Bihar Board Class 10 Model Question Paper 2022 Maths Standard

SECONDARY SCHOOL EXAMINATION – 2022
(ANNUAL)
Model Question Paper ¼vkn”kZ iz”u i=½
Mathematics (COMPULSORY)
xf.kr ¼vfuok;Z½
fo’k; dksM&110@210
le;% 3 ?kaVs15 feuV Subject Code-110/210
Time:3 Hours 15 Minutes

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

Instructions to 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, ijh{kkfFkZ;ksa dks 15 feuV dk vfrfjDr le;
fn;k x;k gSA

1

15 minutes of extra time has been allotted to the candidates to
read the questions carefully.

5- ;g iz”u iqfLrdk nks [k.Mksa esa gS & ,oa A
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 dk gh ewY;kadu
dEI;wVj }kjk 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 OgkbVuj @ rjy inkFkZ
@ CysM @ uk[kwu vkfn dk OMR mÙkj i=d esa iz;ksx djuk euk gS] vU;Fkk
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. 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 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
vfuok;Z gSA izR;sd ds fy, 5 vad fu/kkZfjr gSA

2

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.

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 Qestions

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 nsaA 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. fuEufyf[kr esa dkSu&lh lcls NksVh la;qDr la[;k gS \

(A) 1 (B) 2

(C) 3 (D) 4

Which of the following is the smallest composite number ?

(A) 1 (B) 2

(C) 3 (D) 4

2. 𝜋 gS &

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

(C) ,d iw.kkZad la[;k (D) buesa ls dksbZ ugha

𝜋 is –

(A) an irrational number (B) a rational number

(C) an integer (D) none of these

3. fuEufyf[kr esa dkSu vifjes; la[;k gS \
4

(A) (B)

(C) (D) √2 x √2

Which of the following is an irrational number ?

(A) (B)

(C) (D) √2 x √2

4. nks la[;kvksa ds y0 l0 rFkk e0 l0 cjkcj gS] rks la[;k,¡ gS &

(A) vHkkT; (B) cjkcj

(C) lg&vHkkT; (D) la;qDr

The L.C.M and H.C.F of two numbers are equal. Then the

numbers are –

(A) prime (B) equal

(C) co-prime (D) composite

5. 0.29 =

(A) (B)

(C) (D)

6. 1 - 𝑠𝑖𝑛 60 =

(A) 1 (B)

5

(C) (D) 0

7. fuEufyf[kr esa fdldk n”keyo izlkj lkar gS \

(A) (B)

(C) (D)

Which of the following has terminating decimal expansion ?

(A) (B)

(C) (D)

8. fuEufyf[kr esa fdldk n”keyo izlkj vlkar gS \

(A) (B)

(C) (D)

Which of the following has non-terminating decimal expansion ?

(A) (B)

(C) (D)

9. 400 ds vHkkT; xq.ku[kaM esa 5 dk ?kkrkad gS &

(A) 2 (B) 3
6

(C) 4 (D) 1

The exponent of 5 in the prime factorization of 400 is -

(A) 2 (B) 3

(C) 4 (D) 1

10. ;fn 130 = 15 x 8 + 10 rFkk e0 l0 (130, 15) = e0l0(x, y), rks (x, y) =

(A) (15, 10) (B) (8, 10)

(C) (15, 8) (D) (130, 8)

If 130 = 15 x 8 + 10 and H.C.F(130, 15) = H.C.F(x, y) then (x, y) =

(A) (15, 10) (B) (8, 10)

(C) (15, 8) (D) (130, 8)

11. √125 dk ifjes;hdj.k xq.kkad gS &

(A) √2 (B) √3

(C) √5 (D) 5√5

The rationalizing factor of √125 is -

(A) √2 (B) √3

(C) √5 (D) 5√5

12. fuEufyf[kr esa dkSu&lk ;qXe lgvHkkT; gS \

(A) (9, 16) (B) (15, 20)

(C) (21, 84) (D) (13, 65)

7

Which of the following pair is co-prime ?

(A) (9, 16) (B) (15, 20)

(C) (21, 84) (D) (13, 65)

