Bihar Board Class 10th Model Paper 2024 Maths

Bihar Board

SAMPLE
PAPER

2024

SECONDARY SCHOOL EXAMINATION – 2024

Subject Code – 110 (ANNUAL/ )
MATHEMATICS (COMPULSORY)
xf.kr ¼vfuok;Z½

Total no. of Questions : 100+30+8 = 138 Full Marks : 100

Instructions for the candidates :

1- ijh{kkFkhZ OMR mÙkj i=d ij viuk iz’u iqfLrdk Øekad ¼10 vadksa dk½ vo’; fy[ksAa
Candidates must enter his/her Question Booklet Serial No. (10 Digits) in
the OMR Answer Sheet.

2- ijh{kkFkhZ ;FkklaHko vius 'kCnksa esa gh mÙkj nsAa
Candidates are required to give their answers in own words as far as
practicable.

3- nkfguh vksj gkf’k;s ij fn;s gq, vad iw.kkZad fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.

4- iz’uksa dks /;ku iwoZd i<+us ds fy, 15 feuV dk vfrfjDr le; fn;k x;k gSA
An extra time of 15 minutes has been allotted for the candidates to read
the questions carefully.

5- ;g iz’u iqfLrdk nks [k.Mksa esa gS & ,oa A

1

This question booklet is divided into two sections – Section-A and
Section-B.
6- [k.M&v esa 100 oLrqfu"B iz’u gS]a ftuesa ls fdUgha 50 iz’uksa dk mÙkj nsuk vfuok;Z gSA
ipkl ls vf/kd iz’uksa ds mÙkj nsus ij izFke 50 mÙkjksa dk gh ewY;kadu fd;k tk,xkA
izR;sd iz’u ds fy, 1 vad fu/kkZfjr gSA lgh mÙkj dks miyC/k djk;s x;s OMR mÙkj
i=d esa fn;s x;s lgh fodYi dks uhys@dkys ckWy isu ls izxk<+ djsaA fdlh Hkh izdkj
ds g~okbVuj @ rjy inkFkZ @ CysM @ uk[kwu vkfn dk OMR mÙkj iqfLrdk esa
iz;ksx djuk euk gS] vU;Fkk ijh{kk ifj.kke vekU; gksxkA
In Section-A, there are 100 objective type questions, out of which any 50
questions are to be answered. First 50 answers will be evaluated in case
more than 50 questions are answered. Each question carries 1 mark.
For answering these darken the circle with blue / black ball pen against
the correct option on OMR Answer Sheet provided to you. Do not use
Whitener / liquid / blade / nail etc. on OMR-sheet, otherwise the result
will be treated invalid.

7- [k.M&c esa 30 y?kq mÙkjh; iz’u gSa] ftuesa ls fdUgha 15 iz’uksa dk mÙkj nsuk vfuok;Z
gSA izR;sd iz’u ds fy, 2 vad fu/kkZfjr gSA buds vfrfjDr] bl [k.M esa 8 nh?kZ
mÙkjh; iz’u fn;s x;s gSa] ftuesa ls fdUgha 4 iz’uksa dk mÙkj nsuk gSA izR;sd iz’u ds
fy, 5 vad fu/kkZfjr gSaA
In Section-B, there are 30 short answer type questions, out of which any
15 questions are to be answered. Each question carries 2 marks. Apart
from these, there are 8 long answer type questions, out of which any 4
questions are to be answered. Each question carries 5 marks.

2

8- fdlh izdkj ds bysDVªkWfud midj.k dk iz;ksx iw.kZr;k oftZr gSA
Use of any electronic appliances is strictly prohibited.

3

[k.M & v @ SECTION - A
oLrqfu"B iz’u @ Objective Type Questions

iz’u la[;k 1 ls 100 rd ds iz’u ds lkFk pkj fodYi fn, x, gSa ftuesa ls ,d lgh gSA fdUgha
50 iz’uksa ds mÙkj vius }kjk pqus x, lgh fodYi dks OMR 'khV ij fpfàr djsAa
50x1=50
Question Nos. 1 to 100 have four options, out of which only one is correct.
Answer any 50 questions. You have to mark your selected option on the OMR-
sheet. 50x1=50
1. 2052 ds vHkkT; xq.ku[k.M esa 3 dk ?kkr D;k gS \

(A) 2 (B) 3

(C) 4 (D) 5

What is the power of 3 in the prime factorisation of 2052 ?

(A) 2 (B) 3

(C) 4 (D) 5

2. fuEufyf[kr esa fdldk n’keyo izlkj lkar ugha gS \

(A) (B)

(C) (D)

Which of the following is not a terminating decimal expansion ?

(A) (B)

(C) (D)

3. cgqin x(2x – 5) - 3 ds 'kwU;dksa dk ;ksx gS
4

(A) (B) -

(C) - (D)

The sum of the zeros of the polynomial x(2x – 5) – 3 is

(A) (B) -

(C) - (D)

4. ;fn lehdj.k 2x2 + px – 3 = 0 dk ,d ewy &3 gks rks p dk eku gS

(A) 3 (B) 4

(C) 5 (D) 6

If one root of the equation 2x2 + px – 3 = 0 is – 3 then the value of p is

(A) 3 (B) 4

(C) 5 (D) 6

5. ;fn cgqin ax2 – bx + c ds 'kwU;d 𝛼, 𝛽 gks rks 𝛼. 𝛽 dk eku gS

(A) (B) -

(C) (D) -

If 𝛼, 𝛽 be the zeros of the polynomial ax2 – bx + c then the value of 𝛼. 𝛽 is

(A) (B) -

(C) (D) -

6. fuEufyf[kr esa dkSu l0 Js0 esa gS \

5

(A) 2, 5, 7, … (B) 22, 32, 72, …

(C) x, x2, x3, … (D) 1, 3, 5, …

Which of the following is in an A.P. ?

(A) 2, 5, 7, … (B) 22, 32, 72, …

(C) x, x2, x3, … (D) 1, 3, 5, …

7. (1 + cot2𝜃)(1 – cos2𝜃) =

(A) sin2𝜃 (B) tan2𝜃

(C) cos2𝜃 (D) 1

8. 5x2 – 4x + 2 = 0 dk fofoDrdj gS

(A) 24 (B) &24

(C) 26 (D) &28

The discriminant of 5x2 – 4x + 2 = 0 is

(A) 24 (B) &24

(C) 26 (D) &28

9. 2x2 - 4x + 3 = 0 ds ewyksa dh izd`fr D;k gksxh \

(A) okLrfod ugha (B) okLrfod vkSj leku

(C) okLrfod ,oa vleku (D) buesa ls dksbZ ugha

What will be the nature of roots of 2x2 - 4x + 3 = 0 ?

