CG Board Class 10th Question Bank 2024 Maths

NÙkhlx<+ ek/;fed f'k{kk e.My jk;iqj
}kjk fufeZr ç'u cSad
2023-24

कक्षा 10
गणित (MATHEMATICS)
1

अनक्र
ु मणिका
इकाई अध्याय पृ. क्र.

1 cgqin 3
POLYNOMIALS

nks pjks dk jSf[kd lehdj.k 8

Linear Educations In two Variable

,d pj dk f}?kkr lehdj.k 13

Quadrate Equations in one variable
lekarj Js<+h 19

ARITHMETIC SERIES

vuqikr ,oa lekuqikr 24

Ratio and Proportion
2 funsZ'kkad T;kfefr 29

Co-ordinate geometry
vkys[k 29

Graph
3 okf.kT; xf.kr 34

Commercial Mathematics
4 f=dks.kferh 39

TRIGONOMETRY
5 ज्याणमणि 47
GEOMETRY
6 गणििीय कथनों की जााँच 52
CHECKING MATHEMATICAL STATEMENTS
7 क्षेत्रणमणि 54
MENSURATION
8 साांणययकी 61
STATISTICS
2

bdkbZ 1 (UNIT 1)
cgqin POLYNOMIALS
lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)

1. cgqin 4𝑥 2 + 4√3𝑥 + 3 ds 'kwU;dksa dk ;ksxQy gS &
¼v½ √3
¼c½ −√3
3
¼l½
4
√3
¼n½
4

Sum of zero of polynomial 4𝑥 2 + 4√3𝑥 + 3 is -

(a) √3
(b) −√3
3
(c)
4
√3
(d)
4
2. cgqin 𝑥 2 + 𝑥 − 12 ds 'kwU;dksa dk xq.kuQy gS &
¼v½ 1
¼c½ -1
¼l½ 12
¼n½ -12
Product of zeros of polynomial 𝑥 2 + 𝑥 − 12 is -

(a) 1
(b) -1
(c) 12
(d) -12

3

3. ;fn 𝑓(𝑥) dk Hkktd (𝑥 − 2) gks rc 'ks"kQy gksxk &
(v) 𝑓(2)
(c) 𝑓(−2)
(l) 𝑓(0)
(n) 𝑓(𝑥)
If divisor of 𝑓(𝑥) is (𝑥 − 2) then reminder will be -

(a) 𝑓(2)
(b) 𝑓(−2)
(c) 𝑓(0)
(d) 𝑓(𝑥)

4. cgqin 𝑥 2 − 9 ds 'kwU;d gS &
(v) 3, 3
(c) −3, −3
(l) −3, 3
(n) 9, −9
Zeros of polynomial (𝑥 2 − 9) are -

(a) 3, 3
(b) −3, −3
(c) −3, 3
(d) 9, −9

5. cgqin 𝑥 2 − 16 dk xq.ku[k.M gS &
(v) (𝑥 + 4)(𝑥 − 4)
(c) (𝑥 − 4)(𝑥 − 4)
(l) (𝑥 + 4)(𝑥 + 4)
(n) buesa ls dksbZ ugha

4

Factor of Polynomial 𝑥 2 − 16 is -

(a) (𝑥 + 4)(𝑥 − 4)
(b) (𝑥 − 4)(𝑥 − 4)
(c) (𝑥 + 4)(𝑥 + 4)
(d) None of them

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. HkkT; = ---------------------------------  HkkxQy $ 'ks"kQy
Dividend = ………………………  Quotient + Remainder

2. f}?kkrh; cgqin ds vf/kdre --------------------- 'kwU;d gksrs gSA
The maximum of zeros of quadratic polynomial are …………………………..

3. ;fn 𝑃(𝑥) = 2𝑥 2 + 5𝑥 + 4 gks rks 𝑃(2) dk eku -------------------------- gksrk gSA
If 𝑃(𝑥) = 2𝑥 2 + 5𝑥 + 4] then value of P(2) is ……………….

4. f}?kkrh; cgqin 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ds 'kwU;dksa dk xq.kuQy ------------------------- gksrk gSA
in quadratic polynomial 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, product of zeros is ……………….

5. cgqin 𝑥 2 + 11𝑥 + 30 ds 'kwU;dksa dk ;ksxQy --------------------- gksxkA
In polynomial 𝑥 2 + 11𝑥 + 30, sum of zeros is ………………

5

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. tc 'ks"kQy 'kwU; gks rc Hkktd] HkkT; dk ,d xq.ku[k.M gksrk gSaA
When remainder is zero. then divisior is a factor of dividend.
2. HkkT; = Hkktd  'ks"kQy $ HkkxQy]
Dividend = Divisor  Remainder +quotient.
3. 'ks"kQy dh ?kkr] HkkT; ls de fdUrq HkkxQy ls vf/kd gksrh gSaA
Degree of remainder is less than dividend. but more than quotient.

vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. tk¡p dhft, fd D;k 𝑔(𝑥), 𝑃(𝑥) dk ,d xq.ku[k.M gS ;k ugh\
;fn 𝑔(𝑥 ) = 𝑥 + 4, 𝑃(𝑥 ) = 𝑥 2 + 2𝑥 − 1
Check whether g(𝑥) is a factor of P(𝑥).
If 𝑔(𝑥 ) = 𝑥 + 4, 𝑃(𝑥 ) = 𝑥 2 + 2𝑥 − 1.
2. ;fn Hkktd = 3𝑥 + 1, HkkxQy = 2𝑥 − 1, 'ks"kQy = 4 gks rc HkkT; Kkr dhft,A
If divisior = 3𝑥 + 1, quotient = 2𝑥 − 1, remainder = 4 then find dividend.
3. ;fn (𝑥 − 1)] cgqqin 𝑃(𝑥) = 𝑥 2 + 𝑥 + 𝑘 dk ,d xq.ku[k.M gSa] rc 𝐾 dk eku Kkr
dhft,A
If (𝑥 − 1) is a factor of polynomial 𝑃(𝑥 ) = 𝑥 2 + 𝑥 + 𝑘, then find the value of
K.
4. fl) dhft, fd (2𝑥 2 + 4𝑦 2 + 3𝑦 + 1) dks (𝑦 + 1) ls Hkkx djus ij 'ks"kQy 'kwU; gSA
¼'ks"kQy izes; fof/k ls½
Prove that on dividing 2𝑥 2 + 4𝑦 2 + 3𝑦 + 1 by (y+1) the remainder is zero.
(By Remainder Theorem method.)
5. f}?kkrh; cgqin 𝑥 2 − 8𝑥 + 15 dk xq.ku [k.M Kkr dhft,A
factorize 𝑥 2 − 8𝑥 + 15

6

y?kq mRrjh; iz'u (Short Answer Question): 3 vad (3 Mark)

1. cgqin 𝑥 2 − 𝑥 + 1 dks 𝑥 + 1 ls Hkkx nsdj HkkxQy ,ao 'ks"kQy Kkr dhft,A
Divide the polynomials 𝑥 2 − 𝑥 + 1 by (𝑥 + 1) Also find the quotient and
remainder.
2. tc fdlh cgqin 𝑓(𝑥) dks 𝑥 2 − 9 ls Hkkx fn;k tkrk gS rc 'ks"kQy 3𝑥 + 2 gS ysfdu tc
blh cgqin dks (𝑥 − 3) ls Hkkx fn;k tkrk gS] rc 'ks"kQy D;k gksxk ?
on dividing polynomial f(𝑥) by 𝑥 2 − 9, we get remainder 3𝑥 + 2 , but when
f(𝑥) is divided by (𝑥 − 3), what will be the remainder.?
3. f}?kkrh; cgqin 2𝑥 2 − 7𝑥 − 9 ds 'kwU;dksa dk ;ksxQy o xq.kuQy Kkr dhft,A
find the sum and product of the zeros of following polynomial:- 2𝑥 2 − 7𝑥 − 9
4. 𝑎 dk eku Kkr dhft, tcfd (𝑥 − 1) cgqin 𝑎𝑥 2 − 5𝑥 + 3 dk ,d xq.ku[k.M gSA
Fnd the value of a , where (𝑥 − 1) is a factor of polynomials 𝑎𝑥 2 − 5𝑥 + 3

7

nks pjks dk jSf[kd lehdj.k
Linear Educations In two Variable
lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)
1. lehdj.k fudk; 𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1 rFkk 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2 ds fy, ,d vf}rh; gy gksus
dk izfrca/k gS%&
𝑎1 𝑏 𝑎1 𝑏 𝑐
¼v½ ≠ 1 ¼c½ = 1≠ 1
𝑎2 𝑏2 𝑎2 𝑏2 𝑐2

𝑎1 𝑏 𝑐
¼l½ = 1= 1 ¼n½ buesa ls dksbZ ughA
𝑏1 𝑏2 𝑐2

Condition for unique solution , in system of equation 𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1 and

𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2
𝑎1 𝑏 𝑎1 𝑏 𝑐
¼v½ ≠ 1 ¼c½ = 1≠ 1
𝑎2 𝑏2 𝑎2 𝑏2 𝑐2

𝑎1 𝑏 𝑐
¼l½ = 1= 1 ¼n½ None of these
𝑏1 𝑏2 𝑐2

2. ,d vf}rh; gy gksus ds fy, lehdj.k fudk; 𝑥 − 𝑘𝑦 = 2, 3𝑥 + 2𝑦 = −5 esa 𝐾 dk
2 3
eku ugh gksxk& ¼v½ 𝑘 ≠ ¼c½ 𝑘 ≠
3 2

2 3
¼l½ 𝑘 ≠ − ¼n½ 𝑘 ≠ −
3 2

for a unigue solution, in system of equation 𝑥 − 𝑘𝑦 = 2, 3𝑥 + 2𝑦 = −5 the

value of K should not be equal to –
2 3
¼v½ 𝑘 ≠ ¼c½ 𝑘 ≠
3 2

2 3
¼l½ 𝑘 ≠ ¼n½ 𝑘 ≠
3 2

8

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)

1. jSf[kd lehdj.k fudk; ds vkys[k laikrh gksus ij muds ------------------------- gy gksrs gSA
The lines representing a pair of linear equation are coincident then it has
………….. Solution.
2. jSf[kd lehdj.k fudk; ds vkys[k ,d fcanq ij izfrPNsn djrh gS rks muds ------------------------- gy
gksrs gSA
The lines representing a pair of linear equation intersect at one points then the
linear pair has …………. Solution.
3. ;fn fdlh jSf[kd lehdj.k fudk; dk dksbZ Hkh gy ugh gS rc] mudk vkys[k ----------------- js[kk,a
gksxhA
If a pair of linear equation has no solution then the lines are …………..

