Karnataka SSLC Question Paper 2024 Maths

Karnataka School Examination and Assessment Board

QUESTION
PAPERS
2024

B∆«M•⁄ O⁄}⁄¬° “
CCE RF/PF/RR/PR/NSR/NSPR(A)/666/032

A

Question Paper Serial No.
Jlflo »⁄flfl¶√}⁄ Æ⁄‚¥lV⁄◊⁄ —⁄MSÊ¿ : 16 ]
Total No. of Printed Pages : 16 ]

Jlflo Æ⁄√ÀÊ-V⁄◊⁄ —⁄MSÊ¿ : 38 ] CCE RF/PF/RR/
PR/NSR/NSPR
Total No. of Questions : 38 ]
FULL SYLLABUS
—⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E
Code No. : 81-E
…Œ⁄æ⁄fl : V⁄{}⁄

TEAR HERE TO OPEN THE QUESTION PAPER
Subject : MATHEMATICS

Æ⁄√ÀÊ-Æ⁄~√OÊæ⁄fl´⁄fl-}Ê¡Êæ⁄flƒfl B∆« O⁄}⁄°¬“
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A»Ü¦ì / G®….GÓ….BÃ…. / G®….GÓ….².BÃ….)
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02. 04. 2024
¶´¤MO⁄ : 02. 04. 2024 ] [ Date : 02. 04. 2024
ÓÜÊÜá¿á : ¸æÙÜWæY 10-15 Äí¨Ü ÊÜá«ÝÂÖÜ° 1-30 ÃÜÊÜÃæWæ ] [ Time : 10-15 A.M. to 1-30 P.M.
V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80 ] [ Max. Marks : 80
Cut here /B∆« O⁄}°⁄¬“
General Instructions to the Candidate :
1. This question paper consists of 38 questions in all.
2. This 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 ( Follow the arrow ). Do not cut the left side to open
the paper. Check whether all the pages of the question paper are intact.
3. Follow the instructions given against the questions.
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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.
Tear here

6. Ensure that the Version of the question paper distributed to you and
the Version printed on your admission ticket is the same.

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CCE RF/PF/RR/PR/NSR/NSPR(A)/666/032 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. The product of HCF and LCM of two numbers 15 and 20 is

(A) 15 (B) 20

(C) 300 (D) 35

2. If α and β are the zeroes of the quadratic polynomial

p ( x ) = ax 2 + bx + c, then αβ is

b −b
(A) (B)
a a

−c c
(C) (D)
a a

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4
3. If sin θ = , then the value of 1 − cos 2 θ is
5

16 4
(A) (B)
25 5

5 9
(C) (D)
4 25

4. The probability of a sure event is

(A) 1 (B) 0

(C) –1 (D) 1·5

5. The secant of the circle in the figure, is

(A) MN (B) OE

(C) CD (D) AB

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6. The volume of the frustum of a cone whose base radii are r 1 and

r 2 and height ‘h’ , is

1
(A) π ( r1 + r 2 + r1 .r 2 )h
3

1
(B) π ( r 12 + r 2 2 − r 1 . r 2 ) h
3

1
(C) π ( r 12 + r 2 2 + r 1 . r 2 ) h
3

1
(D) π ( r 12 − r 2 2 − r 1 . r 2 ) h
3

7. If 2, x, 26 are in Arithmetic progression, then the value of x is

(A) 12 (B) 14

(C) 28 (D) 24

8. If tan ( 90° – θ ) = 3 , then the value of cot θ is

1
(A) (B) 1
3

(C) 0 (D) 3

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

9. In the figure, ∆ ADE ~ ∆ ABC and DE : BC = 2 : 3. Find

Area of ∆ ADE
.
Area of ∆ ABC

10. The radii of the base and the height of a cylinder and a cone are

same. If the volume of the cylinder is 27 cubic units, then find

the volume of the cone.

11. If 200 = 2 m × 5 n , then find the values of m and n.

12. Find the number of solutions of the pair of linear equations

2x – 3y + 4 = 0 and 3x + 5y + 8 = 0.

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13. In an Arithmetic progression, sum of the first six terms and sum

of the first five terms are 78 and 55 respectively. Then find the

sixth term of the progression.