(A) not real (B) real and equal

(C) real and unequal (D) none of these
6

10. fuEufyf[kr esa dkSu x2 – 3x + 2 = 0 dk ,d ewy gS \

(A) 0 (B) – 1

(C) - 2 (D) 2

Which of the following is a root of x2 – 3x + 2 = 0 ?

(A) 0 (B) – 1

(C) - 2 (D) 2

11. fuEufyf[kr esa fdldk xq.kuQy ,d ifjes; la[;k gS \

(A) √3 x √5 (B) √7 x √9

(C) √8 x √2 (D) √3 x √6

Which of the following is a product of rational number ?

(A) √3 x √5 (B) √7 x √9

(C) √8 x √2 (D) √3 x √6

12. fuEufyf[kr esa dkSu 154 dk vHkkT; xq.ku[k.M gS \

(A) 22 x 7 x 11 (B) 2 x 7 x 112

(C) 2 x7 x 11 (D) 23 x 7 x 11

Which of the following is the prime factorization of 154 ?

(A) 22 x 7 x 11 (B) 2 x 7 x 112

(C) 2 x7 x 11 (D) 23 x 7 x 11

13. fuEufyf[kr esa dkSu ,d?kkrh; lehdj.k gS \

7

(A) x + 5 = x + 7 (B) (x + 1)2 = x2 + 4x + 2

(C) x2 + 2x + 1 = 0 (D) x2 + 5x + 6 = 0

Which of the following is a linear equation ?

(A) x + 5 = x + 7 (B) (x + 1)2 = x2 + 4x + 2

(C) x2 + 2x + 1 = 0 (D) x2 + 5x + 6 = 0

14. fuEufyf[kr esa ls dkSu&lk f}?kkr lehdj.k gS \

(A) (x+1) (x-2) = (x-1)(x+5) (B) x2 – 2x = x2 + 1

(C) 2x2 + 4x + 2= 0 (D) x2 = x -

Which of the following is a quadratic equation ?

(A) (x+1) (x-2) = (x-1)(x+5) (B) x2 – 2x = x2 + 1

(C) 2x2 + 4x + 2= 0 (D) x2 = x -

15. ;fn lehdj.k 4x2 – 3Kx + 1 = 0 ds ewy cjkcj gks] rks K ¾

(A) ± (B) ±

(C) ± (D) ±

If the equation 4x2 – 3Kx + 1 = 0 has equal roots then K =

(A) ± (B) ±

(C) ± (D) ±

8

16. ;fn fdlh f}?kkr cgqin ds 'kwU;dksa dk ;ksxQy 2 ,oa xq.kuQy &15 gS] rks og f}?kkr

cgqin gksxk

(A) x2 + 2x + 15 (B) x2 + 2x - 15

(C) x2 – 2x + 15 (D) x2 – 2x - 15

If the sum of the zeros of a quadratic polynomial is 2 and their product is

-15 then that quadratic polynomial is

(A) x2 + 2x + 15 (B) x2 + 2x - 15

(C) x2 – 2x + 15 (D) x2 – 2x - 15

17. fuEufyf[kr esa dkSu l0 Js0 esa ugha gS \

(A) 12, 16, 20, … (B) 22, 52, 72, …

(C) 𝑎, 3𝑎, 5𝑎, … (D) 3, 5, 7, …

Which of the following is not in an A.P. ?

(A) 12, 16, 20, … (B) 22, 52, 72, …

(C) 𝑎, 3𝑎, 5𝑎, … (D) 3, 5, 7, …

18. l0 Js0 √2] √8] √18] √32] --- dk lkoZ&varj gS

(A) √3 (B) √2

(C) 2 (D) 3

The common difference of the A. P. √2] √8] √18] √32] --- is

(A) √3 (B) √2

(C) 2 (D) 3
9

19. ;fn x + 2, 3x vkSj 4x + 1 l0 Js0 esa gS rks x dk eku gS

(A) 2 (B) 4

(C) 3 (D) 5

If x + 2, 3x and 4x + 1 are in A.P. then the value of x is

(A) 2 (B) 4

(C) 3 (D) 5

20. tan300.sin300.cot600.cosec300 =

(A) (B)

(C) (D) 1

21. l0 Js0 % 10] 7] 4] --- dk 10ok¡ in gS

(A) -10 (B) -15

(C) -17 (D) -20

The 10th term of an A.P. : 10, 7, 4, … is

(A) -10 (B) -15

(C) -17 (D) -20

22. l0 Js0 % 2] 7] 12] --- dk dkSu lk in 47 gS \

(A) 8ok¡ (B) 10ok¡

(C) 12ok¡ (D) 14ok¡

Which term of the A.P. : 2, 7, 12, … is 47 ?

10

(A) 8th (B) 10th

(C) 12th (D) 14th

23. izFke 10 fo"ke izkd`r la[;kvksa dk ;ksxQy gS

(A) 50 (B) 100

(C) 150 (D) 200

The sum of first 10 odd natural numbers is

(A) 50 (B) 100

(C) 150 (D) 200

24. ;fn fdlh l0 Js0 dk nok¡ in tn = 2n + 5 rks t4 =

(A) 9 (B) 11

(C) 13 (D) 15

If the nth term of an A.P. be tn = 2n + 5 then t4 =

(A) 9 (B) 11

(C) 13 (D) 15

( )
25. p(x) = x5 + 1 esa g(x) = x + 1 ls Hkkx nsus ij dk ?kkr gS
( )

(A) 3 (B) 4

(C) 5 (D) 2

( )
On dividing p(x) = x5 + 1 by g(x) = x + 1 then the degree of is

( )

(A) 3 (B) 4

11

(C) 5 (D) 2

26. ;fn cgqin p(x) dk ,d 'kwU;d &3 gks] rks fuEufyf[kr esa dkSu p(x) dk ,d xq.ku[k.M

gksxk \

(A) x + 3 (b) x - 3

(C) x + √3 (D) x - √3

If one zero of the polynomial p(x) is -3 then which of the following will be a

factor of p(x) ?