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)

1. laikrh js[kk,a iznf'kZr djus okys nks pjksa ds lehdj.k ds varr% vusd gy gksrs gSA
The system of linear equation depicting coincident lines have infinite solution.
2. lehdj.k 7𝑥 − 3𝑦 = 𝑝 esa 𝑥 = 3, 𝑦 = 4 j[kus ij 𝑝 dk eku 5 gSA
On putting 𝑥 = 3, 𝑦 = 4 in the equation 7𝑥 − 3𝑦 = 𝑝 then the value of p is 5.
3. lehdj.k fudk; 𝑎 𝑥 + 𝑏 𝑦 = 𝑐 vkSj 𝑎 𝑥 + 𝑏 𝑦 = 𝑐 esa ;fn izfrca/k 𝑎1 ≠ 𝑏1 gks rks
1 1 1 2 2 2 𝑎 𝑏 2 2

vf}rh; gy izkIr gksxkA
𝑎1
On system of equation 𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1 and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2 if condition ≠
𝑎2
𝑏1
then a unique solution is obtained.
𝑏2

9

vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. K dk eku Kkr dhft, ;fn ,d ljy js[kk 2𝑥 − 𝑘𝑦 = 9 fcanq ¼1] &1½ ls xqtjrh gSA
Find the value of k if the straight line 2𝑥 − 𝑘𝑦 = 9 passes through the point
(1, -1).
2. ;fn 𝑥 = 1, 𝑦 = 1 gS rks lehdj.k 7𝑥 − 4𝑦 = 𝑝 esa 𝑝 dk eku Kkr dhft,A
If 𝑥 = 1, 𝑦 = 1 then find the value of p from the equation 7𝑥 − 4𝑦 = 𝑝
3. dFkuksa dk lehdj.k :Ik esa fyf[k, &
**nks la[;kvksa dk ;ksx 16 rFkk mudk varj 4 gS**
Write the statement in the form of equation "The sum of two number is 16 and
difference is 4"

y?kq mRrjh; iz'u (Short Answer Question): 3 vad (3 Mark)
1. K ds fdl eku ds fy, lehdj.k fudk; dk vf}rh; gy gksxk &
8𝑥 + 5𝑦 = 9
𝑘𝑥 + 10𝑦 = 15
Find the value of k for which the given systems of equation have a unique
solution
8𝑥 + 5𝑦 = 9
𝑘𝑥 + 10𝑦 = 15
2. lehdj.k gy dhft, &
2𝑥 + 𝑦 = 8
𝑥 − 2𝑦 = −1
Solve the equation :
2𝑥 + 𝑦 = 8
𝑥 − 2𝑦 = −1

10

3. lehdj.k fudk; dks foyksiu fof/k ls gy dhft, &
𝑥+𝑦 =7
𝑥 − 𝑦 = −1
Solve the following system of equation by elimination method
𝑥+𝑦 =7
𝑥 − 𝑦 = −1
4. lehdj.k fudk; dks izfrLFkkiu fof/k ls gy dhft, &
𝑥 − 𝑦 = −1
3𝑥 − 2𝑦 = 12
Solve the following system of equation by substitution method
𝑥 − 𝑦 = −1
3𝑥 − 2𝑦 = 12

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)

1. nks la[;kvksa dk varj 14 rFkk muds oxksZ dk varj 448 gSA la[;k,a Kkr dhft,A
The difference of two numbers is 14 and the difference of the sum of square
of the numbers is 448 find the numbers.
2. ,d f=Hkqt ABC esa ∠𝐴 = 𝑥 𝑜 , ∠𝐵 = 3𝑥 𝑜 ,oa ∠𝐶 = 𝑦 𝑜 gSA ;fn 3𝑥 𝑜 − 5𝑦 𝑜 = 30𝑜
gks rc fl) dhft, fd ;g ,d ledks.k f=Hkqt gSA
In a triangle ABC ∠𝐴 = 𝑥 𝑜 , ∠𝐵 = 3𝑥 𝑜 and ∠𝐶 = 𝑦 𝑜 if 3𝑥 𝑜 − 5𝑦 𝑜 = 30𝑜
then prove that, this is a right triangle.
3. nks la[;kvksa dk ;ksx 25 rFkk muds O;qRØeksa dk ;ksx 1⁄ gSA la[;k,a Kkr dhft,A
4
The sum of two numbers is 25 and the sum of their reciprocal is 1⁄4 then Find
the numbers.

11

4. nks la[;kvksa dk xq.kuQy 45 rFkk mudk ;ksx 14 gS] la[;k,a Kkr dhft,A
The product of two numbers is 45 and their sum is 14 find the numbers.
5. nks vadks okyh ,d la[;k dk 7 xquk] vadks dks iyVus ij cuus okyh la[;k ds 4 xquk ds cjkcj
gS rFkk la[;kvksa ds vadks dk ;ksx 3 gSA og la[;k Kkr dhft,A
Seven times a two digit number is equal to 4 times a two digit number obtained
by reserving the digits and sum of the digits is 3 find the number.
6. vkys[kh fof/k ls lehdj.k fudk; gy dhft, &
𝑥 + 𝑦 = 10, −𝑥 + 𝑦 = 4
Solve the system of equation by graphical method
𝑥 + 𝑦 = 10, −𝑥 + 𝑦 = 4
7. ikap o"kZ iwoZ esjh vk;q esjs iq= dh vk;q dh frxquh FkhA nl o"kZ ckn esjh vk;q iq= dh vk;q dh
nqxquh gks tk;sxhA esjh o esjs iq= dh orZeku vk;q Kkr dhft,A
Five years ago, I was 3 times as old as my son. Ten years later, I shall be 2
times as old as my son. Find the present age of mine and my son.

12

,d pj dk f}?kkr lehdj.k
Quadrate Equations in one variable

lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)
1. f}?kkr lehdj.k 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 ds ewyksa dk ;ksxQy gksxk %&
𝑏 𝑏
¼v½ − ¼c½
𝑎 𝑎
𝑎 𝑎
¼l½ ¼n½ −
𝑏 𝑏

The sum of the roots of the Quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 will be
𝑏 𝑏
¼a½ − ¼b ½
𝑎 𝑎
𝑎 𝑎
¼c ½ ¼ d½ −
𝑏 𝑏
2. f}?kkr lehdj.k 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 ds ewy gsrq lw= gSa%&
+ √𝑏 2 −4𝑎𝑐
−𝑏− + √𝑏 2 −4𝑎𝑐
−𝑎−
¼v½ 𝑥 = ¼c½ 𝑥 =
2𝑎 2𝑎
+ √𝑏 2 −4𝑎𝑐
−𝑏− + √𝑏 2 −4𝑎𝑐
−𝑎−
¼l½ 𝑥 = ¼n½ −𝑥 =
2𝑎 2𝑎

The formula for the roots of the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is –
+ √𝑏 2 −4𝑎𝑐
−𝑏− + √𝑏 2 −4𝑎𝑐
−𝑎−
¼v½ 𝑥 = ¼c½ 𝑥 =
2𝑎 2𝑎
+ √𝑎2 −4𝑏𝑐
−𝑏− + √𝑏 2 −4𝑎𝑏
−𝑎−
¼l½ 𝑥 = ¼n½ −𝑥 =
2𝑎 2𝑎
3. ∝ o 𝛽 ewy ds f}?kkr lehdj.k gSa&
¼v½ 𝑥 2 + (∝ +𝛽 )𝑥+∝ 𝛽 = 0
¼c½ 𝑥 2 + (∝ −𝛽 )𝑥+∝ 𝛽 = 0
¼l½ 𝑥 2 − (∝ +𝛽 )𝑥+∝ 𝛽 = 0
¼n½ 𝑥 2 − (∝ +𝛽 )𝑥 − (∝ 𝛽 ) = 0

13

The quadratic equation with roots ∝ and is 𝛽 -

¼a½ 𝑥 2 + (∝ +𝛽 )𝑥+∝ 𝛽 = 0

¼b½ 𝑥 2 + (∝ −𝛽 )𝑥+∝ 𝛽 = 0

¼c½ 𝑥 2 − (∝ +𝛽 )𝑥+∝ 𝛽 = 0

¼d½ 𝑥 2 − (∝ +𝛽 )𝑥 − (∝ 𝛽 ) = 0

4. f}?kkr lehdj.k 3𝑥 2 + 2𝑥 + 7 = 0 ds ewyksa dk xq.kuQy gSa%&
7 3
¼v½ ¼c½
3 7

2 3
¼l½ ¼n½
3 2

The product of the roots of the Quadratic equation 3𝑥 2 + 2𝑥 + 7 = 0
7 3
¼a½ ¼b½
3 7

2 3
¼c½ ¼d½
3 2

5. oxZ lehdj.k 𝑥 2 − 4𝑥 + 2 = 0 dk foHksnd gSa%&

¼v½ 4 ¼c½ 2

¼l½ 8 ¼n½ 24

The discriminant of the quadratic equation 𝑥 2 − 4𝑥 + 2 = 0 is

¼a½ 4 ¼b½ 2

¼c½ 8 ¼d½ 24

14

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. oxZ lehdj.k ds vf/kdare ewyksa dh la[;k --------------------------- gksrh gSaA
The maximum number of roots of the quadratic equation is ……….......

