Bihar Board Class 10th Model Paper 2026 Maths

iz'u iqfLrdk Øekad / Question Booklet Serial No. :

SECONDARY SCHOOL EXAMINATION – 2026
ek/;fed Ldwy ijh{kk & 2026
¼ANNUAL@okf"kZd ½
fo"k; dksM % MODEL QUESTION PAPER Question Booklet Set

Subject Code :
110 MATHEMATICS (COMPULSORY)
Code

xf.kr ¼vfuok;Z½
dqy iz'u % 100 $ 30 $ 8 ¾ 138 dqy eqfnzr i`"B % 43
Total Questions : 100 + 30 +8 = 138 Total Printed Pages : 43
¼le; % 3 ?kaVs 15 feuV½ ¼iw.kkZad % 100½
[Time : 3 Hours 15 Minutes] [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 their 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 their 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 /;kuiwoZd i<+us ds fy, 15 feuV dk vfrfjä le; fn;k x;k gSA
An extra time of 15 minutes has been allotted for thecandidates to read the questions carefully.

5- ;g iz'u i= nks [k.Mksa esa gS & [k.M Þvß ,oa [k.M ÞcßA
This question booklet is divided into two sections – Section “A” and Section “B”.

6- [k.M&v esa 100 oLrqfu"B ç'u gSa] ftuesa ls fdUgha dsoy 50 ç'uksa dk mÙkj nsuk vfuok;Z gS ¼çR;sd ds
fy, 1 vad fu/kkZfjr gS ½A ipkl ls vf/kd ç”uksa ds mÙkj nsus ij ÁFke 50 mÙkjksa dk gh ewY;kadu dEI;wVj
}kjk fd;k tk,xkA lgh mÙkj dks miyC/k djk;s x;s OMR mÙkj&i=d esa fn;s x;s lgh xksys dks
uhys@dkys c‚y isu ls izxk<+ djsaA fdlh Hkh çdkj ds OgkbVuj@rjy inkFkZ@CysM@uk[kwu vkfn dk
mÙkj&iqfLrdk esa ç;ksx djuk euk gS] vU;Fkk ijh{kk ifj.kke vekU; gksxkA

Page 1 of 43

In Section-A, there are 100 objective type questions, out of which any 50 questions are to be
answered (each carrying 1 mark). First 50 answers will be evaluated by the computer in case more
than 50 questions are answered. 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 Þcß esa 30 y?kq mÙkjh; iz'u gSAa ftuesa ls fdUgha 15 iz'uksa ds mÙkj nsuk vfuok;Z gSA izR;sd iz'u ds
fy, 2 vad fu/kkZfjr gSAa buds vfrfjä] 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.

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

Page 2 of 43

[k.M & v @ Section-A
oLrqfu"B iz'u @ Objective Type Questions
iz'u la[;k 1 ls 100 rd ds izR;sd 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 fpfàr
djsaA 50 x 1 = 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. 50 x 1 = 50

1- fuEufyf[kr esa dkSu vifjes; la[;k gS \

(A) √441 (B) √2601

(C) √2209 (D)

Which of the following is an irrational number?

(A) √441 (B) √2601

(C) √2209 (D)

2- ;fn nks /kukRed iw.kkZad ds yŒlŒ rFkk eŒl( leku gks rks la[;k,¡ gksx
a h
(A) vHkkT; (B) lg vHkkT;

(C) cjkcj (D) la;ä
q
If the LCM and HCF of two positive integers are equal then the numbers will be

(A) prime (B) Co-prime

(C) equal (D) Composite

Page 3 of 43

3- 0. 0001 =

(A) (B)

(C) (D)

4- fdlh iw.kkZad m ds fy, fo"ke la[;k dk :i gS &

(A) m (B) m + 1

(C) 6m (D) 6m + 1

For an integer m, the form of an odd number is -

(A) m (B) m + 1

(C) 6m (D) 6m + 1

5- ;fn 𝑎 = 3 × 5, 𝑏 = 3 × 5 × 7 , 𝑐 = 3 × 7 rFkk yŒlŒ (𝑎, 𝑏, 𝑐) = 3 × 5 × 7
rks 𝑝 =

(A) 1 (B) 2

(C) 3 (D) 4

If 𝑎 = 3 × 5, 𝑏 = 3 × 5 × 7 , 𝑐 = 3 × 7 and LCM (𝑎, 𝑏, 𝑐) = 3 × 5 × 7

then p =.

(A) 1 (B) 2

(C) 3 (D) 4

6- √81 dk ifjes;dkjh xq.kd gS &

(A) (B) 3

(C) √9 (D) √3

Page 4 of 43

Rationalising factor of √81 is

(A) (B) 3

(C) √9 (D) √3

7- 36𝑥 𝑦 , 54𝑥 𝑦 rFkk 90𝑥 𝑦 dk eŒlŒ gS ¼tgk¡ x rFkk y vHkkT; la[;k,¡ gSa½ &

(A) 18𝑥 𝑦 (B) 36𝑥 𝑦

(C) 90𝑥 𝑦 (D) 180𝑥 𝑦

The HCF of 36𝑥 𝑦 , 54𝑥 𝑦 and 90𝑥 𝑦 is (where x and y are prime numbers)

(A) 18𝑥 𝑦 (B) 36𝑥 𝑦

(C) 90𝑥 𝑦 (D) 180𝑥 𝑦

8- ;fn 1440 = 2 × 3 × 5 rks 𝑝 + 𝑞 − 𝑟 =
(A) 6 (B) 8

(C) 2 (D) 5

If 1440 = 2 × 3 × 5 then 𝑝 + 𝑞 − 𝑟 =

(A) 6 (B) 8

(C) 2 (D) 5

9- fuEufyf[kr esa fdldk n'keyo izlkj vlkar gS \

(A) (B)

(C) (D)

Page 5 of 43

Which of the following has non-terminating decimal expansion?

(A) (B)

(C) (D)

10- ;fn n ,d izkd`r la[;k gS] rks 9 −4 lnSo foHkkT; gksxk
(A) 13 (B) 5

(C) 5 rFkk 13 nksuksa (D) buesa ls dksbZ ugha

If n is a natural number, then 9 −4 is always divisible by
(A) 13 (B) 5

(C) both 5 and 13 (D) None of these

11- ;fn 65 rFkk 117 dk eŒlŒ 65𝑛 − 117 ds :i esa gS] rks 𝑛 dk eku gS

(A) 4 (B) 3

(C) 2 (D) 1

If the HCF of 65 and 117 is in the form 65𝑛 − 117, then the value of n is -

(A) 4 (B) 3

(C) 2 (D) 1

12- dk n'keyo izlkj] n'keyo ds fdrus LFkkuksa ds ckn lkar gksxk \

(A) 1 (B) 2

(C) 3 (D) 4

Page 6 of 43

How many places after decimal, the decimal expansion of will be terminating?