14. Write the degree of the polynomial p ( x ) = x ( x 2 + 3 ) + 5 x 2 + 7.

15. If the value of discriminant of a quadratic equation is zero, then

write the nature of roots of the quadratic equation.

16. Find the value of θ in the figure.

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

17. Prove that 3 + 2 is an irrational number.

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18. Solve the given pair of linear equations by Elimination method :

2x + y = 8

3x – y = 7

19. Find the sum of first 20 terms of the Arithmetic progression

1, 5, 9, .... using formula.

20. Find the roots of the quadratic equation 2 x 2 – 3x – 1 = 0 using

quadratic formula.

cos θ − sin θ . cos θ cosec θ − 1
21. Prove that = .
cos θ + sin θ . cos θ cosec θ + 1

OR

sin 30 o + cos 60 o
Prove that = sin 90°.
cosec 30 o − cot 45 o

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22. Find the coordinates of the point P and Q in the given graph and

hence find the length of PQ using distance formula.

OR

Find the coordinates of the point which divides the line segment

joining the points ( 4, – 3 ) and ( 8, 5 ) in the ratio 3 : 1

internally.
1
23. A basket contains 36 mangoes. th of them are rotten and
4

others are good. If one mango is drawn at random from the

basket, then find the probability of getting a good mango.

24. Draw a circle of radius 3·5 cm and construct a pair of tangents to

the circle such that the angle between the tangents is 60°.

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

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

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

OR

When the polynomial p ( x ) = x 3 + 4 x 2 + 5x – 2 is divided by the

polynomial g ( x ), the quotient [ q ( x ) ] and remainder [ r ( x ) ]

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

26. Find the mean for the following data :

Class-interval Frequency

2–6 2

7 – 11 4

12 – 16 5

17 – 21 3

22 – 26 1

OR

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Find the mode for the following data :

Class-interval Frequency

1–5 1

5–9 3

9 – 13 7

13 – 17 10

17 – 21 9

27. ‘D’ is a point on the side BC of a ∆ ABC such that
ADC = BAC . Then prove that AC 2 = BC . CD.

OR

In the figure, ∆ ABC and ∆ AMP are right angled triangles, right
CA BC
angled at B and M respectively. Then prove that = .
PA MP

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

to a circle are equal”.
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29. In the figure area of sector AOBPA of radius ‘r’ is 231 cm 2 and

the length of the arc APB is 22 cm. Find the radius of the sector

and angle θ.

OR

In the figure a rectangle ROQP is inscribed in the quadrant of a

circle. If the length and breadth of the rectangle are 16 cm and

12 cm respectively, find the area of the shaded region.

30. Age of mother is twice the square of age of her son. After 8 years

mother’s age becomes 4 years more than the thrice of age of her

son. Find their present ages.

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31. In the figure, ABC is a triangle whose vertices are A ( x, 10 ),

B ( 2, 2 ) and C ( 12, 2 ). If Q ( 9, 6 ) is the mid-point of AC and

area of ∆ APQ is 12 cm 2 , then find the area of quadrilateral

PBCQ.

32. The ages of 100 patients admitted in a hospital are as follows.

Draw a “less than type ogive” for the given data :

Age ( in years ) Number of patients

( cumulative frequency )

Less than 10 6

Less than 20 15

Less than 30 38

Less than 40 46

Less than 50 65

Less than 60 84

Less than 70 100

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33. Construct a triangle with sides 6 cm, 8 cm and 9 cm and then

2
construct another triangle whose sides are of the
3

corresponding sides of the first triangle.

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

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

graphical method :

2x + y = 8

x+y = 5

35. In an Arithmetic progression the sum of first n terms is 210 and

the sum of first ( n – 1 ) terms is 171. If the first term of the

Arithmetic progression is 3, then find the Arithmetic progression

and find its 20 th term.

OR

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The sum of interior angles of a polygon of ‘n’ sides is

( n – 2 ) 180°. If the interior angles of a pentagon are in Arithmetic

progression and its least angle is 72°, then find all the interior

angles of the pentagon.

