Bihar Board 12th Question Paper 2023 Maths (Commerce)

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BIHAR • EXAINATION BOARD • CLASS 12 •
QUESTION PAPER • PDF • DOWNLOAD

12TH INTER
CLASS Examination

Previous Year
YEAR
2023 QUESTION PAPER
View For Free. Download PDF Online at

SECONDARY SCHOOL EXAMINATION – 2023
(ANNUAL)
Sub. Code – 110
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
Candidate must enter his/her Question Booklet Serial No. (of 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, ij fn;s gq, vad iw.kkZad fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.

4- iz’uksa dks /;kuiwoZd i<+us ds fy, 15 feuV dk vfrfjDr le; fn;k x;k gSA
15 minutes of extra time 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 gSa] ftuesa ls fdUgha 50 iz’uksa dk mÙkj nsuk
vfuok;Z gSA 50 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, x;s OMR mÙkj i=d esa fn, x, lgh fodYi dks uhys@dkys
ckWy isu ls izxk<+ djsaA fdlh Hkh izdkj ds âkbVuj @ rjy inkFkZ @ CysM @
uk[kwu vkfn dk OMR mÙkj i=d 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 by the computer 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 gSA
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,

2

out of which any 4 questions are to be answered. Each question
carries 5 marks.

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 fpfUgr djsaA
50x1=50
Question No.s 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. ¼tgk¡ p, q iw.kkZad gSa] q≠0½ ds :Ik esa 0.41 dks fy[kk tk ldrk gS

(A) (B)

(C) (D)

In the form (where p, q are integers, q≠0), 0.41 can be expressed

as

(A) (B)

(C) (D)

2. fuEufyf[kr esa dkSu vifjes; la[;k gS \

(A) √9 (B) √8

(C) √16 (D) √25

Which of the following is an irrational number ?

(A) √9 (B) √8

4

(C) √16 (D) √25

3. cgqin 6x – 21 dk 'kwU;d gS

(A) (B) -

(C) (D) -

The zero of the polynomial 6x – 21 is

(A) (B) -

(C) (D) -

4. cgqin x2 – 10x + 25 dk ?kkr gS

(A) 1 (B) 2

(C) 0 (D) 25

Degree of the polynomial x2 – 10x + 25 is

(A) 1 (B) 2

(C) 0 (D) 25

5. fuEufyf[kr esa dkSu f}?kkr lehdj.k gS \

(A) x2 + 5√𝑥 + 3 = 7 (B) (x – 1)3 = x2

(C) (x + 1)2 = 2(x + 1) (D) (x + 1)3 = 9

Which of the following is a quadratic equation ?

(A) x2 + 5√𝑥 + 3 = 7 (B) (x – 1)3 = x2

(C) (x + 1)2 = 2(x + 1) (D) (x + 1)3 = 9

5

6. ;fn m rFkk n vHkkT; la[;k,¡ gS rks m3n2 vkSj m2n2 dk e0 l0 gksxk

(A) m3n2 (B) m2n

(C) m2n2 (D) mn

If m and n are prime numbers then the H.C.F. of m3n2 and m2n2 is

(A) m3n2 (B) m2n

(C) m2n2 (D) mn

7. ;fn a vkSj 18 dk y0 l0 36 gS rFkk a vkSj 18 dk e0 l0 2 gS rks a =

(A) 3 (B) 4

(C) 2 (D) 1

If the L.C.M. of a and 18 is 36 and the H.C.F. of a and 18 is 2 then

a=

(A) 3 (B) 4

(C) 2 (D) 1

8. nks vifjes; la[;kvksa 2 + √3 rFkk 2 – √3 dk xq.kuQy gS ,d

(A) ifjes; la[;k (B) vifjes; la[;k

(C) (A) ;k (B) (D) buesa dksbZ ugha

The product of two irrational numbers 2 + √3 and 2 – √3 is a/an

(A) rational number (B) irrational number

(C) either A or B (D) none of these

9. f}?kkr cgqin x2 + x – 20 ds 'kwU;d gSa
6

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

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

The zeroes of the polynomial x2 + x – 20 are

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

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

10. ;fn fdlh f}?kkr cgqin p(x) = 2x2 + 5x – 12 ds 'kwU;d 𝛼, 𝛽 gksa rks

𝛼+𝛽=

(A) (B) –

(C) (D)

If 𝛼, 𝛽 are zeros of any quadratic polynomial p(x) = 2x2 + 5x – 12

then 𝛼 + 𝛽 =

(A) (B) –

(C) (D)

11. ;fn 𝛼, 𝛽 cgqin x2 + 4x + 1 ds 'kwU;d gSa rks + ¾
𝛼

(A) 2 (B) 4

(C) -4 (D) 1

If 𝛼, 𝛽 are the zeroes of the polynomial x2 + 4x + 1 then + =

𝛼

(A) 2 (B) 4

7

(C) -4 (D) 1

12. ;fn lehdj.k Kx2 + 6x + 9K = 0 ds ewyksa dk ;ksx vkSj xq.kuQy leku gks rks

K dk eku gS

(A) (B) –

(C) (D) -

If the sum and product of roots of the equation Kx2 + 6x + 9K = 0 are

equal then the value of K is

(A) (B) –

(C) (D) -

13. ;fn x = 1 nksuksa lehdj.kksa x2 + x + a = 0 vkSj bx2 + bx + 3 = 0 ds ewy gksa rks

ab =

(A) 4 (B) 3

(C) -3 (D) -4

If x = 1 is a root of both the equations x2 + x + a = 0 and

bx2 + bx + 3 = 0 then ab =

(A) 4 (B) 3

(C) -3 (D) -4

14. ;fn 2x+y = 2x-y = 2 rks y dk eku gS

(A) (B)
8

(C) 0 (D) buesa ls dksbZ ugha

If 2x+y = 2x-y = 2 then the value of y is

(A) (B)

