HBSE Class 10 Mathematics (B) Question Paper 2018

CLASS : 10th (Secondary) Code No. 3504
Series : Sec. M/2018
Roll No.

xf.kr
MATHEMATICS
( Academic/Open )
[ fgUnh ,oa vaxszth ek/;e ]
[ Hindi and English Medium ]
(Only for Blind Candidates)
(Only for Fresh/Re-appear Candidates)
le; % 4 ?k.Vs ] [ iw.kk±d % 80
Time allowed : 4 hours ] [ Maximum Marks : 80
• d`i;k tk¡p dj ysa fd bl iz'u-i= esa eqfnzr iz'u 17 gSaA
Please make sure that the printed question paper are contains 17 questions.
• iz'u-i= esa nkfgus gkFk dh vksj fn;s x;s dksM uEcj dks Nk= mÙkj-iqfLrdk ds eq[;-i`"B ij
fy[ksaA
The Code No. on the right side of the question paper should be written by
the candidate on the front page of the answer-book.
• d`i;k iz'u dk mÙkj fy[kuk 'kq: djus ls igys] iz'u dk Øekad vo'; fy[ksaA
Before beginning to answer a question, its Serial Number must be written.
• mÙkj-iqfLrdk ds chp esa [kkyh iUuk / iUus u NksMsa+A
Don’t leave blank page / pages in your answer-book.
• mÙkj-iqfLrdk ds vfrfjDr dksbZ vU; 'khV ugha feysxhA vr% vko';drkuqlkj gh fy[ksa vkSj fy[kk
mÙkj u dkVsaA
Except answer-book, no extra sheet will be given. Write to the point and do
not strike the written answer.
• ijh{kkFkhZ viuk jksy ua0 iz'u-i= ij vo'; fy[ksaA
Candidates must write their Roll Number on the question paper.

3504 P. T. O.

(2) 3504
• d`i;k iz'uksa dk mÙkj nsus lss iwoZ ;g lqfuf'pr dj ysa fd iz'u-i= iw.kZ o lgh gS] ijh{kk ds
mijkUr bl lEcU/k esa dksbZ Hkh nkok Lohdkj ugha fd;k tk;sxkA
Before answering the question, ensure that you have been supplied the
correct and complete question paper, no claim in this regard, will be
entertained after examination.

lkekU; funsZ'k %
General Instructions :
(i) LkHkh iz'u vfuok;Z gSaA
All questions are compulsory.
(ii) bl iz'u&i= esa dqy 17 iz'u gSa] tks fd pkj [k.Mksa % v] c] l vkSj n esa ck¡Vs x,
gSa %
[k.M ^v* % bl [k.M esa ,d iz'u gS] ftlesa cgqfodYih; izdkj ds lksyg (i-xvi)
iz'u gS]a izR;sd iz'u 1 vad dk gSA lgh mÙkj viuh mÙkj&iqfLrdk esa
fy[ksAa
[k.M ^c* % bl [k.M esa 2 ls 6 rd dqy ik¡p iz'u gS]a izR;sd iz'u 3 vadksa dk gSA
[k.M ^l* % bl [k.M esa 7 ls 12 rd dqy N% iz'u gSa] izR;sd iz'u 4 vadksa dk gSA
[k.M ^n* % bl [k.M esa 13 ls 17 rd dqy ik¡p iz'u gSa] izR;sd iz'u 5 vadksa dk
gSA
This question paper consists of 17 questions in all, which are
divided into four Sections : A, B, C and D :
Section 'A' : This section consists of one question which has
Sixteen (i-xvi) Multiple type questions, each of 1
mark. Write correct answer in your answer-book.
Section 'B' : This section consists of five questions from 2 to 6,
each of 3 marks.
Section 'C' : This section consists of six questions from 7 to 12,
each of 4 marks.
Section 'D' : This section consists of five questions from 13 to 17,
each of 5 marks.
(iii) lexz :i ls dksbZ fodYi ugha gS] ysfdu [k.M ^n* ds rhu iz'uksa esa vkUrfjd fodYi
fn, x, gSaA ,sls iz'uksa esa ls vkidks dsoy ,d gh iz'u djuk gSA
3504

(3) 3504
There is no overall choice, but in three questions of Section 'D'
internal choices are given. You have to attempt only one of the given
choice in such questions.
[k.M & v
SECTION – A
1. (i) fuEufyf[kr esa ls 156 ds vHkkT; xq.ku[k.M gSa % 1
(A) 2 × 6 × 13 (B) 4 × 3 × 13
(C) 2 × 2 × 3 × 13 (D) buesa ls dksbZ ugha
Which of the following is prime factorization of 156 ?
(A) 2 × 6 × 13 (B) 4 × 3 × 13
(C) 2 × 2 × 3 × 13 (D) None of these
(ii) 42 vkSj 98 dk egÙke lekiorZd (H.C.F.)
gksxk % 1
(A) 26 (B) 14
(C) 13 (D) 91
The H.C.F. of 42 and 98 will be :
(A) 26 (B) 14
(C) 13 (D) 91
(iii) f}?kkr cgqin 2x 2 − 4x − 1 ds 'kwU;dksa dk ;ksx gksrk gS % 1
1 1
(A) (B) −
4 4
1
(C) − (D) 2
2
The sum of zeroes of the quadratic polynomial 2x 2 − 4x − 1 is :
1 1
(A) (B) −
4 4
1
(C) − (D) 2
2