36. In the figure the poles AB and CD of different heights are

standing vertically on a level ground. From a point P on the line

joining the foots of the poles on the level ground, the angles of

elevation to the tops of the poles are found to be complementary.

The height of CD and the distance PD are 20 3 m and 20 m

respectively. If BP is 10 m, then find the length of the pole AB and

the distance AC between the tops of the poles.

37. Prove : “Basic proportionality theorem” or “Thales theorem”.

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

38. An insect control device made of a cylinder, a frustum of a cone

and a hemisphere attached to each other is as shown in the

figure. Sticky liquid is completely filled in the hemispherical part.

If the radii of hemisphere and cylinder are 21 cm and 14 cm

respectively and total height of the device is 60 cm and height of

the cylinder is 15 cm, then calculate the curved surface area of

the device and also find the quantity of the sticky liquid in the

hemisphere.

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CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031

Question Paper Serial No.
A

CÈÉí¨Ü PÜñܤÄÔ
Joár ÊÜáá©ÅñÜ ±ÜâoWÜÙÜ ÓÜíTæ : 16 ]
Total No. of Printed Pages : 16 ]
Joár ±ÜÅÍæ°WÜÙÜ ÓÜíTæ : 38 ] CCE RF/PF/RR/
PR/NSR/NSPR
Total No. of Questions : 38 ]
FULL SYLLABUS
ÓÜíPæàñÜ ÓÜíTæ : 81-K
Code No. : 81-K
ËÐÜ¿á : WÜ~ñÜ
Subject : MATHEMATICS

TEAR HERE TO OPEN THE QUESTION PAPER
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium
ÍÝÇÝ A»Ü¦ì / TÝÓÜX A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ

±ÜÅÍ氱ܣÅPæ¿á®Üá° ñæÃæ¿áÆá CÈÉ PÜñܤÄÔ
A»Ü¦ì / G®….GÓ….BÃ…. / G®….GÓ….².BÃ….
Regular Fresh / Private Fresh / Regular Repeater / Private Repeater / NSR / NSPR
©®ÝíPÜ 02. 04. 2024 ] [ Date : 02. 04. 2024

02. 04. 2024
ÓÜÊÜá¿á ¸æÙÜWæY 10-15 Äí¨Ü ÊÜá«ÝÂÖÜ° 1-30 ÃÜÊÜÃæWæ ] [ Time : 10-15 A.M. to 1-30 P.M.
WÜÄÐÜu AíPÜWÜÙÜá 80 ] [ Max. Marks : 80
±ÜÄàûݦìWÝX ÓÝÊÜޮܠÓÜãaÜ®æWÜÙÜá Cut here/CÈÉ PÜñܤÄÔ
1. D ±ÜÅÍ氱ܣÅPæ¿áá Joár 38 ±ÜÅÍæ°WÜÙÜ®Üá° Öæãí©¨æ.
2. D ¯ÜÅÍ殯ܣÅPæ¿á¬Üá® ×ÊÜáá¾S hÝPæp… ÊÜáãÆPÜ ÊæãÖÜÃÜá ÔàÇ… ÊÜÞvÜÇÝX¨æ. ±ÜÄàûæ
±ÝÅÃÜí»ÜÊÝWÜáÊÜ ÓÜÊÜá¿áPæR ¯ÊÜá¾ ±ÜÅÍ氱ܣÅPæ¿á ŸÆŸ© ±ÝÍÜÌìÊÜ®Üá° PÜñܤÄÔ ¸Ý|ÊÜ®Üá°
A®ÜáÓÜÄÔ , GvÜŸ© ±ÝÍÜÌìÊÜ®Üá° PÜñܤÄÓܸæàw. ±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 AíPÜWÜÙܬÜá® 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.
Tear here

6. ¯ÊÜáWæ ËñÜÄÓÜÇÝXÃÜáÊÜ ±ÜÅÍ氱ܣÅPæ¿á BÊÜ꣤ ( Version ) ÊÜáñÜᤠ¯ÊÜá¾ ±ÜÅÊæàÍÜ ±ÜñÜŨÜÈÉ
ÊÜáá©ÅñÜÊÝXÃÜáÊÜ ±ÜÅÍ氱ܣÅPæ¿á BÊÜ꣤ Jí¨æà BXÃÜáÊÜâ¨Ü®Üá° TÝñÜıÜwÔPæãÚÛ.