(C) 0 (D) none of these

15. cos𝜃 x cosec𝜃 =

(A) tan𝜃 (B) cot𝜃

(C) sec𝜃 (D) cos𝜃

16. ;fn cgqin x2 + ax – b ds 'kwU;d ,d nwljs ds O;qRØe gks rks b =

(A) 2 (B) 1

(C) -1 (D) a

If zeroes of the polynomial x2 + ax – b be reciprocal of each other

then b =

(A) 2 (B) 1

(C) -1 (D) a

17. ;fn lehdj.k ax2+bx+c=0 ds ewy leku gSa rks izR;sd ewy dk eku gS

(A)- (B) -

(C) (D)

If equation ax2+bx+c=0 has equal roots then the value of each root is

(B) - (B) -

9

(D) (D)

18. (x + 2) (x – 2) = 5 ds ewy fuEufyf[kr esa dkSu gSa \

(A) 2, -2 (B) 3, -3

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

Which of the following are the roots of (x + 2) (x – 2) = 5 ?

(A) 2, -2 (B) 3, -3

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

19. ;fn fdlh f}?kkr cgqin p(x) = x2+bx+c ds 'kwU;d 6 vkSj & 2 gks] rks b vkSj c

ds eku Øe’k% gSa

(A) 4, 12 (B) -4, 12

(C) 4, -12 (D) -4, -12

If 6 and -2 be the zeroes of a quadratic polynomial p(x) = x2+bx+c

then the values of b and c are respectively

(A) 4, 12 (B) -4, 12

(C) 4, -12 (D) -4, -12

20. fuEufyf[kr esa dkSu jSf[kd cgqin gS \

(A) 2𝑥 – 9√5 (B) 𝑥 +

(C) 5 - (D) 𝑧 2 -

Which of the following is a linear polynomial ?

10

(A) 2𝑥 – 9√5 (B) 𝑥 +

(C) 5 - (D) 𝑧 2 -

21. ,d f=?kkr cgqin ds 'kwU;dkas dh vf/kdre la[;k gS

(A) 1 (B) 2

(C) 3 (D) 4

The maximum number of zeroes of a cubic polynomial is

(A) 1 (B) 2

(C) 3 (D) 4

22. cgqin m3 – 3m2 – m + 3 ds 'kwU;dksa dk ;ksx gS

(A) -3 (B) 2

(C) 3 (D) 4

The sum of the zeroes of the polynomial m3 – 3m2 – m + 3 is

(A) -3 (B) 2

(C) 3 (D) 4

23. lehdj.k bx2+bx+c=0 dk foospd gS

(A) b2 – 4ac (B) c2 – 4bc

(C) b2 – 4bc (D) c2 – 4b2

The discriminant of the equation bx2+bx+c=0 is

(A) b2 – 4ac (B) c2 – 4bc

11

(C) b2 – 4bc (D) c2 – 4b2

24. ;fn x-y = 2 rFkk x+y = 10 gS rks

(A) x= 4, y = 2 (B) x = 5, y = 3

(C) x = 6, y = 4 (D) x = 7, y = 5

If x-y = 2 and x+y = 10 then

(A) x= 4, y = 2 (B) x = 5, y = 3

(C) x = 6, y = 4 (D) x = 7, y = 5

25. ;fn 5, a, b rFkk 11 lekarj Js<+h esa gSa rks a+b dk eku gS

(A) 7 (B) 9

(C) 11 (D) 16

If 5, a, b and 11 are in A. P. then the value of a+b is

(A) 7 (B) 9

(C) 11 (D) 16

26. lekarj Js<+h % , , , … dk lkoZvarj gS

(A) p (b) –p

(C) -1 (D) 1

The common difference of the A. P. : , , , … is

(A) p (b) –p

(C) -1 (D) 1

12

27. A. P. : 10, 7, 4, … dk 30ok¡ in gS

(A) 97 (B) 77

(C) -77 (D) -97

The 30th term of the A. P. : 10, 7, 4, … is

(A) 97 (B) 77

(C) -77 (D) -97

28. lekarj Js<+h % 5] 8] 11] --- dk dkSu lk in 383 gS \

(A) 124 (B) 125

(C) 126 (D) 127

Which term of A. P. : 5, 8, 11, … is 383 ?

(A) 124 (B) 125

(C) 126 (D) 127

√
29. fuEufyf[kr esa ls dkSu ds cjkcj gS \

(A) sin300 (B) sin600

(C) tan300 (D) cos600

√
Which of the following is equal to ?

(A) sin300 (B) sin600

(C) tan300 (D) cos600

30. ;fn tan𝜃 =1 rks 𝜃 dk eku D;k gksxk \

13

(A) 300 (B) 600

(C) 450 (D) 900

If tan𝜃 =1 then what will be the value of 𝜃 ?

