Karnataka SSLC Question Paper 2023 Maths

Karnataka Board
Question Paper

SSLC
Question Paper

B∆«M•⁄ O⁄}⁄¬° “
2
A
RF(A)/100/3311

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

Jlflo Æ⁄√ÀÊ-V⁄◊⁄ —⁄MSÊ¿ : 38 ]
Total No. of Questions : 38 ] CCE RF
UNREVISED

100
—⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E FULL SYLLABUS
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⁄}⁄°¬“
( BMW«ŒÈ »⁄·¤®⁄¥¿»⁄fl / English Medium )
( À¤≈¤ @∫⁄¥¿£% / Regular Fresh )

¶´¤MO⁄ : 03. 04. 2023 ] [ Date : 03. 04. 2023
—⁄»⁄flæ⁄fl : ∑Ê◊⁄VÊX 10-30 ¬M•⁄ »⁄fl®¤¿‘⁄-1-45 ¡⁄»⁄¡ÊVÊ ]
[ Time : 10-30 A.M. to 1-45 P.M.
V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80 ] [ Max. Marks : 80
General Instructions to the Candidate :
1. This question paper consists of objective and subjective types of
38 questions.
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. Check whether all the pages of the question paper are
intact.
3. Follow the instructions given against both the objective and subjective
types of 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
Tear here

question paper. It includes 15 minutes for reading the question paper.

[ Turn over

2 RF(A)/100/3311 2 CCE RF 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 number of zeroes of the polynomial y = p ( x ) in the given

graph is

(A) 3 (B) 2

(C) 1 (D) 4

2. For an event ‘E’, if P ( E ) = 0·75, then P ( E ) is

(A) 2·5 (B) 0·25

(C) 0·025 (D) 1·25

2 RF(A)/100/3311 3 CCE RF 81-E

3. The total surface area of a right circular cylinder having radius ‘r’

and height ‘h’ is

(A) πr(r+h) (B) 2π rh

(C) 2π r ( r – h ) (D) 2π r ( r + h )

4. The number that represents the remainder when 19 = 6 × 3 + 1

is compared with Euclid’s division lemma a = bq + r is

(A) 3 (B) 6

(C) 1 (D) 19

5. In the given figure, PB is a tangent drawn at the point A to the

circle with centre ‘O’. If AOP = 45°, then the measure of

OPA is

(A) 45° (B) 90°

(C) 35° (D) 65°

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2 RF(A)/100/3311 4 CCE RF 81-E

6. In the figure, if DE || BC, then the correct relation among the

following is

AD AE
(A) =
AB EC
AD EC
(B) =
DB AE
AD AE
(C) =
DB EC
DB AE
(D) =
AD EC

7. The lines represented by the equations 4x + 5y – 10 = 0 and

8x + 10y + 20 = 0 are

(A) intersecting lines

(B) perpendicular lines to each other

(C) coincident lines

(D) parallel lines

2 RF(A)/100/3311 5 CCE RF 81-E

8. The distance of the point ( – 8, 3 ) from the x-axis is

(A) – 8 units

(B) 3 units

(C) – 3 units

(D) 8 units

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

7
9. Express the denominator of in the form of 2n × 5m .
80

10. If the pair of lines represented by the linear equations

x + 2y – 4 = 0 and ax + by – 12 = 0 are coincident lines, then find

the values of ‘a’ and ‘b’.

11. ∆ ABC ~ ∆ PQR. Area of the ∆ ABC is 64 cm 2 and the area of the

∆ PQR is 100 cm 2 . If AB = 8 cm, then find the length of PQ.

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2 RF(A)/100/3311 6 CCE RF 81-E

12. Express the equation x ( 2 + x ) = 3 in the standard form of a

quadratic equation.

13. Find the discriminant of the quadratic equation 2 x 2 – 4x + 3 = 0.

14. Find the coordinates of the mid-point of the line segment joining

the points ( 6, 3 ) and ( 4, 7 ).

15. Write the degree of the polynomial

P ( x ) = 3 x 3 − x 4 + 2x 2 + 5x + 2.

16. Write the formula to find the volume of the frustum of a cone

given in the figure.