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CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 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. 15 ÊÜáñÜᤠ20 D ÓÜíTæÂWÜÙÜ ÊÜá.ÓÝ.A. ÊÜáñÜᤠÆ.ÓÝ.A.WÜÙÜ WÜá|ÆŸœÊÜâ

(A) 15 (B) 20

(C) 300 (D) 35

2. ÊÜWÜìŸÖÜá¯Ü¨æãàQ¤ p ( x ) = ax 2 + bx + c ®Ü ÍÜã®ÜÂñæWÜÙÜá  ÊÜáñÜᤠ BX¨ÝªWÜ

 ÊÜâ

(A) b (B) b
a a

(C) c (D) c
a a

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3. sin  = 4 B¨ÜÃæ, 1  cos2  ¨Ü ¸æÇæ¿áá
5

(A) 16 (B) 4
25 5

(C) 5 (D) 9
4 25

4. Jí¨Üá SbñÜ Zo®æ¿á ÓÜí»ÜÊܯà¿áñæ¿áá

(A) 1 (B) 0

(C) –1 (D) 1·5

5. bñÜŨÜÈÉ ÊÜêñܤ¨Ü dæà¨ÜPÜÊÜâ

(A) MN (B) OE

(C) CD (D) AB

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6. ±Ý¨Ü¨Ü £ÅgÂWÜÙÜá r 1 ÊÜáñÜᤠr 2 ÖÝWÜã GñܤÃÜ ‘h’ BXÃÜáÊÜ ÍÜíPÜá訆 ¼®Ü°PܨÜ
Z®Ü¶ÜÆÊÜâ

(A) 1
3
 r  r  r .r  h
1 2 1 2

(B) 1
3
r r
1
2 2
2  r1 . r 2 h

(C)
3 1 
1  r 2  r 2  r .r
2 1 2 h 

(D)
3 1 
1  r 2  r 2  r .r
2 1 2 h 

7. 2, x, 26 CÊÜâ ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿áÈɨܪÃæ ‘x’ ®Ü ¸æÇæ¿áá

(A) 12 (B) 14

(C) 28 (D) 24

8. tan ( 90° –  ) = 3 BX¨ÝªWÜ cot  ¨Ü ¸æÇæ

1
(A) (B) 1
3

(C) 0 (D) 3

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II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8×1=8

9. bñÜŨÜÈÉ  ADE ~  ABC ÊÜáñÜᤠDE : BC = 2 : 3 B¨ÜÃæ,

ÊÜ®Üá° PÜívÜá×wÀáÄ.

10. ±Ý¨Ü¨Ü £Åg ÖÝWÜã GñܤÃÜWÜÙÜá ÓÜÊÜáÊÝXÃÜáÊÜ ÔÈívÜÃ… ÊÜáñÜᤠÍÜíPÜáWÜÙÜÈÉ

ÔÈívÜÄ®Ü Z®Ü¶ÜÆÊÜâ 27 Z®ÜÊÜÞ®ÜWÜÙݨÜÃæ, ÍÜíPÜá訆 Z®Ü¶ÜÆÊÜ®Üá°

PÜívÜá×wÀáÄ.

11. 200 = 2 m  5 n B¨ÜÃæ, m ÊÜáñÜᤠn ¸æÇæWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

12. 2x – 3y + 4 = 0 ÊÜáñÜᤠ3x + 5y + 8 = 0 D ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜá

Öæãí©ÃÜáÊÜ ±ÜÄÖÝÃÜWÜÙÜ ÓÜíTæ¿á®Üá° PÜívÜá×wÀáÄ.

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13. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ BÃÜá ±Ü¨ÜWÜÙÜ Êæãñܤ ÊÜáñÜᤠÊæã¨ÜÆ I¨Üá
±Ü¨ÜWÜÙÜ ÊæãñܤÊÜâ PÜÅÊÜáÊÝX 78 ÊÜáñÜᤠ55 BXÊæ. ÖÝWݨÜÃæ B ÍæÅà{¿á BÃÜ®æà
±Ü¨ÜÊÜ®Üá° PÜívÜá×wÀáÄ.