(A) 300 (B) 600

(C) 450 (D) 900

31. sec2440 – cot2460 dk eku gksxk

(A) 0 (B) 1

(C) -1 (D) 2

The value of sec2440 – cot2460 will be

(A) 0 (B) 1

(C) -1 (D) 2

32. 2sec2530 – 2cot2370 =

(A) 0 (B) 1

(C) -2 (D) 2

33. cosec2350 – tan2550 dk eku gS

(A) 0 (B) 1

(C) -1 (D) 2

The value of cosec2350 – tan2550 is

(A) 0 (B) 1

(C) -1 (D) 2

14

34. ;fn sin𝜃 = cos𝜃 gks rks 𝜃 dk eku gS

(A) 00 (B) 300

(C) 450 (D) 600

If sin𝜃 = cos𝜃 then the value of 𝜃 is

(A) 00 (B) 300

(C) 450 (D) 600

35. tan𝜃 x cosec𝜃 x sec𝜃 =

(A) 1 (B) cos2𝜃

(C) sec2𝜃 (D) sin2𝜃

36. ;fn 4tan𝜃 = 3 gks rks (cos2𝜃 – sin2𝜃) dk eku gS

(A) (B)

(C) 1 (D) -

If 4tan𝜃 = 3 then the value of (cos2𝜃 – sin2𝜃) is

(A) (B)

(C) 1 (D) -

37. + =

(A) 0 (B) 1

(C) 2 (D) ½

38. nks flDdksa dh mNky esa laHko ifj.kkeksa dh la[;k gS
15

(A) 2 (B) 4

(C) 9 (D) buesa ls dksbZ ugha

The number of possible outcomes in tossing two coins is

(A) 2 (B) 4

(C) 9 (D) none of these

39. ,d ikls dks Qsadus ij vad 6 ugha vkus dh izkf;drk gS

(A) (B)

(C) (D)

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

(A) (B)

(C) (D)

40. ;fn P(E) = 0.07 rks P(E ) cjkcj gS

(A) 0.93 (B) 0.08

(C) 0.07 (D) 0.91

If P(E) = 0.07 then P(E ) is equal to

(A) 0.93 (B) 0.08

(C) 0.07 (D) 0.91

41. ,d FkSys esa 4 lQsn] 6 yky] 7 dkyh vkSj 3 uhyh xsan gSA ,d xsan ;kn`PN;k

fudkyh tkrh gSA blds lQsn gksus dh izkf;drk gS

16

(A) (B)

(C) (D)

A bag contains 4 white, 6 red, 7 black and 3 blue balls. One ball is

taken out at random. The probability that it is white is

(A) (B)

(C) (D)

42. ;fn sin𝜃 = gS rks cos𝜃 =

√
(A) (B)
√

(C) (D)
√

If sin𝜃 = then cos𝜃 =

√
(A) (B)
√

(C) (D)
√

43. ABC ,d ledks.k f=Hkqt gS ftlesa C=900 gS rks cos(A+B) dk eku gS

(A) 0 (B) 1

√
(C) ½ (D)

ABC is a right angled triangle in which C=900 then the value of

cos(A+B) is

17

(A) 0 (B) 1

√
(C) ½ (D)

44. (cosec𝜃 – cot𝜃)2 dk eku gksxk

(A) (B)

(C) (D)

The value of (cosec𝜃 – cot𝜃)2 will be

(A) (B)

(C) (D)

45. cos600.cos300 – sin600.sin300 dk eku gksxk

(A) 0 (B) 1

(C) -1 (D) 2

The value of cos600.cos300 – sin600.sin300 is

(A) 0 (B) 1

(C) -1 (D) 2

46. ,d ehukj ls 100eh nwj fLFkr fcUnq ij ehukj dk mUu;u dks.k 600 gS rks ehukj

dh Å¡pkbZ gS

(A) 100√3 eh (B) eh

√

(C) 50√3 eh (D) eh
√

18

The angle of elevation of a tower at a point 100m away from the

tower is 600 then the height of the tower is

(A) 100√3 m (B) m
√

(C) 50√3 m (D) m
√

47. ;fn 𝛼 = 𝛽 = 450 rks sin(𝛼 + 𝛽) dk eku gS

(A) (B) 1
√

(C) 0 (D)

If 𝛼 = 𝛽 = 450 then the value of sin(𝛼 + 𝛽) is

(A) (B) 1
√

(C) 0 (D)

48. ;fn √2 = 2sinA rks A dk eku gS

(A) 00 (B) 300

(C) 450 (D) 600

If √2 = 2sinA then the value of A is

(A) 00 (B) 300

(C) 450 (D) 600

49. dk eku gS

(A) 0 (B) 1
19

(C) -1 (D) 2

The value of is

(A) 0 (B) 1

(C) -1 (D) 2

50. 42 lseh O;kl okys o`Ùk dh ifjf/k gksxh

(A) 14 lseh (B) 21 lseh

(C) 132 lseh (D) 122 lseh

The circumference of a circle of diameter 42 cm will be

(A) 14 cm (B) 21 cm

(C) 132 cm (D) 122 cm

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

(A) 3:2 (B) 2:3

(C) 4:9 (D) 9:4

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

will be

(A) 3:2 (B) 2:3

(C) 4:9 (D) 9:4

52. ;fn ,d o`Ùk dh ifjf/k 8𝜋 gS rks bldk {ks=Qy gS

(A) 8𝜋 (B) 16𝜋

20

(C) 4𝜋 (D) 32𝜋

If the circumference of a circle is 8𝜋 then its area is

(A) 8𝜋 (B) 16𝜋

(C) 4𝜋 (D) 32𝜋

53. ,d ?kM+h dh feuV okyh lqbZ }kjk 35 feuV iwjk djus esa cuk;k x;k dks.k gS

(A) 450 (B) 600

(C) 900 (D) 2100

The angle constructed by the minute hand of a clock in completing 35

minutes is

(A) 450 (B) 600

(C) 900 (D) 2100

54. fcUnq ¼2] &5½ fdl prqFkkZa’k esa fLFkr gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant is the point (2, -5) located ?