(A) x + 3 (b) x - 3

(C) x + √3 (D) x - √3

27. 2x – 5y = 7 dk ,d gy gS

(A) x = -6, y = 1 (B) x = 3, y = 1

(C) x = -1, y = 1 (D) x = 6, y = 1

One solution of 2x – 5y = 7 is

(A) x = -6, y = 1 (B) x = 3, y = 1

(C) x = -1, y = 1 (D) x = 6, y = 1

28. fuEufyf[kr esa dkSu fcUnq x = 4 ds vkys[k ij gS \

(A) (4, 0) (B) (4, 3)

(C) (4, 4) (D) buesa lHkh

Which of the following point lies on the graph of x = 4 ?

(A) (4, 0) (B) (4, 3)

12

(C) (4, 4) (D) all of these

29. l0 Js0 % 2] 6] 10] 14] --- ds fy, 𝑎20 – 𝑎10 dk eku gS

(A) 50 (B) 40

(C) 20 (D) 10

The value of 𝑎20 – 𝑎10 for A.P. : 2, 6, 10, 14, … is

(A) 50 (B) 40

(C) 20 (D) 10

30. lehdj.kksa 5x -15y = 8 rFkk

3x – 9y = ds vkys[k ,slh nks js[kk,¡ gksxh] tks

(A) laikrh gS

(B) lekarj gS

(C) ijLij ,d fcUnq ij izfrPNsn djrh gSA

(D) buesa ls dksbZ ugha

The graph of the equations 5x -15y = 8 and

3x – 9y = are two lines which are

(A) coincident

(B) parallel

(C) intersecting each other at one point

(D) none of these

13

31. ;fn nks le:Ik f=Hkqtksa ds {ks=Qyksa dk vuqikr 49 % 81 gS] rks buds laxr Hkqtkvksa dk

vuqikr gS

(A) 9 : 7 (B) 7 : 9

(C) 7 : 8 (D) 9 : 8

If the ratio of areas of two similar triangles is 49 : 81 then the ratio of their

corresponding sides is

(A) 9 : 7 (B) 7 : 9

(C) 7 : 8 (D) 9 : 8

32. nks le:Ik f=Hkqtksa dh laxr Hkqtk,¡ 4 % 5 ds vuqikr esa gS bu f=Hkqtksa ds {ks=Qyksa dk

vuqikr gS

(A) 25 : 16 (B) 8: 25

(C) 16 : 25 (D) 4 : 5

The ratio of corresponding sides of two similar triangles is 4 : 5 then ratio of

their areas is

(A) 25 : 16 (B) 8: 25

(C) 16 : 25 (D) 4 : 5

33. ,d leprqHkqZt ds fod.kZ 24 lseh vkSj 32 lseh gSA bldh izR;sd Hkqtk fuEufyf[kr esa

fdlds cjkcj gS

(A) 10 lseh (B) 20 lseh

(C) 30 lseh (D) 35 lseh

14

The diagonals of a rhombus are 24 cm and 32 cm. its each side is equal to

which of the following ?

(A) 10 cm (B) 20 cm

(C) 30 cm (D) 35 cm

34. tan2300 + cot2600 =

(A) (B)

(C) (D)
√ √

35. ∆ABC rFkk ∆DEF esa = ;s nksuksa f=Hkqt le:Ik gksax]s ;fn

(A)  A =  D (B)  B =  E

(C)  C =  D (D)  B =  F

If = in ∆ABC and ∆DEF both of these triangles will be similar if

(A)  A =  D (B)  B =  E

(C)  C =  D (D)  B =  F
36. ;fn ∆ABC~∆PQR vkSj {ks0 ∆ABC % {ks0 ∆PQR ¾ 121 % 169 rks BC : QR dk eku

gS

(A) 121 : 169 (B) 15 : 19

(C) 11 : 13 (D) 13 : 11

If ∆ABC~∆PQR and ar ∆ABC : ar ∆PQR = 121 : 169 then the value of BC :

QR is

15

(A) 121 : 169 (B) 15 : 19

(C) 11 : 13 (D) 13 : 11

37. sin2𝜃 + cos2𝜃 =

(A) 0 (B) 1

(C) -1 (D) 2

38. ;fn ∆ABC~∆PQR,  B=470,  R=830 rks  A=

(A) 500 (B) 600

(C) 700 (D) 800

If ∆ABC~∆PQR,  B=470,  R=830 then  A=

(A) 500 (B) 600

(C) 700 (D) 800

39. ∆ABC~∆PQR vkSj 2AB = PQ rFkk BC = 8 lseh rks QR =

(A) 12 lseh (B) 14 lseh

(C)16 lseh (D) 18 lseh

∆ABC~∆PQR and 2AB = PQ and BC = 8 cm then QR =

(A) 12 cm (B) 14 cm

(C) 16 cm (D) 18 cm

40. ∆ABC esa ADBC, AD = DC = 8 lseh] BC = 14 lseh rks AB dk eku gS

(A) 15 lseh (B) 16 lseh

(C) 10 lseh (D) 19 lseh

16

In ∆ABC, ADBC, AD = DC = 8 cm, BC = 14 cm then the value of AB is

(A) 15 cm (B) 16 cm

(C) 10 cm (D) 19 cm

41. fdlh o`Ùk dh ifjf/k mlds O;kl ls 9 lseh vf/kd gSA o`Ùk dh f=T;k gS

(A) 1-1 lseh (B) 2-1 lseh

(C) 2-5 lseh (D) 3-1 lseh

The circumference of a circle exceeds its diameter by 9 cm. The radius of

the circle is

(A) 1.1 cm (B) 2.1 cm

(C) 2.5 cm (D) 3.1 cm

42. nks o`Ùkksa ds {ks=Qyksa dk varj 100 lseh2 rFkk mudh ifjf/k;ksa dk varj 20 lseh gSA mudh

f=T;kvksa dk ;ksx gS

(A) 10𝜋 lseh (B) 20𝜋 lseh

(C) 10 lseh (D) 10√2 lseh

The difference of areas of two circles is 100cm2 and the difference of their

perimeters is 20 cm. The sum of their radii is

(A) 10𝜋 cm (B) 20𝜋 cm

(C) 10 cm (D) 10√2 cm

17

43. ;fn fcUnq (5, P) rFkk (2, 0) ds chp dh nwjh 5 gks] rks P cjkcj gS

(A) ±3 (B) ±4

(C) ±5 (D) ±6

If the distance between the points (5, P) and (2, 0) is 5, then P is equal to

(A) ±3 (B) ±4

(C) ±5 (D) ±6

44. nks o`Ùkksa ds {ks=Qyksa dk vuqikr 36 % 49 gS] mudh f=T;kvksa dk vuqikr gS

(A) 6 : 8 (B) 6 : 7

(C) 7 : 6 (D) 8 : 7

The areas of two circles are in the ratio 36 : 49. The ratio of their radii is

(A) 6 : 8 (B) 6 : 7

(C) 7 : 6 (D) 8 : 7

45. 7 lseh f=T;k okys ,d o`Ùk dss fcUnq P ls 24 lseh yEch Li’kZ js[kk PT [khph tkrh gSA