13. jSf[kd cgqin esa 'kwU;dksa dh la[;k gksrh gS &

(A) 3 (B) 2

(C) 1 (D) 0

The number of zeros in a linear polynomial is -

(A) 3 (B) 2

(C) 1 (D) 0

14. cgqin x-1 dk 'kwU;d gS &

(A) 1 (B) 0

(C) 2 (D) vifjHkkf"kr

Zero of the polynomial x-1 is –

(A) 1 (B) 0

(C) 2 (D) undefined

15. fdlh f}?kkr cgqin ds 'kwU;d ;fn &4 ,oa 7 gS] rks f}?kkr cgqin gksxk &

(A) x2 + 4x + 7 (B) x2 + 3x – 28

(C) x2 – 3x – 28 (D) x2 + 3x + 28

If zeros of any quadratic polynomial are -4 and 7 then the

quadratic polynomial will be –

8

(A) x2 + 4x + 7 (B) x2 + 3x – 28

(C) x2 – 3x – 28 (D) x2 + 3x + 28

16. ,d f}?kkr cgqin ds 'kwU;dksa dk ;ksx rFkk xq.kuQy Øe”k% 5 rFkk 6 gS] rks

f}?kkr cgqin gksxk &

(A) x2 - 5x - 6 (B) x2 – 5x + 6

(C) x2 +5x – 6 (D) x2 + 5x + 6

The sum and product of the zeros of a quadratic polynomial are 5

and 6 respectively then the quadratic polynomial will be -

(A) x2 - 5x - 6 (B) x2 – 5x + 6

(C) x2 +5x – 6 (D) x2 + 5x + 6

17. ;fn cgqin y2 – 6y + 8 ds 'kwU;d 𝛼 rFkk 𝛽 gS] rks 𝛼. 𝛽 dk eku gksxk &

(A) 8 (B) -8

(C) 6 (D) -6

If 𝛼 and 𝛽 are the zeros of the polynomial y2 – 6y + 8 then the

value of 𝛼. 𝛽 is -

(A) 8 (B) -8

(C) 6 (D) -6

18. cgqin f(x) = 3x2 – 7x + 2 rks f(-1) dk eku gS &

(A) -8 (B) 12

(C) -12 (D) 0

9

The polynomial f(x) = 3x2 – 7x + 2 then the value of f(-1) is –

(A) -8 (B) 12

(C) -12 (D) 0

19. cgqin x2 – 3 ds 'kwU;d gSa &

(A) 3, -3 (B) √3,-√3

(C) √3,√3 (D) 3, 3

The zeros of the polynomial x2 – 3 are –

(A) 3, -3 (B) √3,-√3

(C) √3,√3 (D) 3, 3

20. fuEu esa ls dkSu 6x – 2y = 0 rFkk 10x + 2y = 32 dk gy gS \

(A) x = 6, y = 6 (B) x = 2, y = 6

(C) x= 6, y = 2 (D) x = 2, y = 2

Which of the following is the solution of 6x – 2y = 0 and

10x + 2y = 32 ?

(A) x = 6, y = 6 (B) x = 2, y = 6

(C) x= 6, y = 2 (D) x = 2, y = 2

21. ;fn js[kk,¡ 4x + py = 6 rFkk 12x + 9y = 15 lekukUrj gS] rks p dk eku

gksxk &

10

(A) (B) 3

(C) 6 (D) -3

If the lines 4x + py = 6 and 12x + 9y = 15 are parallel then the

value of p will be –

(A) (B) 3

(C) 6 (D) -3

22. lehdj.k fudk; x + 2y = 3, 5x + Ky = 15ds vufxur gy gksus ds fy, K

dk eku gS &

(A) 5 (B) 20

(C) 6 (D) 10

The value of K for which the system of equations x + 2y = 3,

5x + Ky = 15 has infinitely many solutions is –

(A) 5 (B) 20

(C) 6 (D) 10

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

(A) (x + 2) (x – 2) = x2 – 4x3 (B) (X + 2 )2 = 3(x + 4)

(C) (2x2 + 3) = (5 + x) (2x – 3) (D) 2x + = 4x2

Which of the following is a quadratic equation ?

(A) (x + 2) (x – 2) = x2 – 4x3 (B) (X + 2 )2 = 3(x + 4)

11

(C) (2x2 + 3) = (5 + x) (2x – 3) (D) 2x + = 4x2

24. f}?kkr lehdj.k px2+qx+r = 0, p≠0 ds ewy gaS &

± ±
(A) (B)

± ±
(C) (D)

The roots of quadratic equation px2+qx+r = 0, p≠0 are

± ±
(A) (B)

± ±
(C) (D)

25. f}?kkr lehdj.k 6x2+6x+1 = 0 ds ewyksa dh izd`fr gksxh &

(A) okLrfod vkSj cjkcj (B) okLrfod ugh

(C) okLrfod vkSj vleku (D) buesa ls dksbZ ugha

The nature of roots of the quadratic equation 6x2+6x+1 = 0 is –

(A) real and equal (B) not real

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

26. lehdj.k 3x2-18x+6=0 ds ewyksa dk ;ksx gS &

(A) 3 (B) -3

(C) -6 (D) 6

The sum of the roots of the equation 3x2-18x+6=0 is –

(A) 3 (B) -3

12

(C) -6 (D) 6

27. f}?kkr lehdj.k ax2+bx+c = 0 (a≠0) ds ewy ,d nwljs ds O;qRØe gksaxs] ;fn

(A) b = c (B) c = a

(C) a = b (D) buesa ls dksbZ ugha

The roots of the quadratic equation ax2+bx+c = 0 (a≠0) will be

reciprocal of each other, if

(A) b = c (B) c = a

(C) a = b (D) none of these

28. lekarj Js<+h % , 1, ,2, … dk 15ok¡ in gS &

(A) (B)

(C) 7 (D)

The 15th term of the A.P. : , 1, ,2, … is –

(A) (B)

(C) 7 (D)

29. fuEufyf[kr esa dkSu lk lekarj Js<+h esa ugha gS \

(A) 5, 4 , 4, 3 , … (B) -1, , , ,…

(C) 8, 14, 20, 26, … (D) 4, 10, 15, 20, …

13

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

(A) 5, 4 , 4, 3 , … (B) -1, , , ,…

(C) 8, 14, 20, 26, … (D) 4, 10, 15, 20, …

30. ;fn fdlh lekarj Js<h+ dk igyk in 9 ,oa lkoZvUrj &3 gks rks blds izFke

12 inksa dk ;ksxQy gksxk &

(A) 78 (B) 90

(C) -90 (D) -78

If in an A.P., first term is 9 and common difference is -3 then sum

of its first 12 terms is –

(A) 78 (B) 90

(C) -90 (D) -78

31. Lkekarj Js<+h % 17] 34] 51] ---------- dk dkSu lk in 170 gS \

(A) 9ok¡ (B) 10 ok¡

(C) 11 ok¡ (D) 12 ok¡

Which term of the A.P. : 17, 34,. 51, … is 170 ?

