝÃÜ PÜÅÊÜá˘¿á®Üá°
E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

81-K 7 CCE RF
18. ÊÜiìÓÜáÊÜ Ë«Ý®Ü©í¨Ü ¹wÔ

2x + 3y = 14

2x + y = 10

19. 3, 7, 11, .... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ 30 ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá° ÓÜãñÜÅ

E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

20. x 2 – 7x + 12 = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá° ÊÜWÜìÓÜËáàPÜÃÜ| ÓÜãñÜÅ

E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

21. sin 30° + cos 60° + tan 45° = sec 60° Gí¨Üá ÓݘÔ.

A¥ÜÊÝ

cos A 1  sin A
  2 sec A Gí¨Üá ÓݘÔ.
1  sin A cos A

22. ( 2, 1 ) ÊÜáñÜᤠ( 7, 6 ) ¹í¨ÜáWÜÙÜ®Üá° ÓæàÄÓÜáÊÜ ÃæàTÝSívÜÊÜ®Üá° 3 : 2 ÃÜ

A®Üá±ÝñܨÜÈÉ Ë»ÝXÓÜáÊÜ ¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

[ Turn over

81-K 8 CCE RF
23. Jí¨Üá ±æqrWæ¿áÈÉ 1 Äí¨Ü 15 ÃÜÊÜÃæWæ ®ÜÊÜáã¨ÝXÃÜáÊÜ ¹ÇæÉWÜÚÊæ. ±æqrWæÀáí¨Ü

Jí¨Üá ¹ÇæÉ¿á®Üá° ¿Þ¨ÜêbfPÜÊÝX ÖæãÃÜñæWæ¨ÝWÜ A¨Üá AË»Ýg ÓÜíTæ¿á®Üá°

Öæãí©ÆÉ¨Ü ÓÜí»ÜÊܯà¿áñæ¿á®Üá° PÜívÜá×wÀáÄ.

24. 4 cm £ÅgÂÊÜâÙÜÛ ÊÜêñܤÊÜ®Üá° ÃÜbÔ, D ÊÜêñܤPæR ÓܳÍÜìPÜWÜÙÜ ®ÜvÜá訆 Pæãà®Ü 60°

CÃÜáÊÜíñæ Jí¨Üá hæãñæ ÓܳÍÜìPÜWÜÙÜ®Üá° GÙæÀáÄ.

IV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 9 × 3 = 27

25. p ( x ) = x 4 – 3 x 2 + 4x + 5 ®Üá° g ( x ) = x 2 – 1 Äí¨Ü »ÝXÔ »ÝWÜÆŸœ

[ q ( x ) ] ÊÜáñÜᤠÍæàÐÜ [ r ( x ) ] PÜívÜá×wÀáÄ.

A¥ÜÊÝ

x 3  3x 2  x  2 ®Üá° g ( x ) GíŸ ŸÖÜá±Ü¨æãàQ¤Àáí¨Ü »ÝXÔ¨ÝWÜ ÔWÜáÊÜ

»ÝWÜÆŸœ ÊÜáñÜᤠÍæàÐÜWÜÙÜá PÜÅÊÜáÊÝX ( x – 2 ) ÊÜáñÜᤠ( – 2x + 4 ) B¨ÜÃæ, g ( x )

®Üá° PÜívÜá×wÀáÄ.

26. Jí¨Üá B¿áñÝPÝÃÜ¨Ü gËáà¯®Ü PÜ|ìÊÜâ A¨ÜÃÜ bPÜR ¸ÝÖÜáËXíñÜ 20 m

ÖæaÝcX¨æ. bPÜR ¸ÝÖÜáÊÜâ ¨æãvÜx ¸ÝÖÜáËXíñÜ 10 m PÜwÊæá BX¨ÜªÃæ, B

B¿áñÝPÝÃÜ¨Ü gËáà¯®Ü ¸ÝÖÜáWÜÙÜ AÙÜñæ PÜívÜá×wÀáÄ.

81-K 9 CCE RF
27. P ( 1, 6 ), Q ( 3, 2 ) ÊÜáñÜᤠR ( 10, 8 ) ÍÜêíWܹí¨ÜáWÜÙÜ®Üá° Öæãí©ÃÜáÊÜ

PQR £Å»Üág¨Ü ËÔ¤à|ìÊÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

ABC £Å»Üág¨Ü ÍÜêíWܹí¨ÜáWÜÙÜá A ( 1, 4 ), B ( – 2, – 2 ), C ( 4, – 2 )

BXÊæ, ÊÜáñÜᤠAD ¿áá BC ¿á ÊÜá«ÜÂÃæàTæ¿ÞX¨æ. AD ¿á E¨ÜªÊÜ®Üá°

PÜívÜá×wÀáÄ.

