HBSE Class 10 Question Paper 2025 Math Standard

CLASS : 10th (Secondary) Code No. 2103
Series : Sec/Annual Exam.-2025
Roll No. SET : A

xf.kr ¼ekud½
MATHEMATICS (Standard)
(Academic/Open)
[ fgUnh ,oa vaxzsth ek/;e ]
[ Hindi and English Medium ]
(Only for Fresh/Re-appear/Improvement/Additional Candidates)

le; : 3 ?k.Vs ] [ iw.kk±d : 80
Time allowed : 3 hours ] [ Maximum Marks : 80

• Ñi;k tk¡p dj ysa fd bl iz'u&i= esa eqfnzr i`"B 24 rFkk iz'u 38 gSaA
Please make sure that the printed pages in this question paper are 24 in number
and it contains 38 questions.

• iz'u&i= esa nkfgus gkFk dh vksj fn;s x;s dksM uEcj rFkk lsV dks Nk= mÙkj&iqfLrdk ds eq[;&i`"B ij
fy[ksaA
The Code No. and Set on the right side of the question paper should be written by
the candidate on the front page of the answer-book.

• Ñi;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.

2103/(Set : A) P. T. O.

(2) 2103/(Set : A)
• 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 jksy ua0 ds vfrfjDr iz'u&i= ij vU; dqN Hkh u
fy[ksa vkSj oSdfYid iz'uksa ds mÙkjksa ij fdlh izdkj dk fu'kku u yxk,¡A
Candidates must write their Roll No. on the question paper. Except Roll No. do not
write anything on question paper and don't make any mark on answers of objective
type questions.
• d`i;k iz'uksa ds mÙkj nsus lss iwoZ ;g lqfuf'pr dj ysa fd iz'u&i= iw.kZ o lgh gS] ijh{kk ds mijkUr bl
lEcU/kk esa dksbZ Hkh nkok Lohdkj ugha fd;k tk;sxkA
lEcU/
Before answering the questions, 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) bl ç'u-i= esa 5 [k.M % d] [k] x] ?k vkSj ³ gSaA
There are 5 Sections : A, B, C, D and E in this question paper.

(ii) [k.M – d esa 1 ls 20 rd 1-1 vad ds ç'u gSAa 1 ls 18 rd cgqfodYih; (MCQs), ,d 'kCn
mÙkjh;] fjDr LFkku iwfrZ rFkk ç'u la[;k 19 vkSj 20 vfHkdFku-dkj.k vk/kkfjr ç'u gSaA
Section–A consists of 1 mark questions from 1 to 20. 1 to 18 are Multiple
Choice Questions (MCQs), one word answer, fill in the blank and question
numbers 19 and 20 are Assertion-Reason based questions.

2103/(Set : A)

(3) 2103/(Set : A)
(iii) [k.M – [k esa 21 ls 25 rd vfr y?kq mÙkjh; (VSA) çdkj ds 2-2 vadksa ds ç'u gSaA
Section – B consists of Very Short Answer (VSA) type questions of 2 marks
each from 21 to 25.

(iv) [k.M – x esa 26 ls 31 rd y?kq mÙkjh; (SA) çdkj ds 3-3 vadksa ds ç'u gSasA
Section – C consists of Short Answer (SA) type questions of 3 marks each
from 26 to 31.

(v) [k.M – ?k esa 32 ls 35 rd nh?kZ mÙkjh; (LA) çdkj ds 5-5 vadksa ds ç'u gSaA
Section – D consists of Long Answer (LA) type questions of 5 marks each
from 32 to 35.

(vi) [k.M – ³ esa ç'u la[;k 36 ls 38 rd çdj.k v/;;u vk/kkfjr 4-4 vadksa ds ç'u gSaA

Question Numbers 36 to 38 in Section – E are case study based questions
of 4 marks each.

(vii) lHkh ç'u vfuok;Z gSaA gk¡ykfd [k.M – [k ds 2 ç'uksa esa] [k.M – x ds 2 ç'uksa esa] [k.M – ?k ds
lHkh ç'uksa esa rFkk [k.M – ³ ds 3 ç'uksa esa vkarfjd fodYi dk çko/kku fn;k x;k gSA muesa ls
vkidks ,d ç'u dks pquuk gSA

All questions are compulsory. However, provision of internal choice has
been made in 2 questions of Section–B, 2 questions of Section–C, all
questions of Section–D, 3 questions of Section–E. You have to choose one
question of them.

2103/(Set : A) P. T. O.