;fn O o`Ùk dk dsUnz gS rks OT dh yackbZ gS

(A) 20 lseh (B) 25 lseh

(C) 30 lseh (D) 35 lseh

In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT=

24 cm. If O is the centre of the circle, then length of OT is

18

(A) 20 cm (B) 25 cm

(C) 30 cm (D) 35 cm

46. nks o`Ùkksa dh f=T;kvksa dk vuqikr 2 % 3 gS rks muds {ks=Qyksa dk vuqikr gS

(A) 2 : 3 (B) 4 : 9

(C) 3 : 2 (D) 9 : 4

The ratio of the radii of two circles is 2 : 3 then the ratio of their areas is

(A) 2 : 3 (B) 4 : 9

(C) 3 : 2 (D) 9 : 4

47. Sin520 – Cos380 =

(A) 0 (B) 1

√
(C) -1 (D)

48. fdlh o`Ùk ds dsUnz ls 13 lseh nwj fLFkr fcUnq P ls [khph xbZ Li’kZ js[kk dh yackbZ 12

lseh gS] rks o`Ùk dh f=T;k gS

(A) 6 lseh (B) 12 lseh

(C) 5 lseh (D) 13 lseh

The length of a tangent from a point P at a distance 13 cm from the centre

of a circle is 12 cm, then the radius of the circle is

(A) 6 cm (B) 12 cm

(C) 5 cm (D) 13 cm

19

49. ,d ?kM+h dh feuV dh lwbZ 14 lseh yach gS blds }kjk 15 feuV esa jfpr {ks=Qy gS

(A) 144 lseh2 (B) 154 lseh2

(C) 164 lseh2 (D) 140 lseh2

The minute needle of a watch is 14 cm long. The area covered by it in 15

minutes is

(A) 144 cm2 (B) 154 cm2

(C) 164 cm2 (D) 140 cm2

50. Sec2300 – tan2300 =

(A) 0 (B) 1

(C) -1 (D) 2

51. ∆PQR esa QR dks S fcUnq rd c<+k;k x;k gS ftlls PRS = 1500 rFkk PQR =

800 rks QPR dk eku gS

(A) 500 (B) 700

(C) 800 (D) 1500

In ∆PQR, side QR is extended up to a point S such that PRS = 1500 and

PQR = 800, then the value of QPR is

(A) 500 (B) 700

(C) 800 (D) 1500

52. r f=T;k okys o`Ùk ds fy, 𝜃 dks.k okys f=T;[k.M dss pki dh yEckbZ gS

20

(A) x 𝜋 r2 (B) x 2𝜋r

(C) x 𝜋r (D) x 2𝜋r2

For a circle of radius r the length of an arc of a sector of angle 𝜃 is

(A) x 𝜋r2 (B) x 2𝜋r

(C) x 𝜋r (D) x 2𝜋r2

53. ;fn A = 450 rks sinA + cosA + tanA dk eku gS

(A) √2 (B) √2 + 1

√
(C) (D) 2 + √2
√

If A = 450 then the value of sinA + cosA + tanA is

(A) √2 (B) √2 + 1

√
(C) (D) 2 + √2
√

54. ;fn fdlh o`Ùk dh f=T;k nqxquk dj fn;k tk, rks u, o`Ùk vkSj iqjkus o`Ùk ds {ks=Qyksa dk

vuqikr gS

(A) 1 % 2 (B) 1 % 4

(C) 4 % 1 (D) 2 % 1

If the radius of a circle is doubled then the ratio of the areas of the new

circle and the old circle is

(A) 1 % 2 (B) 1 % 4

21

(C) 4 % 1 (D) 2 % 1

55. nks o`Ùkksa dh ifjf/k;ksa dk vuqikr 2 % 3 gS rks muds {ks=Qyksa dk vuqikr gS

(A) 4 : 9 (B) 2 : 3

(C) 3 : 2 (D) 9 : 4

The circumferences of two circles are in the ratio 2 : 3 then the ratio of their

areas is

(A) 4 : 9 (B) 2 : 3

(C) 3 : 2 (D) 9 : 4

56. 9sec2A – 9tan2A =

(A) 3 (B) 9

(C) (D) 0

57. ;fn ,d o`Ùk dh ifjf/k 2𝜋 ls c<+dj 8𝜋 gks tkrk gS rks bldk u;k {ks=Qy gksxk

(A) pkj xquk (B) vkB xquk

(C) lksyg xquk (D) buesa ls dksbZ ugha

If the circumference of a circle increases from 2𝜋 to 8𝜋 then its new area

will be

(A) four times (B) eight times

(C) sixteen times (D) none of these

22

58. fdlh f=Hkqt ds 'kh"kZ fcUnqvksa ds funsZ’kkad ¼4] 7½ ¼0] 5½ vkSj ¼5] 3½ gSA blds dsUnzd ds

funsZ’kkad gS

(A) (3, 4) (B) (3, 5)

(C) (5, 3) (D) (4, 7)

The co-ordinates of the vertices of a triangle are (4, 7) (0, 5) and (5, 3).

The co-ordinates of its centroid are

(A) (3, 4) (B) (3, 5)

(C) (5, 3) (D) (4, 7)

59. fcUnq ¼&1] &2½ fdl ikn esa fLFkr gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant is the point (-1, -2) situated ?