(A) 9th (B) 10th

(C) 11th (D) 12th

32. Lkekarj Js<+h 51] 48] 45] --- ] 6 esa inksa dh la[;k gS &

(A) 14 (B) 16

(C) 17 (D) 18
14

The number of terms in an A.P. 51, 48, 45, …, 6 is –

(A) 14 (B) 16

(C) 17 (D) 18

33. 0 vkSj 50 ds chp fo"ke la[;kvksa dh la[;k gS &

(A) 25 (B) 26

(C) 24 (D) 27

The number of odd numbers between 0 and 50 is –

(A) 25 (B) 26

(C) 24 (D) 27

34. ledks.k ∆ABC esa  B=900, AB = √15lseh0] BC=1lseh0 rFkk AC =

4lseh0 rks tanA cjkcj gksxk&

√ √
(A) (B)

(C) (D)
√ √

In right ∆ABC, B=900, AB = √15cm, BC=1Cm and AC = 4cm

then tanA will be equal to –

√ √
(A) (B)

(C) (D)
√ √

35. ;fn cosec∅ = K rks cos∅ dk eku gS &

15

√ √
(A) (B)

(C) (D)
√ √

If cosec∅ = K then the value of cos∅ is –

√ √
(A) (B)

(C) (D)
√ √

36. =
∅

(A) tan∅ (B) cos∅

(C) sec∅ (D) cosec∅

37. cotx . tanx =

(A) 1 (B) -1

(C) 0 (D) 2

38. ;fn cos∅ = rks cosec∅ dk eku gS &

(A) 2 (B)
√

√
(C) (D)
√

If cos∅ = then the value of cosec∅ is –

(A) 2 (B)
√

√
(C) (D)
√

16

39. ;fn ∅ = 450 rks sec∅ + cosec∅ =

(A) 1 (B) √2

(C) 2 (D) 2√2

If ∅ = 450 then sec∅ + cosec∅ =

(A) 1 (B) √2

(C) 2 (D) 2√2

40. ;fn √3tanA – 3 = 0 rks A =

(A) 900 (B) 600

(C) 450 (D) 300

If √3tanA – 3 = 0 then A =

(A) 900 (B) 600

(C) 450 (D) 300

41. =

(A) -1 (B) 0

(C) 1 (D) 2

42. fuEufyf[kr esa fdldk eku lcls vf/kd gS \

(A) cos450 (B) sin00

(C) cot450 (D) cos600

17

Which of the following has the maximum value ?

(A) cos450 (B) sin00

(C) cot450 (D) cos600

43. Sin300 =

(A) 1 (B)

√
(C) (D)
√

44. cos2𝜃 - 1 =

(A) –sin2𝜃 (B) sin2𝜃

(C) 0 (D) cot2𝜃

45. (1-sec2𝜃) x cot2𝜃 =

(A) 0 (B) 1

(C) -1 (D) 2

46. Sin𝛼 x cot𝛼 =

(A) 1 (B) cos𝛼

(C) cos2𝛼 (D) sin2𝛼

47. ;fn = 120 rks 3X cosec2𝜃 =

(A) (B)
√

(C) 4 (D) 4√3

18

If = 120 then 3X cosec2𝜃 =

(A) (B)
√

(C) 4 (D) 4√3

48. ;fn sinx + sin2x = 1 rks cos2x + cos4x =

(A) (B)

(C) 1 (D)

If sinx + sin2x = 1 then cos2x + cos4x =

(A) (B)

(C) 1 (D)

49. fuEufyf[kr esa dkSu&lk cosec600 ds cjkcj gS \

(A) sec300 (B) sin600

(C) cot300 (D) tan600

Which of the following is equal to cosec600 ?

(A) sec300 (B) sin600

(C) cot300 (D) tan600

50. ;fn cosec𝜃 – cot𝜃 = 𝑥 rks cosec𝜃 =

(A) (B)

(C) (D)

19

If cosec𝜃 – cot𝜃 = 𝑥 then cosec𝜃 =

(A) (B)

(C) (D)

51. sec2600 – tan2600 + 1 =

(A) 1 (B) 2

(C) -2 (D) 0

52. cot (900 – A) =

(A) cotA (B) secA

(C) cosecA (D) tanA

53. =

(A) sin300 (B) cos600

(C) tan600 (D) sin600

54. ;fn ,d 6 ehVj Å¡ps [kaHks dh Nk;k i`Foh ij 2√3ehVj yEch gS] rks lw;Z dk

mUu;u dks.k gS &

(A) 150 (B) 300

(C) 450 (D) 600

If a pole 6 meter high casts a shadow 2√3 meter long on ground,

then the angle of elevation of sun is –

(A) 150 (B) 300

20

(C) 450 (D) 600

55. fdlh ehukj dh Å¡pkbZ 15 ehVj gSA tc lw;Z dk mUu;u dks.k 600 gks] rks

ehukj dh Nk;k dh yEckbZ tehu ij D;k gksxh \

(A) 15 ehVj (B) 15√3 ehVj

(C) 5√3 ehVj (D) 45√3 ehVj

The height of a tower is 15 meter. When the angle of elevation of

the sun is 600, what will be the length of the tower’s shadow on the

ground ?

(A) 15 m (B) 15√3 m

(C) 5√3 m (D) 45√3 m

56. funsZ”kkad v{kksa ds izfrPNsnu fcUnq dks dgk tkrk gS &

(A) Hkqt (B) dksfV

(C) ewyfcUnq (D) prqFkkZa’k

The point of intersection of the coordinate axes is called –

(A) abscissa (B) ordinate

(C) origin (D) quadrant

57. fcUnq (-6, -8) fdl ikn esa fLFkr gS \

(A) izFke ikn (B) f}rh; ikn

(C) r`rh; ikn (D) prqFkZ ikn

21

In which quadrant does the point (-6, -8) lie ?