(4) 2103/(Set : A)
[k.M – d
SECTION – A

1. ;fn nks /kukRed iw.kkZad X rFkk Y dks bl izdkj O;Dr fd;k tk ldrk gS] fd X = 12 a3b5 vkSj
Y = 14 a4b3] tgk¡ a rFkk b vHkkT; la[;k,¡ gSa] rks L.C.M. (X, Y) gS % 1
(A) 168 a4b3 (B) 84 a4b5 (C) 14 a4b5 (D) 42 a4b3
If two positive integers X and Y can be expressed as X = 12 a3b5 and Y = 14 a4b3,
where a and b are prime numbers, then L.C.M. (X, Y) is :
(A) 168 a4b3 (B) 84 a4b5 (C) 14 a4b5 (D) 42 a4b3

2. ,d vHkkT; la[;k ds dqy xq.ku[kaM gksrs gSa % 1
(A) 0 (B) 1
(C) 2 (D) nks ls vf/kd
The total number of factors of a prime number :
(A) 0 (B) 1
(C) 2 (D) More than two

3. 'kwU;d –7 rFkk 3 okys f}?kkr cgqinksa dh la[;k gksxh % 1
(A) 1 (B) 2
(C) 3 (D) rhu ls vf/kd
The number of quadratic polynomials having zeroes –7 and 3 are :
(A) 1 (B) 2
(C) 3 (D) More than three

4. f}?kkr lehdj.k x 2 − 0.04 = 0 ds ewy gS % 1
(A) ± 0.2 (B) ± 0.02 (C) 0.4 (D) 2
The roots of the quadratic equation x 2 − 0.04 = 0 are :
(A) ± 0.2 (B) ± 0.02 (C) 0.4 (D) 2

2103/(Set : A)

(5) 2103/(Set : A)
5. A.P. : 6, 24 , 54 ......... dk vxyk in gS % 1
(A) 60 (B) 72
(C) 96 (D) 108
The next term of A.P. 6, 24, 54 ......... is :
(A) 60 (B) 72
(C) 96 (D) 108

6. vkÑfr esa EF || PQ] ;fn E Hkqtk PR dks 5 : 4 vuqikr esa foHkkftr djrk gS vkSj PQ = 6.3 cm]
rks EF dh yackbZ gS % R 1

E F

P Q

(A) 5.4 cm (B) 3.5 cm
(C) 2.8 cm (D) buesa ls dksbZ ugha
In figure EF || PQ, if E divides PR in the ratio 5 : 4 and PQ = 6.3 cm, then length
of EF is :
R

E F

P Q

(A) 5.4 cm (B) 3.5 cm
(C) 2.8 cm (D) None of these

2103/(Set : A) P. T. O.

(6) 2103/(Set : A)
7. ml fcUnq ds funsZ'kkad tgk¡ js[kk 4x – 2y + 5 = 0] y-v{k dks izfrPNsn djrh gS] gksaxs % 1
 5  −5  5  −5
(A)  0,  (B)  0,  (C)  0,  (D)  0, 
 2  2   4  4 
The co-ordinates of the point where the line 4x – 2y + 5 = 0 intersect the y-axis
are :
 5  −5  5  −5
(A)  0,  (B)  0,  (C)  0,  (D)  0, 
 2  2   4  4 

8. ;fn sin θ = a gS] rks cos θ cjkcj gS -------------------A 1
b
a
If sin θ = , then cos θ is equal to ………………… .
b

9. ;fn ledks.k ∆ABC esa C ij ledks.k gS] rks cos (A + B ) dk eku gS % 1
1 3
(A) (B) (C) 1 (D) 0
2 2

If ∆ABC is right angled at C, then the value of cos (A + B ) is :
1 3
(A) (B) (C) 1 (D) 0
2 2
2
10. lgh fodYi pqfu, % 1 + cot 2 θ = ………………. 1
1 + tan θ
(A) sec2 θ (B) tan2 θ (C) cot2 θ (D) cosec2 θ
1 + cot2 θ
Choose the correct option : = ……………….
1 + tan2 θ
(A) sec2 θ (B) tan2 θ (C) cot2 θ (D) cosec2 θ

11. fdlh o`Ùk dh Li'kZ js[kk] Li'kZ fcUnq ls xqtjus okyh f=T;k ij --------------- gksrh gSA 1
The tangent to a circle is …………….. to the radius through the point of contact.