(A) 1 (B) 2

(C) 3 (D) 4

13- cgqin 𝑝(𝑥) vkSj 𝑞(𝑥) dk eŒlŒ (6𝑥 − 9) gS] rks 𝑝(𝑥) vkSj 𝑞(𝑥) gks ldrs gSa &

(A) (12𝑥 − 18), 3 (B) 3, (2𝑥 − 3)

(C) 3(2𝑥 − 3), 6(2𝑥 + 3) (D) 3(2𝑥 − 3) , 6(2𝑥 − 3)

The HCF of the polynomials 𝑝(𝑥) and 𝑞(𝑥) is (6𝑥 − 9)then 𝑝(𝑥) and 𝑞(𝑥) may
be -

(A) (12𝑥 − 18), 3 (B) 3, (2𝑥 − 3)

(C) 3(2𝑥 − 3), 6(2𝑥 + 3) (D) 3(2𝑥 − 3) , 6(2𝑥 − 3)

14- fuEufyf[kr esa dkSu chth; O;atd cgqin ugha gS \

(A) 𝑦 + 2𝑦 + √5𝑦 + 8 (B)

(C) 5𝑦2 + 4 𝑦 − 11 (D)

Which of the following algebraic expression is not polynomial?

(A) 𝑦 + 2𝑦 + √5𝑦 + 8 (B)

(C) 5𝑦2 + 4 𝑦 − 11 (D)

Page 7 of 43

15- ;fn cgqin 5𝑦 − 𝑦 + 25 ds 'kwU;d 𝛼, 𝛽 gSa rks 𝛼𝛽. =

(A) −25 (B) −1

(C) −5 (D) 25

If 𝛼, 𝛽 are the zeroes of the polynomial 5𝑦 − 𝑦 + 25 then 𝛼𝛽. =

(A) −25 (B) −1

(C) −5 (D) 25

16- ;fn cgqin 𝑝(𝑥) dk ,d 'kwU;d 5 gks rks fuEufyf[kr esa dkSu 𝑝(𝑥) dk ,d xq.ku[k.M

gS \
(A) 𝑥 − 5 (B) 2𝑥 + 5

(C) 𝑥 − 4 (D) 𝑥 + 5

If one of the zeroes of the polynomial 𝑝(𝑥) is 5 then which of the following is a factor

of 𝑝(𝑥) ?

(A) 𝑥 − 5 (B) 2𝑥 + 5

(C) 𝑥 − 4 (D) 𝑥 + 5

17- ;fn f}?kkr cgqin 2𝑥 + 7𝑥 + 𝑝 ds 'kwU;d 𝛼 rFkk 𝛽 bl izdkj gS fd 𝛼 + 𝛽 +
𝛼𝛽 = rks 𝑝 =

(A) 7 (B) −7

(C) 14 (D) −14

If 𝛼, 𝛽 be the zeroes of the quadratic polynomial 2𝑥 + 7𝑥 + 𝑝 such that 𝛼 +
𝛽 + 𝛼𝛽 = then 𝑝 =
(A) 7 (B) −7
(C) 14 (D) −14

Page 8 of 43

18- cgqin (𝑥 − 2)(𝑥 + 2) + (𝑥 − 1)(𝑥 + 1) dk ?kkr gS &

(A) 4 (B) 3

(C) 2 (D) 5

The degree of the polynomial (𝑥 − 2)(𝑥 + 2) + (𝑥 − 1)(𝑥 + 1) is

(A) 4 (B) 3

(C) 2 (D) 5

19- cgqin 𝑎𝑦 + 𝑏𝑦 + 𝑐, 𝑎 ≠ 0 dk vkys[k gksrk gS

(A) ijoy; (B) vfrijoy;

(C) o`Ùk (D) ljy js[kk

The graph of the polynomial 𝑎𝑦 + 𝑏𝑦 + 𝑐, 𝑎 ≠ 0 is

(A) parabola (B) hyperbola

(C) circle (D) straight line

20- ;fn fdlh f}?kkr cgqin 𝑟(𝑥) = 𝑥 + 𝑏𝑥 + 𝑐 ds 'kwU;d 6 vkSj −2 gksa rks 𝑏 rFkk 𝑐

ds eku Øe'k% gS
(A) −4, −12 (B) −4, 12

(C) 4, 12 (D) 4, −12

If 6 and −2 be the zeroes of a quadratic polynomial 𝑟(𝑥) = 𝑥 + 𝑏𝑥 + 𝑐 then the
values of b and c are respectively

(A) −4, −12 (B) −4, 12

(C) 4, 12 (D) 4, −12

Page 9 of 43

21- ;fn 3 =3 = √27 gS] rks 𝑦 dk eku gS

(A) 1 (B) 0

(C) (D) s

If 3 =3 = √27 then the value of 𝑦 is

(A) 1 (B) 0

(C) (D) s

22- ;fn js[kk,¡ 5𝑥 + 𝑘𝑦 = 8 rFkk 15𝑥 + 6𝑦 = 40 lekukUrj gS] rks 𝑘 dk eku gksxk

(A) 2 (B) 3

(C) 4 (D) 5

If the lines 5𝑥 + 𝑘𝑦 = 8 and 15𝑥 + 6𝑦 = 40 are parallel then the value of 𝑘 will be

(A) 2 (B) 3

(C) 4 (D) 5

23- lehdj.k ;qXe 6𝑥 − 9𝑦 + 6 = 0 vkSj 18𝑥 − 27𝑦 + 15 = 0 dk

(A) ,d vkSj dsoy ,d gy gS (B) vufxur gy gSa

(C) dksbZ gy ugha gS (D) buesa ls dksbZ ugha

The pair of equations 6𝑥 − 9𝑦 + 6 = 0 and 18𝑥 − 27𝑦 + 15 = 0 has

(A) one and only one solution (B) infinitely many solutions

(C) no solution (D) None of these

Page 10 of 43

24- jSf[kd lehdj.k 𝑦 = 𝑥 dk vkys[k fuEufyf[kr esa ls fdl fcUnq ls gksdj xqtjrk gS \

(A) , (B) 0,

(C) , (D) (2,2)

The graph of the linear equation 𝑦 = 𝑥 passes through which of the following point?

(A) , (B) 0,

(C) , (D) (2,2)

25- ;fn lekarj Js<+h dk 𝑛ok¡ in 3𝑛 + 8 gS rks mldk lkoZvUrj gksxk

(A) 3 (B) 2

(C) 4 (D) 5

If the 𝑛 term of an A.P. is 3𝑛 + 8 then its Common difference will be

(A) 3 (B) 2

(C) 4 (D) 5

26- lekarj Js<+h 81, 72, 63 … … … … dk dkSu lk in 0 gS \

(A) 8 ok¡ (B) 9 ok¡

(C) 7 ok¡ (D) 10 ok¡

Which term of the A.P. 81, 72,63 ……… is 0 ?