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

ÊÜWÝìíñÜÃÜ BÊÜ꣤

0 — 10 4

10 — 20 6

20 — 30 17

30 — 40 13

40 — 50 7

50 — 60 3

A¥ÜÊÝ

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81-K 10 CCE RF
D PæÙÜX®Ü ¨ÜñݤíÍÜWÜÚWæ ŸÖÜáÆPÜÊÜ®Üá° PÜívÜá×wÀáÄ

ÊÜWÝìíñÜÃÜ BÊÜ꣤

1—5 1

5 — 10 2

10 — 15 13

15 — 20 15

20 — 25 7

25 — 30 2

29. Jí¨Üá WÝÅÊÜá¨Ü 100 ÖæãÆWÜÙÜÈÉ ±ÜÅ£ ÖæPæràÃ…Wæ Eñݳ©ÓÜáÊÜ »Üñܤ¨Ü
CÙÜáÊÜÄ¿á®Üá° PæÙÜX®Ü PæãàÐÜrPÜÊÜâ ¯àvÜᣤ¨æ. D ¨ÜñݤíÍÜWÜÚWæ A˜PÜ Ë«Ý®Ü¨Ü
KiàÊ… ÃÜbÔ
Eñݳ¨Ü®Ý CÙÜáÊÜÄ ÖæãÆWÜÙÜ ÓÜíTæÂ
Pæi / ÖæPæràÃ…WÜÙÜÈÉ ÓÜíbñÜ BÊÜ꣤
50 A¥ÜÊÝ 50 QRíñÜ A˜PÜ 100

55 A¥ÜÊÝ 55 QRíñÜ A˜PÜ 98

60 A¥ÜÊÝ 60 QRíñÜ A˜PÜ 90

65 A¥ÜÊÝ 65 QRíñÜ A˜PÜ 77

70 A¥ÜÊÝ 70 QRíñÜ A˜PÜ 49

75 A¥ÜÊÝ 75 QRíñÜ A˜PÜ 15

81-K 11 CCE RF
30. bñÜŨÜÈÉ BAC  ADB , BC = 8 cm ÊÜáñÜᤠAB = 6 cm BX¨æ,

16
= Gí¨Üá ÓݘÔ.
9

31. ¸ÝÖÜ ¹í¨Üá˯í¨Ü ÊÜêñܤPæR GÙæ¨Ü ÓܳÍÜìPÜWÜÙÜ E¨ÜªÊÜâ ÓÜÊÜáÊÝXÃÜáñܤÊæ Gí¨Üá

ÓݘÔ.

32. 5 cm, 6 cm ÊÜáñÜᤠ9 cm ¸ÝÖÜáWÜÚÃÜáÊÜ Jí¨Üá £Å»ÜágÊÜ®Üá° ÃÜbÔ, ®Üí ñÜÃÜ

ÊÜáñæã¤í¨Üá £Å»ÜágÊÜ®Üá°, A¨ÜÃÜ ±ÜÅ£Áãí¨Üá ¸ÝÖÜáÊÜâ Êæã¨ÜÆá ÃÜbÔ¨Ü

£Å»Üág¨Ü A®ÜáÃÜã±Ü ¸ÝÖÜáWÜÙÜ 32 ÃÜÑrÃÜáÊÜíñæ ÃÜbÔ.

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81-K 12 CCE RF
33. bñÜŨÜÈÉ ‘O’ Pæàí¨ÜÅÊÝXÃÜáÊÜ ÊÜêñܤ¨Ü £Åg 5 cm ÊÜáñÜᤠAPB ¿áá 8 cm

¸ÝÖÜáÊÜâÙÜÛ ÓÜÊÜá¸ÝÖÜá £Å»ÜágÊÝX¨æ. AP ÊÜáñÜᤠBP WÜÙÜá ÓܳÍÜìPÜWÜÙÜá. ÖÝWݨÜÃæ

dÝÀáàPÜêñÜ »ÝWÜ¨Ü ËÔ¤à|ìÊÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ ABCD Jí¨Üá B¿áñÜ ÊÜáñÜᤠAPB ¿áá A«Üì

ÊÜêñܤÊÝX¨æ. B¿áñÜ¨Ü E¨ÜªÊÜâ ( BC ) A«ÜìÊÜêñܤ £Åg嬆 3 ÃÜÑr¨æ ÊÜáñÜá¤

APBCDA ¿á ±Üä|ì ËÔ¤à|ì 371 cm 2 BX¨ÜªÃæ, A«ÜìÊÜêñݤPÝÃÜ¨Ü PÜíÓܨÜ

E¨ÜªÊÜ®Üá° PÜívÜá×wÀáÄ.