2 RF(A)/100/3311 7 CCE RF 81-E

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

17. Show that 5 + 3 is an irrational number.

OR

Find the H.C.F. of 72 and 120 by using Euclid’s division

algorithm.

18. Solve the given pair of linear equations :

3x + y = 12

x+y = 6

19. Find the 20 th term of the Arithmetic progression 4, 7, 10, .....

by using formula.

20. Find the roots of the equation 2 x 2 – 5x + 3 = 0 by using

‘quadratic formula’.

OR

Find the roots of the equation 5 x 2 – 6x – 2 = 0 by the method of

completing the square.

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2 RF(A)/100/3311 8 CCE RF 81-E

21. In the given figure, if ABC = 90°, then find the values of sin θ

and cos α.

22. A box contains cards which are numbered from 9 to 19. If one

card is drawn at random from the box, find the probability that it

bears a prime number.

23. In the given figure, ABCD is a trapezium in which AB || DC, and

BC ⊥ DC. If AB = 6 cm, CD = 10 cm and AD = 5 cm, then find the

distance between the parallel lines.

24. Draw a circle of radius 4 cm and construct a pair of tangents to

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

2 RF(A)/100/3311 9 CCE RF 81-E

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

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

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

OR

Find the zeroes of the quadratic polynomial p ( x ) = x 2 + 7x + 10,

and verify the relationship between zeroes and the coefficients.

26. Prove that
1 + cos A
= cosec A + cot A
1 − cos A

OR

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

27. Find the mean for the following data :

Class-interval Frequency

1–5 4

6 – 10 3

11 – 15 2

16 – 20 1

21 – 25 5

OR

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2 RF(A)/100/3311 10 CCE RF 81-E

Find the mode for the following data :

Class-interval Frequency

1–3 6

3–5 9

5–7 15

7–9 9

9 – 11 1

28. Find the ratio in which the line segment joining the points

A ( – 6, 10 ) and B ( 3, – 8 ) is divided by the point ( – 4, 6 ).

OR

Find the area of a triangle whose vertices are A ( 1, – 1 ),

B ( – 4, 6 ) and C ( – 3, – 5 )

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

to a circle are equal”.

2 RF(A)/100/3311 11 CCE RF 81-E

30. In the given figure, ‘O’ is the centre of a circle and OAB is an

equilateral triangle. P and Q are the mid-points of OA and OB

respectively. If the area of ∆ OAB is 36 3 cm 2 , then find the

area of the shaded region.

31. Construct a triangle with sides 5 cm, 6 cm and 8 cm and then
3
construct another triangle whose sides are of the
4

corresponding sides of the first triangle.

32. The distance between two cities ‘A’ and ‘B’ is 132 km. Flyovers

are built to avoid the traffic in the intermediate towns between

these cities. Because of this, the average speed of a car travelling

in this route through flyovers increases by 11 km/h and hence,

the car takes 1 hour less time to travel the same distance than

earlier. Find the current average speed of the car.

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2 RF(A)/100/3311 12 CCE RF 81-E

33. A life insurance agent found the following data for distribution of

ages of 100 policy holders. Draw a “Less than type ogive” for the

given data :

Age ( in years ) Number of policy holders

( cumulative frequency )

Below 20 2

Below 25 6

Below 30 24

Below 35 45

Below 40 78

Below 45 89

Below 50 100

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

34. The sum of 2nd and 4th terms of an arithmetic progression is 54

and the sum of its first 11 terms is 693. Find the arithmetic

progression. Which term of this progression is 132 more than its

54th term ?

OR

2 RF(A)/100/3311 13 CCE RF 81-E

The first and the last terms of an arithmetic progression are 3

and 253 respectively. If the 20th term of the progression is 98,

then find the arithmetic progression. Also find the sum of the last

10 terms of this progression.

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

graphical method :

2x + y = 8

x–y = 1

36. Prove that “If in two triangles, corresponding angles are equal,

then their corresponding sides are in the same ratio ( or

proportion ) and hence the two triangles are similar”.