2. f}?kkr lehdj.k esa pj jkf'k dh vf/kdre ?kkr --------------------- gksrh gSaA
In a quadratic equation the highest power of the variable is ……………...

3. ;fn foHksnd 𝐷 < 𝑂 gks rks f}?kkr lehdj.k ds ewyksa dh izd`fr -------------------- gksxa hA

If in a quadratic equation the Discriminant D < 0 then the nature of the roots

of the equation will be ----------------

4. f}?kkr lehdj.k 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 ds fy, foHksnd ¼fofoDydj½ dk eku --------------- gSaA

Discriminant of the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is --------------

5. oxZ lehdj.k 2𝑥 2 − 4𝑥 + 3 = 0 ds ewyksa dk ;ksxQy -------------------- gksxkA

The sum of the roots of the quadratic equation 𝑎2 − 4𝑥 + 3 = 0 is ------------

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. oxZ lehdj.k 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 dk foHksnd 𝐷 = 𝑏2 − 4𝑎𝑐 gSaA
In a quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 the Discriminant 𝐷 = 𝑏 2 − 4𝑎𝑐.
2. oxZ lehdj.k dk foHksnd dk eku 'kwU; gks rks ewy okLrfod ,ao leku gksrs gSA
In a quadratic equation, the value of the Discriminant is zero then its roots will
be real and equal.
3. oxZ lehdj.k ds vf/kdre nks ewy gksrs gSaA
The maximum number of roots of a Quadratic equation is two.

15

4. oxZ lehdj.k 𝑥 2 − 7𝑥 = 0 dk gy 0 vkSj 7 gSaA
The solution of the Quadratic equation 𝑥 2 − 7𝑥 = 0 is 0 and 7
5. oxZ lehdj.k 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 ds ewyksa dk xq.kuQy 𝑐 gksrk gSaA
𝑎
𝑐
Product of the Roots of the Quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is
𝑎

vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. K ds fdl eku ds fy, lehdj.k 𝐾𝑥 2 + 4𝑥 + 1 = 0 ds ewy okLrfod ,ao cjkcj gksxa s ?
For what value of k of the quadratic equation 𝑘𝑥 2 + 4𝑥 + 1 = 0 the roots are
real and eaual.
2. oxZ lehdj.k cukb, ftlds ewy 7 vkSj 4 gS?
Form a Quadratic equation whose roots are 7 and 4.
3. f}?kkr lehdj.k 3𝑥 2 + 2𝑥 + 7 = 0 ds ewyksa dk ;ksxQy ,ao xq.kuQy Kkr dhft,A
Find the sum and the product of the roots of the Quadratic equation 3𝑥 2 +
2𝑥 + 7 = 0
4. (2𝑥 + 3)(3𝑥 − 7) = 0 ds ewy Kkr dhft,A
Find the roots of the quadratic equation (2𝑥 + 3)(3𝑥 − 7) = 0
5. f}?kkr lehdj.k 𝑥 2 + 16𝑥 + 64 = 0 dk foHksnd Kkr dhft,A
Find the Discriminant of the Quadratic equation 𝑥 2 + 16𝑥 + 64 = 0

vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. f}?kkr lehdj.k 𝑥 2 − 4𝑥 + 4 = 0 ds ewyksa dh izd`fr Kkr dhft,A
find the nature of roots of quadratic equation 𝑥 2 − 4𝑥 + 4 = 0
2. f}?kkr lehdj.k 9𝑥 2 − 7𝑥 − 2 = 0 dks gy dhft,A ¼lw= fof/k ls½
Solve the quadratic equation by formula method 9𝑥 2 − 7𝑥 − 2 = 0

16

3. f}?kkr lehdj.k cukb, ftuds ewyksa dk ;ksxQy 5 vkSj xq.kuQy 6 gSA
form a quadratic equation whose sum of roots is 5 and product is 6
4. f}?kkr lehdj.k 3𝑥 2 − 11𝑥 + 10 = 0 dks xq.ku[k.M fof/k }kjk gy dhft,A
Solve the quadratic equation 3𝑥 2 − 11𝑥 + 10 = 0 by factor nation method.
5. f}?kkr lehdj.k 𝑥 2 − 6𝑥 + 5 = 0 dks iw.kZ oxZ fof/k ls gy dhft,A
Solve the quadratic equation 𝑥 2 − 6𝑥 + 5 = 0 by perfect square method.
6.
√6 + √6 + √6 +− − − − − dks gy dhft,A

Solve √6 + √6 + √6 +− − − − −
7. f}?kkr lehdj.k cukb, ftuds ewy 6 + √5 o 6 − √5 gSaA
form a quadratic equation whose roots are 6 + √5 and 6 − √5

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)
1
1. ;fn ,d la[;k vkSj mlds O;qRdzeksa dk ;ksx 2 30 gSA rks la[;k,W Kkr dhft,A
1
if sum of a number and its reciprocal is 2 then find the numbers.
30
2. nks dzekxr izkd`r la[;kvksas ds oxkZs dk ;ksx 85 gS] la[;k, Kkr dhft,A
The sum of square of two consecutive natural number is 85 then find the
number.
3. ,d O;fDr dh orZeku vk;q] mlds iw= dh orZeku vk;q ds oxZ ds cjkcj gSA ;fn 1 o"kZ igys
ml O;fDr dh vk;q mlds iq= dh vk;q dh 8 xquh Fkh rks nksuksa dh orZeku vk;q Kkr dhft,A
If Father is present, age is equal to the square of his son's present age. If one
year ago his age is 8 times his son's age then find their present age.

17

4. ,d vk;rkdkj [ksr dk ifjeki 82 ehVj gSa rFkk mldk {ks=Qy 400 oxZ ehVj gSaA [ksr dh yEckbZ
o pkSMkbZ Kkr dhft,A
The perimeter of a rectangular field is 82 meter and its area is 400 sq. meter
find the length and width of the field.
5. esjh 5 o"kZ iwoZ dh vk;q rFkk 8 o"kZ iwoZ dh vk;q dk xq.kuQy 40 gSaA rc esjh orZeku vk;q Kkr
dhft,A
The product of my age 5 years ago and 8 years ago is 40 then find my present
age.

6. lehdj.k gy dhft,&
𝑥+1 𝑥−1 5
− = , 𝑥 ≠ 1,1
𝑥−1 𝑥+1 6
Solve the equation –
𝑥+1 𝑥−1 5
− = , 𝑥 ≠ 1,1
𝑥−1 𝑥+1 6

18

lekarj Js<+h
ARITHMETIC SERIES

lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)
1. izFke 10 izkd`r la[;kvksa dk ;ksxQy gksxk &
¼v½ 65 ¼c½ 45
¼l½ 5-5 ¼n½55
The sum of first 10 natural number will be –
¼a½ 65 ¼b½ 45
¼c½ 5-5 ¼d½55
2. √3 + 1 vkSj √3 − 1 dk lekarj ek/; gS&
¼v½ 1 ¼c½ √3
¼l½ 3√2 ¼n½ 2√3
The arithmetic mean of √3 + 1 and √3 − 1 is
¼a½ 1 ¼b½ √3
¼c½ 3√2 ¼d½ 2√3
3. Js<+h 7]13]19 ----------- dk 5 oka in gksxk &
¼v½ 38 ¼c½ 33
¼l½ 31 ¼n½ 37
The 5𝑡ℎ term of series 7,13,19 ………………… will be
¼a½ 38 ¼b½ 33
¼c½ 31 ¼d½ 37
4. lekarj Js<+h 5]11]17 ----------- dk lokZrj gSa&
¼v½ 7 ¼c½ 5
¼l½ &6 ¼n½ 6

19

common difference of A.P. 5,11,17 ………. is
¼a½ 7 ¼b½ 5
¼c½ &6 ¼d½ 6
5. ;fn fdlh lekarj Js<+h dk izFke in a rFkk lkoZvarj d gS rc n os in dk eku gksxk &

¼v½ 𝑎 + (𝑛 + 1)𝑑 ¼c½ 𝑎 + (𝑛 − 1)𝑑

¼l½ 𝑑 + (𝑛 + 𝑎)𝑎 ¼n½ 𝑑 + (𝑛 − 𝑎)𝑎

𝑛𝑡ℎ there of an A.P. whose first term is a and common difference is d is

¼a½ 𝑎 + (𝑛 + 1)𝑑 ¼b½ 𝑎 + (𝑛 − 1)𝑑

¼c½ 𝑑 + (𝑛 + 𝑎)𝑎 ¼d½ 𝑑 + (𝑛 − 𝑎)𝑎

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. a rFkk b dk lekarj ek/; -------------------- gSA
The arithmetic mean of a and b is …………… .
2. Js<+h dk n ok¡ in (3𝑛 − 1) gS rks 21 ok¡ in ------------------- gksxkA
If 𝑛𝑡ℎ tterm of the series is (3𝑛 − 1) then 21st term will be …………… .
3. (𝑥 + 7) rFkk (𝑥 − 7) dk lekarj ek/; ---------------------- gSA
Arithmetic mean of (𝑥 + 7) and (𝑥 − 7) is …………… .
4. Js<+h 7]12]17]22------- dk 12 oka in ---------------------- gksxkA
12th term of the series 7, 12, 17, 22…… will be …………
5. ;fn izFke a in rFkk vafre in 𝑙 gks rks n inksa dk ;ksxQy --------------------- gksxkA
If first term a , and last term is 𝑙 then sum of n terms will be …………. .