(A) first quadrant (B) second quadrant

(C) third quadrant (D) fourth quadrant

58. ewy fcUnq ls fcUnq P(sin𝜃, cos𝜃) dh nwjh gS &

(A) 0 bdkbZ (B) 1 bdkbZ

(C) 4 bdkbZ (D) 2 bdkbZ

The distance of the point P(sin𝜃, cos𝜃) from the origin is –

(A) 0 unit (B) 1 unit

(C) 4 units (D) 2 units

59. fcUnqvksa R(1, 1) vkSj S(-1, 5) ds chp dh nwjh gS &

(A) 4√5 bdkbZ (B) 3√5 bdkbZ

(C) 2√5 bdkbZ (D) 5 bdkbZ

The distance between the points R(1, 1) and S(-1, 5) is –

(A) 4√5 units (B) 3√5 units

(C) 2√5 units (D) 5 units

60. js[kk y=8 dk vkys[k fuEu esa ls fdl fcUnq ls gksdj ugha xqtjsxh &

(A) (1, 8) (B) (4, 8)

(C) (0, 8) (D) (8, 1)

22

The graph of the line y=8 does not passes through which of the

following points ?

(A) (1, 8) (B) (4, 8)

(C) (0, 8) (D) (8, 1)

61. fuEufyf[kr esa ls dkSu&lk fcUnq izFke prqFkkZa”k esa gS \

(A) (10, 4) (B) (-4, 3)

(C) (-4, -3) (D) (4, -3)

Which of the following points lie in the first quadrant ?

(A) (10, 4) (B) (-4, 3)

(C) (-4, -3) (D) (4, -3)

62. fcUnq A(5, 6) dh y &v{k ls yacor~ nwjh gS &

(A) 5 bdkbZ (B) 6 bdkbZ

(C) 11 bdkbZ (D) √61 bdkbZ

The perpendicular distance of the point A(5, 6) from the y-axis is –

(A) 5 units (B) 6 units

(C) 11 units (D) √61 units

63. fcUnqvksa M(-2, 10) vkSj N(-8, 14) dks feykusokyh js[kk[kaM ds e/;fcUnq ds

fu;ked gSa &

(A) (-10, 24) (B) (6, 1)

(C) (-5, 12) (D) (8, 6)

23

The coordinates of the midpoint of the line segment joining the

points M(-2, 10) and N(-8, 14) are –

(A) (-10, 24) (B) (6, 1)

(C) (-5, 12) (D) (8, 6)

64. ;fn fdlh o`Ùk ds O;kl ds Nksjksa ds funsZ”kkad ¼6] &8½ vkSj ¼&6] 8½ gS rks o`Ùk ds

dsUnz ds funsZ”kkad gSa &

(A) (6, -6) (B) (-8, 8)

(C) (0, 0) (D) (12, 16)

If (6, -8) and (-6, 8) are the coordinates of the ends points of the

diameter of a circle, then the coordinates of the centre of the circle

are –

(A) (6, -6) (B) (-8, 8)

(C) (0, 0) (D) (12, 16)

65. fcanq, A(3, 4) vkSj B(p, 1) ewy fcUnq ls lenwjLFk gS] rks p dk eku gS &

(A) 2 (B) ±√6

(C) ±6 (D) ±2√6

Points A(3, 4) and B(p, 1) are equidistant from the origin then the

value of p is –

(A) 2 (B) ±√6

(C) ±6 (D) ±2√6

24

66. fcUnqvksa A(0, 8), B(6, 12) vkSj C(-16, -4) ls cus f=Hkqt dk {ks=Qy gS &

(A) 48 oxZ bdkbZ (B) 8 oxZ bdkbZ

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

The area of the triangle formed by points A(0, 8), B(6, 12) and

C(-16, -4) is –

(A) 48 sq. units (B) 8 sq. units

(C) 6 sq. units (D) 4 sq. units

67. fcanq A(4, -3) rFkk B(9, 7) dks feykus okys js[kk[kaM dks 3%2 ds vuqikr esa vUr%

foHkkftr djus okys fcUnq dk y &funsZ”kkad gS &
( ) [ ( )]
(A) (B)
( ) ( )

[ ( )] [ ( )]
(C) (D)
( ) ( )

The y-coordinates of a point which divides the line segment joining

A(4, -3) and B(9, 7) in the ratio 3:2 internally is –

( ) [ ( )]
(A) (B)
( ) ( )

[ ( )] [ ( )]
(C) (D)
( ) ( )

68. ∆ABC esa ;fn A + C =1200 ;fn rks B =

(A) 500 (B) 600

(C) 700 (D) 900

25

In ∆ABC if A + C =1200 then B =

(A) 500 (B) 600

(C) 700 (D) 900

69. ;fn ∆ABC rFkk ∆DEF esa = rks ;s le:Ik gksaxs tc &

(A) A = D (B) B = E

(C) A = F (D) B = D

If in ∆ABC and ∆DEF, = then they will be similar when –

(A) A = D (B) B = E

(C) A = F (D) B = D

(∆ )
70. ;fn ∆ABC~∆PQR vkSj = rks ¾
(∆ )

(A) 25 (B) 5

(C) (D)

(∆ )
If ∆ABC~∆PQR and = then =
(∆ )

(A) 25 (B) 5

(C) (D)

71. tan(900 – A) =

(A) tanA (B) cotA

(C) - cotA (D) – tanA

26

A

72. nh xbZ vkd`fr esa DEIIBC.