(A) first (B) second

(C) third (D) fourth

60. ;fn fdlh js[kk[k.M ds ,d Nksj ds funsZ’kkad ¼4] &5½ vkSj e/; fcUnq ¼0] 0½ gks] rks nwljs

Nksj ds funsZ’kkad gS -

(A) (-4, -5) (B) (-4, 5)

(C) (4, 5) (D) (5, 4)

If the co-ordinates of one end of a line segment are (4, -5) and its middle

point is (0, 0) then the co-ordinates of the other end are

23

(A) (-4, -5) (B) (-4, 5)

(C) (4, 5) (D) (5, 4)

61. ;fn A = 300 rFkk B = 600 rks sinA.cosB + cosA.sinB dk eku gS

(A) - 1 (B) 1

√
(C) (D)

If A = 300 and B = 600 then the value of sinA.cosB + cosA.sinB is

(A) - 1 (B) 1

√
(C) (D)

62. ;fn O(0,0) ewy fcUnq gks vkSj fcUnq P ds funsZ’kkad ¼&3] &4½ gks rks nwjh OP gS

(A) 3 bdkb;k¡ (B) 4 bdkb;k¡

(C) 5 bdkb;k¡ (D) 2 bdkb;k¡

If O(0,0) be the origin and the co-ordinates of point P be (-3, -4) then the

distance OP is

(A) 3 units (B) 4 units

(C) 5 units (D) 2 units

63. x&v{k ij fdlh fcUnq ds funsZ’kkad dk :Ik gksrk gS

(A) (x, o) (B) (o, x)

(C) (o, y) (D) buesa ls dksbZ ugha

A point on x-axis has co-ordinates of the form

24

(A) (x, o) (B) (o, x)

(C) (o, y) (D) none of these

64. fcUnq (10sin600, 0) vkSj (0, 10sin300) ds chp dh nwjh gS

(A) 6 bdkb;k¡ (B) 8 bdkb;k¡

(C) 10 bdkb;k¡ (D) 12 bdkb;k¡

The distance between the points (10sin600, 0) and (0, 10sin300) is

(A) 6 units (B) 8 units

(C) 10 units (D) 12 units

65. ;fn fcUnq,¡ (0, 0), (1, 2) rFkk (x, y) lajs[k gS] rks

(A) x = y (B) 2x = y

(C) x = 2y (D) 2x = -y

If the points (0, 0), (1, 2) and (x, y) are collinear then

(A) x = y (B) 2x = y

(C) x = 2y (D) 2x = -y

66. A(0, 1), B(0, 5) rFkk C(3, 4) ls cus f=Hkqt dk {ks=Qy gS

(A) 16 oxZ bdkbZ (B) 8 oxZ bdkbZ

(C) 6 oxZ bdkbZ (D) 4 oxZ bdkbZ

The area of the triangle formed by A(0, 1), B(0, 5) and C(3, 4) is

(A) 16 sq unit (B) 8 sq unit

(C)6 sq unit (D) 4 sq unit

25

67. fuEufyf[kr esa fdlesa nksuksa fod.kZ cjkcj gksrs gSa \

(A) leprqHkqZt (B) lekarj prqHkqZt

(C) vk;r (D) buesa ls dksbZ ugha

In which of the following both the diagonals are equal ?

(A) rhombus (B) parallelogram

(C) rectangle (D) none of these

68. 2x + 4y = 10, 3x + 6y = 12 }kjk fu:fir js[kk,¡ dSlh gksxh \

(A) izfrPNsnh (B) laikrh

(D) lekarj (D) buesa ls dksbZ ugha

The lines represented by 2x + 4y = 10, 3x + 6y = 12 are of which type?

(A) intersecting (B) coincident

(C) parallel (D) none of these

69. ;fn fdlh ?ku ds ,d fod.kZ dh yackbZ 8√3 lseh gS] rks bldk izR;sd fdukjk gS

(A) 8√3 lseh (B) 8 lseh

(C) 4√3 lseh (D) √3 lseh

If the length of a diagonal of a cube is 8√3 cm then its each edge is

(A) 8√3 cm (B) 8 cm

(C) 4√3 cm (D) √3 cm

26

70. ,d ?ku dk i`"B&{ks=Qy 96 lseh2 gS rks bldk izR;sd fdukjk gS

(A) 4 lseh (B) 5 lseh

(C) 6 lseh (D) 16 lseh

The surface area of a cube is 96cm2 then its each edge is

(A) 4 cm (B) 5 cm

(C) 6 cm (D) 16 cm

71. xksys dk i`"Bh; {ks=Qy gksrk gS

(A) 2𝜋r2 (B) 3𝜋r2

(C) 4𝜋r2 (D) 𝜋r2

The surface area of a sphere is

(A) 2𝜋r2 (B) 3𝜋r2

(C) 4𝜋r2 (D) 𝜋r2

72. nks v/kZxksys ds vk;ruksa dk vuqikr 27 % 64 gS rks muds oØ i`"B ds {ks=Qyksa dk vuqikr

gS

(A) 2 % 3 (B) 9 % 16

(C) 16 % 9 (D) 3 % 2

The ratio of volumes of two hemisphere is 27 : 64 then the ratio of their

curved surface areas is

(A) 2 % 3 (B) 9 % 16

(C) 16 % 9 (D) 3 % 2
27

73. nks csyuksa dh f=T;kvksa dk vuqikr 3 % 2 vkSj mudh Å¡pkb;ksa dk vuqikr 5 % 7 gS] rks

mudss vk;ruksa dk vuqikr gS

(A) 28 : 45 (B) 45 : 28

(C) 15 : 14 (D) 14 : 15

The radii of two cylinders are in the ratio 3 : 2 and their heights are in the

ratio 5 : 7 then the ratio of their volumes is

(A) 28 : 45 (B) 45 : 28

(C) 15 : 14 (D) 14 : 15

74. sin2𝜃(1 + cot2𝜃) =

(A) cos2𝜃 (B) sec2𝜃

(C)1 (D) 0
75. ;fn fdlh 'kadq esa Å¡pkbZ h ¾ 12 eh0 vkSj f=T;k r ¾ 5 eh0 gks rks fr;Zd Å¡pkbZ 𝑙 dk

eku gS

(A) 11 eh0 (B) 12 eh0

(C) 13 eh0 (D) 14 eh0

If in any cone height h = 12 m and radius r = 5 m then the value of slant

height 𝑙 is

(A) 11 m (B) 12 m

(C) 13 m (D) 14 m

28

76. fdlh yaco`Ùkh; csyu ds vk/kkj dk {ks=Qy ;fn 𝜋𝑎2 lseh2 gks vkSj Å¡pkbZ b lseh gks] rks

mlds oØ i`"B dk {ks=Qy gS

(A) 2𝜋𝑎 2 lseh2 (B) 2𝜋𝑎b lseh2

(C) 2𝜋𝑎2b lseh2 (D) buesa ls dksbZ ugha

If the area of a base of a right circular cylinder be 𝜋𝑎2 cm2 and height is b

cm then the area of its curved surface is

(A) 2𝜋𝑎2 cm2 (B) 2𝜋𝑎b cm2

(C) 2𝜋𝑎2b cm2 (D) none of these

77. ;fn 𝑥tan450 x cos600 = sin600 x cot600 rks 𝑥 cjkcj gS

(A) √3 (B)