(A) 8th (B) 9th

(C) 7th (D) 10th

Page 11 of 43

27- ;fn A. P. dk 17ok¡ in 11osa in ls 36 vf/kd gS rks lkoZvUrj dk eku gS

(A) 5 (B) 6

(C) 7 (D) 8

If the 17th term of an A.P. is 36 more than the 11th term then the value of Common
difference is

(A) 5 (B) 6

(C) 7 (D) 8

28- 1 + 3 + 5 + ⋯ 𝑛 inksa rd =

(A) 𝑛 (B) 2𝑛

( )
(C) (D)

1 + 3 + 5 + ⋯ up to n terms =

(A) 𝑛 (B) 2𝑛

( )
(C) (D)

29- ;fn 𝑥, 9, 𝑦, 25 lekarj Js<h+ esa gSa rks 𝑥 rFkk 𝑦 dk eku gS

(A) 𝑥 = 1, 𝑦 = 17 (B) 𝑥 = 1, 𝑦 = 16

(C) 𝑥 = 2, 𝑦 = 16 (D) 𝑥 = 1, 𝑦 = 15

If 𝑥, 9, 𝑦, 25 are in A.P. then the value of 𝑥 𝑎𝑛𝑑 𝑦 is

(A) 𝑥 = 1, 𝑦 = 17 (B) 𝑥 = 1, 𝑦 = 16

(C) 𝑥 = 2, 𝑦 = 16 (D) 𝑥 = 1, 𝑦 = 15

Page 12 of 43

30- ;fn lekarj Js<+h ds 𝑛 inksa dk ;ksx 𝑆 gS rks 𝑛ok¡ in gksxk
(A) Sn+Sn-1 (B) Sn+Sn+1

(C) Sn−Sn-1 (D) buesa ls dksbZ ugha

If sum of 𝑛 terms of an A.P. is 𝑆 then 𝑛 term will be

(A) Sn+Sn-1 (B) Sn+Sn+1

(C) Sn−Sn-1 (D) None of these

31- lekarj Js<+h √7, √28, √63, … dk vxyk in gS

(A) √84 (B) √98

(C) √112 (D) √70

The next term of the A.P. √7, √28, √63, … is

(A) √84 (B) √98

(C) √112 (D) √70

32- 6 + 6 + √6 + … dk eku gS

(A) 3 (B) 4

(C) 6 (D) 3.5

The value of 6 + 6 + √6 + … is

(A) 3 (B) 4

(C) 6 (D) 3.5

Page 13 of 43

33- ;fn lehdj.k 𝑥 + 𝑝𝑥 + 12 = 0 dk ,d ewy 2 rFkk lehdj.k 𝑥 + 𝑝𝑥 + 𝑞 = 0 dk ewy

leku gks rks 𝑞 =
(A) −8 (B) 8

(C) −16 (D) 16

If 2 is a root of the equation 𝑥 + 𝑝𝑥 + 12 = 0 and the equation 𝑥 + 𝑝𝑥 + 𝑞 = 0
has equal roots then q =

(A) −8 (B) 8

(C) −16 (D) 16

34- ;fn lehdj.k 𝑥 − 𝑥 = 𝑘(4𝑥 − 1) ds ewyksa dk ;ksx 'kwU; gS rks 𝑘 =

(A) 4 (B) −4

(C) − (D)

If the sum of the roots of the equation 𝑥 − 𝑥 = 𝑘(4𝑥 − 1) is zero then k =

(A) 4 (B) −4

(C) − (D)

35- lehdj.k 5√5𝑥 + 10𝑥 + √5 = 0 dk foospd gS

(A) 25 (B) 0

(C) 50 (D) 20

The discriminant of equation 5√5𝑥 + 10𝑥 + √5 = 0 is

(A) 25 (B) 0

(C) 50 (D) 20

Page 14 of 43

36- ;fn lehdj.k 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 ds ewy 𝑠𝑖𝑛𝜃 rFkk 𝑐𝑜𝑠𝜃 gSa rks 𝑏 =

(A) 𝑎 + 2𝑐𝑎 (B) 𝑎 + 𝑐𝑎

(C) 𝑎 − 2𝑐𝑎 (D) 𝑎 − 𝑎𝑐

If 𝑠𝑖𝑛𝜃 and 𝑐𝑜𝑠𝜃 are roots of the equation 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 then 𝑏 =

(A) 𝑎 + 2𝑐𝑎 (B) 𝑎 + 𝑐𝑎

(C) 𝑎 − 2𝑐𝑎 (D) 𝑎 − 𝑎𝑐

37- ;fn lehdj.k 3(𝑥 + 2) − 6 = 0 ds ewy 𝛼 rFkk 𝛽 gksa rks + =

(A) −2 (B) 2

(C) −1 (D) −3

If 𝛼 and 𝛽 are the roots of the equation 3(𝑥 + 2) − 6 = 0 then + =

(A) −2 (B) 2

(C) −1 (D) −3

38- ;fn 𝑠𝑖𝑛𝛼 = rFkk 𝑐𝑜𝑠𝛽 = rks 𝛼 + 𝛽 =
(A) 90 (B) 60

(C) 30 (D) 0

If 𝑠𝑖𝑛𝛼 = and 𝑐𝑜𝑠𝛽 = then 𝛼 + 𝛽 =

(A) 90 (B) 60

(C) 30 (D) 0

Page 15 of 43

39- 𝑐𝑜𝑠𝑒𝑐 77 − tan 13 + 1 =

(A) 0 (B) 1

(C) −1 (D) 2

40- ;fn 𝐴 = 𝐵 = 45 rks 𝑠𝑖𝑛𝐴 + 𝑠𝑖𝑛𝐵 =
(A) (B) √2
√

√
(C) 1 (D)

If 𝐴 = 𝐵 = 45 then 𝑠𝑖𝑛𝐴 + 𝑠𝑖𝑛𝐵 =
(A) (B) √2
√

√
(C) 1 (D)

41- 𝑐𝑜𝑠𝑒𝑐50 =

(A) 𝑠𝑒𝑐50 (B) 𝑠𝑖𝑛40

(C) 𝑐𝑜𝑡40 (D) 𝑠𝑒𝑐40

42- ;fn = 20 rks 𝑠𝑒𝑐 3𝐴 =

(A) (B) 2

(C) 4 (D) 1

If = 20 then 𝑠𝑒𝑐 3𝐴 =

(A) (B) 2

(C) 4 (D) 1

Page 16 of 43

43- 𝑠𝑖𝑛15 . 𝑠𝑒𝑐75 + 𝑐𝑜𝑠15 . 𝑐𝑜𝑠𝑒𝑐75 =

(A) 2 (B) 0

(C) −1 (D) 1

44- (𝑐𝑜𝑡 45 + 1) =

(A) 2 (B) 4

(C) 9 (D)

45- ;fn √2𝑐𝑜𝑠2𝜃 − 1 = 0 rks 𝜃 =
1𝑜 1𝑜
(A) 22 (B) 20
2 2

(C) 45 (D) 30

If √2𝑐𝑜𝑠2𝜃 − 1 = 0 then 𝜃 =
1𝑜 1𝑜
(A) 22 (B) 20
2 2

(C) 45 (D) 30

46- × 𝑐𝑜𝑡60 =

(A) (B) 1

(C) 2 (D) 3

47- ;fn 𝑠𝑖𝑛𝜙 + sin 𝜙 = 1 rks cos 𝜙 + cos 𝜙 =

(A) 1 (B) 2

(C) (D) 3

Page 17 of 43

If 𝑠𝑖𝑛𝜙 + sin 𝜙 = 1 then cos 𝜙 + cos 𝜙 =
(A) 1 (B) 2

(C) (D) 3

48- 3(sin 𝛼 + sin (90 − 𝛼)) =
(A) 1 (B) 0

(C) 3 (D) −1

49- ;fn 𝑠𝑖𝑛𝐴 = rks 𝑡𝑎𝑛𝐴 =

(A) (B)
√ √

√ √
(C) (D)

If 𝑠𝑖𝑛𝐴 = then 𝑡𝑎𝑛𝐴 =

(A) (B)
√ √

√ √
(C) (D)