;fn AD = 5 lseh0] DB = 6 lseh0 D E

rFkk AE = 8 lseh0 rks EC = B C

(A) 5-6 lseh0 (B) 7-6 lseh0

(C) 9-6 lseh0 (D) 11-6 lseh0
A
In the given figure DEIIBC.

If AD = 5 cm, DB = 6 cm and AE = 8 cm then EC =

(A) 5.6 cm (B) 7.6 cm D E

(C) 9.6 cm (D) 11.6 cm B C

73. ^nks lekudksf.kd f=Hkqtksa esa mudh laxr Hkqtkvksa dk vuqikr lnSo leku jgrk

gSA* fdlus dgk \

(A) FksYl (B) U;wVu

(C) vk;ZHkV~V (D) ikbFkkxksjl

Who said, ‘the ratio of any two corresponding sides in two

equiangular triangle is always the same’ ?

(A) Thales (B) Newton

(C) Aryabhatta (D) Pythagoras

74. ,d cká fcUnq ls ,d o`Ùk ij fdruh Li”kZ js[kk,¡ [khaph tk ldrh ga S \

27

(A) 1 (B) 2

(C) 3 (D) vufxur

How many tangents can be drawn to a circle from an external

point ?

(A) 1 (B) 2

(C) 3 (D) infinite

75. nh xbZ vkd`fr esa PA rFkk PB nks Li”kZ js[kk,¡ O dsUnz okys o`Ùk ij bl izdkj

gS] fd APO = 400 rks OAB =

(A) 400 (B) 300

(C) 200 (D) 900

A

P O

B

In the given figure PA and PB are two tangents to the circle with

centre O such that APO = 400 then OAB =

(A) 400 (B) 300

(C) 200 (D) 900
A

P 28 O

B

76. nh xbZ vkd`fr esa QOR o`Ùk dk O;kl gS rFkk PQ=PR gS] rks PQR =

(A) 900 (B) 600

(C) 450 (D) 300
P

Q R
O

In the given figure QOR is a diameter of a circle and PQ=PR, then

PQR =

(A) 900 (B) 600

(C) 450 (D) 300
P

Q R
O

77. ;fn O dsUnz okys o`Ùk esa AB vkSj AC nks Li”kZ js[kk;sa bl izdkj gS fd

BOC=1100 rks CAO=

(A) 600 (B) 450

(C) 350 (D) 1500

29

If AB and AC are two tangents to a circle with centre O such that

BOC=1100 then CAO=

(A) 600 (B) 450

(C) 350 (D) 1500

78. fcUnq T ls fdlh o`Ùk dh Li”kZ js[kk dh yEckbZ 12 lseh0 gS vkSj dsUnz ls T dh

nwjh 13 lseh0 gS] rc o`Ùk dh f=T;k gS &

(A) 6 lseh0 (B) 12-5 lseh0

(C) 25 lseh0 (D) 5 lseh0

From a point T the length of the tangent to a circle is 12cm and the

distance of T from the centre is 13cm. The radius of circle is –

(A) 6 cm (B) 12.5 cm

(C) 25 cm (D) 5 cm

79. nh xbZ vkd`fr esa O o`Ùk dk dsUnz gS rFkk QPR=600 rks OQR =

(A) 600 (B) 300

(C) 1200 (D) 400
P

O

Q R

In the given figure, O is the centre of the circle and QPR=600

then OQR =

30

(A) 600 (B) 300

(C) 1200 (D) 400

80. vk¡dM+s 15] 20] 84] 29] 41] 45] 60] 65 dk ifjlj gS &

(A) 69 (B) 65

(C) 84 (D) 15

The range of the data 15, 20, 84, 29, 41, 45, 60, 65 is –

(A) 69 (B) 65

(C) 84 (D) 15

81. 1] 2] 3] ---] n dk lekarj ek/; gS &

(A) +1 (B)

(C) (D)

The arithemetic mean of 1, 2, 3, …, n is –

(A) +1 (B)

(C) (D)

82. ;fn 3 ek/;d ¾ cgqyd $ 𝑥 ek/; rks 𝑥 dk eku gS

(A) (B)

(C) 2 (D)

If 3 Median = Mode + 𝑥 Mean then the value of 𝑥 is

31

(A) (B)

(C) 2 (D)

83. izs{k.k 12] 13] 11] 14] 12] 13] 15] 11] 13 dk cgqyd gS &

(A) 11 (B) 12

(C) 13 (D) 15

The mode of the observation 12, 13, 11 ,14, 12, 13, 15, 11, 13, is–

(A) 11 (B) 12

(C) 13 (D) 15

84. ;fn 7, 9, 13, 15, 𝑥 dk ek/; 10 gks rks 𝑥 dk eku gksxk &

(A) 6 (B) 9

(C) 11 (D) 13

If 10 is mean of 7, 9, 13, 15, 𝑥, then the value of 𝑥 will be –

(A) 6 (B) 9

(C) 11 (D) 13

85. fdlh ?kVuk dh izkf;drk dk vf/kdre eku gksrk gS &

(A) 1 (B) 0

(C) 2 (D)

The maximum value of probability of any event is –

(A) 1 (B) 0

32

(C) 2 (D)

86. fuEufyf[kr esa dkSu lh ,d ?kVuk dh izkf;drk ugha gks ldrh gS \

(A) 50% (B) 0.12

(C) (D)

Which of the following cannot be probability of an event ?