2103/(Set : A)

(7) 2103/(Set : A)
12. nh xbZ vkÑfr es]a O o`Ùk dk dsUnz gS] AB ,d thok gS vkSj A ij Li'kZ js[kk PA, AB ds 40° dk
dks.k cukrh gS] rks ∠AOB cjkcj gS % A 1
P
40°°

O

B
(A) 75° (B) 80° (C) 85° (D) 100°
In the given figure, O is the centre of circle, AB is a chord and the tangent PA at
A makes an angle of 40° with AB, then ∠AOB is equal to :
A
P
40°°

O

B
(A) 75° (B) 80° (C) 85° (D) 100°

13. nh xbZ vkÑfr es]a o`Ùk[k.M PRQ dk {ks=Qy gS % 1
Q
R
r
90°°
O r P

r2 r2
(A) (π − 2) (B) (π − 1)
4 4

r2 r2
(C) (π + 2) (D) (π + 1)
4 4

2103/(Set : A) P. T. O.

(8) 2103/(Set : A)
In the given figure, the area of segment PRQ is :
Q
R
r
90°°
O r P

r2 r2 r2 r2
(A) (π − 2) (B) (π − 1) (C) (π + 2) (D) (π + 1)
4 4 4 4

14. ,d dkj ds ifg;s dk O;kl 42 cm gSA 132 km pyus esa ;g fdrus pDdj yxk,xk \ 1
4 5 6 3
(A) 10 (B) 10 (C) 10 (D) 10
The diameter of a car wheel is 42 cm. The number of complete revolutions it will
make in moving 132 km is :
(A) 104 (B) 105 (C) 106 (D) 103

15. nks 'kadqvksa dh Å¡pkb;ksa dk vuqikr 1 : 3 gS rFkk f=T;kvksa dk vuqikr 3 : 1 gSA muds vk;ruksa dk
vuqikr D;k gS \ 1
(A) 1:3 (B) 1:9
(C) 9:1 (D) 3 : 1
Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What
is the ratio of their volumes ?
(A) 1:3 (B) 1:9
(C) 9:1 (D) 3 : 1

16. ;fn dqN vk¡dM+ksa dk cgqyd rFkk ek/; Øe'k% 12.6 rFkk 10.5 gks] rks mu vk¡dM+ksa dk ek/;d gS % 1
(A) 10.8 (B) 11.2 (C) 11.6 (D) 12
If Mode and Mean of a data is 12.6 and 10.5 respectively, then median of data is :
(A) 10.8 (B) 11.2 (C) 11.6 (D) 12

2103/(Set : A)

(9) 2103/(Set : A)
17. fuEufyf[kr vk¡dM+ksa ij fopkj djsa % 1
vad 10 ls de 20 ls de 30 ls de 40 ls de 50 ls de 60 ls de
Nk=ksa dh la[;k 3 12 27 57 75 80
cgqyd oxZ gS %
(A) 10-20 (B) 20-30
(C) 30-40 (D) 40-50
Consider the following data :

Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60

No. of Students 3 12 27 57 75 80

The Mode class is :
(A) 10-20 (B) 20-30
(C) 30-40 (D) 40-50

18. 52 iÙkksa dh ,d xìh esa ls 2 iÙks iku ds rFkk 4 iÙks gqdqe ds xk;c gSaA 'ks"k xìh esa ls ,d dkyk iÙkk
fudkyus dh izkf;drk gS % 1
11 11
(A) (B)
26 23

6 12
(C) (D)
13 23

2 cards of hearts and 4 cards of spades are missing from a pack of 52 cards.
What is the probability of getting a black card from the remaining pack ?
11 11
(A) (B)
26 23

6 12
(C) (D)
13 23

2103/(Set : A) P. T. O.

( 10 ) 2103/(Set : A)
iz'u 19 vkSj 20 ds fy, fn'kk-funsZ'k % iz'u la[;k 19 vkSj 20 esa vfHkdFku (A) ds ckn dkj.k (R) dk
dFku gSA (A), (B), (C) vkSj (D) esa ls lgh fodYi pqusa tSlk fd uhps fn;k x;k gS %
Directions for questions 19 and 20 : In question no. 19 and 20 a statement of
Assertion (A) is followed by a statement of Reason (R). Choose the correct options
from (A), (B), (C) and (D) as given below :

19. vfHkdFku (A) : fdlh Hkh izkÑfrd la[;k n ds fy, la[;k 8n vad 0 ij lekIr ugha gks ldrhA 1
dkj.k (R) : fdlh izkÑfrd la[;k n ds fy, la[;k 12n vad 5 ij lekIr gks ldrh gSA
(A) vfHkdFku (A) vkSj dkj.k (R) nksuksa lgh gSa vkSj dkj.k (R), vfHkdFku (A) dh lgh O;k[;k djrk
gSA
(B) vfHkdFku (A) vkSj dkj.k (R) nksuksa lgh gSa] ysfdu dkj.k (R), vfHkdFku (A) dh lgh O;k[;k ugha
djrk gSA
(C) vfHkdFku (A) lgh gS] ysfdu dkj.k (R) xyr gSA
(D) vfHkdFku (A) xyr gS] ysfdu dkj.k (R) lgh gSA

Assertion (A) : The number 8n cannot end with the digit 0 for any natural
number n.