(A) first (B) second

(C) third (D) fourth

55. y-v{k ls fcUnq ¼7] 3½ dh nwjh gS

(A) 3 (B) 7

21

(C) 0 (D) 10

The distance of the point (7, 3) from y-axis is

(A) 3 (B) 7

(C) 0 (D) 10

56. ewy fcUnq ls ¼&5] 5½ dh nwjh D;k gS \

(A) 5 (B) 5√2

(C) 10 (D) 10√2

What is the distance of the point (-5, 5) from the origin ?

(A) 5 (B) 5√2

(C) 10 (D) 10√2

57. fcUnqvksa P(2, 3) vkSj Q(4, 1) ds chp dh nwjh gS

(A) 2 (B) 2√2

(C) 4 (D) 4√2

The distance between the points P(2, 3) and Q(4, 1) is

(A) 2 (B) 2√2

(C) 4 (D) 4√2

58. cosec2𝜃 x tan2𝜃 x cos2𝜃 =

(A) 1 (B) -1

(C) 0 (D) 2

22

59. ,d gh f=T;k okys ,d xksys dk i`"B {ks=Qy rFkk ,d v)Zxksys ds dqy i`"B
{ks=Qy dk vuqikr gS
(A) 4:9 (B) 2:1
(C) 3:2 (D) 4:3
The ratio of the total surface area of a sphere and that of a
hemisphere of the same radius is
(A) 4:9 (B) 2:1
(C) 3:2 (D) 4:3
60. ;fn dksbZ fcUnq A y-v{k ds nk¡;h vksj 5 bdkbZ nwjh ij x-v{k ij fLFkr gks rks A
ds funsZ’kkad gksaxs
(A) (0, 5) (B) (-5, 0)

(C) (5, 0) (D) (0, -5)

If a point A is situated on the x-axis at a distance of 5 units on the

right side of y-axis then the co-ordinates of A will be

(A) (0, 5) (B) (-5, 0)

(C) (5, 0) (D) (0, -5)

61. + + + … 9 inksa rd dk ;ksx gS

(A) - (B)

(C) (D) buesa ls dksbZ ugha

The sum of + + + … up to 9 terms is

(A) - (B)

23

(C) (D) none of these

62. cos2A(1 + tan2A) =

(A) sin2A (B) cotA

(C) 1 (D) cosec2A

63. ;fn fcUnqvksa A(6, 5) vkSj B(4, y) dks feykus okys js[kk[k.M dk e/; fcUnq (x, 6)

gks rks y dk eku gS

(A) 5 (B) 6

(C) 8 (D) 7

If (x, 6) is the midpoint of the line segment joining the point A(6, 5)

and B(4, y) then the value of y is

(A) 5 (B) 6

(C) 8 (D) 7

64. fdlh f=Hkqt ds 'kh"kZ fcUnqvksa ds funsZ’kkad (4, 6) (0, 4) vkSj (5, 5) gSaA blds

dsUnzd ds funsZ’kkad gksaxs

(A) (5, 2) (B) (3, 5)

(C) (1, 3) (D) (4, 4)

The co-ordinates of the vertices of a triangle are (4, 6) (0, 4) and

(5, 5). The co-ordinates of its centroid will be

(A) (5, 2) (B) (3, 5)

(C) (1, 3) (D) (4, 4)

24

65. fdlh ?ku dk fdukjk nqxquk gks tkus ij u;s ?ku ds dqy i`"B {ks=Qy rFkk igys

okys ?ku ds dqy i`"B {ks=Qy dk vuqikr gS

(A) 2:1 (B) 4:1

(C) 6:1 (D) 8:1

If the edge of a cube is doubled then the ratio of total surface area of

the new cube to the total surface area of the old cube is

(A) 2:1 (B) 4:1

(C) 6:1 (D) 8:1

66. ;fn fdlh f=Hkqt ds 'kh"kksZa ds funsZ’kkad (a, 0), (0,0) ,oa (0, b) gks rks f=Hkqt dk