81-K 13 CCE RF
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+y=4

2x + y = 7

35. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿áÈÉ 20 ±Ü¨ÜWÜÚÊæ. ÍæÅà{¿á Êæã¨ÜÆ ±Ü¨Ü ÊÜáñÜᤠ6 ®æà

±Ü¨ÜWÜÙÜ ÊæãñܤÊÜâ Óæã®æ°¿ÞX¨æ. ÍæÅà{¿á 4 ®æà ÊÜáñÜᤠ5 ®æà ±Ü¨ÜWÜÙÜá PÜÅÊÜáÊÝX

2 ÊÜáñÜᤠ6 BX¨æ, ÖÝWݨÜÃæ ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á®Üá° PÜívÜá×wÀáÄ ÖÝWÜã

A¨ÜÃÜ GÐÜr®æà ±Ü¨ÜÊÜâ 62 BX¨æ

36. JŸº ÊÜÂQ¤¿áá AD PÜorvÜ¨Ü ÊæáàÇæ ‘A’ ¹í¨ÜáË®ÜÈÉ ¯í£¨Ýª®æ. PÜorvܨÜ

±Ý¨Ü©í¨Ü ÖæãÃÜvÜáÊÜ ®æàÃÜÊÝ¨Ü ÃÜÓ椿á ÊæáàÇæ ‘C’ ¹í¨ÜáË®ÜÈÉÃÜáÊÜ PÝÃÜ®Üá°

ÊÜÂQ¤¿áá ËàüÓÜáñݤ®æ. PÝÃÜá PÜorvÜ¨Ü PÜvæWæ 500 m ¨ÜãÃÜÊÜ®Üá° aÜÈÔ

‘B’ ¹í¨ÜáÊÜ®Üá° ñÜÆᲨÝWÜ ÊÜÂQ¤¿áá ‘A’ ¹í¨Üá˯í¨Ü PÝÃÜ®Üá° ËàüÓÜáñݤ®æ. D

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81-K 14 CCE RF
GÃÜvÜã ÓÜí¨Ü»ÜìWÜÙÜÈÉ EípÝ¨Ü AÊÜ®ÜñÜ Pæãà®ÜWÜÙÜá ±ÜÃÜÓܳÃÜ ±ÜäÃÜPÜ

Pæãà®ÜWÜÙÝXÊæ. PÝÃÜá ‘C’ ¹í¨Üá˯í¨Ü ‘D’ ¹í¨ÜáËWæ 100 m/min gÊܨÜÈÉ

aÜÈÔ 9 ¯ËáÐÜWÜÙÜÈÉ ‘D’ ¹í¨ÜáÊÜ®Üá° ñÜÆᲨÜÃæ PÜorvÜ¨Ü GñܤÃÜÊÜ®Üá°

PÜívÜá×wÀáÄ.

A¥ÜÊÝ

 ABC ¿áÈÉ AD  BC BX¨æ, A B C = 60°, ACB = 30° ÊÜáñÜá¤

BC = 36 cm B¨ÜÃæ, AB, AC ÊÜáñÜᤠAD WÜÙÜ AÙÜñæWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

37. ÊÜáãÆ ÓÜÊÜÞ®Üá±ÝñÜñæ¿á ±ÜÅÊæáà¿á ¥æàÇ…Õ ±ÜÅÊæáà¿á ÊÜ®Üá° ÓݘÔ.

81-K 15 CCE RF
VI. PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×5=5

38. Jí¨Üá ±ÜÅ®ÝÙÜÊÜ®Üá° bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ Jí¨Üá ÔÈívÜÃ… ÊÜáñÜá¤

A«ÜìWæãàÙÜ©í¨Ü ÊÜÞvÜÇÝX¨æ. A«ÜìWæãàÙÜ¨Ü ÊÝÂÓÜ 3·5 cm ÊÜáñÜᤠ±ÜÅ®ÝÙܨÜ

Joár GñܤÃÜ 17·5 cm B¨ÜÃæ, ±ÜÅ®ÝÙÜ¨Ü ÊÜPÜÅ ÊæáàÇ澄 ËÔ¤à|ì ÊÜáñÜá¤

A«ÜìWæãàÙÝPÝÃÜ¨Ü »ÝWܨÜÈÉ ÓÜí±Üä|ìÊÝX ñÜáퟟÖÜá¨Ý¨Ü ¨ÝÅÊÜ|¨Ü

±ÜÅÊÜÞ|ÊÜ®Üá° PÜívÜá×wÀáÄ.

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81-K 16 CCE RF