Reason (R) : The number 12n can end with the digit 5 for some natural
number n.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct
explanation of Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct
explanation of Assertion (A).

(C) Assertion (A) is true, but Reason (R) is false.

(D) Assertion (A) is false, but Reason (R) is true.

2103/(Set : A)

( 11 ) 2103/(Set : A)
20. vfHkdFku (A) : ;fn fdlh o`Ùk ds O;kl dk ,d fljk (2, 3) gS vkSj dsUnz (–2, 5) gS] rks O;kl dk
nwljk fljk (–6, 7) gSA 1
dkj.k (R) : fdlh o`Ùk dk dsUnz mlds O;kl dk e/; fcUnq gksrk gSA
(A) vfHkdFku (A) vkSj dkj.k (R) nksuksa lgh gSa vkSj dkj.k (R), vfHkdFku (A) dh lgh O;k[;k djrk
gSA
(B) vfHkdFku (A) vkSj dkj.k (R) nksuksa lgh gSa] ysfdu dkj.k (R), vfHkdFku (A) dh lgh O;k[;k ugha
djrk gSA
(C) vfHkdFku (A) lgh gS] ysfdu dkj.k (R) xyr gSA
(D) vfHkdFku (A) xyr gS] ysfdu dkj.k (R) lgh gSA
Assertion (A) : If one end of a diameter of a circle is (2, 3) and centre is (–2, 5),
then other end is (–6, 7).
Reason (R) : Centre of any circle is mid point of its diameter.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct
explanation of Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct
explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

[k.M – [k
SECTION – B

21. fuEufyf[kr jSf[kd lehdj.kksa ds ;qXe dks gy djsa % 2
px + qy = p – q
qx − py = p + q
Solve the following pair of linear equations :
px + qy = p – q
qx − py = p + q

2103/(Set : A) P. T. O.

( 12 ) 2103/(Set : A)
vFkok
OR
nks la[;k,¡ 5 : 6 ds vuqikr esa gSaA ;fn izR;sd la[;k esa ls 7 ?kVk fn;k tk,] rks vuqikr 4 : 5 gks tkrk
gSA la[;k,¡ Kkr dhft,A
Two numbers are in the ratio 5 : 6. If 7 is subtracted from each of the numbers,
the ratio becomes 4 : 5. Find the numbers.

22. nh xbZ vkÑfr esa ∠A = ∠B = 53° gSA x dks a, b rFkk c ds :i esa O;Dr djs]a tgk¡ a, b vkSj c
Øe'k% AD, AB vkSj BP dh yEckb;k¡ gSaA 2
D

a
C
x

53°° 53°°
A b B c P

In the given figure, ∠A = ∠B = 53°. Express x in terms of a, b and c, where a, b
and c are lengths of AD, AB and BP respectively.
D

a
C
x

53°° 53°°
A b B c P

23. ;fn 5 cot θ = 7 gS] rks 7 sin θ + 5 cos θ dk eku Kkr dhft,A 2
5 sin θ + 7 cos θ
7 sin θ + 5 cos θ
If 5 cot θ = 7, then find the value of .
5 sin θ + 7 cos θ

2103/(Set : A)

( 13 ) 2103/(Set : A)
vFkok
OR

fl) dhft, % (1 + tan2 θ) (1 − sin θ) (1 + sin θ) = 1
Prove that : (1 + tan2 θ) (1 − sin θ) (1 + sin θ) = 1

vFkok
OR

;ksX;rk vk/kkfjr iz'u %
Competency Based Question :

;fn sin θ – 3 cos θ = 0 vkSj 0° < θ < 90° gS] rks θ dk eku Kkr dhft,A

If sin θ – 3 cos θ = 0 and 0° < θ < 90°, find the value of θ.

24. fdlh o`Ùk ftldk dsUnz O gS] ds fcUnq P ij Li'kZ js[kk PQ gSA ;fn ∆OPQ ,d lef}ckgq f=Hkqt gS] rks
∠OQP Kkr dhft,A 2
P

Q O

PQ is a tangent to a circle with centre O at point P. If ∆OPQ is an isosceles
triangle, then find ∠OQP.
P

Q O

2103/(Set : A) P. T. O.