{ks=Qy gksxk

(A) ab (B) ab

(C) b2 (D) a2b2

If the co-ordinates of the vertices of a triangle are (a, 0), (0,0) and

(0, b) then the area of the triangle is

(A) ab (B) ab

(C) b2 (D) a2b2

67. ;fn ∆ABC rFkk ∆DEF le:Ik gSa rFkk AB = 1.2 lseh] DE = 1.4 lseh rks

∆ABC vkSj ∆DEF ds {ks=Qyksa dk vuqikr gS

(A) 16:25 (B) 49:81

25

(C) 36:49 (D) 9:16

If ∆ABC and ∆DEF are similar and AB = 1.2 cm, DE = 1.4 cm then

the ratio of areas of ∆ABC and ∆DEF is

(A) 16:25 (B) 49:81

(C) 36:49 (D) 9:16

68. ;fn ∆ABC rFkk ∆DEF le:Ik gS rFkk ∆ABC dk ifjeki = 30 lseh] ∆DEF

dk ifjeki = 20 lseh vkSj AB = 12 lseh rks DE cjkcj gS

(A) 12 lseh (B) 8 lseh

(D) 16 lseh (D) 10 lseh

If ∆ABC and ∆DEF are similar and perimeter of ∆ABC = 30 cm,

perimeter of ∆DEF = 20 cm and AB = 12 cm then DE is equal to

(A) 12 cm (B) 8 cm

(C) 16 cm (D) 10 cm

69. fdlh cká fcUnq P ls fdlh o`Ùk ij PQ vkSj PR nks Li’kZ js[kk,¡ gSA ;fn PQ =

6 lseh rks PR cjkcj gS

(A) 12 lseh (B) 3 lseh

(C) 6 lseh (D) 5 lseh

PQ and PR are two tangents to a circle from an external point P. If

PQ = 6 cm then PR is equal to

(A) 12 cm (B) 3 cm
26

(C) 6 cm (D) 5 cm

70. ;fn fdlh o`Ùk dk O;kl 14 lseh gks rks o`Ùk dk {ks=Qy gksxk

(A) 616 lseh2 (B) 221 lseh2

(C) 154 lseh2 (D) 77 lseh2

If the diameter of a circle is 14cm then the area of the circle is

(A) 616 cm2 (B) 221 cm2

(C) 154 cm2 (D) 77 cm2

71. ;fn fdl o`Ùk dk O;kl 14 lseh gks rks o`Ùk dh ifjf/k gksxh

(A) 7 lseh (B) 22 lseh

(C) 44 lseh (D) 66 lseh

If the diameter of a circle is 14cm then the circumference of the circle

is

(A) 7 cm (B) 22 cm

(C) 44 cm (D) 66 cm

72. ,d o`Ùk dh ifjf/k rFkk O;kl dk ;ksx 29 lseh gSA o`Ùk dk O;kl gksxk

(A) 3-5 lseh (B) 7 lseh

(C) 14 lseh (D) 21 lseh

The sum of the circumference and diameter of a circle is 29 cm. The

diameter of the circle is

(A) 3.5 cm (B) 7 cm
27

(C) 14 cm (D) 21 cm

73. 8 lseh f=T;k okys o`Ùk ds vanj ls 6 lseh O;kl okys o`Ùk dks dkVdj gVkus ij

cps Hkkx dk {ks=Qy gS

(A) 28 lseh2 (B) 2𝜋 lseh2

(C) 55𝜋 lseh2 (D) 14𝜋 lseh2

The area of the left part after cutting a circle of diameter 6 cm from a

circle of radius 8 cm is

(A) 28 cm2 (B) 2𝜋 cm2

(C) 55𝜋 cm2 (D) 14𝜋 cm2

74. ;fn ,d leckgq f=Hkqt dk {ks=Qy 4√3 lseh2 gS rks bldh izR;sd Hkqtk gksxh

(A) 9 lseh (B) 4√3 lseh

(C) 4 lseh (D) 2√3 lseh
If the area of an equilateral triangle is 4√3cm2 then its each side will
be
(A) 9 cm (B) 4√3 cm
(C) 4 cm (D) 2√3 cm

75. ;fn sin3A = cos(A – 100) rFkk 3A U;wudks.k gS rks A =

(A) 200 (B) 250

(C) 300 (D) 350

If sin3A = cos(A – 100) and 3A is acute angle then A =
28

(A) 200 (B) 250

(C) 300 (D) 350

76. f=Hkqt ABC esa AC2 = AB2 + BC2 rks B =

(A) 900 (B) 600

(C) 450 (D) buesa dksbZ ugha

In ∆ABC, AC2 = AB2 + BC2 then B =

(A) 900 (B) 600

(C) 450 (D) none of these

77. ,d fcUnq Q ls ,d o`Ùk ij Li’kZ js[kk dh yackbZ 24 lseh rFkk Q dh dsUnz ls nwjh

25 lseh gSA o`Ùk dh f=T;k gS %

(A) 7 lseh (B) 12 lseh

(C) 15 lseh (D) 24 lseh

From a point Q, the length of the tangent to a circle is 24 cm and the

distance of Q from the centre is 25 cm. The radius of circle is

(A) 7 cm (B) 12 cm

(C) 15 cm (D) 24 cm

78. ,d flDds dks mNkyk tkrk gSA fpr vkus dh izkf;drk gS

(A) 0 (B)

(C) (D) 1

29

A coin is tossed. The probability of getting head is

(A) 0 (B)

(C) (D) 1

79. fdlh ijh{kk esa lkSE;k ds ikl gksus dh izkf;drk gSA lkSE;k ds Qsy gksus dh

izkf;drk gS

(A) (B)

(C) (D)

The probability that Saumya passes in an examination is . The

probability that Saumya fails is

(A) (B)

(C) (D)

80. - - cos00 =

(A) 1 (B) -1

(C) 2 (D) 0

81. 2] 5] 8] 3] 8 dk ek/;d gS

(A) 3 (B) 2

(C) 5 (D) 8
30

The median of 2, 5, 8, 3, 8 is

(A) 3 (B) 2

(C) 5 (D) 8

82. 2] 5] 8] 3] 8] 3] 2] 3] 3 dk cgqyd gS

(A) 3 (B) 5

(C) 8 (D) 2

The mode of 2, 5, 8, 3, 8, 3, 2, 3, 3 is

(A) 3 (B) 5

(C) 8 (D) 2

83. oxZ varjky 10&20 dk oxZ fpà gS

(A) 10 (B) 15

(C) 20 (D) 30

The class mark of the class interval 10-20 is

(A) 10 (B) 15

(C) 20 (D) 30

84. oxZ varjkyksa 5&10] 10&15] 15&20 ------ ds oxksZ dh yEckbZ gS

(A) 5 (B) 10

(C) 15 (D) 20

The length of the class intervals of the classes 5-10, 10-15, 15-20 .. is

(A) 5 (B) 10

31

(C) 15 (D) 20

85. (secA + tanA)(1 – sinA) =

(A) cosA (B) sinA

(C) cosecA (D) secA

86. ik¡p la[;kvksa dk ek/; 18 gSA ;fn ,d la[;k gVk nh tk, rks ek/; 16 gSA gVkbZ

xbZ la[;k gS

(A) 16 (B) 18

(C) 24 (D) 26

The mean of five numbers is 18. If one number is removed then the

mean is 16. The removed number is

(A) 16 (B) 18

(C) 24 (D) 26

87. cgqyd lkekU;r% og pj&eku gksrk gS ftldh ckjackjrk gksrh gS

(A) lcls de (B) lokZf/kd

(C) ckjackjrkvksa dk dqy ;ksx (D) buesa dksbZ ugha

Mode is usually a variable value whose frequency is

(A) lowest (B) highest

(C) sum of total frequencies (D) none of these

88. fuEukafdr vk¡dM+s dks c<+rs Øe esa fy[kk x;k gS %

24] 27] 28] 31] 34] x] 37] 40] 42] 45- ;fn ef/;dk 35 gks rks x dk eku gS

32

(A) 35 (B) 34.5

(C) 35.5 (D) 36

The following data has been written in ascending order :