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2 RF(A)/100/3311 14 CCE RF 81-E

37. In the given figure, a rope is tightly stretched and tied from the

top of a vertical pole to a peg on the same level ground such that

the length of the rope is 20 m and the angle made by it with the

ground is 30°. A circus artist climbs the rope, reaches the top of

the pole and from there he observes that the angle of elevation of

the top of another pole on the same ground is found to be 60°. If

the distance of the foot of the longer pole from the peg is 30 m,

then find the height of this pole. ( Take 3 = 1·73 )

2 RF(A)/100/3311 15 CCE RF 81-E

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

38. A wooden solid toy is made by mounting a cone on the circular

base of a hemisphere as shown in the figure. If the area of base of

the cone is 38·5 cm 2 and the total height of the toy is 15·5 cm,

then find the total surface area and volume of the toy.

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2 RF(A)/100/3311 16 CCE RF 81-E

2
A
RF(A)/100/3310

CÈÉí¨Ü PÜñܤÄÔ
Question Paper Serial No.
Joár ÊÜáá©ÅñÜ ±ÜâoWÜÙÜ ÓÜíTæ : 16 ]
Total No. of Printed Pages : 16 ]

Joár ±ÜÅÍæ°WÜÙÜ ÓÜíTæ : 38 ] CCE RF
Total No. of Questions : 38 ] UNREVISED
ÓÜíPæàñÜ ÓÜíTæ : 81-K FULL SYLLABUS

100
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 / Regular Fresh)

±ÜÅÍ氱ܣÅPæ¿á®Üá° ñæÃæ¿áÆá CÈÉ PÜñܤÄÔ
©®ÝíPÜ : 03. 04. 2023 ] [ Date : 03. 04. 2023
ÓÜÊÜá¿á : ¸æÙÜWæY 10-30 Äí¨Ü ÊÜá«ÝÂÖÜ° 1-45 ÃÜÊÜÃæWæ ] [ Time : 10-30 A.M. to 1-45 P.M.
WÜÄÐÜu AíPÜWÜÙÜá : 80 ] [ Max. Marks : 80

±ÜÄàûݦìWÝX ÓÝÊÜޮܠÓÜãaÜ®æWÜÙÜá :
1. D ±ÜÅÍ氱ܣÅPæ¿áá ÊÜÓÜ᤯ÐÜu ÊÜáñÜᤠËÐÜ¿á¯ÐÜu ÊÜÞ¨ÜÄ¿á Joár 38 ±ÜÅÍæ°WÜÙÜ®Üá°
Öæãí©¨æ.
2. D ±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. ÊÜÓÜ᤯ÐÜu ÊÜáñÜᤠËÐÜ¿á¯ÐÜu ÊÜÞ¨ÜÄ¿á ±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
Tear here

¯WÜ©±ÜwÓÜÇÝ¨Ü ÓÜÊÜá¿áÊÜ®Üá° ±ÜÅÍ氱ܣÅPæ¿á ÊæáàÇݽWܨÜÈÉ ¯àvÜÇÝX¨æ.

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2 RF(A)/100/3310 2 CCE RF 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. PæãqrÃÜáÊÜ ®Üûæ¿áÈÉ y = p ( x ) ŸÖÜá±Ü¨æãàQ¤¿áá Öæãí©ÃÜáÊÜ ÍÜã®ÜÂñæWÜÙÜ

ÓÜíTæ¿áá,

(A) 3 (B) 2

(C) 1 (D) 4

2. Jí¨Üá Zo®æ "E ' Wæ P ( E ) = 0·75 B¨ÜÃæ, P ( E ) ¿áá

(A) 2·5 (B) 0·25

(C) 0·025 (D) 1·25

2 RF(A)/100/3310 3 CCE RF 81-K
3. £Åg "r ' ÖÝWÜã GñܤÃÜ "h ' BXÃÜáÊÜ ®æàÃÜ ÊÜêñܤ±Ý¨Ü ÔÈívÜÄ®Ü ±Üä|ì ÊæáàÇ澄
ËÔ¤à|ìÊÜâ,

(A) πr(r+h) (B) 2π rh

(C) 2π r ( r – h ) (D) 2π r ( r + h )