20

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. 1 ls 10 rd izkd`r la[;kvksa dk ;ksxQy 57 gksrk gSA
The sum of natural number from 1 to 10 is 57.
2. 1 4 7
, , ds 10 osa in dk eku
28
gSA
9 9 9 9
1 4 7 28
10th term of series , , is
9 9 9 9
3. 1 1
rFkk − dk lekarj ek/; 'kwU; gksxkA
2 2
1 1
Arithmetic mean of and − will be zero
2 2
4. m rFkk n dk lekarj ek/; 𝑚+𝑛 gksxkA
2
𝑚+𝑛
Arithmetic mean of m and n will be
2
5. 2]4]6]8----- ds 10 ossa in dk eku 20 gSA
vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. lekarj Js<+h 4]7]10]13------------ dk 10 oka in Kkr dhft,A
In A.P. 4,7,10,13…….. find 10th term.
2. lekarj Js<+h 2]6]10------ dk m oka in Kkr dhft,A
Find the mth term of an A.P. 2,6,10…..
3. lekarj Js<+h 9]5]1]&3------------- dk 11 oka in Kkr dhft,A
Find the 11th term of an A.P. 9,5,1,-3……..
4. 4𝑥 rFkk 6𝑥 dk lekarj ek/; Kkr dhft,A
find the arithmetic mean of 4𝑥 and 6𝑥
5. 100]70]40------ dk 51 oka in Kkr dhft,A
Find the 51st term of 100, 70, 40….

21

y?kq mRrjh; iz'u (Short Answer Question): 3 vad (3 Mark)
1. Js<+h 9]12]15------ ds 10 inksa dk ;ksxQy Kkr dhft,A
Find the sum of 10 term of the series 9, 12, 15…….
2. Js<+h 27]24]21------- dk dkSu lk in 'kwU; gksxk ?
Which term will be zero of the series 27, 24, 21……?
3. Js<+h 3]8]13------------253 esa vafre ls 10 oka in Kkr dhft,A
Find the 10th term from the last of the series 3, 8, 13…….253.
4. 0 ls 50 ds e/; leLr fo"ke la[;kvksa dk ;ksxQy Kkr dhft,A
Find the sum of all odd number between 0 to 50.

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)
1. 100 vkSj 200 ds chp dh fo"ke la[;kvkssa dk ;ksxQy Kkr dhft,A
Find the sum of all odd number between 100 and 200.

2. fdlh Js<+h ds izFke 7 inksa dk ;ksx 49 gS rFkk izFke 17 inksa dk ;ksx 289 gS rc Js<+h ds 𝑛 inksa

dk ;ksxQy Kkr dhft,A

The sum of first 7 terms is 49 of an A.P. and sum of first 17 term is 289 then

find the sum of n terms.

3. ;fn fdlh Js<+h dk izFke] f}rh; rFkk vafre in Øe'k% a, b vkSj 2a gS rks Js<+h ds n inksa dk

;ksxQy Kkr dhft,A

If first, second and last terms of an A.P are a, b and 2a respectively, then find

the sum of the n term of the series.

22

4. ;fn a, b, c fdlh lekarj Js<+h ds Øe'k% p os]a q osa] r osa in gS] rks fl} dhft, fd &

𝑎(𝑞 − 𝑟) + 𝑏(𝑟 − 𝑝) + 𝑐 (𝑝 − 𝑞 ) = 0

If a, b, c are Pth, Qth and rth term of an A.P. then prove that ;

𝑎(𝑞 − 𝑟) + 𝑏(𝑟 − 𝑝) + 𝑐 (𝑝 − 𝑞 ) = 0

5. ;fn ,d lekarj Js.kh dk P oka in q rFkk q oka in p gks rks fl) dhft, (𝑝 + 𝑞) oka in 'kwU;

gksxkA

If Pth term is q, qth term is p of an A.P. then prove that (𝑝 + 𝑞) term will be

zero.

23

vuqikr ,oa lekuqikr
RATIO AND PROPORTION

lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)
1. ;fn a:b::c:d gks rks fuEufyf[kr esa ls dkSu lk lac/a k lR; gS\
¼v½ ad=bc
¼c½ ab=cd
¼c½ ac=bd
¼n½ buesa ls dksbZ ugha A
If a:b::c:d then which of the following is correct?

(a) ad=bc
(b) ab=cd
(c) ac=bd
(d) None of the these
2. a:b:c gks rks e/;kuqikrh D;k gksxk\

¼v½ 𝑏 2 = 𝑎𝑐
¼c½ 𝑐 2 = 𝑎𝑏
¼c½ 𝑎2 = 𝑏𝑐
¼n½ buesa ls dksbZ ugha A
If a:b:c then the mean proportional will be?

(a) 𝑏 2 = 𝑎𝑐

(b) 𝑐 2 = 𝑎𝑏

(c) 𝑎2 = 𝑏𝑐
(d) None of the these

24

3. ;fn 6: 𝑥: : 𝑥: 54 gks rks 𝑥 dk eku gksxk\
¼v½ 3
¼c½ 6
¼c½ 9
¼n½ 18
If 6: 𝑥: : 𝑥: 54 then value of 𝑥 will be?

(a) 3

(b) 6

(c) 9
(d) 18

4. 6: 10 ∷ 𝑥: 25 esa 𝑥 dk eku gksxk\
¼v½ 20
¼c½ 15
¼c½ 25
¼n½ 30
The value of 𝑥 is in
(a) 20

(b) 15

(c) 25
(d) 30
5. nks rqY; vuqikrksa dk lac/a k gS&
¼v½ lekuqikr ¼c½ oxkZuqikr ¼l½ izfryksekuqikr ¼n½ ?kukuqikr
The relation between two equal ratio is –
¼a½ proportion ¼b½ Duplicate ratio ¼c½ inverse ratio ¼d½ Triplicate ratio

25

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. 12 rFkk 6 dk r`rh;uqikrh ------------------------ gksxkA
The third proportional of 12 and 6 will be ……….
2. vk/kk ehVj ,ao 50 ls-eh- esa --------------------- vuqikr gksxkA
Ratio between half meter and 50 cm is ………….
3. 8]14]16 dk prqFkkZuqikrh -------------------------- gksxkA
Forth proportion of 8, 14, 16 is ……………..
4. 7]3]21 dk prqFkkZuqikrh ------------------------ gksxkA
Fourth term of 7, 3, 21 will be …………
5. ;fn 𝑎: 𝑏: 𝑐 gks rks 𝑏 dks ----------------------- dgrs gSaA
If 𝑎: 𝑏: 𝑐 then b is called …………

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. 5 rFkk 10 dk r`rh;kuqikrh 20 gksxkA
Third proportion of 5 and 10 will be 20.
2. nks rqY; vuqikrksa dh rqyuk lekuqikr ugh dgykrh gSaA
The ratio between two equal ratios is not known as proportion.
3. ;fn 𝑎 = 𝑏 gks rks 𝑏 = 𝑎𝑐 gksxkA
𝑏 𝑐
𝑎 𝑏
If = then b = ac
𝑏 𝑐
4. 2𝑥𝑦, 𝑥 2 , 𝑦 2 dk prqFkkZuqikrh 𝑥𝑦 gSA
2
𝑥𝑦
forth proportion of 2𝑥𝑦, 𝑥 2 , 𝑦 2 is
2

26

vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. 75 lseh- yacs ,d js[kk[kaM dks 3%5%7 ds vuqikr esa rhu Hkkx djus ij izR;sd Hkkx dh yackbZ fdruh
gksxh\
If a 75 cm. long, line segment is divided into three parts in the ratio 3:5:7 then what will
be the length of each part.
2. 6 vkSj 54 dk e/;kuqikrh Kkr dhft,A
Find the mean proportion of 6 & 54.

3. ;fn 14%35%%16%𝑥 gks rks 𝑥 dk eku Kkr dhft,A
If 14: 35 ∷ 16: 𝑥 then find the mean value of 𝑥.

4. ;fn 29 iqLrdksa dk ewY; 783 #- gks rks 2214 #- esa fdruh iqLrdsa feysxhA
If the cost of 29 book is Rs 783 then how many books can be purchased in Rs 2214

y?kq mRrjh; iz'u (Short Answer Question): 3 vad (3
Mark)
1. ;fn a vkSj b dk e/;kuqikrh b gks rks fl) dhft, fd
𝑎2 + 𝑏 2 𝑎 + 𝑐
=
𝑎𝑏 𝑏
If b is the mean proportional of a and c then prove that
𝑎2 + 𝑏 2 𝑎 + 𝑐
=
𝑎𝑏 𝑏
2. la[;k, 10] 18] 22] 38 esa ls izR;sd la[;k esa D;k tksM+k tk, fd ;s la[;k, lekuqikrh gks tk,\
What should be added to 10, 18, 22, 38 so that they become proportional?