3504 P. T. O.

(4) 3504
(iv) cgqin 4x − 8x ds 'kwU;d gksaxs %
2
1

(A) 0, 2 (B) 4, 2

(C) 4, 0 (D) 0, –2

The zeroes of polynomial 4x 2 − 8x will be :

(A) 0, 2 (B) 4, 2

(C) 4, 0 (D) 0, –2

(v) jSf[kd lehdj.k ;qXe x + 2y − 4 = 0 rFkk 2x + 4y − 12 = 0 ds gy gksaxs % 1

(A) vf}rh; gy
(B) dksbZ gy ugha
(C) vifjfer :i ls vusd gy
(D) buesa ls dksbZ ugha
The solution of pair of linear equations x + 2y − 4 = 0 and
2x + 4y − 12 = 0 will be :

(A) Unique solution

(B) No solution

(C) Infinitely many solution

(D) None of these

(vi) f}?kkr lehdj.k 2x 2 + x − 6 = 0 dk fofoDrdj gksxk %1
(A) 49 (B) 14
(C) 48 (D) 13

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(5) 3504
2
The discriminant of quadratic equation 2x + x − 6 = 0 will be :
(A) 49 (B) 14
(C) 48 (D) 13
(vii) lekarj Js.kh (A.P.) 1, 7, 13, ………….. , 199 esa fdrus in gSa \ 1

(A) 32 (B) 34
(C) 36 (D) 38
How many terms are in A.P. 1, 7, 13, ………….. , 199 ?
(A) 32 (B) 34
(C) 36 (D) 38

(viii) A.P. 2, 8, 14, ….. dk 15oka in gksxk % 1

(A) 82 (B) 84
(C) 86 (D) 90
15th term of an A.P. 2, 8, 14, ….. will be :
(A) 82 (B) 84
(C) 86 (D) 90

(ix) fcanv
q ksa (0, 0) rFkk (p, q) ds chp dh nwjh gksxh % 1

(A) 2 pq (B) p2 + q 2

(C) p2 − q 2 (D) p2 + q 2

The distance between points (0, 0) and (p, q) will be :

(A) 2 pq (B) p2 + q 2

(C) p2 − q 2 (D) p2 + q 2

3504 P. T. O.

(6) 3504
(x) fcUnqvksa (4, –1) rFkk (–2, –3) dks tksM+us okys js[kk[k.M ds e/;fcanq ds funsZ'kkad gksaxs
% 1
(A) (1, –2) (B) (2, –4)
(C) (–8, 3) (D) (12, 2)
The mid point of the line segment joining the points (4, –1) and (–
2, –3) will be :
(A) (1, –2) (B) (2, –4)
(C) (–8, 3) (D) (12, 2)
(xi) sec(90° − A ) cjkcj gS % 1
(A) cos A (B) sec A
(C) cosec A (D) cot A
sec(90° − A ) is equals to :
(A) cos A (B) sec A
(C) cosec A (D) cot A

(xii) cosec 59° – sec 31° dk eku gksxk % 1
(A) 1 (B) 0
(C) 2 (D) –1

The value of cosec 59° – sec 31° will be :
(A) 1 (B) 0
(C) 2 (D) –1
(xiii) fdlh o`Ùk dk O;kl 14 lseh0 gS] bldk {ks=Qy gksxk % 1
2
(A) 44 lseh0 (B) 44 lseh0
2
(C) 154 lseh0 (D) 154 lseh0
The diameter of a circle is 14 cm. Its area will be :

3504

(7) 3504
(A) 44 cm2 (B) 44 cm

(C) 154 cm (D) 154 cm2
(xiv) ;fn fdlh o`Ùk dk ifjeki o {ks=Qy la[;kRed :i ls cjkcj gS] rks o`Ùk dh f=T;k
gksxh % 1

(A) 7 ek=d (B) 2 ek=d

(C) π ek=d (D) 4 ek=d
If the perimeter and the area of a circle are numerically equal,
then the radius of the circle is :
(A) 7 units (B) 2 units
(C) π units (D) 4 units

(xv) fuEufyf[kr esa ls dkSu-lh la[;k fdlh ?kVuk dh izkf;drk ugha gks ldrh \ 1
1
(A) (B) –10%
3
(C) 18% (D) 0.5