4. 19 = 6 × 3 + 1 C¨Ü®Üá° ¿áãQÉv…®Ü »ÝWÝPÝÃÜ A®Üá±ÜÅÊæáà¿á a = bq + r Wæ

ÖæãàÈÔ¨ÝWÜ ÍæàÐÜÊÜ®Üá° ÓÜãbÓÜáÊÜ ÓÜíTæ¿áá,

(A) 3 (B) 6

(C) 1 (D) 19

5. bñÜŨÜÈÉ "O ' Pæàí¨ÜÅÊÝXÃÜáÊÜ ÊÜêñܤPæR A ¹í¨ÜáË®ÜÈÉ PB ÓܳÍÜìPÜÊÜ®Üá°
GÙæ¿áÇÝX¨æ. AOP = 45° B¨ÜÃæ, OPA ¿á AÙÜñæ¿áá,

(A) 45° (B) 90°

(C) 35° (D) 65°

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2 RF(A)/100/3310 4 CCE RF 81-K
6. bñÜŨÜÈÉ DE || BC B¨ÜÃæ, D PæÙÜX®ÜÊÜâWÜÙÜÈÉ ÓÜÄ¿Þ¨Ü ÓÜíŸí«ÜÊÜâ,

AD AE AD EC
(A) = (B) =
AB EC DB AE

AD AE DB AE
(C) = (D) =
DB EC AD EC

7. 4x + 5y – 10 = 0 ÊÜáñÜᤠ8x + 10y + 20 = 0 D ÓÜËáàPÜÃÜ|WÜÙÜ®Üá°

±ÜÅ£¯◊ÓÜáÊÜ ÃæàTæWÜÙÜá,

(A) dæà©ÓÜáÊÜ ÃæàTæWÜÙÜá

(B) ±ÜÃÜÓܳÃÜ ÆíŸÃæàTæWÜÙÜá

(C) IPÜÂÊÝWÜáÊÜ ÃæàTæWÜÙÜá

(D) ÓÜÊÜÞíñÜÃÜ ÃæàTæWÜÙÜá

8. x&AûÜ©í¨Ü ( – 8, 3 ) ¹í¨ÜáËWæ CÃÜáÊÜ ¨ÜãÃÜÊÜâ

(A) – 8 ÊÜÞ®ÜWÜÙÜá (B) 3 ÊÜÞ®ÜWÜÙÜá

(C) – 3 ÊÜÞ®ÜWÜÙÜá (D) 8 ÊÜÞ®ÜWÜÙÜá

2 RF(A)/100/3310 5 CCE RF 81-K
II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ D}⁄°¬“ : 8×1=8

7 ÃÜ dæà¨ÜÊÜ®Üá° n
9. 2 × 5 m ÃÜã±Ü¨ÜÈÉ ÊÜÂPܤ±ÜwÔ.
80

10. x + 2y – 4 = 0 ÊÜáñÜᤠax + by – 12 = 0 D ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ®Üá°

±ÜÅ£¯◊ÓÜáÊÜ ÃæàTæWÜÙÜ hæãàw¿áá ±ÜÃÜÓܳÃÜ IPÜÂWæãÙÜáÛÊÜ ÃæàTæWÜÙݨÜÃæ, a ÊÜáñÜá¤

b WÜÙÜ ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

11. ∆ ABC ~ ∆ PQR BX¨æ. ∆ ABC ¿á ËÔ¤à|ìÊÜâ 64 cm 2 , ∆ PQR ®Ü

ËÔ¤à|ìÊÜâ 100 cm 2 ÊÜáñÜᤠAB = 8 cm B¨ÝWÜ, PQ ®Ü E¨ÜªÊÜ®Üá°

PÜívÜá×wÀáÄ.

12. x ( 2 + x ) = 3 D ÓÜËáàPÜÃÜ|ÊÜ®Üá° ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü B¨ÜÍÜìÃÜã±Ü¨ÜÈÉ

ÊÜÂPܤ±ÜwÔ.

13. 2 x 2 – 4x + 3 = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü Íæãà«ÜPÜÊÜ®Üá° PÜívÜá×wÀáÄ.