( 14 ) 2103/(Set : A)
25. ,d ?kM+h dh feuV dh lwbZ ftldh yackbZ 12 lseh gSA bl lwbZ }kjk 35 feuV esa jfpr {ks=Qy Kkr
dhft,A 2
The minute hand of a clock is 12 cm long. Find the area swept by the minute
hand in 35 minutes.
[k.M – x
SECTION – C

26. fl) dhft, 5 ,d vifjes; la[;k gSA 3
Prove that 5 is an irrational number.

27. ;fn α vkSj β] cgqin x 2 − P (x + 1) − K ds 'kwU;d gSa] rc K dk eku Kkr dhft,] tgk¡ (α + 1) (β
3
+ 1) = gSA 3
2
If α and β are zeroes of the polynomial x 2 − P (x + 1) − K , then find the value of K,
3
where (α + 1) (β + 1) = .
2

28. a vkSj b ds fdu ekuksa ds fy, fuEu jSf[kd lehdj.kksa ds ;qXe ds vifjfer vusd gy gksaxs % 3
3x + 4y = 12
(a + b)x + 2(a – b)y = 5a – 1
For what values of a and b the following pair of linear equations have infinitely
many solutions :
3x + 4y = 12
(a + b)x + 2(a – b)y = 5a – 1
vFkok
OR
,d O;fDr us vius /ku dk ,d Hkkx 10% izfro"kZ rFkk 'ks"k Hkkx 15% izfro"kZ dh nj ls m/kkj fn;kA
mldh okf"kZd vk; ` 1,900 gSA ;fn mlus nks jkf'k;ksa ij C;kt dh nj dks vkil esa cny fn;k gksrk]
rks mls ` 200 vf/kd izkIr gksrsA izR;sd ekeys esa m/kkj nh xbZ jkf'k Kkr dhft,A
A man lent a part of money at 10% p.a. and rest at 15% p.a. His annual income
is ` 1,900. If he had interchanged the rate of interest on two sums, he would
have earned ` 200 more. Find the amount lent in each case.

2103/(Set : A)

( 15 ) 2103/(Set : A)
29. fl) dhft, fd fdlh o`Ùk ds ifjxr lekUrjprqHkZqt leprqHkZqt gksrk gSA 3
Prove that the parallelogram circumscribing a circle is a rhombus.
30. fl) dhft, % 3
tan θ cot θ
+ = 1 + sec θ . cosec θ
1 − cot θ 1 − tan θ
Prove that :
tan θ cot θ
+ = 1 + sec θ . cosec θ
1 − cot θ 1 − tan θ
vFkok
OR
fl) dhft, %
1 1 1 1
2
+ 2
+ 2
+ =2
1 + sin θ 1 + cos θ 1 + sec θ 1 + cosec 2 θ
Prove that :
1 1 1 1
+ + + =2
1 + sin θ 1 + cos θ 1 + sec θ 1 + cosec 2 θ
2 2 2

31. cPpksa ds ,d [ksy esa 8 f=Hkqtsa] ftlesa ls 3 uhys vkSj 'ks"k yky gSaA lkFk gh bl [ksy esa 10 oxZ gSa
ftlesa ls 6 uhys vkSj 'ks"k yky gSaA buesa ls ,d VqdM+k ;kn`fPNd :i ls [kks tkrk gSA bl VqdM+s ds
fuEufyf[kr gksus dh izkf;drk Kkr dhft, % 3
(i) oxZ
(ii) yky jax dk oxZ
(iii) uhys jax dk f=Hkqt
A child's game has 8 triangles of which 3 are blue and rest are red and 10
squares of which 6 are blue and rest are red. One piece is lost at random. Find
the probability that it is a :
(i) Square
(ii) Square of red colour
(iii) Triangle of blue colour

2103/(Set : A) P. T. O.

( 16 ) 2103/(Set : A)
[k.M – ?k
SECTION – D

32. ,d jsyxkM+h 63 km dh nwjh fdlh fuf'pr vkSlr pky ls r; djrh gS vkSj fQj 72 km dh nwjh
izkjafHkd pky ls 6 km/h vf/kd vkSlr pky ls r; djrh gSA ;fn ;g iwjh ;k=k 3 ?k.Vs esa r; dh
xbZ gS] rks izkjafHkd vkSlr pky D;k Fkh \ 5
A train travels at a certain average speed for a distance of 63 km and then
travels a distance of 72 km at an average speed of 6 km/h more than its original
speed. If it takes 3 hours to complete the total journey. What is its original
average speed ?
vFkok
OR
,d eksVj cksV ftldh xfr fLFkj ty esa 18 km/h gS] mls /kkjk ds izfrdwy 24 km tkus esa mlh
LFkku ij okil vkus ls 1 ?k.Vk vf/kd le; yxrk gSA /kkjk dh xfr Kkr dhft,A
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go
24 km upstream than to return to the same spot. Find the speed of the stream.