24, 27, 28, 31, 34, x, 37, 40, 42, 45. If the median is 35 then the value

of x is

(A) 35 (B) 34.5

(C) 35.5 (D) 36

89. +1=

(A) sin2A (B) sec2A

(C) tan2A (D) cot2A

90. lecsyu dk iw.kZ i`"B {ks=Qy gksxk

(A) 2𝜋rh (B) 𝜋 r2h

(C) 2𝜋𝑟(r + h) (D) 𝜋 r2h

The total surface area of a right circular cylinder is

(A) 2𝜋rh (B) 𝜋 r2h

(C) 2𝜋𝑟(r + h) (D) 𝜋 r2h

91. 5 lseh fdukjs okys ,d ?ku dks 1 lseh fdukjs okys ?kuksa esa foHkDr fd;k x;k gSA

?kuksa dh la[;k gS

(A) 5 (B) 50

33

(C) 125 (D) 625

A cube of side 5cm is cut into cubes of side 1cm. The number of

cubes formed is

(A) 5 (B) 50

(C) 125 (D) 625

92. nks csyuksa dh f=T;kvksa ds oxksZa dk vuqikr 2%5 gS vkSj mudh Å¡pkb;ksa dk vuqikr

5%2 gSA muds vk;ruksa dk vuqikr gksxk

(A) 1:1 (B) 2:5

(C) 5:2 (D) 2:3

The ratio of square of the radii of two cylinders is 2:5 and the ratio of

their heights is 5:2. The ratio of their volumes will be

(A) 1:1 (B) 2:5

(C) 5:2 (D) 2:3

93. =
.

(A) √3 (B)
√

(C) 1 (D) 2

94. ;fn fcUnq (4, P) rFkk (1, 0) ds chp dh nwjh 5 gks rks P cjkcj gS

(A) ± 4 (B) ±5

(C) ±3 (D) 1

34

If the distance between (4, P) and (1, 0) is 5 then P is equal to

(A) ± 4 (B) ±5

(C) ±3 (D) 1

95. ,d 'kadq ds vk/kkj dh f=T;k 7 lseh gS rFkk Å¡pkbZ 15 lseh gS rks bldk vk;ru

gksxk

(A) 665 lseh3 (B) 770 lseh3

(C) 880 lseh3 (D) 429 lseh3

The radius of the base of a cone is 7 cm and its height is 15 cm then

its volume is

(A) 665 cm3 (B) 770cm3

(C) 880 cm3 (D) 429 cm3

96. ;fn fcUnq,¡ A(1, 2), O(0, 0) rFkk C(a, b) lajs[k gS rks

(A) a = b (B) a = 2b

(C) 2a = b (D) a + b = 0

If the points A(1, 2), O(0, 0) and C(a, b) are collinear then

(A) a = b (B) a = 2b

(C) 2a = b (D) a + b = 0

97. ,d f=Hkqt ftlds 'kh"kZ (5, 0), (8, 0) rFkk (8, 4) gSa] dk {ks=Qy ¼oxZ bdkbZ esa½ gS

(A) 6 (B) 12

(C) 16 (D) 20

35

The area of a triangle (in square units) with vertices (5, 0), (8, 0) and

(8, 4) is

(A) 6 (B) 12

(C) 16 (D) 20

98. ;fn fcUnq (5, a) js[kk 2x – 3y = 5 ij fLFkr gks rks a dk eku gksxk

(A) (B)

(C) - (D) -

If the point (5, a) lies on the line 2x – 3y = 5 then the value of a is

(A) (B)

(C) - (D) -

99. ;fn acos𝜃 + bsin𝜃 = 4 rFkk asin𝜃 – bcos𝜃 = 3 gks rks a2 + b2 =

(A) 7 (B) 16

(C) 25 (D) 36

If acos𝜃 + bsin𝜃 = 4 and asin𝜃 – bcos𝜃 = 3 then a2 + b2 =

(A) 7 (B) 16

(C) 25 (D) 36

100. 'kCn “mobile” ls ;kn`PN;k ,d v{kj pquk tkrk gSA ml v{kj ds Loj pqus tkus

dh izkf;drk gS

(A) (B)

36

(C) (D)

A letter is chosen at random from the word “mobile”. The probability

that the letter chosen is a vowel is

(A) (B)

(C) (D)

[k.M&c @ Section-B

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

iz'u la[;k 1 ls 30 y?kq mÙkjh; iz’u gSaA fdUgha 15 iz’uksa ds mÙkj nsaA izR;sd ds fy, 2

vad fu/kkZfjr gSA 15x2=30

Question Nos. 1 to 30 are short Answer Type. Answer any 15 questions.

Each question carries 2 marks. 15x2=30

1- vHkkT; xq.ku[kaM }kjk 126 rFkk 156 dk e0 l0 ,oa y0 l0 Kkr djsaA

Find the HCF and LCM of 126 and 156 by prime factorization

method.