50- ;fn 𝑡𝑎𝑛𝛼 = rks 𝑠𝑒𝑐𝛼 =
√

(A) (B) 2

6
(C) (D)
5 √

Page 18 of 43

If 𝑡𝑎𝑛𝛼 = then 𝑠𝑒𝑐𝛼 =
√

(A) (B) 2

6
(C) (D)
5 √

51- ∆𝐴𝐶𝐵 esa ;fn ∠𝐶 = 90 rFkk 𝑐𝑜𝑡𝐵 = rks 𝑡𝑎𝑛𝐴 =

(A) (B)

(C) (D)

In ∆𝐴𝐶𝐵 if ∠𝐶 = 90 and 𝑐𝑜𝑡𝐵 = then 𝑡𝑎𝑛𝐴 =

(A) (B)

(C) (D)

52- ;fn 𝑡𝑎𝑛𝑥 + 𝑐𝑜𝑡𝑥 = 7 rks tan 𝑥 + cot 𝑥 =

(A) 49 (B) 47

(C) 45 (D) 50

If 𝑡𝑎𝑛𝑥 + 𝑐𝑜𝑡𝑥 = 7 then tan 𝑥 + cot 𝑥 =
(A) 49 (B) 47

(C) 45 (D) 50

53- ;fn 𝑠𝑖𝑛3𝐵 = cos (𝐵 − 10 ) rFkk 3𝐵 U;wudks.k gS rks ∠𝐵 =

(A) 25 (B) 20

(C) 35 (D) 45

Page 19 of 43

If 𝑠𝑖𝑛3𝐵 = cos (𝐵 − 10 ) and 3𝐵 is acute angle then ∠𝐵 =

(A) 25 (B) 20

(C) 35 (D) 45

54- ;fn sin 𝜙 − cos 𝜙 = 𝑇 rks sin 𝜙 − cos 𝜙 =

(A) 𝑇 (B) 𝑇

(C) 𝑇 (D) 𝑇

If sin 𝜙 − cos 𝜙 = 𝑇 then sin 𝜙 − cos 𝜙 =
(A) 𝑇 (B) 𝑇

(C) 𝑇 (D) 𝑇

55- ∆𝐴𝐵𝐶 esa tan =

(A) cot (B) tan

(C) 𝑐𝑜𝑡𝐵 (D) 𝑡𝑎𝑛𝐵

In ∆𝐴𝐶𝐵 tan =

(A) cot (B) tan

(C) 𝑐𝑜𝑡𝐵 (D) 𝑡𝑎𝑛𝐵

56- 7𝑐𝑜𝑠𝑒𝑐 30 − 7𝑐𝑜𝑡 30 =

(A) 0 (B) 7

(C) 14 (D) 5

Page 20 of 43

57- 6𝑐𝑜𝑠1 ∙ 𝑐𝑜𝑠2 ∙ 𝑐𝑜𝑠3 ∙∙∙ 𝑐𝑜𝑠180 =

(A) 6 (B) 1

(C) 0 (D) 3

jes'k ns[krk gS fd fnu ds le; ,d Hkou dh ijNkbZ dh yackbZ le;&le; ij cnyrh
jgrh gSA og ;g Hkh voyksdu djrk gS fd ;g vyx&vyx le; ij lw;Z dh fLFkfr cnyus
ds dkj.k gks jgk gSA bl tkudkjh ds vk/kkj ij iz'u la[;k 58 ,oa 59 dk mÙkj nsAa
Ramesh observes that the length of the shadow of a building is variable at different
timings in daytime. He also observes that it is due to position of the sun at different
timings. On the basis of this, answer the question number 58 and 59.

58- ;fn Hkou dh Å¡pkbZ vkSj mldh ijNkbZ dh yackbZ dk vuqikr √3: 1 gS rks lw;Z dk

mUu;u dks.k gS &
(A) 30 (B) 45

(C) 60 (D) buesa ls dksbZ ugha

If the ratio of the height of the building and the length of its shadow is √3: 1, then
the angle of elevation of the sun is -

(A) 30 (B) 45

(C) 60 (D) None of these

59- ;fn lw;Z dk mUu;u dks.k 45 gS rks Hkou dh Å¡pkbZ vkSj mldh ijNkbZ dh yackbZ dk
vuqikr gS &
(A) 1 ∶ 1 (B) 1 ∶ √3

(C) 1 ∶ 3 (D) √3 ∶ 1

Page 21 of 43

If the angle of elevation of the Sun is 45o, then the ratio of the height of the tower
to the length of its shadow is -

(A) 1 ∶ 1 (B) 1 ∶ √3

(C) 1 ∶ 3 (D) √3 ∶ 1

60- fcUnq − , fdl prqFkkZa'k esa gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

The point − , lies in which quadrant?

(A) first (B) second

(C) third (D) fourth

61- fcUnqvksa (𝑝 𝑐𝑜𝑠𝜙 + 𝑞𝑠𝑖𝑛𝜙, 0) rFkk (0, 𝑝𝑠𝑖𝑛𝜙 − 𝑞𝑐𝑜𝑠𝜙) ds chp dh nwjh gS

(A) (𝑝 + 𝑞) bdkbZ (B) (𝑝 − 𝑞 ) bdkbZ

(C) 𝑝 +𝑞 bdkbZ (D) (𝑝 + 𝑞 ) bdkbZ

The distance between the points (𝑝 𝑐𝑜𝑠𝜙 + 𝑞𝑠𝑖𝑛𝜙, 0) and (0, 𝑝𝑠𝑖𝑛𝜙 − 𝑞𝑐𝑜𝑠𝜙) is

(A) (𝑝 + 𝑞) units (B) (𝑝 − 𝑞 ) units

(C) 𝑝 +𝑞 units (D) (𝑝 + 𝑞 ) units

62- ewy fcUnq ls fcUnq 𝑃(2√2, −2√2) dh nwjh gS

(A) 2 bdkbZ (B) 4 bdkbZ

(C) √2 bdkbZ (D) 8 bdkbZ

Page 22 of 43

The distance of the point 𝑃(2√2, −2√2) form the origin is

(A) 2 units (B) 4 units

(C) √2 units (D) 8 units

63- lehdj.k 4𝑥 + 3𝑦 = 12 dk vkys[k] 𝑦 − v{k dks ftl fcUnq ij dkVrk gS] og gS

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

(C) (0, 4) (D) (4, 0)

The graph of the equation 4𝑥 + 3𝑦 = 12 cuts the y- axis at the point

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

(C) (0, 4) (D) (4, 0)

64- ;fn fcUnqvksa 𝑃(0, 0) rks 𝑄 (𝑥, 5) ds chp dh nwjh 15 bdkbZ gS] rks 𝑥 dk eku gksxk &

(A) ±10 bdkbZ (B) ±10√2 bdkbZ

(C) ±5 bdkbZ (D) ±5√2 bdkbZ

If the distance between the points 𝑃(0, 0) and 𝑄 (𝑥, 5) is 15 units then the value
of x is -
(A) ±10 units (B) ±10√2 units

(C) ±5 units (D) ±5√2 units

65- ;fn fdlh js[kk[k.M ds ,d Nksj ds funsZ'kkaad (8, −6) vkSj e/; fcUnq (4, 6) gksa rks

nwljs Nksj ds funsZ'kkad gSa
(A) (0, 18) (B) (18, 0)

(C) (6, 0) (D) (0, 12)

Page 23 of 43

If the co-ordinates of one end of a line segment are (8, -6) and its middle point is
(4, 6) then the co-ordinates of the other end are