(C) 1 (D)
√

If 𝑥tan450 x cos600 = sin600 x cot600 then 𝑥 is equal to

(A) √3 (B)

(C) 1 (D)
√

78. ;fn cot3x = sin450.cos450 + cos600 rks x dk eku gS

(A) 300 (B) 450

(C)150 (D) 200

If cot3x = sin450.cos450 + cos600 then the value of x is

(A) 300 (B) 450

29

(C)150 (D) 200

79. ;fn cos𝜃 + sec𝜃 = 4 rks cos2𝜃 + sec2𝜃 dk eku gS

(A) 12 (B) 13

(C) 14 (D) 16

If cos𝜃 + sec𝜃 = 4 then the value of cos2𝜃 + sec2𝜃 is

(A) 12 (B) 13

(C) 14 (D) 16

80. ;fn secA + tanA = 2𝑎 vkSj secA – tanA = b rks 𝑎b dk eku gS

(A) 1 (B) -1

(C) (D)

If secA + tanA = 2𝑎 and secA – tanA = b then the value of 𝑎b is

(A) 1 (B) -1

(C) (D)

81. sec600 =

(A) 1 (B) 2

(C) (D) 0

82. ;fn A + B = 900 rks cotA cjkcj gS

(A) sinB (B) cosB

(C) tanB (D) secB

30

If A + B = 900 then cotA is equal to

(A) sinB (B) cosB

(C) tanB (D) secB

83. ;fn cos𝜃 = rks cosec𝜃 cjkcj gS

(A) (B)

(C) (D)

If cos𝜃 = then cosec𝜃 is equal to

(A) (B)

(C) (D)

84. ;fn 4tan𝜃 = 3 rks cos2𝜃 + sin2𝜃 dk eku gS

(A) -1 (B) 1

(C) 2 (D)

If 4tan𝜃 = 3 then the value of cos2𝜃 + sin2𝜃 is

(A) -1 (B) 1

(C) 2 (D)

85. cot2𝜃(sec𝜃 – 1) (sec𝜃 + 1) = x rks x dk eku gS

(A) -1 (B) 1

(C) 2 (D) -2

31

cot2𝜃(sec𝜃 – 1) (sec𝜃 + 1) = x then the value of x is

(A) -1 (B) 1

(C) 2 (D) -2

86. fdlh ehukj dh Å¡pkbZ 35 eh0 gSA tc lw;Z dk mUu;u dks.k 450 gks] rks ehukj dh Nk;k

dh yackbZ tehu ij D;k gksxk \

(A) 30 eh0 (B) 35√3 eh0

(C) 35 eh0 (D) 30√3 eh0

The height of a tower is 35 m. when the angle of elevation of the sun is 45 0,

what will be the length of the tower’s shadow on the ground ?

(A) 30 m (B) 35√3 m

(C) 35 m (D) 30√3 m

87. ;fn fdlh m/okZ/kj LraHk dh Å¡pkbZ tehu ij ml LraHk dh Nk;k dh yackbZ dh xquh
√

gS rks lw;Z dk mUu;u dks.k gksxk

(A) 300 (B) 450

(C) 600 (D) 750

If the height of a vertical pole is times the length of its shadow on the
√

ground then the angle of elevation of the sun is

(A) 300 (B) 450

(C) 600 (D) 750
32

88. (cosec𝜃 – sin𝜃) (sec𝜃 – cos𝜃) (tan𝜃 + cot𝜃) =

(A) 0 (B) 1

(C) -1 (D) 2

89. dk eku gS

(A) 1 (B) -1

(C) 2 (D) -2

The value of is

(A) 1 (B) -1

(C) 2 (D) -2

90. ;fn sin500 + cos400 = 2sinA rks A dk eku gS

(A) 400 (B) 500

(C) 900 (D) 300

If sin500 + cos400 = 2sinA then the value of A is

(A) 400 (B) 500

(C) 900 (D) 300

91. ;fn 2𝜃 = rks sec𝜃 dk eku gksxk

(A) (B) √2
√

√
(C) (D)

33

If 2𝜃 = then the value of sec𝜃 will be

(A) (B) √2
√

√
(C) (D)

92. + 𝑠𝑖𝑛 59 + cos59 .sin31 =

(A) 3 (B) 2

(C) 1 (D) 0

93. ,d o"kZ esa ,d eghuk ;n`PN;k pquk tkrk gSA blds vizSy ;k twu gksus dh izkf;drk

(A) (B)

(C) (D)

A month is selected at random in a year. The probability of it being April or

June is

(A) (B)

(C) (D)

94. ,d ikls dh Qsd
a esa la[;k 4 ;k 5 vkus dh izkf;drk gS

(A) (B)

(C) (D)

34

The probability of getting a number 4 or 5 in throwing a die is

(A) (B)

(C) (D)

95. ,d ikls dks Qsd
a us ij vad 6 ugha vkus dh izkf;drk gS

(A) (B)

(C) (D) buesa ls dksbZ ugha

In throwing a die, the probability of not getting 6 is

(A) (B)

(C) (D) none of these

96. pkj flDdksa dh mNky esa laHko ifj.kkeksa dh la[;k gS

(A) 4 (B) 24

(C) 43 (D) buesa ls dksbZ ugha

The number of possible outcomes in tossing four coins is

(A) 4 (B) 24

(C) 43 (D) none of these

97. ,d ikls dh ,d Qsd esa 3 vkSj 6 ds chp dk vad izkIr gksus dh izkf;drk gS

(A) 0 (B)

(C) (D) 1

35

The probability of getting a digit between 3 and 6 in throwing a die once is

(A) 0 (B)

(C) (D) 1

98. 4, 5, 6, 7, 7, 5, 4, 5 dk cgqyd D;k gS \

(A) 4 (B) 5

(C) 6 (D) 7

What is the mode of 4, 5, 6, 7, 7, 5, 4, 5 ?