2- ;fn p(x) = 2x2+3x-5 gks rks p(3) vkSj p(-3)ds eku Kkr djsaA

If p(x) = 2x2+3x-5 then find the values of p(3) and p(-3).

37

3- ,d f}?kkr cgqin Kkr djsa ftlds 'kwU;dksa dk ;ksx rFkk xq.kuQy Øe’k% √2

vkSj gS

Find a quadratic polynomial, the sum and product of whose zeroes

are √2 and respectively.

4- f}?kkr lehdj.k 3√3x2 + 10x + √3 = 0 dk foospd Kkr djsaA

Find the discriminant of the quadratic equation

3√3x2 + 10x + √3 = 0.

5- fn[kk,¡ fd x = -3, x2 + 6x + 9 = 0 dk ,d gy gSA

Show that x = -3 is a solution of x2 + 6x + 9 = 0.

6- f}?kkr lehdj.k 2x2 – 6x + 3 = 0 ds ewyksa dh izd`fr Kkr djsaA

Find the nature of the roots of the quadratic equation

2x2 – 6x + 3 = 0

7- lekarj Js<+h % 18] 15 ] 13] --- dk dkSu lk in&47 gS \

Which term of the A. P. : 18, 15 , 13, … is -47 ?

8- K ds eku Kkr djsa ftlds fy, lehdj.k 2x2 – 10x + K = 0 ds ewy

okLrfod gSaA

Find the values of K for which the equation 2x2 – 10x + K = 0 has

real roots.

38

9- lekarj Js<h+ 25 $ 28 $ 31 $ ---- $ 100 dk ;ksxQy Kkr djsaA

Find the sum of the A. P. 25 + 28 + 31 + … + 100.

10- K ds fdl eku ds fy, fuEukafdr jSf[kd lehdj.kksa ds ;qXe dk dksbZ gy ugha

gS\

3x + y = 1, (2K – 1)x + (K – 1)y = 2K + 1

For which value of K the following pair of linear equations have no

solution ?

3x + y = 1, (2K – 1)x + (K – 1)y = 2K + 1

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

vlaxr %

2x – 3y = 8, 4x – 6y = 9

By comparing the ratios , and find whether following pair is

consistent or inconsistent :

2x – 3y = 8, 4x – 6y = 9

12- 6x2 – 7x – 3 ds 'kwU;d Kkr djsa ,oa 'kwU;dksa vkSj xq.kkadks ds chp ds lac/k dh

tk¡p djsaA

Find the zeroes of 6x2 – 7x – 3 and verify the relation between

zeroes and coefficients.

39

13- x vkSj y esa laca/k Kkr djsa tcfd fcUnq (x, y) fcUnqvksa (7, 1) vkSj (3, 5) ls

lenwjLFk gS

Find the relation between x and y such that the point (x, y) is

equidistant from the points (7, 1) and (3, 5).

14- ABC ,d lef}ckgq f=Hkqt gS ftlesa AC = BCA ;fn AB2 = 2AC2 gS rks

fl) djsa fd ABC ,d ledks.k f=Hkqt gS

ABC is an isosceles triangle in which AC = BC. If AB2 = 2AC2 then

prove that ABC is a right angled triangle.

15- fl) djsa fd ,d cká fcUnq ls o`Ùk ij [khaph xbZ Li’kZ js[kkvksa dh yackb;k¡

cjkcj gksrh gSA

Prove that the lengths of tangents drawn from an external point to

a circle are equal.

16- ;fn 30 eh0 Å¡ph ,d ehukj Hkwfe ij 10√3 eh0 yach Nk;k cukrh gS rks lw;Z

dk mUu;u dks.k Kkr djsaA

If a tower 30m high casts a shadow 10√3 m long on the ground

then find the angle of elevation of the sun.

17- nks le:Ik f=Hkqtksa ABC vkSj DEF ds {ks=Qy Øe’k% 25 lseh2 vkSj 49 lseh2

gSA ;fn EF = 9.8 lseh] rks BC Kkr djsaA

40

Areas of two similar triangles ABC and DEF are 25cm2 and 49 cm2

respectively . If EF = 9.8 cm then find BC.

18- fdlh o`Ùk ds dsUnz ls 5 lseh nwj fLFkr fcUnq A ls [khaph xbZ Li’kZ js[kk dh

yackbZ 4 lseh gSA o`Ùk dh f=T;k Kkr djsaA

The length of a tangent drawn from a point A at distance 5 cm

from the centre of the circle is 4 cm. Find the radius of the circle.

19- ,d o`Ùk dk {ks=Qy Kkr djsa ftldh ifjf/k 66 lseh gSA

Find the area of a circle whose circumference is 66 cm.

20- ;fn ,d v)Zo`Ùkkdkj pk¡n dk ifjeki 72 lseh0 gS rks bldk O;kl Kkr djsaA

If the perimeter of a semicircular protractor is 72 cm then find its

diameter.

21- ;fn A = 300 gks rks fl) djsa fd tan2A =

If A = 300 then prove that tan2A =

22- ;fn A = 900 rFkk B = 450 gks rks fl) djsa fd

Sin(A – B) = sinAcosB – cosAsinB

If A = 900 and B = 450 then prove that

Sin(A – B) = sinAcosB – cosAsinB

23- ml f=Hkqt dk {ks=Qy Kkr djsa ftlds 'kh"kksZa ds funsZ’kkad (1, -1), (-4, 6) vkSj

(-3, -5) gSaA

41

Find the area of the triangle whose coordinates of the vertices are

(1, -1), (-4, 6) and (-3, -5).