(A) (0, 18) (B) (18, 0)

(C) (6, 0) (D) (0, 12)

66- fdlh f=Hkqt ds 'kh"kksaZ ds funsZ'kkaad (8, 12), (0, 8) vkSj (10, 10) gSa rks blds dsUnzd dk

funsZ'kkad gksxk &
(A) (3, 5) (B) (6, 10)

(C) (9, 15) (D) (10, 6)

The co-ordinates of the vertices of a triangle are (8, 12), (0, 8) and (10, 10), then
the co-ordinates of its centroid will be

(A) (3, 5) (B) (6, 10)

(C) (9, 15) (D) (10, 6)

67- fcUnq,¡ 𝑃(0, 6), 𝑄(4, 10) vkSj 𝑅(−12, −2) ls cus f=Hkqt dk {ks=Qy gS

(A) 6 oxZ bdkbZ (B) 12 oxZ bdkbZ

(C) 8 oxZ bdkbZ (D) 10 oxZ bdkbZ

The area of the triangle formed by points 𝑃(0, 6), 𝑄(4, 10) and 𝑅(−12, −2) is
(A) 6 square units (B) 12 square units

(C) 8 square units (D) 10 square units

68- ;fn fcUnq,¡ 𝐴(𝑘 + 1, 2𝑘), 𝐵(3𝑘, 2𝑘 + 3) rFkk 𝐶(5𝑘 − 1, 5𝑘) lajs[k gks rks 𝑘 =

1 1
(A) −2, (B) 2,
2 2
−1 −1
(C) −2,
2
(D) 2,
2
Z

Page 24 of 43

If the points 𝐴(𝑘 + 1, 2𝑘), 𝐵(3𝑘, 2𝑘 + 3) and 𝐶(5𝑘 − 1, 5𝑘) are collinear then
𝑘 =
1 1
(A) −2, (B) 2,
2 2

−1 −1
(C) −2,
2
(D) 2,
2
Z

69- ;fn f=Hkqt ds 'kh"kksaZ ds fu;ked (0, 8), (0, 0) rFkk (6, 0) gSa rks f=Hkqt dk ifjeki gksxk

&
(A) 48 bdkbZ (B) 14 bdkbZ

(C) 28 bdkbZ (D) 24 bdkbZ

If the co-ordinates of the vertices of a triangle are (0, 8), (0, 0) and (6, 0) then the
perimeter of the triangle will be -
(A) 48 units (B) 14 units

(C) 28 units (D) 24 units

70- 𝑥 − v{k ls fcUnq 𝑅(6, 12) dh nwjh gS

(A) 6 bdkbZ (B) 12 bdkbZ

(C) 18 bdkbZ (D) 9 bdkbZ

The distance of the point 𝑅(6, 12) from the x-axis is -
(A) 6 units (B) 12 units

(C) 18 units (D) 9 units

71- fuEufy[kr esa ls dkSu vkys[k }kjk fu/kkZfjr ugha fd;k tk ldrk gS \

(A) cgqyd (B) ek/;

(C) ekf/;dk (D) buesa ls dksbZ ugha

Page 25 of 43

Which of the following cannot be determined graphically?

(A) Mode (B) Mean

(C) Median (D) None of these

72- ;fn 2, 7, 6, 15 rFkk 𝑥 dk ek/; 10 gS rks 𝑥 =

(A) 10 (B) 20

(C) 15 (D) 50a

If the mean of 2, 7, 6, 15 and x is 10 then 𝑥 =

(A) 10 (B) 20

(C) 15 (D) 50

73- 7, 5, 4, 7, 5, 4, 5, 7, 6 rFkk 7 dk cgqyd gS

(A) 5 rFkk 7 nksuksa (B) flQZ 7

(C) 4 rFkk 7 nksuksa (D) 4] 5] ;k 7

The Mode of 7, 5, 4, 7, 5, 4, 5, 7, 6 and 7 is

(A) both 5 and 7 (B) only 7

(C) both 4 and 7 (D) 4, 5, 𝑜𝑟 7

74- ;fn izFke 𝑛 izkd`r la[;kvksa dk ek/; gS rks 𝑛 =

(A) 3 (B) 4

(C) 7 (D) 14

Page 26 of 43

If the mean of first 𝑛 natural numbers is then 𝑛 =

(A) 3 (B) 4

(C) 7 (D) 14

75- 7,11,5, 10, 6, 9, 8 dh ekf/;dk gS

(A) 6 (B) 7

(C) 8 (D) 9

The Median of 7,11,5, 10, 6, 9, 8 is

(A) 6 (B) 7

(C) 8 (D) 9

76- rhu fu"i{k flDdksa dks ,d ckj mNkyk tkrk gSA de&ls&de ,d fpÙk vkus dh izkf;drk

gS &

(A) (B)

(C) (D)

Three unbiased coins are tossed. The probability of getting at least one head is

(A) (B)

(C) (D)

77- ,d ikls dks ,d ckj Qsadk tkrk gSA 8 ls NksVh la[;k izkIr gksus dh izkf;drk gS

(A) 1 (B) 0

(C) (D)

Page 27 of 43

A die is thrown once. The probability of getting a number less than 8 is

(A) 1 (B) 0

(C) (D)

78- fuEufyf[kr esa ls dkSu ?kVuk dh izkf;drk ugha gks ldrh gS \

(A) 0.6 (B) 20%

(C) (D)

Which of the following cannot be the probability of an event?

(A) 0.6 (B) 20%

(C) (D)

79- ;fn 𝑃(𝐸) = rks 𝑃(𝐸 ) =

(A) (B)

(C) (D)

If 𝑃(𝐸) = then 𝑃(𝐸 ) =

(A) (B)

(C) (D)

80- ,d FkSys esa 9 yky] 10 gjk vkSj 7 lQsn xsansa gSaA ,d xasn ;n`PN;k fudkyk tkrk gSA
blds u yky] u lQsn gksus dh izkf;drk gS
(A) (B)

(C) (D)

Page 28 of 43

A bag contains 9 red, 10 green and 7 white balls. One ball is taken out at random.
The probability that it is neither red nor white is -

(A) (B)

(C) (D)

81- 3√2 lsehŒ Hkqtk okyh leckgq f=Hkqt dk {ks=Qy gS&

(A) √3 lsehŒ2 (B) 18√3 lsehŒ2

(C)9√2 lsehŒ2 (D)36 lsehŒ2

The area of an equilateral triangle with side 3√2 cm. is

(A) √3𝑐𝑚. (B) 18√3𝑐𝑚.

(C) 9√2 𝑐𝑚. (D) 36 𝑐𝑚.

82- 𝑑 + rFkk 𝑑 fod.kksaZ okyh leprqHkqt
Z dk ifjeki gS
(A) (𝑑1 + 𝑑2 ) bdkbZ (B) 2 × (𝑑1 + 𝑑2 ) bdkbZ

2 2 2 2
(C) 2 × 𝑑1 + 𝑑2 bdkbZ (D) 𝑑1 + 𝑑2 bdkbZ

The perimeter of a rhombus with diagonals 𝑑 and 𝑑 is

(A) (𝑑 + 𝑑 ) units (B) 2 × (𝑑1 + 𝑑2 ) units

2 2 2 2
(C) 2 × 𝑑1 + 𝑑2 units (D) 𝑑1 + 𝑑2 units

83- ;fn ,d xksys dk i`"Bh; {ks=Qy 154 lsehŒ2 gS rks mldk O;kl gS

(A) 7 lsehŒ (B) 3-5 lsehŒ

(C) 14 lsehŒ (D) 49 lsehŒ

Page 29 of 43

If the surface area of a sphere is 154cm.2 then its diameter is

(A) 7 cm. (B) 3.5 cm.