(A) 4 (B) 5

(C) 6 (D) 7

99. 18] 13] 17] 12] 16] 19 dh ekf/;dk gS

(A) 15 (B) 16

(C) 16.5 (D) 17

The median of 18, 13, 17, 12, 16, 19 is

(A) 15 (B) 16

(C) 16.5 (D) 17

100. ;fn 1, 4, x, 5, y dk ek/; 7 gS rks x vkSj y esa lac/a k gS

(A) x + y = 15 (B) x + y = 20

(C) x + y = 25 (D) x – y = 25

If the mean of 1, 4, x, 5, y is 7 then the relation between x and y is

36

(A) x + y = 15 (B) x + y = 20

(C) x + y = 25 (D) x – y = 25

[k.M&c @ SECTION-B

y?kq mÙkjh; iz’u @ Short Answer Type Questions

iz'u la[;k 1 ls 30 rd y?kq mÙkjh; iz’u gSaA buesa ls fdUgha 15 iz’uksa ds mÙkj nsAa izR;sd iz’u

ds fy, 2 vad fu/kkZfjr gSA 15x2=30

Question Nos. 1 to 30 are Short Answer Type Questions. Answer any 15

questions. Each question carries 2 marks. 15x2=30

1- f}?kkr cgqin 2x2 – 8x + 6 ds 'kwU;dksa dk ;ksx ,oa xq.kuQy Kkr djsAa 2

Find the sum and product of the zeros of the quadratic polynomial

2x2 – 8x + 6.

2- f}?kkr cgqin 4x2 - 3 ds 'kwU;d Kkr djsAa 2

Find the zeros of the quadratic polynomial 4x2 – 3.

3- ;fn 3x + ay = 1 vkSj bx – 2y = 3 dk gy x = 1, y = 2 gks rks a vkSj b dk eku

Kkr djsaA 2

If the solution of 3x + ay = 1 and bx – 2y = 3 is x = 1, y = 2 then find the

values of a and b.

4- 4x2 + 5x + 2 dks x + 3 ls Hkkx nsaA 2

37

Divide 4x2 + 5x + 2 by x + 3.

5- ;wfDyM foHkktu ,Yxksfjn~e dk iz;ksx dj 135 vkSj 225 dk e0 l0 Kkr djsaA 2

Using Euclid’s division algorithm find the H.C.F. of 135 and 225.

6- f}?kkr lehdj.k 3y2 – 2y + = 0 ds foospd Kkr djsa ,oa ewyksa dh izd`fr crk,aA 2

Find the discriminant of the quadratic equation 3y2 – 2y + = 0 and find

the nature of roots.

7- 2 $ 7 $ 12 $ --- $ 122 dk ;ksxQy Kkr djsAa 2

Find the sum of 2 + 7 + 12 + … + 122.

8- fl) djsa fd ¾1- 2

Prove that ¾1-

9- ∆ABC esa DE𝐼𝐼BC, AD = 1.5 lseh] DB = 3 lseh] AE = 1 lseh gks rks EC dk

eku Kkr djsAa 2

In ∆ABC DE𝐼𝐼BC, AD = 1.5 cm, DB = 3 cm, AE = 1 cm then find the

value of EC.

10- ,d f=Hkqt ABC dh Hkqtk BC ij ,d fcUnq D bl izdkj fLFkr gS fd ADC =

BAC gS] fl) djsa fd CA2 = CB.CD. 2

38

D is a point on the side BC of a triangle ABC such that ADC = BAC,

Prove that CA2 = CB.CD.

11- vuqikrksa , ,oa dh rqyuk dj Kkr djsa fd fuEukafdr ;qXe lehdj.k laxr gS

;k vlaxr % 2

x + 2y + 3 = 0, 3x + 6y + 9 = 0.

By comparing the ratios , and find out whether the following pair

of equations are consistent or inconsistent :

x +2y + 3 = 0, 3x + 6y + 9 = 0.

12- fdlh o`Ùk dk dsUnz ¼2] &3½ gS vkSj mlds O;kl AB dk ,d Nksj B(1, 4) gS rks A ds

funsZ’kkad ,oa o`Ùk dh f=T;k Kkr djsAa 2

The centre of a circle is (2, -3) and one end of its diameter AB is B(1, 4).

Find the coordinates of A and the radius of the circle.

13- f=Hkqt dk {ks=Qy Kkr djsa ftlds 'kh"kksZa ds funsZ’kkad ¼3] 4½ ¼&4] 3½ vkSj ¼8] 6½ gSA 2

Find the area of the triangle, the co-ordianates of whose vertices are

(3, 4), (-4, 3) and (8, 6).

14- fcUnqvksa ¼4] 6½ vkSj ¼2] 4½ dks feykusokyh js[kk[k.M ds e/;&fcUnq dh nwjh fcUnq ¼8]6½

ls Kkr djsAa 2

39

Find the distance of the mid-point of the line segment joining the points

(4, 6) and (2, 4) from the point (8, 6).

15- 8 lseh f=T;k okys ,d Bksl xksys ls 1 lseh f=T;k okyh fdruh Bksl xksykdkj

xksfy;k¡ cukbZ tk ldrh gS \ 2

How many spherical solid bullets each of radius 1 cm can be made from

a solid sphere of radius 8 cm ?

16- ,d v)Zo`Ùkkdkj [ksr dh ifjfefr fudkysa ;fn bldk O;kl 24 lseh gksA 2

Find the perimeter of a semi circular field if its diameter is 24 cm.

17- ,d o`Ùk dh ifjf/k 30-8 lseh gS] mldk {ks=Qy Kkr djsAa 2

The circumference of a circle is 30.8 cm find its area.

18- fl) djsa fd = tan𝜃. 2

Prove that = tan𝜃.

19- fl) djsa fd ¾ - 2

Prove that ¾ -

20- fl) djsa fd (1 + sin𝜃) (1 – sin𝜃) = . 2

Prove that (1 + sin𝜃) (1 – sin𝜃) = .

21- gy djsa 3x + 4y = 10, 2x- 2y = 2. 2

Solve 3x + 4y = 10, 2x- 2y = 2.

40

22- y-v{k ij ,d ,slk fcUnq Kkr djsa tks fcUnq,¡ ¼2] 3½ vkSj ¼&4] 1½ ls lenwjLFk gksA 2

Find a point on the y-axis which is equidistant from the points (2, 3) and

(-4, 1).