14. p ( x ) = x ( x 2 + 3 ) + 5 x 2 + 7 D ŸÖÜá¯Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá [ÝñÜÊܬÜá®

wXÅ ŸÃæÀáÄ.

15. Jí¨Üá ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü Íæãà«ÜPÜ¨Ü ¸æÇæ¿áá ÍÜã®ÜÂÊÝX¨ÝªWÜ A¨ÜÃÜ ÊÜáãÆWÜÙÜ
ÓÜÌ»ÝÊÜÊܬÜá® ŸÃæÀáÄ.

16. bñÜŨÜÈÉ  ¨Ü ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

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

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

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18. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á®Üá° ÊÜiìÓÜáÊÜ Ë«Ý®Ü©í¨Ü
¹wÔ

2x + y = 8

3x – y = 7

19. 1, 5, 9, .... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ 20 ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá° ÓÜãñÜÅ

E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

20. 2 x 2 – 3x – 1 = 0 D ÊÜWÜì ÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá° ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü

ÓÜãñÜÅ ÊÜ®Üá° E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

cos   sin  . cos  cosec   1
21.
cos   sin  . cos 

cosec   1
Gí¨Üá ÓݘÔ.

A¥ÜÊÝ

sin 30  cos 60
= sin 90° Gí¨Üá ÓݘÔ.
cosec 30  cot 45

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22. bñÜŨÜÈÉ PæãqrÃÜáÊÜ ®Üûæ¿á ÓÜÖÝ¿á©í¨Ü P ÊÜáñÜᤠQ ¹í¨ÜáWÜÙÜ
¯¨æàìÍÝíPÜWÜÙÜ®Üá° PÜívÜá×w¨Üá, PQ ®Ü E¨ÜªÊÜ®Üá° ¨ÜãÃÜ ÓÜãñÜÅ E±ÜÁãàXÔ
PÜívÜá×wÀáÄ.

A¥ÜÊÝ

( 4, – 3 ) ÊÜáñÜᤠ( 8, 5 ) ¹í¨ÜáWÜÙÜ®Üá° ÓæàÄÓÜáÊÜ ÃæàTÝSívÜÊÜ®Üá°

BíñÜÄPÜÊÝX 3 : 1 A®Üá±ÝñܨÜÈÉ Ë»ÝXÓÜáÊÜ ¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜ®Üá°
PÜívÜá×wÀáÄ.

23. Jí¨Üá Ÿáqr¿áÈÉ 36 ÊÜÞ訆 ÖÜ|á¡WÜÚÊæ. AÊÜâWÜÙÜ 14 »ÝWܨÜÐÜár ÖÜ|á¡WÜÙÜá
PæãÙæ£Êæ ÊÜáñÜᤠEÚ¨ÜÊÜâ aæ¬Ý®XÊæ. ŸáqrÀáí¨Ü ¿Þ¨ÜêbfPÜÊÝX Jí¨Üá ÊÜÞË®Ü
ÖÜ|¡®Üá° ÖæãÃÜñæWæ¨ÝWÜ A¨Üá aæ®Ý°XÃÜáÊÜ ÓÜí»ÜÊܯà¿áñæ¿á®Üá° PÜívÜá×wÀáÄ.

24. 3·5 cm £ÅgÂËÃÜáÊÜ ÊÜêñܤÊÜ®Üá° ÃÜbÔ ÊÜáñÜᤠÓܳÍÜìPÜWÜÙÜ ®ÜvÜá訆 Pæãà®Ü 60°

CÃÜáÊÜíñæ ÊÜêñܤPæR Jí¨Üá hæãñæ ÓܳÍÜìPÜWÜÙÜ®Üá° ÃÜbÔ.