(C) 14 cm. (D) 49 cm.

84- f=T;k vkSj 4𝑙 fr;Zd Å¡pkbZ okys ,d 'kadq dk dqy i`"Bh; {ks=Qy gS &

(A) 2𝜋𝑟(𝑙 + 𝑟) (B) 𝜋𝑟(𝑙 + )

(C) 4𝜋𝑟𝑙 (D) 𝜋𝑟(𝑟 + 𝑙)

The total surface area of a cone of radius and slant height 4𝑙 is -

(A) 2𝜋𝑟(𝑙 + 𝑟) (B) 𝜋𝑟(𝑙 + )

(C) 4𝜋𝑟𝑙 (D) 𝜋𝑟(𝑟 + 𝑙)

85- ,d ?ku dk i`"B {ks=Qy 486 lsehŒ2 gS rks bldk vk;ru gS &

(A)216 lsehŒ3 (B) 629 lsehŒ3

(C)512 lsehŒ3 (D) 729 lsehŒ3

The surface area of a cube is 486𝑐𝑚. then its volume is -

(A) 216 𝑐𝑚. (B) 629 𝑐𝑚.

(C) 512 𝑐𝑚. (D) 729 𝑐𝑚.

86- leku Å¡pkbZ ds nks csyuksa ds oØi`"Bksa ds {ks=Qyksa dk vuqikr 5 % 3 gSA buds vk;ruksa

dk vuqikr gS
(A) 25 9 (B) 5 3
(C) 9 25 (D) 3 5

Page 30 of 43

The ratio of the curved surface area of two cylinders of the same height is 5: 3. The
ratio of their volume is

(A) 25: 9 (B) 5: 3

(C) 9: 25 (D) 3: 5

87- ;fn ,d csyu vkSj ,d 'kadq ds O;kl vkSj Å¡pkbZ leku gksa rks muds vk;ruksa dk vuqikr

gksxk
(A) 1: 3 (B) 3: 1

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

If the diameter and height of a cylinder and a cone be equal then the ratio of their
volumes will be.

(A) 1: 3 (B) 3: 1

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

88- R f=T;k okyh ,d xksykdkj xsan dks fi?kykdj 𝑅 f=T;k okyh 27 ubZ leku xsna sa cukbZ
tkrh gSa rks 𝑅: 𝑅 =
(A) 3: 1 (B) 1: 3

(C) 1: 9 (D) 9: 1

A spherical ball of radius R is melted to make 27 new identical balls each of radius
𝑅 then 𝑅: 𝑅 =

(A) 3: 1 (B) 1: 3

(C) 1: 9 (D) 9: 1

Page 31 of 43

89- nks le:i f=Hkqtksa ds {ks=Qyksa dk vuqikr 289 121 gS rks buds laxr Hkqtkvksa dk vuqikr
gS
(A) 17: 11 (B) 11: 17

(C) 17: 9 (D) 11: 13

If the ratio of the areas of two similar triangles is 289 : 121 then the ratio of their corresponding
sides is

(A) 17: 11 (B) 11: 17

(C) 17: 9 (D) 11: 13

90- ;fn ∆ 𝐴𝐵𝐶~∆𝐷𝐸𝐹, ∠𝐵 = 47 , ∠𝐹 = 53 rks ∠𝐷 =
(A) 80O (B) 40O

(C) 50O (D) 53O

If ∆ 𝐴𝐵𝐶~∆𝐷𝐸𝐹, ∠𝐵 = 47 , ∠𝐹 = 53 then ∠𝐷 =

(A) 80O (B) 40O

(C) 50O (D) 53O

.
91- ;fn ∆ 𝐴𝐵𝐶 esa 𝐷𝐸 ∥ 𝐵𝐶, = rFkk 𝐴𝐸 = 1lsehŒ rks 𝐸𝐶 =

(A) 1 lsehŒ (B) 2.5 lsehŒ

(C) 2 lsehŒ (D) 3 lsehŒ

.
In ∆ 𝐴𝐵𝐶 if 𝐷𝐸 ∥ 𝐵𝐶, = and 𝐴𝐸 = 1𝑐𝑚. then 𝐸𝐶 =

(A) 1𝑐𝑚. (B) 2.5 𝑐𝑚.

(C) 2 𝑐𝑚. (D) 3 𝑐𝑚.

Page 32 of 43

92- ;fn ∆ 𝐴𝐵𝐶 esa AB = 6lsehŒ] BC = 12lsehŒ vkSj 𝐴𝐶 = 6√3lsehŒ rks ∠𝐴 =

(A) 90O (B) 60O

(C) 45O (D) 30O

In ∆ 𝐴𝐵𝐶 if AB=6cm., BC=12cm. and AC= 6√3cm. then ∠𝐴 =

(A) 90O (B) 60O

(C) 45O (D) 30O

93- ∆ 𝐴𝐵𝐶 esa AD, ∠𝐵𝐴𝐶 dk lef}Hkktd gSA ;fn AB= 8lsehŒ] BD= 6lsehŒ rFkk DC=

3lsehΠrks AC =
(A) 6 lsehŒ (B) 8 lsehŒ

(C) 4 lsehŒ (D) 3 lsehŒ

In a ∆ 𝐴𝐵𝐶, 𝐴𝐷 is the bisector of ∠𝐵𝐴𝐶 . If AB=8cm., BD=6cm. and DC=3cm.
then AC =

(A) 6 𝑐𝑚. (B) 8 𝑐𝑚.

(C) 4 𝑐𝑚. (D) 3 𝑐𝑚.

94- ;fn nks f=Hkqtksa ABC rFk DEF esa ∠𝐴 = ∠𝐸, ∠𝐵 = ∠𝐹 rks fuEufyf[kr esa dkSu lgh

ugha gS \

(A) = (B) =

(C) = (D) =

If in two triangles ABC and DEF, ∠𝐴 = ∠𝐸, ∠𝐵 = ∠𝐹 then which of the following is not
true?

(A) = (B) =

(C) = (D) =

Page 33 of 43

95- ;fn O dsUnz okys o`Ùk ij nks Li'kZ js[kk,¡ PA rFkk PB bl izdkj gS fd ∠𝐴𝑃𝐵 = 100

rks ∠𝑂𝐴𝐵 =
(A) 80O (B) 40O

(C) 60O (D) 50O

If two tangents PA and PB to a circle with centre O are such that

∠𝐴𝑃𝐵 = 100 then ∠𝑂𝐴𝐵 =

(A) 80O (B) 40O

(C) 60O (D) 50O

96- nh xbZ vkd`fr esa BOC o`Ùk dk O;kl rFkk AB=AC rks ∠𝐴𝐶𝐵 =

(A) 45O (B) 60O

(C) 30O (D) 90O

In the given figure BOC is a diameter of a Circle and AB=AC then ∠𝐴𝐶𝐵 =

(A) 45O (B) 60O

(C) 30O (D) 90O

97- ;fn 5 lsehŒ f=T;k okys ,d o`Ùk ij] ijLij 600 dk dks.k cukrh gqbZ nks Li'kZ js[kk,¡

[khaph xbZ gksa] rks izR;sd Li'kZ&js[kk dh yackbZ gS
√
(A) 5√3 lsehŒ (B) lsehŒ

(C) 10 lsehŒ (D) lsehŒ
√

Page 34 of 43

If two tangents inclined at an angle of 60o are drawn to a circle of radius 5cm., then
length of each tangent is

√
(A) 5√3 𝑐𝑚. (B) 𝑐𝑚.