3. fdlh dke dks iwjk djus esa 15 O;fDr;ksa dks 16 fnu yxrs gSA fdrus O;fDRk ml dke ds pkSFkkbZ
Hkkx dks 15 fnu esa iwjk dj ldrs gS\
15 men complete a Task in 16 days. How many men will be needed to complete 1/4rth of
the task in 15 days?

27

4. ;fn 11 edfM+;k¡ 11 fnuksa esa 11 tkys cukrh gS rks crkb, 1 edM+h 1 tky cukus esa fdrus fnu
ysxhA
If 11 spider spin 11 webs in 11 days then how many webs, will one spider spin in one
day?

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)
1. ;fn a:b::c:d gks rks fl) dhft, fd &
𝑎2 − 𝑐 2 𝑎𝑐
=
𝑏 2 − 𝑑2 𝑏𝑑
If a:b::c:d then prove that -
𝑎2 − 𝑐 2 𝑎𝑐
=
𝑏 2 − 𝑑2 𝑏𝑑

2. nks uy A vkSj B ,d Vadh dks Øe'k% 30 vkSj 40 feuV esa Hkj ldrs gSaA rhljk uy C ml Vadh
dks 60 feuV esa [kkyh dj ldrk gSA ;fn rhuksa uy ,d lkFk [kksy fn, tk,a rks Vadh dks Hkjus
esa fdruk le; yxsxkA
Two taps A and B can fill a tank in 30 minutes and 40 minutes respectively. A third tap
C can empty the tank in60 minutes. If all there taps are together. How long will it take the
to fill up?

28

bdkbZ 2 (UNIT 2)
funsZ'kkad T;kfefr
Co-ordinate geometry
vkys[k
Graph
lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)
1. fcanq (-4, 7) funsZ'kkad lery ds fdl prqFkkZa'k esa gksxk &
¼v½ izFke prqFkkZa'k
¼c½ f}rh; prqFkkZa'k
¼l½ r`rh; prqFkkZa'k
¼n½ prqFkZ prqFkkZa'k
Point (-4, 7) lies in which quadrant -

(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
2. fcanq A(4, 5) dh Y-v{k ls yEcor~ nwjh gS &
¼v½ 4
¼c½ 5
¼l½ 0
¼n½ buesa ls dksbZ ugha
The perpendicular distance of the point A(4, 5) from Y-axis is -

(a) 4
(b) 5
(c) 0
(d) None of the above

29

3. ewy fcanq ds funsZ'kkad gS &
¼v½ (0, 0)
¼c½ (𝑥, 0)
¼l½ (𝑦, 0)
¼n½ (𝑥, 𝑦)
Co-ordinate of the origin is -

(a) (0, 0)
(b) (𝑥, 0)
(c) (𝑦, 0)
(d) (𝑥, 𝑦)
4. js[kk 𝑦 = 7𝑥 − 5 dh <ky gS &
¼v½ 7
¼c½ 1@7
¼l½ &5
¼n½ &1@5
Slope of the line 𝑦 = 7𝑥 − 5 is -

(a) 7
(b) 1/7
(c) -5
(d) -1/5

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. fcanq (0, 5) …………………….. prqFkkZa'k esa fLFkr gSA
Point (0, 5) lies on …………………………… axis.

2. {kSfrt js[kk dh izo.krk ----------------------------------- gksrh gSA
The Gradient of the horizontal line is ………………………………

30

3. X- v{k ij fLFkr fdlh fcanq dk y funsZ'kkad ------------------------- gksrk gSA

The Y Coordinate of a point situated on X-axis is ……………………..

4. nksuksa v{kksa ds dVku fcanq dks --------------------------- dgrs gSA
The Intersection point of both the axis is called ……………………….

5. ;fn fdlh fcanq ds funsZ'kkad (-4, 0) gks rks ;g fcanq --------------------------------------- v{k ij fLFkr gksxkA
If the Coordinate of a point is (-4, 0) then this point will be situated on
……………………… axis.

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. Y-v{k ij fLFkr fdlh fcanq dh dksfV 'kwU; gksrh gSA

The Ordinate of any point situated on Y-axis is Zero.

2. X-v{k rFkk Y-v{k ijLij yEcor gksrs gS &

X-axis and Y-axis are mutually perpendicular.

3. ,d ljy js[kk ftldh <ky m vkSj y v{k ls dkVk x;k var%[kaM c gS rks lehdj.k 𝑦 = 𝑚𝑥 +
𝑐 gksxkA
The situation of the straight line whose slope is m and the intercept cut on Y-axis is 𝑦 =

𝑚𝑥 + 𝑐
4. ;fn fdlh fcanq ds funsZ'kkad (8, 0) gS rks ;g fcanq X-v{k Ikj fLFkr gksxkA
Of the Coordinate of a point is (8, 0) then this point is situated on X-axis.

5. ewy fcanw ds funsZ'kkad (𝑥, 𝑦) gSA
The Coordinate of origin is (𝑥, 𝑦)

31

y?kq mRrjh; iz'u (Short Answer Question): 3 vad (3 Mark)
1. fcanq (0, 0) rFkk (5, 3) dh chp dh nwjh Kkr dhft,A
Find the distance between of the point (0,0) and (5, 3)

2. ewy fcanw ls gksdj tkus okyh ml js[kk dh izo.krk Kkr dhft, tks fcanq (−3, 5) ls Hkh tkrh
gSA
Find the slope of the the straight line which passes through the origin and also pasees
through the point (-3, 5)
3. ,d js[kk fcanq ¼1] 2½ vkSj ¼5] 10½ ls xqtjrh gS bldh <ky Kkr dhft,A
Find the gradient of the line, which passes through (1, 2) & (5, 10).

4. ljy js[kk 5𝑥 + 6𝑦 = 7 dks 𝑦 = 𝑚𝑥 + 𝑐 ds :Ik esa fyf[k, rFkk js[kk dh <ky rFkk
𝑦 −v{k ls var%[k.M Kkr dhft,A
Express Straight line 5𝑥 + 6𝑦 = 7 in the form 𝑦 = 𝑚𝑥 + 𝑐 find the gradient and g
intercept.

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)

1. 𝑦 −v{k ij ,d ,slk fcanq Kkr dhft, tks fcanqvksa 𝐴(6, 5) vkSj 𝐵(−4, 3) ls lenwjLFk gksA

Find the coordinate of any point on Y-axis which is equidistant from the point A(6, 5) and

B(-4, 3)

2. ewy/ku 300 :Ik;s ij 5 izfr'kr okf"kZd C;kt dh nj ls 1] 2] 3] 4 o 5 o"kZ ds fy, lk/kkj.k
C;kt fuEu lkj.kh esa iznf'kZr gSA

le; ¼o"kZ es½a 0 1 2 3 4 5

lk/kkj.k C;kt ¼#i;s es½a 0 15 30 45 60 75

le; vkSj lk/kkj.k C;kt ds chp vkys[k [khfpa,A

32

The simple interest on principal Rs 300 at the rate 5 per annum for 1, 2, 3, 4 and 5 years

respectively are shown in the following table

Time (In year) 0 1 2 3 4 5

Simple Interest (In Rs) 0 15 30 45 60 75

Draw a graph between time and simple interest.

3. ,d ifjokj esa 5 lIrkg rd mi;ksx fd, x, I;kt dh ek=k fd-xzk- esa fuEu lkj.kh esa nh xbZ
gS&
lIrkg 1 2 3 4 5

I;kt dh ek=k ¼fd- xzk-½ 1 2 3 4 5

LkIrkg rFkk mi;ksx fd, x, I;kt dh ek=k ds chp vkys[k [khfp,A

Quantity of onion (In Kg) used by a person for 5 weeks are given in following table

Week 1 2 3 4 5

Quantity of Onion (In Kg) 1 2 3 4 5

Draw a graph between week and quantity of onion used.

4. oxksZa dh ,d Hkqtk dh eki o muds oxksZa ds ifjeki dks lkj.kh esa iznf'kZr fd;k x;k gS &
oxZ dh Hkqtk ¼lseh- esa½ 1 2 3 4 5 6 7

oxZ dh ifjeki ¼lseh- esa½ 4 8 12 16 20 24 28

oxZ dh Hkqtk o ifjeke ds chp vkys[k [khfp,A

Length of side of squares and perimeters are shown below in the table

Side of square (In cm) 1 2 3 4 5 6 7

Perimeter of square (In cm) 4 8 12 16 20 24 28

Drawn a graph between side of square and its perimeter.