14. ( 6, 3 ) ÊÜáñÜᤠ( 4, 7 ) D ¹í¨ÜáWÜÙÜ®Üá° ÓæàÄÓÜáÊÜ ÃæàTÝSívܨÜ

ÊÜá«Ü¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

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2 RF(A)/100/3310 6 CCE RF 81-K
15. P ( x ) = 3 x 3 − x 4 + 2x 2 + 5x + 2 D ŸÖÜá±Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá

[ÝñÜÊÜ®Üá° (wXÅ) ŸÃæÀáÄ.

16. bñÜŨÜÈÉ PæãqrÃÜáÊÜ ÍÜíPÜá訆 ¼®Ü°PÜ¨Ü Z®Ü¶ÜÆÊÜ®Üá° PÜívÜá×w¿ááÊÜ ÓÜãñÜÅÊÜ®Üá°

ŸÃæÀáÄ.

III. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ D}⁄°¬“ : 8 × 2 = 16

17. 5 + 3 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ÓÝ◊Ô.

A¥ÜÊÝ

72 ÊÜáñÜᤠ120 ÃÜ ÊÜá.ÓÝ.A.ÊÜ®Üá° ¿áãQÉv…®Ü »ÝWÝPÝÃÜ PÜÅÊÜáË◊¿á®Üá°

E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

2 RF(A)/100/3310 7 CCE RF 81-K
18. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á®Üá° ¹wÔ :

3x + y = 12

x+y = 6

19. 4, 7, 10, ..... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á 20 ®æà ±Ü¨ÜÊÜ®Üá° ÓÜãñÜÅ E±ÜÁãàXÔ

PÜívÜá×wÀáÄ.

20. 2 x 2 – 5x + 3 = 0 D ÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá° "ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü ÓÜãñÜÅ'

E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

A¥ÜÊÝ

5 x 2 – 6x – 2 = 0 D ÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá° ÊÜWÜì±Üä|ìWæãÚÓÜáÊÜ
˫ݮܩí¨Ü PÜívÜá×wÀáÄ.

21. bñÜŨÜÈÉ ABC = 90° B¨ÜÃæ, sin θ ÖÝWÜã cos α CÊÜâWÜÙÜ ¸æÇæ
PÜívÜá×wÀáÄ.

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2 RF(A)/100/3310 8 CCE RF 81-K
22. Jí¨Üá ±æqrWæ¿áÈÉ 9 Äí¨Ü 19 ÃÜ ÊÜÃæX®Ü ÓÜíTæÂWÜÙÜ®Üá° ®ÜÊÜáã©ÔÃÜáÊÜ

PÝv…ìWÜÚÊæ. ±æqrWæÀáí¨Ü ¿Þ¨ÜêbfPÜÊÝX Jí¨Üá PÝvÜì®Üá° ñæWæ¨ÝWÜ A¨Üá

Jí¨Üá AË»Ýg ÓÜíTæ¿ÞXÃÜáÊÜ ÓÜí»ÜÊܯà¿áñæ¿á®Üá° PÜívÜá×wÀáÄ.

23. bñÜŨÜÈÉ ABCD Jí¨Üá ñÝŲgÂ, AB || DC ÊÜáñÜᤠBC ⊥ DC BX¨æ.

AB = 6 cm, CD = 10 cm ÊÜáñÜᤠAD = 5 cm B¨ÜÃæ, ÓÜÊÜÞíñÜÃÜ ¸ÝÖÜáWÜÙÜ

®ÜvÜá訆 ¨ÜãÃÜÊÜ®Üá° PÜívÜá×wÀáÄ.