33. ,d f=Hkqt ABC dh Hkqtk,¡ AB vkSj BC rFkk ekf/;dk AD ,d vU; f=Hkqt PQR dh Øe'k% Hkqtkvksa
PQ vkSj QR rFkk ekf/;dk PM ds lekuqikrh gSA n'kkZb, fd ∆ABC ~ ∆PQR. 5
P
A

B D C Q R
M
Sides AB and BC and median AD of a triangle ABC are respectively proportional
to sides PQ and QR and median PM of ∆PQR. Show that ∆ABC ~ ∆PQR.
P
A

B D C Q R
M
2103/(Set : A)

( 17 ) 2103/(Set : A)
vFkok
OR

,d f=Hkqt ABC dh Hkqtk BC ij ,d fcUnq D bl izdkj fLFkr gS fd ∠ADC = ∠BAC gSA n'kkZb,
fd CA2 = CB.CD gSA
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show
that CA2 = CB.CD.

34. ,d Bksl f[kykSuk ,d v/kZ xksys ds vkdkj dk gS ftl ij ,d yaco`Ùkh; 'kadq vkjksfir gSA bl 'kadq dh
Å¡pkbZ 3 cm gS vk/kkj dk O;kl 4 cm gSA bl f[kykSus dk vk;ru Kkr dhft,A ;fn ,d yaco`Ùkh;
csyu bl f[kykSus ds ifjxr gks] rks csyu vkSj f[kykSus ds vk;ruksa dk vUrj Kkr dhft,A 5
A solid toy is in the form of a hemisphere surmounted by a right circular cone.
The height of the cone is 3 cm and the diameter of the base is 4 cm. Determine
volume of the toy. If a right circular cylinder circumscribes the toy, find the
difference of the volumes of the cylinder and the toy.

vFkok
OR
,d rEcw ,d csyu ds vkdkj dk gS ftl ij ,d 'kadq v/;kjksfir gSA ;fn csyukdkj Hkkx dh Å¡pkbZ
vkSj O;kl Øe'k% 2.1 m vkSj 5 m gS rFkk 'kadq dh fr;Zd Å¡pkbZ 2.8 m gS] rks bl rEcw dks cukus esa
iz;qDr dSuol dk {ks=Qy Kkr dhft,A lkFk gh ` 600 izfr m2 dh nj ls blesa iz;qDr dSuol dh
ykxr Kkr dhft,A
A tent is in the shape of a cylinder surmounted by a conical top. If the height
and diameter of the cylindrical part are 2.1 m and 5 m respectively and slant
height of the top is 2.8 m, find the area of the canvas used for making the tent.
Also find the cost of the canvas of the tent at the rate of ` 600 per m2.
35. fuEufyf[kr vk¡dM+ksa dk ek/;d Kkr dhft, % 5

ekfld [kir 85 ls de 105 ls de 125 ls de 145 ls de 165 ls de 185 ls de 205 ls de
¼bdkbZ;ksa esa½
miHkksDrkvksa dh la[;k 4 9 22 42 56 64 68

2103/(Set : A) P. T. O.

( 18 ) 2103/(Set : A)
Find the median of the following data :
Monthly Consumption Below 85 Below 105 Below 125 Below 145 Below 165 Below 185 Below 205
(in Units)

No. of Consumers 4 9 22 42 56 64 68

vFkok
OR

fuEufyf[kr caVu fdlh eksgYys ds 68 cPpksa ds nSfud tsc [kpZ n'kkZrk gSA ek/; tsc [kpZ ` 18 gSA
x rFkk y dk eku Kkr dhft,A
nSfud tsc [kpZ ¼#0 esa½ 11-13 13-15 15-17 17-19 19
19-21 21-23 23-25
cPpksa dh la[;k 7 6 x 13 y 5 4
The following distribution show
shows the daily pocket allowance of 68 children of a
locality. The mean pocket allowance is ` 18. Find the value of x and y.
Daily pocket allowance (in `) 11-13 13-15 15-17 17-19 19-21 21-23 23-25

Number of Children 7 6 x 13 y 5 4

vFkok
OR
;ksX;rk vk/kkfjr iz'u %
Competency Based Question :
uhps fn[kk;k x;k csyukdkj dsd eerk vius csVs ds tUefnu ds fy, cuk jgh gSA dsd 21 lseh yack gS
vkSj bldh f=T;k 15 lseh gS og dsd ds vUnj tSEl ¼VkWfQ;k¡½ Hkjdj vius csVs dks vk'p;Zpfdr djuk
pkgrh gSA ,slk djus ds fy, og fn[kk, x, vuqlkj chp ls dsd dk ,d csyukdkj Hkkx fudkyrh gSA
fudkyk x;k VqdM+k 21 lseh yack gSA