Which of the following cannot be the probability of an event ?
1
(A) (B) –10%
3
(C) 18% (D) 0.5

(xvi) 52 iÙkksa dh vPNh izdkj ls QsaVh xbZ ,d xìh esa ls ,d iÙkk fudkyk tkrk gSA ,d
bZaV dh csxe izkIr djus dh izkf;drk gksrh gS % 1

(A) 1 (B) 1
13 52

(C) 1 (D) 1
26 3

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(8) 3504
One card is drawn from a well shuffled deck of 52 cards. The
probability of getting a queen of diamonds is :

(A) 1 (B) 1
13 52

(C) 1 (D) 1
26 3
[k.M & c
SECTION – B

2. vHkkT; xq.ku[k.M fof/k }kjk 92 vkSj 510 dk HCF Kkr dhft, vkSj fQj budk LCM
Kkr dhft,A 3

Find the HCF of 92 and 510 by the prime factorization method. Hence
find their LCM.

3. f}?kkr cgqin 3x 2 − x − 4 ds 'kwU;d Kkr dhft, vkSj 'kwU;dksa rFkk xq.kkadksa ds chp ds laca/k
dh lR;rk dh tk¡p dhft,A 3

Find the zeroes of the quadratic polynomial 3x 2 − x − 4 and verify the
relationship between the zeroes and the coefficients.

4. ,sls izFke 40 /ku iw.kkZadksa dk ;ksx Kkr dhft, tks 6 ls foHkkT; gSaA 3

Find the sum of the first 40 positive integers divisible by 6.

5. ;fn sin (A + B) = 1 vkSj cos(A − B) = 3 ; 0o < A + B ≤ 90o ; A > B , rks A rFkk B dk
2
eku Kkr dhft,A 3

3 o
If sin (A + B ) = 1 and cos (A − B ) = ; 0 < A + B ≤ 90° ; A > B , then find the
2
values of A and B.

3504

(9) 3504
6. fuEufyf[kr lkj.kh 35 uxjksa dh lk{kjrk nj (% esa) n'kkZrh gSA ek/; lk{kjrk nj Kkr
dhft, % 3

lk{kjrk nj ¼% esa½ 45-55 55-65 65-75 75-85 85-95
uxjksa dh la[;k 3 10 11 8 3

The following table gives the literacy rate (in %) of 35 cities. Find the
mean literacy rate :

Literacy rate (in %) 45-55 55-65 65-75 75-85 85-95

Number of Cities 3 10 11 8 3

[k.M & l
SECTION – C

7. lehdj.kksa ds fuEu ;qXe dks gy dhft, % 4

2 3 5 4
+ = 13, − = −2
x y x y

Solve the pair of equations :
2 3 5 4
+ = 13, − = −2
x y x y

9
8. ;fn fdlh fHkUu ds va'k vkSj gj nksuksa esa 2 tksM+ fn, tk, rks og gks tkrh gSA ;fn
11
va'k vkSj gj nksuksa esa 3 tksM+ fn;k tk, rks og 5 gks tkrh gSA og fHkUu Kkr dhft,A 4
6

3504 P. T. O.

( 10 ) 3504
9
A fraction becomes , if 2 is added to both the numerator and the
11
denominator. If 3 is added to both the numerator and the denominator
5
it becomes . Find the fraction.
6
9. x vkSj y esa ,slk lEcU/k Kkr dhft, fd fcUnq (x , y ) fcanqvksa (3, 6) vkSj (−3, 4) ls
lenwjLFk gksA 4
Find a relation between x and y such that the point (x , y ) is equidistant
from the point (3, 6) and (−3, 4) .

10. ,d leprqHkqZt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ] blh Øe esa (3, 0), (4, 5), (−1, 4) vkSj
(−2, − 1) gSaA 4
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and
(−2, − 1) taken in order.

11. 15 lseh0 f=T;k okys ,d o`Ùk dh dksbZ thok dsUnz ij 60° dk dks.k varfjr djrh gSA
fuEufyf[kr ds {ks=Qy Kkr dhft, % 4
(i) laxr y?kq o`Ùk[kaM
(ii) laxr nh?kZ f=T;k[k.M
[π = 3.14 vkSj 3 = 1.73 dk iz;ksx dhft,]
A chord of a circle of radius 15 cm subtends an angle of 60° at the
centre. Find the area of the corresponding :
(i) Minor segment
(ii) Major sector
[use π = 3.14 and 3 = 1.73]

12. ,d [ksy esa ,d #i;s ds flDds dks rhu ckj mNkyk tkrk gS vkSj izR;sd ckj dk ifj.kke
fy[k fy;k tkrk gSA rhuksa ifj.kke leku gksus ij] vFkkZr~ rhu fpr ;k rhu iV izkIr gksus ij
fl)kUr [ksy esa thr tk,xk] vU;Fkk og gkj tk,xkA fl)kUr ds [ksy esa gkj tkus dh
izkf;drk Kkr dhft,A 4