(A) 50% (B) 0.12

(C) (D)

87. ,d U;k;laxr iklk Qsadk x;k rks 2 dk xq.kt vkus dh izkf;drk gksxh &

(A) (B)

(C) (D) a

An unbiased die is thrown then the probability of getting a multiple

of 2 is –

(A) (B)

(C) (D) a

88. vPNh izdkj ls QsVh xbZ ,d rk”k dh xM~Mh esa ls ,d iÙkk ;kn`PN;k fudkyk

tkrk gS] rks blds yky jax dk iÙkk gksus dh izkf;drk gS &

(A) (B)

(C) (D) buesa ls dksbZ ugha

33

A card is drawn at random from a well shuffled deck of playing

cards, then the probability of getting a red card is –

(A) (B)

(C) (D) None of these

89. nks U;k;laxr flDds mNkys tkrs gS rks de&ls&de ,d 'kh"kZ vkus dh izkf;drk

gS &

(A) (B)

(C) (D)

Two unbiased coins are tossed then the probability of getting at

least one head is –

(A) (B)

(C) (D)

90. rhu flDdksa dh mNky esa laHko ifj.kkeksa dh la[;k gS &

(A) 3 (B) 10

(C) 6 (D) 8

The number of possible outcomes by tossing three coins is –

(A) 3 (B) 10

(C) 6 (D) 8

34

91. 8 lseh0 f=T;k okys o`Ùk ds vUr%oxZ dk {ks=Qy gksxk &

(A) 64 lseh02 (B) 84 lseh02

(C) 100 lseh02 (D) 128 lseh02

The area of a square inscribed in a circle of radius 8 cm is –

(A) 64 cm2 (B) 84 cm2

(C) 100 cm2 (D) 128 cm2

92. nks o`Ùkksa ds ifjf/k;ksa dk vuqikr 4%5 gSa] rks buds {ks=Qyksa dk vuqikr gksxk &

(A) 4:5 (B) 5:4

(C) 16:25 (D) 25:16

The ratio of circumferences of two circles is 4:5 then the ratio of

their areas is –

(A) 4:5 (B) 5:4

(C) 16:25 (D) 25:16

93. ;fn ,d leckgq f=Hkqt dk {ks=Qy 9√3 lseh02 gS rks bldh Hkqtk gS &

(A) 6 lseh0 (B) 3√3 lseh0

(C) 3 lseh0 (D) √3 lseh0

If the area of an equilateral triangle is 9√3 cm2 then its side is –

(A) 6cm (B) 3√3 cm

(C) 3 cm (D) √3 cm

35

94. ;fn R1 rFkk R2 f=T;kvksa okys nks o`Ùkksa ds ifjf/k;ksa dk ;ksx R f=T;k okys o`Ùk

ds ifjf/k ds cjkcj gks] rks

(A) R1 + R2 > R (B) R1 + R2 = R

(C) R1 + R2 < R (D) R1 + R2 =

If the sum of the circumferences of two circles with radii R1 and R2

is equal to the circumference of a circle of radius R then

(A) R1 + R2 > R (B) R1 + R2 = R

(C) R1 + R2 < R (D) R1 + R2 =

95. ;fn 'kadq dh Å¡pkbZ vkSj f=T;k rhu xquh gks tkrh gS] rks 'kadq dk vk;ru gks

tk,xk &

(A) 8 xquk (B) 16 xquk

(C) 6 xquk (D) 27 xquk

If the height and radius of the cone are thriced then volume of the

cone will be –

(A) 8 times (B) 16 times

(C) 6 times (D) 27 times

96. r f=T;k okys v)Zxksyk ds dqy i`"B dk {ks=Qy gS &

(A) 4𝜋r2 (B) 3𝜋r2

(C) 2𝜋r2 (D) 𝜋r2

36

The total surface area of a hemisphere with radius r is –

(A) 4𝜋r2 (B) 3𝜋r2

(C) 2𝜋r2 (D) 𝜋r2

97. 'kadq ds fNUud dk vk;ru gksxk &

(A) 𝜋h(R2 + r2 + Rr) (B) 𝜋 h(R2 + r2)

(C) 𝜋(R2 + r2) (D) 𝜋R2r2h

The volume of a frustum of a cone is –

(A) 𝜋h(R2 + r2 + Rr) (B) 𝜋 h(R2 + r2)

(C) 𝜋 (R2 + r2) (D) 𝜋R2r2h

98. ml lekarj Js<+h dk lkoZvUrj D;k gksxk ftlesa a18 – a14 = 32 ?

(A) 4 (B) -4

(C) 8 (D) -8

What is the common difference of an A.P in which a18 – a14 = 32?

(A) 4 (B) -4

(C) 8 (D) -8

99. xksyk dk O;kl 42 lseh0 gS rks bldk i`"B {ks=Qy gksxk &

(A) 1386 lseh02 (B) 4158 lseh02

(C) 5544 lseh02 (D) 2772 lseh02

The diameter of sphere is 42cm then its surface area is –

37

(A) 1386 cm2 (B) 4158 cm2

(C) 5544 cm2 (D) 2772 cm2

100. ,d csyu dh Å¡pkbZ 14 lseh0 gS vkSj csyu ds oØ lrg dk {ks=Qy 264

lseh02 gSA bldk vk;ru gS &

(A) 369 lseh03 (B) 503 lseh03

(C) 296 lseh03 (D) 396 lseh03

The height of a cylinder is 14cm and its curved surface area is

264cm2. The volume of the cylinder is –

(A) 369 cm3 (B) 503 cm3

(C) 296 cm3 (D) 396 cm3

[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- ;fn n ,d fo"ke iw.kkZad gS rks n”kkZb,¡ fd n2 -1] 8 ls foHkkT; gSA