;fn dsd dk Hkkj 0.5 xzke izfr?ku lseh gS rFkk dsUnzh; Hkkx dks gVkus ds ckn cps gq, dsd dk Hkkj
6600 xzke gS] rks dVs gq, dsUnzh; Hkkx dh f=T;k Kkr dhft,A

2103/(Set : A)

( 19 ) 2103/(Set : A)
Shown below is a cylindrical shaped cake that Mamta is baking for her son's
birthday. The cake is 21 cm tall and has a radius of 15 cm. She wants to
surprise
prise her son by filling gems inside the cake. In order to do that she removes
a cylindrical portion of cake out of the centre
entre as shown. The piece that is
removed is 21 cm tall.

If the cake weights 0.5 g per cubic cm and the weight of the cake that is left
l after
removing the central portion is 6600 gms, find the radius of the central portion
that is cut.

[k.M – ³
SECTION – E
36. euizhr dkSj xksykQsad [ksy eas efgykvksa ds fy, jk"Vªh; fjdkWMZ /kkjd gSA 2017 eas ,f'k;kbZ xzaSM fizDl esa
mudk 18.86 ehVj dk Fkzks fdlh Hkkjrh; efgyk ,,sFk~yhV ds fy, vf/kdre nwjh gSA mUgsa viuk vkn'kZ

ekurs gq,] fleju us ,d fnu vksyafid eas Lo.kZ ind thrus dk n`<+ ladYi fd;kA 'kq:vkr eas mudk Fkzks
dsoy 7.65 ehVj rd gh igq¡pk FkkA Ldwy eas ,,sFk~yhV yhV gksus ds ukrs] og fu;
fu;fer :i ls lqcg vkSj 'kke
nksukas le; vH;kl djrh Fkh vkSj gj g¶rs 9 lseh dh nwjh eas lq/kkj djus eas l{ke FkhA 15 fnuksa ds
fo'ks"k f'kfoj ds nkSjku mUgksaus 40 Fkzks ls 'kq:vkr dh vkSj gj fnu bl mYys[kuh; izxfr dks izkIr djus
ds fy, Fkzks dh la[;k esa 12 dh d o`f) djrh jghA

mijksDr tkudkjh ds vk/kkj ij fuEufyf[kr iz'uksa ds mÙkj nhft, %
(i) f'kfoj eas 12osa fnu fleju us fdrus Fkzks dk vH;kl fd;k \ 1
(ii) 6 lIrkg ds var esa fleuj us fdruh nwjh rd Fkzks fd;k \ 2

2103/(Set : A) P. T. O.

( 20 ) 2103/(Set : A)
vFkok
og 11.25 ehVj dh nwjh rd Fkzks fdrus lIrkg esa dj ik;sxh \
(iii) 15 fnuksa ds iwjs f'kfoj ds nkSjku mlus dqy fdrus Fkzks fd;s \ 1
Manpreet Kaur is the national record holder for women in the shot shot-put
discipline. Her throw of 18.86 m at the Asian Grand Prix in 2017 is the
maximum distance for an Indian female athlete. Keeping her as a role model,
Simran is determined to earn gold in Olympics one day. Initially her throw
reached 7.65 m only. Being an athlete in school, she regularly practiced both in
the mornings and in the evenings and was able to improve the distance by 9 cm
every week. During the special camp for 15 days, she started with 40 throws
and everyday kept increasing the number of throws by 12 to achieve this
remarkable progress.

Based on the above information answer the following questions :
(i) How many throws Simran practiced on 12th day of the camp ?
(ii) What would be Simran's throw distance at the end of the 6 weeks ?
OR
When will she be able to achieve a throw of 11.25 m ?
(iii) How many throws did she do during the entire camp of 15 days ?
37. txnh'k ds ikl ,d [ksr gS tks ledks.k f=Hkqt AQC ds vkdkj dk gSA og [ksr ds vUnj ,d oxZ
PQRS ds :i eas txg NksM+uk pkgrk gS ftlesa xsgw¡ dh [ksrh gks vkSj ckdh txg lfCt;k¡ mxkus fy, gks
¼tSlk fd fp= eas fn[kk;k x;k gS½ [ksr esa] O ¼ewy fcUnw½ ls fpfUgr ,d [kaHkk gSA
2103/(Set : A)