33

bdkbZ 3&okf.kT; xf.kr
UNIT 3 - Commercial Mathematics

lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)
1. ;fn lkof/k tek [kkrk esa C;kt dh x.kuk NSekgh vk/kkj ij dh tkrh gS rks okf"kZd nj dks fy;k
tkuk pkfg, %&
¼v½ nqxquh
¼c½ frxquh
¼l½ vk/kk
¼n½ pkj xquk
If in fix deposit the intersect is calculated half yearly the rate of interest will be taken :-

(a) Double
(b) Three times
(c) half
(d) four times
2. ns; vk;dj 3000# ij f'k{kk midj 3% dh nj ls gksxk &
¼v½ 600#
¼c½ 500#
¼l½ 60#
¼n½ 90#
What will be the education cess 3% in taxable income is

(a) Rs600
(b) Rs900
(c) Rs60
(d) Rs90

34

3. ;fn lkof/k tek [kkrk esa le; dks fr%ekgh fd;k tkrk gS rks okf"kZd C;kt dh nj dks fy;k tkuk
pkfg, &
¼v½ vk/kk
¼c½ ,d pkSFkkbZ
¼l½ pkj xquk
¼n½ nqxquk
If in Fixed Deposit times is taken four times then annual Rate of interest will be taken :-

(a) Half
(b) One fourth
(c) Four times
(d) Double

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. vk;dj foHkkx }kjk izR;sd O;fDr laLFkk ;k daiuh dks nh xbZ igpkj la[;k dks ------------------------------
--------- dgrs gSA
Every individual, farm and company has given an identification number by income tax
department is knows as ………………………..
2. f'k{kk midj] ns;dj ij ---------------------------- izfr'kr ls yxk;k tkrk gSA
Educational cess is calculated on ………………….. on tax payable.

3. ;fn cSadks esa ,d fuf'pr le; ds fy, ,deq'r jkf'k tek dh tkrh gS rks mls --------------------------------
---- tek [kkrk dgrs gSA
The account in which the amount deposited for a fixed time is called is …………………..

4. vkorhZ tek [kkrs esa C;kt dh x.kuk dk lw= ------------------------------ gSA
The formula of calculating interest in recovering deposit account is ………………..

35

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. ftl jkf'k ij C;kt dh x.kuk dh tkrh gS mls feJ/ku dgrs gSA
The sum of money in which interest is calculated is known as amount.

2. foÙkh; o"kZ dk izkjaHk 1 vizSy ls gksrk gSA
Financial year starts on 1st April.

3. foÙkh; o"kZ dh lekfIr 31 ekpZ dks gksrh gSA
Financial year ends on 31st march.

4. cSadks esa [kkrk [kksyus ds fy, PAN dk gksuk vfuok;Z gSA
To open an account on a Bank PAN is must.

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)
1. in~euh us ftyk lgdkjh cSad esa 100 :Ik;s izfrekg dk 10 Ok"kZ ds fy, vkorhZ tek [kkrk [kksykA
;fn bUgsa cSad }kjk C;kt dh jkf'k 3025 #- iznku dh tkrh gS] rks C;kt dh nj fdrus izfr'kr
okf"kZd gksxh\
Padamani opened a recurring deposit account in district cooperative Bank for ten year and
her monthly installment is Rs 100. If on maturity, she gets Rs 3025 as interest. What is
the rate of interest per annum?
2. js'kek us iatkc us'kuy cSad esa 200 # izfrekg dh nj ls 5 o"kZ ds fy, vkorhZ tek [kkrk [kksykA
;fn C;kt dh nj 6% okf"kZd gks rks 5 o"kZ i'pkr~ mls fdruh /kujkf'k izkIr gksxh \
Reshma opened a recurring deposit account in panjab national Bank for the rate of interest
is 6% per annum then how much money will she get after 5 years.
3. Jhjke us 20000 # 1 Ok"kZ ds fy, lkof/k tek [kkrs esa tek djk;kA ;fn C;kt dh nj 10%
okf"kZd gks rFkk C;kt Nekgh la;ksftr gksrk gS rks fu;r frfFk i'pkr feyus okyh /kujkf'k fdruh
gksxh\
Sri Ram deposited Rs 20000 for 1 year in a fixed deposit account. If the annual rate of
interest is 10% per annum and the interest is compounded every six month then what
amount will Sri Ram get after due date.

36

4. fuf[ky xzkeh.k cSad esa 1 o"kZ ds fy, 10000 # lkof/k tek [kkrs esa tek djrk gSA ;fn C;kt dh
nj 8% izfro"kZ gS rFkk mldk la;kstu v)Z okf"kZd gks] rks fuf[ky ds lkof/k tek [kkrs esa tek
jkf'k dk ifjiDork ewY; Kkr dhft,A
Nikhil deposited Rs 10000 for 1 year in a fixed deposit account in rural Bank . If the rate
of interest is 8% per annum and compounded semiannually, then find the maturity amount
in nikhil's account.
5. foÙkh; Ok"kZ 2013&14 esa ,d 'kkldh; deZpkjh dh dqy okf"kZd vk; 360000# FkhA mlus 20000
# thou chek ikfylh dk okf"kZd izhfe;e RkFkk 4000# izfrekg lkekU; Hkfo"; fuf/k esa tek fd;kA
ns; vk;dj dh x.kuk dhft,A
;fn vk;dj x.kuk ds iwoZ lkekU; Hkfo"; fuf/k ,oa thou chek vkfn esa fu;ksftr jkf'k dk
vf/kdre 100000 # dj eqDr gksA
vk;dj dh njsa fuEukuqlkj gS &
Ø dj ;ksX; lhek vk;dj dh nj
1 200000# rd dksbZ vk;dj ugha
2 200001 # 500000# rd 10%
3 500001# ls 1000000# rd 20%
The income of a government employee in the financial year 2013-2014 was Rs 360000.
She deposited R20000 as premium on life insurance policy and R 4000 every month in
general provident fund. Calculate the payable tax
Also, a maximum of R100000 of savings under provident fund, life insurance and national
savings certificate are exempted from tax.
the rates of tax are as follows
S. Tax Limits Rate of tax
No.
1 Upto Rs 200000 NIL
2 Rs 200001 to Rs 500000 10%
3 Rs 500001 to Rs 1000000 20%

37

6. foÙkh; o"kZ 2012&13 l{ke dh dqy okf"kZd vk; 5,25000 # gSA og lkekU; Hkfo"; fuf/k esa 8000#
izfrekg tek djrk gS rFkk 8000# vius Hkkjrh; thou chek dk okf"kZd izhfe;e nsrk gSA vk;dj
esa NwV lHkh cpr i=ksa dk 100% ¼vf/kdre lhek 100000 #½ gks rks l{ke ds }kjk ns; vk;dj
dh x.kuk dhft,A
vk;dj dh njsa fuEukuqlkj gS &
Ø dj ;ksX; lhek vk;dj dh nj
1 200000# rd dksbZ vk;dj ugha
2 200001 # 500000# rd 10%
3 500001# ls 1000000# rd 20%
blds vfrfjDr ns; vk;dj ij 3% f'k{kk midj yxrk gSA
Sakshams total annual income in the financial year 2012-2013 was Rs 525000. He
deposited Rs 8000 every month in GPF and of Rs 8000 annual on life insurance policy.
Maximum savings permissible is 100% under all schemes (up to Rs 100000) then
calculated the tax payable by saksham at the end of the year where the educational sub
tax is 3% of the payable tax
S. No. Tax Limits Rate of tax
1 Upto Rs 200000 NIL
2 Rs 200001 to Rs 500000 10%
3 Rs 500001 to Rs 1000000 20%

38

bdkbZ 4 & f=dks.kferh
UNIT 4 – TRIGONOMETRY

lgh fodYi pqudj fyf[k, (Choose the correct option): 1 vad (1 Mark)

1. 1 + 𝑡𝑎𝑛2 𝜃 dk eku gS %&
(a) 𝑠𝑖𝑛2 𝜃
(b) 𝑐𝑜𝑠 2 𝜃
(c) 𝑠𝑒𝑐 2 𝜃
(d) 𝑐𝑜𝑠𝑒𝑐𝜃
The value of 1 + 𝑡𝑎𝑛2 𝜃 is :-
(a) 𝑠𝑖𝑛2 𝜃
(b) 𝑐𝑜𝑠 2 𝜃
(c) 𝑠𝑒𝑐 2 𝜃
(d) 𝑐𝑜𝑠𝑒𝑐𝜃

2. 𝑠𝑖𝑛2 36° + 𝑐𝑜𝑠 2 36° dk eku gksxk %&
(a) √2
(b) 1
(c) √3
(d) 4
The value of 𝑠𝑖𝑛2 36° + 𝑐𝑜𝑠 2 36 will be –
(a) √2
(b) 1
(c) √3
(d) 4

39

3. 𝑠𝑖𝑛63°
dk eku gS &
𝑐𝑜𝑠27°
(a) 0
(b)−1
(c) 1
(d) 2
𝑠𝑖𝑛63°
The value of is -
𝑐𝑜𝑠27°
(a) 0
(b)−1
(c) 1
(d) 2

4. 2 𝑠𝑖𝑛30° dk eku gksxk &
𝑐𝑜𝑠60°
(a) 4
(b) 3
(c) 2
(d) 1
𝑠𝑖𝑛30°
The value of 2 will be : -
𝑐𝑜𝑠60°
(a) 4
(b) 3
(c) 2
(d) 1

5. sin(90 − 𝜃) . 𝑐𝑜𝑠𝑒𝑐(90 − 𝜃) dk eku gksxk %&
(a) 1
(b) 0
(c) 2
(d) -1

40

The value of sin(90 − 𝜃) . 𝑐𝑜𝑠𝑒𝑐(90 − 𝜃) will be
(a) 1
(b) 0
(c) 2
(d) -1

fjDr LFkku dh iwfrZ dhft, (Fill in th blanks): 1 vad (1 Mark)
1. 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 =…………………………. ;fn 0° ≤ 𝜃 ≤ 90°

𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 =…………………………. If 0° ≤ 𝜃 ≤ 90°
2. 1 + 𝑐𝑜𝑡 2 𝜃 =……………………………….
1 + 𝑐𝑜𝑡 2 𝜃 =……………………………….
3. 3 𝑡𝑎𝑛15° dk eku ---------------------------------------------------- gksxkA
𝑐𝑜𝑡75°
𝑡𝑎𝑛15°
Value of 3 is ……………………………………
𝑐𝑜𝑡75°
4. cos(90° − 67°)dk eku ---------------------------------------- gksxkA
The value of cos(90° − 67°) is …………………………..
5. 𝑐𝑜𝑠𝜃 × 𝑠𝑒𝑐𝜃 dk eku ---------------------------------------- gksxkA
Value of 𝑐𝑜𝑠𝜃 × 𝑠𝑒𝑐𝜃 will be ……………………………

lR;@vlR; fyf[k, (Write True / false): 1 vad (1 Mark)
1. (1 − 𝑐𝑜𝑠𝜃)(1 + 𝑐𝑜𝑠𝜃) = 𝑐𝑜𝑠 2 𝜃
(1 − 𝑐𝑜𝑠𝜃 )(1 + 𝑐𝑜𝑠𝜃 ) = 𝑐𝑜𝑠 2 𝜃
2. tan(90 − 𝜃) = 𝑐𝑜𝑡𝜃
tan(90 − 𝜃 ) = 𝑐𝑜𝑡𝜃

41

3. 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 2
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 2
4. 𝑠𝑖𝑛25°. 𝑐𝑜𝑠65° + 𝑐𝑜𝑠25°. 𝑠𝑖𝑛65° = 2
𝑠𝑖𝑛25°. 𝑐𝑜𝑠65° + 𝑐𝑜𝑠25°. 𝑠𝑖𝑛65° = 2
5. 𝑠𝑖𝑛30° dk eku 𝑐𝑜𝑠60° ds eku ds cjkcj gksrk gSA

The value of 𝑠𝑖𝑛30° is equal to 𝑐𝑜𝑠60°
vfr y?kq mRrjh; iz'u (Very short Answer question): 2 vad (2 Mark)
1. eku Kkr dhft, &
𝑐𝑜𝑠80° 𝑠𝑖𝑛31°
+
𝑠𝑖𝑛10° 𝑐𝑜𝑠59°
Find the value -
𝑐𝑜𝑠80° 𝑠𝑖𝑛31°
+
𝑠𝑖𝑛10° 𝑐𝑜𝑠59°
2. eku Kkr dhft, &
𝑠𝑖𝑛2 35° + 𝑠𝑖𝑛2 55°
Find the value -
𝑠𝑖𝑛2 35° + 𝑠𝑖𝑛2 55°
3. eku Kkr dhft, &
𝑡𝑎𝑛40°
3
𝑐𝑜𝑡50°
Find the value -
𝑡𝑎𝑛40°
3
𝑐𝑜𝑡50°
4. eku Kkr dhft, &
3𝑐𝑜𝑠80°. 𝑐𝑜𝑠𝑒𝑐10° + 2𝑐𝑜𝑠59°. 𝑐𝑜𝑠𝑒𝑐31°
Find the value -
3𝑐𝑜𝑠80°. 𝑐𝑜𝑠𝑒𝑐10° + 2𝑐𝑜𝑠59°. 𝑐𝑜𝑠𝑒𝑐31°

42

5. fl) dhft, &
𝑠𝑖𝑛63°. 𝑐𝑜𝑠27° + 𝑐𝑜𝑠63°. 𝑠𝑖𝑛27° = 1
Prove that -
𝑠𝑖𝑛63°. 𝑐𝑜𝑠27° + 𝑐𝑜𝑠63°. 𝑠𝑖𝑛27° = 1
6. fl) dhft, &
𝑐𝑜𝑡𝜃 + 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑠𝑒𝑐𝜃. 𝑠𝑒𝑐𝜃
Prove that -
𝑐𝑜𝑡𝜃 + 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑠𝑒𝑐𝜃. 𝑠𝑒𝑐𝜃
7. fl) dhft, &
𝑠𝑒𝑐 2 𝜃 + 𝑐𝑜𝑠𝑒𝑐2 𝜃 = 𝑠𝑒𝑐 2 𝜃. 𝑐𝑜𝑠𝑒𝑐 2 𝜃
Prove that -
𝑠𝑒𝑐 2 𝜃 + 𝑐𝑜𝑠𝑒𝑐2 𝜃 = 𝑠𝑒𝑐 2 𝜃. 𝑐𝑜𝑠𝑒𝑐 2 𝜃
8. fl) dhft, &
𝑠𝑖𝑛4 𝜃 + 𝑐𝑜𝑠 4 𝜃 = 𝑠𝑒𝑐 2 𝜃 − 𝑐𝑜𝑠 2 𝜃
Prove that -
𝑠𝑖𝑛4 𝜃 + 𝑐𝑜𝑠 4 𝜃 = 𝑠𝑒𝑐 2 𝜃 − 𝑐𝑜𝑠 2 𝜃

nh?kZ mÙkjh; iz'u (Long Answer Question): 6 vad (6 Mark)
1. fl) dhft, &

1 − 𝑐𝑜𝑠𝜃
√ = 𝑐𝑜𝑠𝑒𝑐𝜃 − 𝑐𝑜𝑡𝜃
1 + 𝑐𝑜𝑠𝜃

Prove that -

1 − 𝑐𝑜𝑠𝜃
√ = 𝑐𝑜𝑠𝑒𝑐𝜃 − 𝑐𝑜𝑡𝜃
1 + 𝑐𝑜𝑠𝜃

43

2. fl) dhft, fd &
𝑡𝑎𝑛𝜃
sin(90° − 𝜃 ) . cos(90° − 𝜃 ) =
1 + 𝑐𝑜𝑡 2 (90° − 𝜃)
Prove that -
𝑡𝑎𝑛𝜃
sin(90° − 𝜃 ) . cos(90° − 𝜃 ) =
1 + 𝑐𝑜𝑡 2 (90° − 𝜃)
3. fl) dhft, fd &
𝑐𝑜𝑠𝜃 𝑠𝑖𝑛(90° − 𝜃 )
+ = 2𝑡𝑎𝑛𝜃
sec(90° − 𝜃 ) + 1 𝑐𝑜𝑠𝑒𝑐𝜃 − 1
Prove that
𝑐𝑜𝑠𝜃 𝑠𝑖𝑛(90° − 𝜃 )
+ = 2𝑡𝑎𝑛𝜃
sec(90° − 𝜃 ) + 1 𝑐𝑜𝑠𝑒𝑐𝜃 − 1
4. fl) dhft, fd
𝑠𝑖𝑛𝜃 1 + 𝑐𝑜𝑠𝜃
+ = 2𝑐𝑜𝑠𝑒𝑐𝜃
1 + 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
Prove that
𝑠𝑖𝑛𝜃 1 + 𝑐𝑜𝑠𝜃
+ = 2𝑐𝑜𝑠𝑒𝑐𝜃
1 + 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
7. ;fn 𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃 = √2𝑠𝑖𝑛𝜃 gks rks fl) dhft, fd 𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃 = √2𝑐𝑜𝑠𝜃
If 𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃 = √2𝑠𝑖𝑛𝜃 then, Prove that 𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃 = √2𝑠𝑖𝑛𝜃
8. ;fn 𝑥 = 𝑎𝑐𝑜𝑠𝑒𝑐𝜃 rFkk 𝑦 = 𝑏𝑐𝑜𝑡𝜃 gks rks fl) dhft, fd
𝑥 2 𝑦2
− =1
𝑎2 𝑏 2
If 𝑥 = 𝑎𝑐𝑜𝑠𝑒𝑐𝜃 and 𝑦 = 𝑏𝑐𝑜𝑡𝜃 then prove that
𝑥 2 𝑦2
− =1
𝑎2 𝑏 2

44

9. lehdj.k gy dhft, ;fn 0° ≤ 𝜃 ≤ 90°

2𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠𝜃 =1

Solve the trigonometric equation if 0° ≤ 𝜃 ≤ 90°

2𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠𝜃 =1

10. f=dks.kferh; lehdj.k gy dhft, ;fn 0° ≤ 𝜃 ≤ 90°
𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
+ =4
1 − 𝑠𝑖𝑛𝜃 1 + 𝑠𝑖𝑛𝜃

Solve the trigonometric equation : if 0° ≤ 𝜃 ≤ 90°

𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
+ =
1 − 𝑠𝑖𝑛𝜃 1 + 𝑠𝑖𝑛𝜃

11. f=dks.kferh; lehdj.k gy dhft,] ;fn 0° ≤ 𝜃 ≤ 90°

𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
+ =2
𝑐𝑜𝑠𝑒𝑐𝜃 + 1 𝑐𝑜𝑠𝑒𝑐𝜃 − 1

Solve the trigonometric equation

𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
+ =2
𝑐𝑜𝑠𝑒𝑐𝜃 + 1 𝑐𝑜𝑠𝑒𝑐𝜃 − 1

12. ,d ehukj ds vk/kkj ls ,d ljy js[kk esa 𝑎 vkSj 𝑏 nwjh ij fLFkr nks fcanqvksa ls ehukj

ds f'k[kj dk mUu;u dks.k iwjd dks.k gSA rks fl) dhft, fd ehukj dh Å¡pkbZ √𝑎𝑏
gksxhA
The angle of elevation of the top of a tower from the points at distances a and b meters

from the base and in the same straight line with it are complementary. Prove that the

height of the tower is √𝑎𝑏 meters.