02. 04. 2024 8 of 16

CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 81-K
IV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 9 × 3 = 27

25. ŸÖÜá¯Ü¨æãàQ¤ p ( x ) = x 3 + 3 x 2 + 4x + 5 ®Üá° g ( x ) = x 2 – x + 1

Äí¨Ü »ÝXÔ ÊÜáñÜᤠ»ÝWÜÆŸœ [ q ( x ) ] ÖÝWÜã ÍæàÐÜ [ r ( x ) ] WÜÙÜ®Üá°

PÜívÜá×wÀáÄ.

A¥ÜÊÝ

ŸÖÜá¯Ü¨æãàQ¤ p ( x ) = x 3 + 4 x 2 + 5x – 2 ®Üá° ŸÖÜá¯Ü¨æãàQ¤ g ( x ) ¯í¨Ü

»ÝXÔ¨ÝWÜ »ÝWÜÆŸœ [ q ( x ) ] ÊÜáñÜᤠÍæàÐÜ [ r ( x ) ] WÜÙÜá PÜÅÊÜáÊÝX

x 2 – x + 2 ÊÜáñÜᤠ4 BXÊæ. ÖÝWݨÜÃæ g ( x ) ®Üá° PÜívÜá×wÀáÄ.

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

ÊÜWÝìíñÜÃÜ BÊÜ꣤

2—6 2

7 — 11 4

12 — 16 5

17 — 21 3

22 — 26 1

A¥ÜÊÝ

02. 04. 2024 9 of 16

CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 81-K
D PæÙÜX¬Ü ¨ÜñݤíÍÜWÜÚWæ ŸÖÜáÆPÜÊܬÜá® PÜívÜá×wÀáÄ

ÊÜWÝìíñÜÃÜ BÊÜ꣤

1—5 1

5—9 3

9 — 13 7

13 — 17 10

17 — 21 9

27.  ABC ¿áÈÉ ADC  BAC BWÜáÊÜíñæ ‘D’ ¿áá BC Êæáà騆 Jí¨Üá
¹í¨ÜáÊÝX¨æ. ÖÝWݨÜÃæ AC 2 = BC . CD Gí¨Üá ÓݘÔ.

A¥ÜÊÝ

bñÜŨÜÈÉ  ABC ÊÜáñÜᤠ AMP WÜÙÜá PÜÅÊÜáÊÝX B ÊÜáñÜᤠM ¹í¨ÜáWÜÙÜÈÉ
ÆíŸPæãà¬ÜWÜÙܬÜá® Öæãí©ÃÜáÊÜ ÆíŸPæãà¬Ü £Å»ÜágWÜÙÝXÊæ. ÖÝWݨÜÃæ,
CA BC
 Gí¨Üá ÓݘÔ.
PA MP

28. ¸ÝÖÜ ¹í¨Üá˯í¨Ü ÊÜêñܤPæR GÙæ¨Ü ÓܳÍÜìPÜWÜÙÜ E¨ÜªÊÜâ ÓÜÊÜáÊÝXÃÜáñܤ¨æ Gí¨Üá
ÓݘÔ.

02. 04. 2024 10 of 16

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29. bñÜŨÜÈÉ ‘r’ £ÅgÂËÃÜáÊÜ £ÅhÝÂíñÜÃÜ SívÜ AOBPA ¿á ËÔ¤à|ìÊÜâ 231 cm 2
ÖÝWÜã APB PÜíÓÜ¨Ü E¨ÜªÊÜâ 22 cm WÜÙÝXÊæ. £ÅhÝÂíñÜÃÜ SívÜ¨Ü £Åg ÊÜáñÜá¤
Pæãà®Ü ‘’ ¨Ü ¸æÇæWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

bñÜŨÜÈÉ ROQP B¿áñÜÊÜâ ÊÜêñܤ aÜñÜá¥ÜìPܨÜÈÉ AíñÜÓܧÊÝX¨æ. B¿áñÜ¨Ü E¨Üª
ÊÜáñÜᤠAWÜÆWÜÙÜá PÜÅÊÜáÊÝX 16 cm ÊÜáñÜᤠ12 cm WÜÙݨÜÃæ dÝÀáàPÜêñÜ »ÝWܨÜ
ËÔ¤à|ìÊÜ®Üá° PÜívÜá×wÀáÄ.