23- fl) djsa fd fcUnq ¼2] &2½] ¼&3] 8½ vkSj ¼&1] 4½ lajs[kh; gSaA 2

Prove that the points (2, -2), (-3, 8) and (-1, 4) are collinear.

24- 8-4 lseh f=T;k okys o`Ùk dh ifjf/k ,oa {ks=Qy Kkr djsAa 2

Find the circumference and area of a circle with radius 8.4 cm.

25- f=T;k 4 lseh okys ,d o`Ùk ds f=T;[k.M dk {ks=Qy Kkr djsa ftldk dsUnzh; dks.k

300 gSA 2

Find the area of the sector of a circle with radius 4 cm and of central

angle 300.

26- ∆PQR dh Hkqtk,¡ PR vkSj QR ij Øe’k% fcUnq S vkSj T fcUnq,¡ bl izdkj gS fd

P = RTS fl) djsa fd ∆RPQ~∆RTS. 2

S and T are points on sides PR and QR respectively of ∆PQR such that

P = RTS prove that ∆RPQ~∆RTS.

27- nks le:Ik f=Hkqtksa ds {ks=Qy 121 lseh2 vkSj 64 lseh2 gSA ;fn igys f=Hkqt dh

ekf/;dk 12-1 lseh gS] rks nwljs f=Hkqt dh laxr ekf/;dk Kkr djsAa 2

41

The areas of two similar triangle are 121 cm2 and 64 cm2. If the median

of first triangle is 12.1 cm, find the corresponding median of the other

triangle.

28- ;fn 30] 35] 20] 15 vkSj x dk lekUrj ek/; 24 gks rks x dk eku Kkr djsAa 2

If the arithmetic mean of 30, 35, 20, 15 and x is 24 then find the value of

x.

29- nks vadksa dh la[;k esa ngkbZ dk vad bdkbZ ds vad dk frxquk gSA ;fn ml la[;k esa

ls 54 ?kVk nas rks mlds vad cny tk;sxsAa bu dFkuksa ds lehdj.k fy[ksAa 2

In a two digit number, the ten’s digit is three times of unit’s digit. When

the number is decreased by 54, the digit are reversed. Write the

equations for these statements.

30- lehdj.k fudk; 3x – 5y = 4 ,oa 9x – 2y = 7 dks otzxq.ku fof/k ls gy djsAa 2

Solve the system of equations 3x – 5y = 4 and 9x – 2y = 7 by cross

multiplication method.

Long Answer Type Questions.

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; iz’u gSAa buesa ls fdUgha 4 iz’uksa ds mÙkj nsAa izR;sd iz’u ds

fy, 5 vad fu/kkZfjr gSA 4x5=20

Question Nos 31 to 38 are Long Answer Type Questions. Answer any 4

questions. Each question carries 5 marks. 4x5=20

42

31- jSf[kd lehdj.k ;qXe x – 4y + 14 = 0 rFkk 3x + 2y - 14 = 0 dk vkys[k [khpsa vkSj

gy djsaA 5

Draw the graphs of the pair of linear equations x – 4y + 14 = 0 and 3x +

2y - 14 = 0 and solve them.

32- vk;r ABCD ds vanj fLFkr O dksbZ fcUnq gS rks fl) djsa fd 5

OB2 + OD2 = OA2 + OC2.

Inside a rectangle ABCD, O is any point then Prove that

OB2 + OD2 = OA2 + OC2.

33- ,d eksVj oksV] ftldh fLFkj ty esa pky 18 fd0eh0@?kaVk gS] 24 fd0eh0 /kkjk ds

izfrdwy tkus esa] ogh nwjh /kkjk ds vuqdwy ykSVus dh vis{kk 1 ?kaVk vf/kd ysrh gSA

/kkjk dh pky Kkr djsaA 5

A motot boat whose speed is 18 km/h in still water takes 1 hour more to

go 24 km up stream than to return down stream to the same spot. Find

the speed of the stream.

34- ,d ehukj ds ikn ls xqtjus okyh lh/kh js[kk ij ikn ls Øe’k% 𝑎 ehVj rFkk b ehVj

dh nwfj;ksa ij fLFkr nks fcUnqvksa ls ehukj ds f’k[kj ds mUu;u dks.k ,d nwljs ds iwjd

gSaA fl) djsa fd ehukj dh Å¡pkbZ √𝑎𝑏 ehVj gSA 5

The angles of elevation of the top of a tower from two points on the

ground at distances 𝑎 metres and b metres from the base of the tower

43

and in the same straight line are complementary. Prove that the height

of the tower is √𝑎𝑏 metres.

35- ,d ledks.k f=Hkqt ABC vafdr djsa ftlesa AB = 8 lseh] BC = 6 lseh rFkk B =

900, B ls AC ij yac BD dk ikn D gSA ∆BCD dk ifjo`r [khpasA 5

Draw a right angled triangle ABC in which AB = 8cm, BC = 6 cm and B

= 900, D is the foot of perpendicular BD from B on AC. Draw circum

circle of ∆BCD.

36- fl) djsa fd + = 1 + sec𝜃.cosec𝜃. 5

Prove that + = 1 + sec𝜃.cosec𝜃.

37- fuEufyf[kr cVu dk cgqyd Kkr djsa % 5

izkIrkad 0&10 10&20 20&30 30&40 40&50 50&60 60&70 70&80

lap;h 25 40 60 75 95 130 195 245
ckjackjrk

Find the mode of following distribution :

Marks 0&10 10&20 20&30 30&40 40&50 50&60 60&70 70&80
obtained
Cumulative 25 40 60 75 95 130 195 245
frequency

44

38- /kkrq dh pknj ls cuk vkSj Åij ls [kqyk ,d crZu 'kadq ds ,d fNUud ds vkdkj dk

gS ftldh Å¡pkbZ 16 lseh gS rFkk fupys vkSj Åijh fljksa dh f=T;k,¡ Øe’k% 8 lseh

vkSj 20 lseh gS :0 50 izfr yhVj dh nj ls bl crZu dks iwjk Hkjus okys nw/k dk ewY;

Kkr djsaA ¼𝜋 ¾ 3-14½

A container opened from the top and made up of a metal sheet is in the

form of a frustum of a cone of height 16 cm with radii of its lower and

upper ends as 8 cm and 20 cm respectively. Find the cost of the milk

which can completely fill the container, at the rate of Rs. 50 per litre. ¼𝜋

¾ 3-14½

45

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