24- izfrLFkkiu fof/k ls lehdj.k x + y = 5 vkSj 5x – 3y = 4 dk gy fudkysaA

Find the solution of the equations x + y = 5 and 5x – 3y = 4 by

substitution method.

25- dk eku Kkr djsaA

Find the value of .

26- 2x2 + 3x + 1 dks x+2 ls Hkkx nsa vkSj foHkktu ,Yxksfjn~e dks lR;kfir djsaA

Divide 2x2 + 3x + 1 by x+2 and verify division algorithm.

27- ,d 14 lseh0 yacs o`Ùkkdkj lecsyu dk vk;ru 11 lseh0 ds fdukjs ij cuk,

x, ?ku ds vk;ru ds cjkcj gSA csyu dh f=T;k Kkr djsaA

The volume of a right circular cylinder of length 14 cm is equal to

the volume of a cube of side 11 cm. Find the radius of the cylinder.

28- fl) djsa fd + sin𝜃.cos𝜃 = 1

Prove that + sin𝜃.cos𝜃 = 1

29- fn[kk,¡ fd fcUnq A(a, b+c), B(b, c+a) rFkk C(c, a+b) lajs[k gSA

Show that points A(a, b+c), B(b, c+a) and C(c, a+b) are collinear.

30- nks Øfed izkd`r la[;kvksa ds oxksZ dk ;ksx 421 gSA la[;k,¡ Kkr djsaA

42

The sum of the squares of two consecutive natural numbers is

421. Find the numbers.

Long Answer Type Questions.

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

31- lehdj.k ;qXe 2(x+y) = 1 vkSj 2y – 3x = 6 dk xzkQ [khpsa rFkk gy djsaA

Draw the graph of pair of equations 2(x+y) = 1 and 2y – 3x = 6 and

solve them.

32- nks la[;kvksa dk ;ksx 15 gSA ;fn muds O;qRØeksa dk ;ksx gks rks la[;k,¡ Kkr

djsaA

The sum of two numbers is 15. If the sum of their reciprocals is

then find the numbers.

33- fuEukafdr ckjackjrk lkj.kh ls ek/; vkSj ek/;d Kkr djsaA

oxZ 30&40 40&50 50&60 60&70 70&80 80&90

Ckkjackjrk 5 22 63 74 30 6

Find the mean and median from the following frequency table

43

Class 30&40 40&50 50&60 60&70 70&80 80&90

Frequency 5 22 63 74 30 6

34- 1-2 eh0 yEch ,d yM+dh Hkwfe ls 88-2 eh0 dh Å¡pkbZ ij ,d {kSfrt js[kk esa

gok esa mM+ jgs xqCCkkjs dks ns[krh gSA fdlh Hkh {k.k yM+dh dh vk¡[k ls xqCckjs

dk mUu;u dks.k 600 gSA dqN le; ckn mUu;u dks.k ?kVdj 300 gks tkrk gSA

bl varjky ds nkSjku xqCckjs }kjk r; dh xbZ nwjh Kkr djssaA

A 1.2 m tall girl spots a balloon moving with the wind in a

horizontal line at a height of 88.2 m from the ground. The angle of

elevation of the balloon from the eyes of the girl at any instant is

600. After some time the angle of elevation reduces to 300. Find

the distance travelled by the balloon during the interval.

35- 8 lseh yach ,d js[kk[kaM AB [khapsA A dks dsUnz ekudj 4 lseh0 f=T;k dk ,d

o`Ùk rFkk B dks dsUnz ekudj 3 lseh f=T;k dk ,d vU; o`Ùk [khpsaA izR;sd o`Ùk

ij nwljs o`Ùk ds dsUnz ls Li’kZ js[kkvksa dh jpuk djsaA

Draw a line segment AB of length 8 cm. Taking A as centre draw a

circle of radius 4 cm and taking B as centre, draw another circle of

radius 3 cm. Construct tangents to each circle from the centre of

the other circle.

44

36- ,d crZu ,d mYVs 'kadq ds vkdkj dk gSA bldh Å¡pkbZ 8 lseh gS vkSj blds

mijh fljs ¼tks [kqyk gqvk gS½ dh f=T;k 5 lseh gSA ;g Åij rd ikuh ls Hkjk

gqvk gSA tc bl crZu esa lhls dh dqN xksfy;k¡ ftuesa izR;sd 0-5 lseh okyk

,d xksyk gS] Mkyh tkrh gS rks bles ls Hkjs gq, ikuh dk ,d pkSFkkbZ Hkkx ckgj

fudy tkrk gSA crZu esa Mkyh xbZ lhls dh xksfy;ksa dh la[;k Kkr djsaA

A vessel is in the form of an inverted cone. Its height is 8 cm and

the radius of its top, which is open, is 5 cm. It is filled with water up

to the brim. When lead shots, each of which is a sphere of radius

0.5 cm, are dropped into the vessel, one fourth of the water flows

out. Find the number of lead shots dropped in the vessel.

37- fl) djsa fd

Sin2100+sin2200+sin2300+sin2400+sin2500+sin2600+sin2700+sin2800

+ sin2900 = 5.

Prove that

Sin2100+sin2200+sin2300+sin2400+sin2500+sin2600+sin2700+sin2800

+ sin2900 = 5.

38- ml f=Hkqt ds 'kh"kksZa ds funsZ’kkad Kkr djsa ftldh Hkqtkvksa ds e/; fcUnqvksa ds

funsZ’kkad (3, 3), (1, 4) rFkk (2, 1) gSaA

Find the co-ordinates of the vertices of the triangle whose co-

ordinates of the midpoints of the sides are (3, 3), (1, 4) and (2, 1).

45

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