24. 4 cm £ÅgÂËÃÜáÊÜ ÊÜêñܤÊÜ®Üá° ÃÜbÔÄ ÊÜáñÜᤠÓܳÍÜìPÜWÜÙÜ ®ÜvÜá訆 Pæãà®Ü 60°

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

IV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ D}⁄°¬“ : 9 × 3 = 27

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

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

A¥ÜÊÝ

2 RF(A)/100/3310 9 CCE RF 81-K
p ( x ) = x 2 + 7x + 10 D ÊÜWÜì ŸÖÜá±Ü¨æãàQ¤¿á ÍÜã®ÜÂñæWÜÙÜ®Üá°

PÜívÜá×wÀáÄ ÖÝWÜã ÍÜã®ÜÂñæWÜÙÜá ÊÜáñÜᤠÓÜÖÜWÜá|PÜWÜÙÜ ®ÜvÜá訆 ÓÜíŸí«ÜÊÜ®Üá°

ñÝÙæ ®æãàw.

1 + cos A
26. = cosec A + cot A Gí¨Üá ÓÝ◊Ô.
1 − cos A

A¥ÜÊÝ

sin A 1 + cos A
+ = 2 cosec A Gí¨Üá ÓÝ◊Ô.
1 + cos A sin A

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

ÊÜWÝìíñÜÃÜ BÊÜ꣤

1—5 4

6 — 10 3

11 — 15 2

16 — 20 1

21 — 25 5

A¥ÜÊÝ

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D PæÙÜX®Ü ¨ÜñݤíÍÜWÜÚWæ ŸÖÜáÆPÜÊÜ®Üá° PÜívÜá×wÀáÄ :

ÊÜWÝìíñÜÃÜ BÊÜ꣤

1—3 6

3—5 9

5—7 15

7—9 9

9 — 11 1

28. A ( – 6, 10 ) ÊÜáñÜᤠB ( 3, – 8 ) ¹í¨ÜáWÜÙÜ®Üá° ÓæàÄÓÜáÊÜ ÃæàTÝSívÜÊÜâ

( – 4, 6 ) ¹í¨Üá˯í¨Ü ¿ÞÊÜ A®Üá±ÝñܨÜÈÉ Ë»ÝXÓÜƳvÜáñܤ¨æ Gí¨Üá

PÜívÜá×wÀáÄ.

A¥ÜÊÝ

ÍÜêíWÜ ¹í¨ÜáWÜÙÜá A ( 1, – 1 ), B ( – 4, 6 ) ÊÜáñÜᤠC ( – 3, – 5 ) BXÃÜáÊÜ

£Å»Üág¨Ü ËÔ¤à|ìÊÜ®Üá° PÜívÜá×wÀáÄ.

2 RF(A)/100/3310 11 CCE RF 81-K
29. ""¸ÝÖÜ ¹í¨Üá˯í¨Ü ÊÜêñܤPæR GÙæ¨Ü ÓܳÍÜìPÜWÜÙÜ E¨ÜªÊÜâ ÓÜÊÜáÊÝXÃÜáñܤ¨æ'' Gí¨Üá

ÓÝ◊Ô.

30. bñÜŨÜÈÉ "O ' ÊÜêñܤPæàí¨ÜÅ ÊÜáñÜᤠOAB Jí¨Üá ÓÜÊÜá¸ÝÖÜá £Å»ÜágÊÝX¨æ. P ÊÜáñÜá¤

Q WÜÙÜá PÜÅÊÜáÊÝX OA ÊÜáñÜᤠOB WÜÙÜ ÊÜá«Ü¹í¨ÜáWÜÙÝXÊæ. ∆ OAB ¿á

ËÔ¤à|ìÊÜâ 36 3 cm 2 B¨ÜÃæ, dÝÁáWæãÚst »ÝWÜ¨Ü ËÔ¤à|ìÊÜ®Üá°

PÜívÜá×wÀáÄ.

31. 5 cm, 6 cm ÊÜáñÜᤠ8 cm ¸ÝÖÜáWÜÚÃÜáÊÜ Jí¨Üá £Å»ÜágÊÜ®Üá° ÃÜbÔ, ®ÜíñÜÃÜ

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

£Å»Üág¨Ü A®ÜáÃÜã±Ü ¸ÝÖÜáWÜÙÜ 34 ÃÜÑrÃÜáÊÜíñæ ÃÜbÔ.