If n is an odd integer then show that n2 -1 is divisible by 8. 2

2- ;qfDyM ds foHkktu ,YxksfjFe dk iz;ksx dj 960 rFkk 432 dk e0 l0 Kkr

djsaA
38

Using Euclid’s division algorithm, find the HCF of 960 and 432. 2

3- og lcls cM+h la[;k Kkr djsa tks 438 vkSj 606 dks foHkkftr djrh gS vkSj

izR;sd n”kk esa 6 'ks"kQy nsrh gSA

Find the largest number which divides 438 and 606, leaving

remainder 6 in each case. 2

4- f}?kkr cgqin 5y2 – 13y – 6ds 'kwU;dksa dks Kkr djsaA

Find the zeros of the polynomial 5y2 – 13y – 6. 2

5- f}?kkr cgqin x2 + 9x + 20 ds 'kwU;dksa rFkk xq.kkadksa ds chp ds laca/k dh tk¡p

djsaA

Verify the relation between the zeroes and the coefficient of the

quadratic polynomial x2 + 9x + 20. 2

6- K dk og eku Kkr dhft, ftlds fy, (y – 1), cgqin (2y3 + 9y2 + y + K)

dk ,d xq.ku[kaM gSA

Find the value of K for which (y – 1), is a factor of polynomial

(2y3 + 9y2 + y + K) 2

7- gy djsa %

x – y = 3, + =6

Solve :

x – y = 3, + =6 2

39

8- f}?kkr lehdj.k 16x2 – 8x + 1 = 0 ds ewyksa dh izd`fr Kkr djsaA

Find the nature of roots of the quadratic equation

16x2 – 8x + 1 = 0. 2

9- t ds og eku Kkr djsa] ftlds fy, f}?kkr lehdj.k 2y2 + ty = -3 ds nksuksa

ewy cjkcj gksA

Find the values of t for which the quadratic equation 2y2 + ty = -3

has both equal roots. 2

10- ml lekarj Js<+h ds izFke 51 inksa dk ;ksx Kkr djsa] ftlds nwljs vkSj rhljs

in Øe”k% 14 vkSj 18 gSA

Find the sum of first 51 terms of an A.P. whose second and third

terms are 14 and 18 respectively. 2

11- ;fn , K, 2 fdlh lekarj Js<+h ds rhu Øekxr in gS] rks K dk eku Kkr

djsaA

If , K, 2 are three consecutive terms of an A.P. then find the

value of K. 2

12- ;fn cosecA = gks rks cosA vkSj tanA dk eku Kkr djsa

If cosecA = then find the values of cosA and tanA 2

13- Ekku Kkr djsa &

40

4(𝑠𝑖𝑛 60 + 𝑐𝑜𝑠 60 )
𝑡𝑎𝑛 45 − 𝑐𝑜𝑠 90 + 𝑠𝑖𝑛90

Find the value :

( )
2

14- fl) djsa fd &

(sec2A – 1) (1 – cosec2A) = -1

Prove that

(sec2A – 1) (1 – cosec2A) = -1 2

15- fl) djsa fd

tan10.tan20.tan30 … tan890 = 1

Prove that :

tan10.tan20.tan30 … tan890 = 1 2

16- ;fn A vkSj B U;wudks.k gksa rFkk sinA = cosB gks rks fl) djsa fd

A+B=900.

If A and B are acute angles and sinA = cosB then prove that

A+B=900. 2

17- ;fn 3cotA = 4] rks dk eku Kkr djsaA

If 3cotA = 4, then find the value of 2

41

18- fcanqvksa A(-3, 5) vkSj B(4, -9) dks tksM+us okys js[kk[kaM dks fcanq P(2, -5) fdl

vuqikr esa foHkkftr djrk gS \

Find the ratio in which the line segment joining the points A(-3, 5)

and B(4, -9) is divided by P(2, -5). 2

19- ;fn fcUnq P(a, o), Q(o, b) rFkk R(2, 2) lajs[k gS rks fl) djsa fd + =

If points P(a, o), Q(o, b) and R(2, 2) are collinear then prove that

+ = . 2

20- nh xbZ vkd`fr esa = rFkk APQ = ACB rks fl) djsa fd ABC ,d

lef}ckgq f=Hkqt gSA
A

P Q

B C

In the given figure = and APQ = ACB then prove that

ABC is an isosceles triangle. 2

21- f=Hkqt ABC esa BC ij AD yac gS rks fl) djsa fd

AB2 – BD2 = AC2 – CD2.

In triangle ABC, AD is perpendicular to BC then prove that

42

AB2 – BD2 = AC2 – CD2. 2

22- ;fn O dsUnz okys o`Ùk ds ckg~; fcUnq T ls o`Ùk ij nks Li”kZ js[kk,¡ TA rFkk

TB [khaph xbZ gS rks fl) djsa fd TA = TB.