(C) 10 𝑐𝑚. (D) 𝑐𝑚.
√

98- ,d o`r ds ,d gh [k.M ds nks dks.k gksrs gSa

(A) laiwjd (B) cjkcj

(C) iwjd (D) buesa ls dksbZ ugha

Two angles in the same segment of a circle are

(A) Supplementary (B) equal

(C) Complementary (D) None of these

99- ,d o`r ds ifjxr ,d prqHkqZt PQRS gSA ;fn PQ = 6lseh( QR = 7lseh rFkk RS =

4lseh gS rks PS dh yackbZ gS

(A) 7 lsehŒ (B) 3 lsehŒ

(C) 4 lsehŒ (D) 6 lsehŒ

PQRS is quadrilateral circumscribing a circle. If PQ = 6cm., QR = 7cm. and
RS = 4cm. then the length of PS is

(A) 7 𝑐𝑚. (B) 3 𝑐𝑚.

(C) 4 𝑐𝑚. (D) 6 𝑐𝑚.

Page 35 of 43

100- Ckká :i ls Li'kZ djus okys nks o`Ùkks ds vf/kdre mHk;fu"B Li'kZ js[kkvksa dh la[;k gSa

(A) 1 (B) 2

(C) 4 (D) 3

The maximum number of common tangents of two circles touching eternally is

(A) 1 (B) 2

(C) 4 (D) 3

Section-B

Short Answer Type Questions
iz'u la[;k 1 ls 30 rd y?kq mÙkjh; iz'u gSaA bueas ls fdUgha 15 iz'uksa ds mÙkj nsaA izR;sd
iz'u ds fy, 2 vad fu/kkZfjr gSaA 15 x 2 = 30
Question Nos 1 to 30 are Short Answer Type Questions. Answer any 15 questions. Each question
carries 2 marks. 15 x 2 = 30

1- ;wfDyM foHkktu ,YxksfjFe dk iz;ksx dj 180] 252 rFkk 324 dk eŒl Kkr djsaA 2
Using Euclid’s division algorithm find the H.C.F. of 180, 252 and 324.

2- fl) djsa fd √3 + √5 ,d vifjes; la[;k gSA 2
Prove that √3 + √5 is an irrational number.

3- O;k[;k djsa fd 7 × 11 × 13 + 13 vkSj 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 HkkT;
la[;k,¡ D;ksa gS \ 2
Explain why 7 × 11 × 13 + 13 and 7×6×5×4×3×2×1+5 are
composite numbers?

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4- f}?kkr cgqin 4𝑥 + 5√2 𝑥 − 3ds 'kwU;dksa dks Kkr djsAa 2
Find the zeroes of the polynomial 4𝑥 + 5√2 𝑥 − 3.

5- ;fn cgqin 𝑝(𝑥) ds 'kwU;d 𝛼 rFkk 𝛽 bl izdkj gks fd 𝛼 + 𝛽 = 5 rFkk 𝛼𝛽 = −30

rks 𝑝(𝑥) Kkr djsaA 2
If 𝛼 𝑎𝑛𝑑 𝛽 are the zeroes of the polynomial 𝑝(𝑥) such that 𝛼 + 𝛽 = 5, and 𝛼𝛽 =
−30 then find 𝑝(𝑥).

6- ;fn 𝑃 = rks 𝑃 + Kkr djsaA 2

If 𝑃 = then find 𝑃 + .

7- gy djsa %

s−t=3 2

Solve :
s−t=3

8- vuqikrksa , rFkk dh rqyuk dj Kkr djsa fd ;qXe lehdj.k 2𝑥 − 3𝑦 + 4 = 0,

𝑥 + 2𝑦 − 5 = 0 laxr gS ;k vlaxrA 2

By comparing the ratios , and find out whether the pair of linear equations

2𝑥 − 3𝑦 + 4 = 0, 𝑥 + 2𝑦 − 5 = 0 are consistent or inconsistent.

9- gy djsa 2
4𝑥 − 4𝑎𝑥 + (𝑎 − 𝑏 ) = 0.
Solve :
4𝑥 − 4𝑎𝑥 + (𝑎 − 𝑏 ) = 0.

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10- 𝑘 ds fdl eku ds fy, f}?kkr lehdj.k 𝑘𝑥(𝑥 − 3) + 9 = 0 ds ewy leku gksaxsa \ 2
For what value of k will the roots of the equation 𝑘𝑥(𝑥 − 3) + 9 = 0 are equal?

11- lekarj Js<+h 6, 11, 16, 21, ∙∙∙ ds izFke 30 inksa dk ;ksx Kkr djsaA 2
Find the sum of the first 30 terms of the A.P. 6, 11, 16, 21,∙∙∙.

12- 10 vkSj 250 ds chp esa 4 ds fdrus xq.kt gS \ 2
How many multiples of 4 lie between 10 and 250.

13- lekarj Js<+h 213] 205] 197] ∙∙∙ 37 dk e/; in Kkr djsaA 2
Find the middle term of the A.P. 213, 205, 197 ∙∙∙,37.

14- og lekarj Js<h+ Kkr djsa ftldk rhljk in 5 vkSj lkrok in 9 gSA 2
Find the A.P. whose third term is 5 and seventh term is 9.

15- ;fn 𝑐𝑜𝑠𝜃 = gks rks 𝑡𝑎𝑛𝜃 + 𝑠𝑖𝑛𝜃 dk eku Kkr djsaA
If 𝑐𝑜𝑠𝜃 = then find the value of 𝑡𝑎𝑛𝜃 + 𝑠𝑖𝑛𝜃.

16- fl) djsa fd 2
= 𝑡𝑎𝑛60 .
Prove that

= 𝑡𝑎𝑛60 .

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17- fl) djsa fd 2
+ = 𝑠𝑖𝑛𝐴 + 𝑐𝑜𝑠𝐴.
Prove that

+ = 𝑠𝑖𝑛𝐴 + 𝑐𝑜𝑠𝐴.

18- ;fn 𝑠𝑖𝑛𝜙 − 𝑐𝑜𝑠𝜙 = 0 rks sin 𝜙 + cos 𝜙 dk eku Kkr djsaA

If 𝑠𝑖𝑛𝜙 − 𝑐𝑜𝑠𝜙 = 0 then find the value of sin 𝜙 + cos 𝜙.

19- ;fn fcUnqvksa 𝐴(7, −2), 𝐵(5, 1) rFkk 𝐶(3, 𝑝) ls cus f=Hkqt dk {ks=Qy 'kwU; gks rks 𝑝

dk eku Kkr djsaA 2
If the area of the triangle formed by the points 𝐴(7, −2), 𝐵(5, 1) and 𝐶(3, 𝑝) be
zero then find the value of p.

20- x vkSj y esa ,d laca/k Kkr djsa rkfd fcUnq (𝑥, 𝑦), fcUnqvksa (3, 6) vkSj (−3, 4)ls
lenwjLFk gksaA 2
Find a relation between x and y such that the point (x, y) is equidistant from the
points (3, 6) and (-3, 4).