30. ñÝÀá¿á ÊÜ¿áÓÜáÕ ÊÜáWÜ®Ü ÊÜ¿áÔÕ®Ü ÊÜWÜì¨Ü GÃÜvÜÃÜÑr¨æ. 8 ÊÜÐÜìWÜÙÜ ®ÜíñÜÃÜ
ñÝÀá¿á ÊÜ¿áÓÜáÕ ÊÜáWÜ®Ü ÊÜ¿áÔÕ®Ü ÊÜáãÃÜÃÜÐÜrQRíñÜ 4 ÊÜÐÜìWÜÙÜá ÖæaÝcWÜáñܤ¨æ.
AÊÜÃÜ DX®Ü ÊÜ¿áÓÜáÕWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

02. 04. 2024 11 of 16

CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 81-K
31. bñÜŨÜÈÉ  ABC ¿á ÍÜêíWܹí¨ÜáWÜÙÜá A ( x, 10 ), B ( 2, 2 ) ÊÜáñÜá¤

C ( 12, 2 ) BXÊæ. AC ¿á ÊÜá«Ü¹í¨Üá Q ( 9, 6 ) BX¨Üáª,  APQ ®Ü

ËÔ¤à|ìÊÜâ 12 cm 2 WÜÙݨÜÃæ aÜñÜá»Üáìg PBCQ ®Ü ËÔ¤à|ìÊÜ®Üá°
PÜívÜá×wÀáÄ.

32. Jí¨Üá BÓܳñæÅ¿áÈÉ ¨ÝSÇÝXÃÜáÊÜ 100 g®Ü ÃæãàXWÜÙÜ ÊÜ¿áÓÜáÕWÜÙÜ ËÊÜÃÜ D
PæÙÜX®Üí£¨æ. D ¨ÜñݤíÍÜWÜÚWæ PÜwÊæá Ë«Ý¬Ü¨Ü KiàÊ… ÃÜbÔ
ÊÜ¿áÓÜáÕ ÊÜÐÜìWÜÙÜÈÉ ÃæãàXWÜÙÜ ÓÜíTæÂ
ÓÜíbñÜ BÊÜ꣤
10 QRíñÜ PÜwÊæá 6

20 QRíñÜ PÜwÊæá 15

30 QRíñÜ PÜwÊæá 38

40 QRíñÜ PÜwÊæá 46

50 QRíñÜ PÜwÊæá 65

60 QRíñÜ PÜwÊæá 84

70 QRíñÜ PÜwÊæá 100

02. 04. 2024 12 of 16

CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 81-K
33. 6 cm, 8 cm ÊÜáñÜᤠ9 cm ¸ÝÖÜáWÜÚÃÜáÊÜ Jí¨Üá £Å»ÜágÊÜ®Üá° ÃÜbÔ, ®ÜíñÜÃÜ

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

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

V. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 4 × 4 = 16

34. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á ±ÜÄÖÝÃÜÊÜ®Üá° ®Üûæ¿á

˫ݮܩí¨Ü PÜívÜá×wÀáÄ

2x + y = 8

x+y = 5

35. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ ‘n’ ±Ü¨ÜWÜÙÜÊÜÃæX®Ü Êæãñܤ 210 ÊÜáñÜá¤

Êæã¨ÜÆ ( n – 1 ) ±Ü¨ÜWÜÙÜÊÜÃæX®Ü Êæãñܤ 171 BXÊæ. B ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á

Êæã¨ÜÆ ±Ü¨Ü 3 B¨ÜÃæ, B ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á®Üá° PÜívÜá×wÀáÄ ÊÜáñÜᤠB

ÍæÅà{¿á 20 ®æà ±Ü¨ÜÊÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

02. 04. 2024 13 of 16

CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 81-K
‘n’ »ÜágWÜÙܬÜá® Öæãí©ÃÜáÊÜ Jí¨Üá ŸÖÜá»ÜáhÝPÜꣿá GÇÝÉ JÙÜPæãà¬ÜWÜÙÜ