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32. "A' ÊÜáñÜᤠ"B ' GíŸ GÃÜvÜá ®ÜWÜÃÜWÜÙÜ ®ÜvÜá訆 ¨ÜãÃÜÊÜâ 132 km BX¨æ. D
®ÜWÜÃÜWÜÙÜ ÊÜÞWÜì ÊÜá«Ü¨ÜÈÉ ŸÃÜáÊÜ ±Üor|WÜÙÜÈÉ EípÝWÜáÊÜ ÓÜíaÝÃÜ
¨ÜorOæ¿á®Üá° PÜwÊæá ÊÜÞvÜÆá, ÊæáàÆá ÓæàñÜáÊæWÜÙÜ®Üá° ¯ËáìÓÜÇÝX¨æ. D
PÝÃÜ|©í¨ÝX, ÊæáàÆáÓæàñÜáÊæWÜÙÜ ÊÜáãÆPÜ D ÊÜÞWÜì¨ÜÈÉ aÜÈÓÜáÊÜ Jí¨Üá
PÝÄ®Ü ÓÜÃÝÓÜÄ gÊÜÊÜâ 11 km/h ÖæaÝcWÜáñܤ¨æ ; B¨ÜªÄí¨Ü C¨æà ¨ÜãÃÜÊÜ®Üá°
PÜÅËáÓÜÆá PÝÃÜá Êæã¨ÜÈXíñÜ 1 WÜípæ PÜwÊæá ÓÜÊÜá¿áÊÜ®Üá° ñæWæ¨ÜáPæãÙÜáÛñܤ¨æ.
ÖÝWݨÜÃæ PÝÄ®Ü DX®Ü ÓÜÃÝÓÜÄ gÊÜÊÜ®Üá° PÜívÜá×wÀáÄ.

33. JŸº iàÊÜËÊÜÞ Hhæío®Üá ±Üvæ¨Ü 100 ±ÝÈÔ¨ÝÃÜÃÜ ÊÜ¿áÓÜáÕWÜÙÜ ËñÜÃÜOæ¿á
¨ÜñݤíÍÜWÜÙÜá D PæÙÜX®Üíñæ CÊæ. D ¨ÜñݤíÍÜWÜÚWæ ""PÜwÊæá Ë«Ý®Ü¨Ü KiàÊ…''
GÙæÀáÄ :

±ÝÈÔ¨ÝÃÜÃÜ ÓÜíTæÂ
ÊÜ¿áÓÜáÕ (ÊÜÐÜìWÜÙÜÈÉ )
(ÓÜíbñÜ BÊÜ꣤)
20 QRíñÜ PÜwÊæá
2

25 QRíñÜ PÜwÊæá
6

30 QRíñÜ PÜwÊæá
24

35 QRíñÜ PÜwÊæá
45

40 QRíñÜ PÜwÊæá
78

45 QRíñÜ PÜwÊæá
89

50 QRíñÜ PÜwÊæá
100

2 RF(A)/100/3310 13 CCE RF 81-K
V. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ D}⁄°¬“ : 4 × 4 = 16

34. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á 2 ®æà ÊÜáñÜᤠ4 ®æà ±Ü¨ÜWÜÙÜ Êæãñܤ 54 ÖÝWÜã A¨ÜÃÜ

Êæã¨ÜÆ 11 ±Ü¨ÜWÜÙÜ Êæãñܤ 693 B¨ÜÃæ, ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á®Üá° PÜívÜá×wÀáÄ

ÊÜáñÜᤠD ÍæÅà{¿á GÐÜr®æà ±Ü¨ÜÊÜâ A¨ÜÃÜ 54 ®æà ±Ü¨ÜQRíñÜ 132 ÖæaÝcXÃÜáñܤ¨æ ?

A¥ÜÊÝ

Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ ÊÜáñÜᤠPæã®æ¿á ±Ü¨ÜWÜÙÜá PÜÅÊÜáÊÝX 3 ÊÜáñÜá¤

253 BXÊæ. ÍæÅà{¿á 20 ®æà ±Ü¨ÜÊÜâ 98 B¨ÜÃæ, ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á®Üá°

PÜívÜá×wÀáÄ ÖÝWÜã D ÍæÅà{¿á Pæã®æ¿á 10 ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá°

PÜívÜá×wÀáÄ.