21- og vuqikr Kkr djsa ftlesa fcUnq 𝑃 , , fcUnqvksa 𝑅 , rFkk 𝑆(2, −5) dks
feykusokys js[kk[k.M dks vUr% foHkkftr djrk gSA
Find the ratio in which the point 𝑃 , , divides internally the line segment joining the points

𝑅 , and 𝑆(2, −5).

22- ;fn fcUnq,¡ 𝑃(−2, 1), 𝑄(𝑥, 𝑦) rFkk 𝑅(4, −1) lajs[kh gS rFkk 𝑥 − 𝑦 = 1] rks 𝑥 rFkk 𝑦
ds eku Kkr djsaA 2
If the points 𝑃(−2, 1), 𝑄(𝑥, 𝑦) and 𝑅(4, −1) are collinear and 𝑥 − 𝑦 = 1 then
find the value of 𝑥 and y.

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23- nh xbZ vkd`fr esa 𝑂𝐴. 𝑂𝐵 = 𝑂𝐶. 𝑂𝐷 2

rks fl) djsa fd ∠𝐴 = ∠𝐶 rFkk ∠𝐵 = ∠𝐷.

In the given figure 𝑂𝐴. 𝑂𝐵 = 𝑂𝐶. 𝑂𝐷

then prove that ∠𝐴 = ∠𝐶and ∠𝐵 = ∠𝐷.

24- nks le:i f=Hkqtksa dk {ks=Qy Øe'k% 144lsehŒ2 vkSj 81lsehŒ2 gSA ;fn izFke f=Hkqt dh
ekf/;dk 16lsehΠgks] rks nwljsa f=Hkqt dh laxr ekf/;dk Kkr djsAa 2
The areas of two similar triangles are 144cm.2 and 81cm.2 respectively. If the
median of first triangle is 16cm. find the corresponding median of the other.

25- 5lsehŒ f=T;k okys ,d o`Ùk ij o`Ùk ds dsUnz ls 20lsehŒ nwjh ij fLFkr fcUnq R ls [khaph
xbZ Li'kZ js[kk dh yackbZ Kkr djsaA 2
Find the length of the tangent drawn to a circle of radius 5cm from a point R
situated at a distance of 20cm from the centre of the circle.

26- rhu cPpksa ds ifjokj esa de&ls&de ,d yM+dk gksus dh izkf;drk Kkr djsaA 2
In a family of three children, find the probability of having at least one boy.

27- ,d vko`fr forj.k dk ek/; vkSj cgqyd Øe'k% 30 rFkk 18 gSaA bldk ekf/;dk Kkr
djsaA 2
The mean and mode of a frequency distribution are 30 and 18 respectively. Find
its median.

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28- ,d v)Zxksyk dk vk;ru 2425 lsehŒ3 gSA bldk oØi`"B dk {ks=Qy fudkysAa

The volume of a hemisphere is 2425 𝑐𝑚. . Find its curved surface area.

29- ,d o`Ùk dh ifjf/k 8lsehŒ gSA ml f=T;[kaM dk {ks=Qy Kkr djsa ftldk dsUnzh; dks.k
720 gSA
The circumference of a circle is 8cm. Find the area of the sector whose central
angle is 72o.

30- ,d lkbfdy dk ifg;k 11fdehΠpyus esa 5000 pDdj yxkrk gSA ifg;s dk f=T;k Kkr
djsaA
A bicycle wheel makes 5000 revolutions in moving 11km. Find the radius of the
wheel.

Long Answer Type Questions

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; gaSA buesa ls fdUgha 4 iz'uksa dk mÙkj nsaA izR;sd iz'u ds
fy, 5 vad fu/kkZfjr gaSA 4 x 5 = 20
Question Nos 31 to 38 are Long Answer Type. Answer any 4 Questions. Each question carries 5
marks. 4 x 5 = 20

31- 2 fdyksŒ lsc vkSj 1 fdyksŒ vaxwj dk ewY; fdlh fnu ₹400 FkkA ,d eghus ckn 4
fdyksŒ lsc vkSj 2 fdykssŒ vaxwj dk ewY; ₹600 gks tkrk gSA bl fLFkfr dks chtxf.krh;
rFkk T;kferh; :iksa esa O;ä djsaA 5
The Cost of 2kg. of apples and 1kg of grapes on a day was found to be ₹400. After
a month, the cost of 4kg. of apples and 2kg of grapes is ₹600. Represent the
situation algebraically and geometrically.

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32- fl) djsa fd ,d ledks.k f=Hkqt esa d.kZ dk oxZ] vU; nks Hkqtkvksa ds oxksZa ds ;ksx ds
cjkcj gksrk gSA 5
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of
the squares of the other two sides.

33- ,d jsyxkM+h ,d leku pky ls 360 fdehΠdh nwjh r; djrh gSA ;fn ;g pky 5
fdehŒ@?kaVk vf/kd gksrh] rks og mlh ;k=k esa 1 ?kaVk de le; ysrhA jsyxkM+h dh pky
Kkr djsaA 5
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it
would have taken 1 hour less for the same journey. Find the speed of the train.

34- 8 ehVj Å¡ps ,d Hkou ds f'k[kj ls ,d dscy ehukj ds f'k[kj dk mUu;u dks.k 60 0
rFkk blds ikn dk vouu dks.k 450 gSA ehukj dh Å¡pkbZ Kkr djsAa 5
Find the top of a 8meter high building, the angle of elevation of the top of a cable
tower is 60o and the angle of depression of its foot is 450. Find the height of the
tower.

35- fl) djsa fd 5
+ = 1 + 𝑠𝑒𝑐𝜃𝑐𝑜𝑠𝑒𝑐𝜃.
Prove that

+ = 1 + 𝑠𝑒𝑐𝜃𝑐𝑜𝑠𝑒𝑐𝜃.

36- ,d f=Hkqt PQR dh jpuk djsa] ftlesa PQ = 5lsehŒ] QR = 12lsehŒ rFkk PR = 13lsehŒ
gSA PQR f=Hkqt ds le:i ,d nwljs f=Hkqt 𝑃′𝑄𝑅′ dh jpuk djsa] ftldh Hkqtk,¡ ∆𝑃𝑄𝑅
dh laxr Hkqtkvksa ds xquk gSA 5

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Construct a triangle PQR in which PQ = 5cm, QR = 12cm and PR = 13cm.
Construct another triangle 𝑃’𝑄𝑅’ similar to PQR, whose sides are of the

corresponding sides of ∆𝑃𝑄𝑅.

37- ,d 'kadq ds fNUud] tks 45lsehŒ Å¡pk gSa] ds fljksa dh f=T;k,¡ 28lsehŒ vkS 7 lsehŒ gSaA
bldk laiw.kZ i`"Bh; {ks=Qy Kkr djsAa 5

The radii of the ends of a frustum of a cone 45cm. high are 28cm. and 7cm. Find
its total surface area.

38- fl) djsa fd o`Ùk dh nks lekuakrj Li'kZ js[kkvksa ds chp ,d Li'kZ js[kk dk var%[kaM dsUnz
ij ledks.k varfjr djrk gSA 5
Prove that the intercept of a tangent between two parallel tangents to a circle
subtends a right angle at the centre.

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