Êæãñܤ ( n – 2 ) 180°. Jí¨Üá ±ÜíaÜ»ÜáhÝPÜꣿá JÙÜPæãà®ÜWÜÙÜá ÓÜÊÜÞíñÜÃÜ

ÍæÅà{¿áÈɨÜ᪠A¨ÜÃÜ AñÜÂíñÜ bPÜR Pæãà®ÜÊÜâ 72° B¨ÜÃæ, B ±ÜíaÜ»ÜáhÝPÜꣿá

GÇÝÉ JÙÜPæãà®ÜWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

36. bñÜŨÜÈÉ Jí¨Üá ÓÜÊÜáñÜpÝr¨Ü ®æÆ¨Ü ÊæáàÇæ ¸æàÃæ ¸æàÃæ GñܤÃÜWÜÙÜ®Üá° Öæãí©ÃÜáÊÜ

AB ÊÜáñÜᤠCD PÜíŸWÜÙÜá ¬æàÃÜÊÝX ­í£Êæ. D GÃÜvÜã PÜíŸWÜÙÜ ¯Ý¨ÜWÜÙܬÜá®

hæãàwÓÜáÊÜ ÃæàTæ¿á Êæáà騆 Jí¨Üá ¹í¨Üá ‘P’ ­í¨Ü GÃÜvÜá PÜíŸWÜÙÜ

ÊæáàÆᤩWæ EípÝ¨Ü E®Ü°ñÜ Pæãà®ÜWÜÙÜá ±ÜÃÜÓܳÃÜ ±ÜäÃÜPÜ Pæãà®ÜWÜÙÝXÊæ.

CD PÜíŸ¨Ü E¨Üª 20 3 m ÊÜáñÜᤠPD ¿á E¨Üª 20 m WÜÙÝXÊæ. BP = 10 m

WÜÙݨÜÃæ AB PÜíŸ¨Ü E¨Üª ÊÜáñÜᤠPÜíŸWÜÙÜ ÊæáàÆᤩWÜÙÜ ¬ÜvÜáË¬Ü E¨Üª

AC ¿á®Üá° PÜívÜá×wÀáÄ.

37. ÊÜáãÆÓÜÊÜÞ¬Üá¯ÝñÜñæ¿á ¯ÜÅÊæáà¿á A¥ÜÊÝ ¥æàÇ…Õ¬Ü ¯ÜÅÊæáà¿áÊܬÜá® ÓݘÔ.

02. 04. 2024 14 of 16

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VI. PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×5=5

38. Jí¨Üá ÔÈívÜÃ…, Jí¨Üá ÍÜíPÜáË¬Ü ¼¬Ü®PÜ ÊÜáñÜᤠJí¨Üá A«ÜìWæãàÙÜWÜÙܬÜá®
bñÜŨÜÈÉÃÜáÊÜíñæ hæãàwÔ Jí¨Üá Qào ¯¿áíñÜÅPÜ ÓÝ«Ü®ÜÊÜ®Üá° ñÜ¿ÞÄÓÜÇÝX¨æ.
A«ÜìWæãàÙÜ¨Ü »ÝWܨÜÈÉ Aíoá ¨ÜÅÊÜÊÜ®Üá° ñÜáí¹ÓÜÇÝX¨æ. A«ÜìWæãàÙÜ ÊÜáñÜá¤
ÔÈívÜÃ…WÜÙÜ £ÅgÂWÜÙÜá PÜÅÊÜáÊÝX 21 cm ÊÜáñÜᤠ14 cm WÜÙÝX¨Üáª, ±Üä|ì
ÓÝ«Ü®Ü¨Ü Joár GñܤÃÜ 60 cm ÊÜáñÜᤠÔÈívÜÃ… »ÝWÜ¨Ü GñܤÃÜ 15 cm B¨ÜÃæ,
ÓÝ«Ü®Ü¨Ü ÊÜPÜÅÊæáàÇ澄 ËÔ¤à|ì ÊÜáñÜᤠA«ÜìWæãàÙܨÜÈÉ¬Ü Aíoá¨ÜÅÊܨÜ
±ÜÅÊÜÞ|ÊÜ®Üá° PÜívÜá×wÀáÄ.

02. 04. 2024 15 of 16

CCE RF/PF/RR/PR/NSR/NSPR(A)/666/031 81-K

02. 04. 2024 16 of 16

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