If TA and TB are two tangents drawn from the external point T to a

circle with centre at O then prove that TA = TB. 2

23- f=T;k 28 lseh0 rFkk dsUnzh; dks.k 450 okys o`Ùk ds f=T;[kaM dk {ks=Qy Kkr

djsaA

Find the area of a sector of circle of radius 28 cm and central

angle 450. 2

24- ,d ifg, dh f=T;k 50 lseh- gSA ;fn ;g ifg;k 66 fdeh0 izfr ?kaVs dh pky

ls py jgk gS] rks 20 feuV esa ;g ifg;k fdrus pDdj yxkrh gS \

The radius of a wheel is 50cm. How many complete revolutions

does the wheel make in 20 minutes when the wheel is travelling at

a speed of 66 km per hour ? 2

25- ,d ?ku dk lEiw.kZ i`"B dk {ks=Qy 726 lseh0 2 gSA bldk vk;ru Kkr djsaA

The total surface area of a cube is 726 cm2. Find its volume. 2

43

26- ,d csyukdj [kaEHks dk oØi`"B dk {ks=Qy 264 eh02 rFkk bldk vk;ru 924

eh03 gS] rks [kEHks dh Å¡pkbZ Kkr djsaA

The curved surface area of a cylindrical pillar is 264m2 and its

volume is 924 m3 then find the height of the pillar. 2

27- vPNh izdkj ls QsaVh xbZ 52 iÙkks dh ,d xM~Mh esa ls ,d iÙkk fudkyk tkrk

gSA yky jax dk jkuh izkIr djus dh izkf;drk Kkr djsaA

One card is drawn from a well-shuffled deck of 52 cards. Find the

probability of getting a queen of red colour. 2

28- vaxzsth o.kZekyk dk ,d v{kj ;kn`PN;k pquk tkrk gS] rks blds O;atu v{kj

gksus dh izkf;drk Kkr djsa

A letter of english alphabet is chosen at random then find the

probability that letter is a consonant. 2

29- fuEu forj.k ls ekf/;dk dh x.kuk djsa &

oxZ&varjky 0&10 10&20 20&30 30&40 40&50

ckjackjrk 5 15 30 8 2

Calculate the median from the following distribution 2

Class- 0&10 10&20 20&30 30&40 40&50

Interval

Frequency 5 15 30 8 2

44

30- fuEu caVu dk ek/; Kkr djsa &

oxZ&varjky 0&2 2&4 4&6 6&8 8&10 10&12

ckjackjrk 1 2 1 5 6 5

Find the mean of the following distribution 2

Class- 0&2 2&4 4&6 6&8 8&10 10&12

Interval

Frequency 1 2 1 5 6 5

nh?kZ mÙkjh; iz”u @ 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. Answer any 4 questions.

Each question carries 5 marks. 4x5=20

31- vkys[kh; fof/k ls gy djsa %

x – y = 1, 2x + 3y = 12.

Solve graphically : x – y = 1, 2x + 3y = 12. 5

32- ,d ehukj ds ikn ls ,d Hkou ds f”k[kj dk mUu;.k dks.k 300 gS Hkou ds ikn

ls ehukj ds f”k[kj dk mUu;u dks.k 600 gSA ;fn ehukj dh Å¡pkbZ 60 eh0 gks

rks Hkou dh Å¡pkbZ Kkr djsaA
45

The angle of elevation of the top of a building from the foot of the

tower is 300 and the angle of elevation of the top of the tower from

the foot of the building is 600. If the tower is 60m high, find height of

the building. 5

33- fl) djsa fd ;fn fdlh f=Hkqt dh ,d Hkqtk ds lekarj ,d js[kk [khaph tk,]

rks vU; nks Hkqtk,¡ ,d gh vuqikr esa foHkkftr gks tkrh gSaA

Prove that if a line is drawn parallel to one side of a triangle, the

other two sides are divided in the same ratio. 5

34- fn[kkb, fd fcUnq (-1, -2), (1, 0), (-1, 2) vkSj (-3, 0) ,d oxZ ds 'kh"kZ gSaA

Show that the points (-1, -2), (1, 0), (-1, 2) and (-3, 0) are the

vertices of a square. 5

35- ,d fØdsV Vhe ds dksp us 7 cYys rFkk 6 xsansa 3800 :0 esa [kjhnhaA ckn esa

mlus 3 cYys rFkk 5 xsansa 1750 :0 esa [kjhnhaA izR;sd cYys vkSj izR;sd xsan dk

ewY; Kkr djsaA

The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800.

Later he buys 3 bats and 5 balls for Rs. 1750. Find the cost of

each bat and each ball. 5

36- ;fn sec𝜃 + tan𝜃 = m rks fl) djsa fd sin𝜃 =

If sec𝜃 + tan𝜃 = m then prove that sin𝜃 = 5

46

37- 4 lseh0 f=T;k ds ,d o`Ùk ij ,slh nks Li”kZ js[kk,¡ [khaps tks ijLij 600 ds

dks.k ij >qdh gksaA

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

inclined to each other at an angle of 600. 5

38- ,d Bksl ftlesa 120 lseh0 Å¡pkbZ vkSj 60 lseh0 f=T;k okyk ,d 'kadq

lfEefyr gS] tks 60 lseh0 f=T;k okys ,d v)xksys ij vkjksfir gSA bl Bksl

dks ikuh ls Hkjs gq, ,c yac o`Ùkh; csyu esa bl izdkj yacor Mky fn;k tkrk gS

fd ;g csyu dh ryh dks Li”kZ djsaA ;fn csyu dh f=T;k 60 lseh0 gS vkSj

Å¡pkbZ 180 lseh0 gS rks csyu esa 'ks"k cps ikuh dk vk;ru Kkr djsaA

A solid consisting of a right circular cone of height 120 cm and

radius 60 cm standing on a hemisphere of radius 60 cm is placed

upright in a right circular cylinder full of water such that it touches

the bottom. Find the volume of water left in the cylinder, if the

radius of the cylinder is 60 cm and its height is 180 cm. 5

47