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

˫ݮܩí¨Ü PÜívÜá×wÀáÄ :

2x + y = 8

x–y = 1

36. ""GÃÜvÜá £Å»ÜágWÜÙÜÈÉ A®ÜáÃÜã±Ü Pæãà®ÜWÜÙÜá ÓÜÊÜáÊݨÜÃæ AÊÜâWÜÙÜ A®ÜáÃÜã±Ü
¸ÝÖÜáWÜÙÜ A®Üá±ÝñÜWÜÙÜá ÓÜÊÜá (A¥ÜÊÝ ÓÜÊÜÞ®Üá±ÝñܨÜÈÉÃÜáñܤ¨æ) B¨ÜªÄí¨Ü B

£Å»ÜágWÜÙÜá ÓÜÊÜáÃÜã±ÜÊÝXÃÜáñܤÊæ'' Gí¨Üá ÓÝ◊Ô.

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37. bñÜŨÜÈÉ ÓÜÊÜáñÜpÝr¨Ü ®æÆ¨Ü ÊæáàÈÃÜáÊÜ ®æàÃÜÊÝ¨Ü PÜíŸ¨Ü ñÜá©Àáí¨Ü ®æƨÜ

Êæáà騆 Jí¨Üá WÜãoPæR 20 m ËáàoÃ… E¨ÜªËÃÜáÊÜíñæ Jí¨Üá ÖÜWÜYÊÜ®Üá° ¹X¨Üá

PÜorÇÝX¨æ. ÖÜWÜYÊÜâ ®æƨæãí©Wæ 30° Pæãà®ÜÊÜ®Üá° EíoáÊÜÞw¨æ. JŸº ÓÜPÜìÓ…

PÜÇÝ˨ܮÜá D ÖÜWÜYÊÜ®Üá° ÖÜ£¤ PÜíŸ¨Ü ñÜá©¿á®Üá° ñÜÆá², AÈÉí¨Ü AÊÜ®Üá A¨æà

®æÆ¨Ü ÊæáàÇæ ®æàÃÜÊÝX ¯í£ÃÜáÊÜ ÊÜáñæã¤í¨Üá PÜíŸ¨Ü ñÜá©¿á®Üá° ËàüÔ¨ÝWÜ

EípÝWÜáÊÜ E®Ü°ñÜ Pæãà®ÜÊÜâ 60° BXÃÜáñܤ¨æ. ®æÆ¨Ü Êæáà騆 WÜão©í¨Ü ¨æãvÜx

PÜíŸ¨Ü ±Ý¨ÜPæR CÃÜáÊÜ ¨ÜãÃÜÊÜâ 30 m B¨ÜÃæ, D PÜíŸ¨Ü GñܤÃÜÊÜ®Üá°

PÜívÜá×wÀáÄ.

( 3 = 1·73 Gí¨Üá ñæWæ¨ÜáPæãÚÛ)

2 RF(A)/100/3310 15 CCE RF 81-K
VI. OÊ◊⁄W´⁄ Æ⁄√ÀÊ-Wæ D}⁄°¬“ : 1×5=5

38. A«ÜìWæãàÙÝPÝÃÜ¨Ü ÊÜêñܤ ±Ý¨Ü¨Ü ÊæáàÇæ Jí¨Üá ÍÜíPÜáÊÜ®Üá° hæãàwÔ bñÜŨÜÈÉ

ñæãàÄÔÃÜáÊÜíñæ Jí¨Üá ÊÜáÃÜ¨Ü Z®Ü BqPæ¿á®Üá° ñÜ¿ÞÄÓÜÇÝX¨æ. ÍÜíPÜáË®Ü

±Ý¨Ü¨Ü ËÔ¤à|ìÊÜâ 38·5 cm 2 ÊÜáñÜᤠBqPæ¿á Joár GñܤÃÜ 15·5 cm B¨ÜÃæ,

BqPæ¿á ±Üä|ì ÊæáàÇ澄 ËÔ¤à|ì ÖÝWÜã Z®Ü¶ÜÆÊÜ®Üá° PÜívÜá×wÀáÄ.

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