3504

( 11 ) 3504
A game consists of tossing a one rupee coin 3 times and noting
its outcome each time. Siddhant wins if all the tosses give the same
result i.e. three heads or three tails, and loses otherwise. Find the
probability that Siddhant will lose the game.
[k.M & n
SECTION – D

13. ,d ledks.k f=Hkqt dk 'kh"kZyac blds vk/kkj ls 7 lseh0 de gSA ;fn d.kZ dh yEckbZ 13
lseh0 gks] rks vU; nks Hkqtkvksa dh yEckbZ Kkr dhft,A 5

The altitude of a right triangle is 7 cm less than its base. If the
hypotenuse is 13 cm, find the other two sides.

14. ml A.P. ds 51 inksa dk ;ksx Kkr dhft, ftlds nwljs vkSj rhljs in Øe'k% 14 vkSj 18
gSaA 5
Find the sum of first 51 terms of an A.P. whose second and third terms
are 14 and 18 respectively.
vFkok
OR

fdlh A.P. dk rhljk o ukSaok in Øe'k% 4 vkSj –8 gSA Kkr dhft, fd bl A.P. dk
dkSu-lk in 'kwU; gS \ 5

If the 3rd and 9th terms of an A.P. are 4 and –8 respectively. Which term
of this A. P. is zero ?

15. 7 eh0 Å¡ps Hkou ds f'k[kj ls ,d dscy VkWoj ds f'k[kj dk mUu;u dks.k 60° gS vkSj blds
ikn dk voueu dks.k 45° gSA VkWoj dh Å¡pkbZ Kkr dhft,A 5
From the top of a 7 m high building, the angle of elevation of the top of
a cable tower is 60° and the angle of depression of its foot is 45°.
Determine the height of the tower.

3504 P. T. O.

( 12 ) 3504
16. O;kl 3 eh0 dk ,d dq¡vk 14 eh0 dh xgjkbZ rd [kksnk tkrk gSA blls fudyh gqbZ feêh dks
dq¡, ds pkjksa vksj 4 eh0 pkSM+k ,d o`Ùkkdkj oy; cukrs gq,] leku :i ls QSykdj ,d
izdkj dk ck¡/k cuk;k tkrk gSA bl ck¡/k dh Å¡pkbZ Kkr dhft,A5
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has
been spread evenly all around it in the shape of a circular ring of width
4 m to form an embankment. Find the height of the embankment.
vFkok
OR
Å¡pkbZ 220 lseh0 vkSj vk/kkj O;kl 24 lseh0 okys ,d csyu] ftl ij Å¡pkbZ 60 lseh0
vkSj f=T;k 8 lseh0 okyk ,d vU; csyu vkjksfir gS] ls yksgs dk ,d LraHk cuk gSA bl
LraHk dk nzO;eku Kkr dhft,] tcfd fn;k gS % 1 lseh03 yksgs dk nzO;eku yxHkx 8 xzke gksrk
gSA (π = 3.14 yhft,½ 5
A solid iron pole consists of a cylinder of height 220 cm and base
diameter 24 cm, which is surmounted by another cylinder of height 60
cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron
has approximately 8 g mass. (use π = 3.14)

17. fuEufyf[kr ckjackjrk caVu fdlh eksgYys ds 68 miHkksDrkvksa dh fctyh dh ekfld [kir
n'kkZrk gSA bu vk¡dM+ksa dk ek/;d Kkr dhft, % 5
ekfld [kir 65-85 85-105 105-125 125-145 145-165 165-185 185-205
¼bdkb;ksa es½a
miHkksDrkvksa dh 4 5 13 20 14 8 4
la[;k
The following frequency distribution gives the monthly consumption of
electricity of 68 consumers of a locality. Find the median of the data :
Monthly 65-85 85-105 105-125 125-145 145-165 165-185 185-205
Consumption
(in units)
Number of 4 5 13 20 14 8 4
Consumers

3504

( 13 ) 3504
vFkok
OR
fuEufyf[kr caVu Hkkjr ds mPprj ek/;fed Ldwyksa esa] jkT;ksa ds vuqlkj f'k{kd-fo|kFkhZ
vuqikr dks n'kkZrk gSA bu vk¡dM+ksa dk cgqyd Kkr dhft, % 5
izfr f'k{kd fo|kfFkZ;ksa dh la[;k jkT;@la?kh; {ks=ksa dh la[;k
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2
The following distribution gives the state-wise teacher-student ratio in
higher secondary schools of India. Find the mode of this data :
Number of Students Number of States/U. T.
per teacher
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2

S
3504 P. T. O.