( 21 ) 2103/(Set : A)

Y

A
800

700

600 B

500

S 400
R
300

200

100

–600 –500 –400 –300 –200 –100 0 100 200 300 400
00
X' X
C P(–200, 0) Q(200, 0)

mijksDr tkudkjh ds vk/kkj ij fuEufyf[kr ç'uksa ds mÙkj nhft, %
(i) O dks ewy fcUnq ekurs gq,] P ds funsZ'kkad (–200, 0) vkSj Q ds funsZ'kkad (200, 0) gSA PQRS
,d oxZ gS] R vkSj S ds funsZ'kkad Kkr dhft,A 1

(ii) oxZ PQRS dk {ks=Qy
=Qy D;k gS \ 1

vFkok
oxZ PQRS eas fod.kZ PR dh yackbZ fdruh gS \
(iii) ;fn fcUnq S js[kk CA dks K : 1 eas foHkkftr djrk gks] rks K dk eku Kkr dhft,A 2
Jagdish has a field which is in the shape of a right angled triangle AQC. He
wants to leave a space in the form of a square PQRS inside the field for growing
wheat and the remaining for growing vegetables (as shown in the figure). In the
field,
d, there is a pole marked as O (origin).

2103/(Set : A) P. T. O.

( 22 ) 2103/(Set : A)

Y

A
800

700

600 B

500

S 400
R
300

200

100

–600 –500 –400 –300 –200 –100 100 200 300 400
X' – 0 00
X
C P(–200, 0) Q(200, 0)

Based on the above information, answer the following questions :
(i) Taking O as origin, Co
Co-ordinates of P are (–200,
200, 0) and Q are (200, 0). PQRS
being a square, what are the co
co-ordinates of R and S ?
(ii) What is the area of square PQRS ?
OR
What is the length of diagonal PR in square PQRS ?
(iii) If S divides CA in the ratio K : 1, what is the value of K ?

38. csyk vius ifjokj ds lkFk igyh ckj fnYyh xbZ vkSj Å¡ph bekjrksa vkSj Vkojkas dks ns[kdj gSjku jg xbZA
vius gksVy dh bekjr ds vk/kkj ls mlus lkeus ,d Vkoj ds 'kh"kZ dks ns[kk vkSj mlh 'kke gksVy ds
'kh"kZ ls mlus Vkoj ds 'kh"kZ dks ns[kkA gksVy ds 'kh"kZ ls Vkoj ds 'kh"kZ dk mUu;u dks.k 30º gS vkSj
gksVy ds vk/kkj ls Vkoj ds 'kh"kZ dk mUu;u dks.k 60º gS vkSj Vkoj dh Å¡pkbZ 50 ehVj gSA
2103/(Set : A)

( 23 ) 2103/(Set : A)

HOTEL

50 m

Hotel Building
Bu

Tower

mijksDr tkudkjh ds vk/kkj ij fuEufyf[kr iz'uksa ds mÙkj nhft, %
(i) mi;qZDr tkudkjh ds vk/kkj ij ,d vPNh rjg ls ukekafdr vkd`fr cukb,A 1

(ii) gksVy rFkk Vkoj ds chp dh nwjh dh x.kuk dhft,A 1

vFkok
gksVy ds vk/kkj ls Vkoj ds 'kh"kZ dh nwjh Kkr dhft,A
(iii) ;fn csyk Vkoj vkSj gksVy dh bekjr ds chp lM+d ij [kM+h gS] rks ml fcUnq ls gksVy dh bekjr
ds 'kh"kZ dk mUu;u dks.k 45º gS] rks gksVy ds vk/kkj ls mldh nwjh Kkr dhft,A 2

Bela went to Delhi first time with her family and surprised to see tall buildings,
and towers. From the base of her hotel building, she observed the top of a tower
and in the same evening from the top of her hotel she observed the top of the
tower. The angle of elevation of top of towe
towerr from the top of the hotel building is
30º and angle of elevation of top of the tower from the foot of the hotel building
is 60 º and height of tower is 50 m.

2103/(Set : A) P. T. O.

( 24 ) 2103/(Set : A)

HOTEL

50 m

Hotel Building
Bu

Tower

Based on the above information, answer the following questions :

(i) Draw a well-labelled
labelled figure based on the above information.
(ii) Calculate the distance between hotel building and the tower.
OR
What is the distance between foot of the hotel building and the top of the
tower ?
(iii) If Bela is standing on the road in between tower and hotel building from
that point she observed the angle of elevation of top of hotel building is 45º.
Find her distance from the foot of the hotel building.

S
2103/(Set : A)