45

13. ,d eafnj dk f'k[kj rFkk ml ij yxk >.Mk Hkwfe ds fdlh fcanq ij Øe'k% 30° vkSj
60° dk dks.k varfjr djrs gS ;fn eafnj dh Å¡pkbZ 10 ehVj gks rks >.Ms dh Å¡pkbZ

Kkr dhft,A
There is a flagstaff on a tower of temple of height 10m. At a point on a ground the angle

of elevation of the foot and top of the flag are 30° and 60°. Find the height of flagstaff.

14. rst gok ls VwVs ,d isM+ dk fljk >qd dj isM+ ds ikn ls 6 ehVj dh nwjh ij tehu dks Nwrk

gSA ;g fgLlk tehu ls 60° dk dks.k cukrk gSA isM+ dh Å¡pkbZ Kkr dhft,A
A tree breaks due to storm and the broken part bends, So that the top of the tree touches

the ground making an angle of 60° with the ground the distance between the feet of the

tree to the point where the top touches the ground is 6m. Find the height of the tree.

46

इकाई 5 – ज्याणमणि
UNIT 5 - GEOMETRY

सही णिकल्प का चयन कीणजए (Choose the correct option:): 1 अांक (1 Mark)
1. दो समरूप त्रिभजु के क्षेिफलों का अनपु ात 25:49 है तो उनकी संगत भजु ाओ ं का अनपु ात है:
अ. 5:7 ब. 7:5
स. 25:49 द. 49:25
If the ratio of the areas of two similar triangles is 25:49, then the ratio of
their corresponding sides is:
A. 5:7 B. 7:5
C. 25:49 D. 49:25
2. चक्रीय चतभु ुज के सम्मख
ु कोणों का योग होता है:
अ. 90° ब. 180°
स. 270° द. 360°
The sum of the angles in a cyclic quadrilateral is:
A. 90° B. 180°
C. 270° D. 360°
3. समरूप त्रिभजु की संगत भुजाएँ होती है:
अ. समान ब. समांतर
स. समानपु ात्रतक द. लंबवत
Similar triangles have:
A. Equal B. Parallel
C. Proportional D. Perpendicular

47

4. यत्रद ∆ABC ~ ∆PQR और AB:PQ= 4/5, तब BC:QR है:
4 16
अ. 5 ब. 25
2 5
स. 3 द. 4
If ∆ABC ~ ∆PQR and AB:PQ= 4/5, then BC:QR is:
4 16
A. 5 B. 25
2 5
C. 3 D. 4
5. एक वृत्त के अंतगुत सम पंचभजु खींचा गया है | सम पंचभजु की प्रत्येक भजु ा कें द्र पर कोण
बनाती है:
अ. 30° ब. 60°
स. 72° द. 90°
A regular pentagon is inscribed in a circle. Each side of the regular
pentagon forms an angle at the center of the circle is:
A. 30° B. 60°
C. 72° D. 90°
रिक्त स्थानों की पूणिि कीणजए (Fill in the blanks): 1 अांक (1 Mark)
1 सभी सवाांगसम बहुभजु _________ होते हैं।
All congruent polygons are _________.
2
वृत्त का व्यास= _________ x वृत्त की त्रिज्या।
The diameter of a circle = _________ x radius of the circle.
3 वृत्त के कें द्र से जीवा पर डाला गया लंब जीवा को _____________ करता है।
The perpendicular drawn from the centre of a circle to a chord ___________
the chord.

48

4 चक्रीय चतभु ुज के सम्मख
ु कोणों का योग _______________ होता है।
The sum of opposite angles of a cyclic quadrilateral is _______________.
5 वृत्त की सबसे बडी जीवा वृत्त का __________________ होता है।
The longest chord of a circle is the __________________ of the circle.
सत्य/असत्य णिणिए (Write true/false:): 1 अांक (1 Mark)
1 दो समरूप आकृ त्रतयों के माप में त्रवशेष अनपु ात होता है त्रजसे “स्के ल गणु क” कहते हैं।
There is a special ratio in the measurements of two similar shapes called "scale
factor."
2 तीन असमरे ख त्रबदं ओ
ु ं से होकर एक ओर के वल एक वृत्त खींचा जा सकता है।
By connecting three non-collinear points, only one circle can be drawn.
3 वृत्त के एक ही खडं में बने कोण आपस में बराबर होते हैं।
In the same segment of a circle, the angles formed are equal to each other.
4 आयत का प्रत्येक कोण 180° का होता है।
Each angle of a rectangle is equal to 180°.
5 स्पशु त्रबदं ु से खींची गई त्रिज्या वृत्त की स्पशु रे खा पर समातं र होती है।
A radius drawn from the center of a circle to the point of contact is parallel to
the tangent line.
िघु उत्तिीय प्रश्न-4 अांक (Short Answers Questions-4 Marks):
1 त्रसद्ध कीत्रजए समकोण त्रिभजु में कणु का वगु शेष दो भजु ाओ ं के वगों के योग के बराबर होता है।
Prove that in a right triangle, the square of the hypotenuse is equal to the sum
of the squares of the other two sides.
2 त्रसद्ध कीत्रजए वृत्त के कें द्र से जीवा पर डाला गया लंब जीवा को समत्रिभात्रजत करता है।
Prove that the perpendicular drawn from the center of a circle to the chord
bisects the chord.

49

3 एक वृत्त की त्रिज्या 5 सेंटीमीटर है तो वृत्त के कें द्र से 3 सेंटीमीटर की दरू ी पर त्रस्ित जीवा की लंबाई
ज्ञात कीत्रजए।
If the radius of a circle is 5 centimetres, determine the length of a line segment
located 3 centimetres away from the center.
4 त्रसद्ध कीत्रजए चक्रीय चतुभुज के सम्मख ु कोणों का योग 180° होता है।
Prove that the sum of the opposite angles of a cyclic quadrilateral is 180°.
5 त्रसद्ध कीत्रजए त्रकसी बाह्य त्रबंदु से वृत पर खींची गई स्पशु रे खाओ ं की लंबाईया बराबर होती है।
Prove that the lengths of tangents drawn from an external point to a circle are
equal.
6 यत्रद PAB वृत्त की छे दक रे खा है, जो वृत्त को A और B पर प्रत्रतच्छे द करती है और PT स्पशु
रे खाखडं है तो त्रदखाइए त्रक
PA x PB = PT2
If PAB is a chord of a circle that intersects the circle at points A and B, and PT
is a tangent to the circle at point P, then prove that
PA x PB = PT2

7 एक त्रिभजु ABC में AD⊥BC है, तो त्रसद्ध कीत्रजए त्रक
AB2 + CD2 = BD2 + AC2
ABC is a triangle, in which AD⊥BC then prove that
AB2 + CD2 = BD2 + AC2
8
एक त्रिभजु ABC त्रजसमें ∠C समकोण है। भजु ाओ ं CA और CB पर क्रमशः त्रबंदु D और E
त्रस्ित है त्रसद्ध कीत्रजए:
AE2 + BD2 = AB2 + DE2
In triangle ABC with ∠C is right angle and points D and E on sides CA and
CB respectively, prove
AE2 + BD2 = AB2 + DE2

50

दीघि उत्तिीय प्रश्न-5 अांक (Long Answers Questions-5 Marks):
1
∆ABC के पररगत वृत्त की रचना कीत्रजए, जहां AB=3 सेंटीमीटर, BC= 4 सेंटीमीटर और ∠B=
90°, रचना के पद भी त्रलत्रखए।
Construct the circumcircle of ∆ABC, where AB = 3 centimetres, BC = 4
centimetres, and ∠B = 90°. Also, write the construction of steps also.
2
∆ABC के अंतः वृत्त की रचना कीत्रजए, AB=BC=CA= 6 सेंटीमीटर, रचना के पद भी त्रलत्रखए।
Construct the incircle in ∆ABC, where AB = BC = CA = 6 centimetres. Also,
write the construction of steps.
3
∆ABC के पररवृत्त की रचना कीत्रजए, जहां BC=7 सेंटीमीटर, ∠B= 45°, ∠A=105°, रचना
के पद भी त्रलत्रखए।
Construct the circumcircle of triangle ABC, where BC = 7 centimetres, ∠B =
45°, ∠A = 105°. Also, write the steps of construction.
4
∆ABC के पररगत वृत्त की रचना कीत्रजए, जहां BC= 6 सेंटीमीटर, ∠B= 70° और AB=5
सेंटीमीटर, रचना के पद भी त्रलत्रखए।
Construct the circumcircle of triangle ABC, where BC = 6 centimetres, ∠B =
70°, and AB = 5 centimetres. Also, write the construction of steps.
5
एक त्रिभजु ABC बनाइए, त्रजसमें BC=5.5 सेंटीमीटर, ∠ABC= 75° और ∠ACB=45° इस
5
त्रिभजु के समरूप एक त्रिभुज XYZ बनाइए, त्रजसमें YZ= 4 BC हो।
Construct a triangle ABC where BC = 5.5 centimetres, ∠ABC = 75°, and
5
∠ACB=45°. Also, construct a similar triangle XYZ where YZ = 4 BC.

51

इकाई 6 - गणििीय कथनों की जााँच
UNIT 6 - CHECKING MATHEMATICAL STATEMENTS

सही णिकल्प का चयन कीणजए (Select the correct option:): 1 अांक (1 Mark)
1. चतभु ुज के आंतररक कोणों का योग होता है:
अ. 90° ब. 180°
स. 360° द. 540°
The sum of the interior angles of a quadrilateral is:
A. 90° B. 180°
C. 360° D. 540°
2. दो सम सख्ं याओ ं का जोड सदैव होता है:
अ. सम संख्या ब. त्रवषम संख्या
स. सम और त्रवषम संख्या द. इनमें से कोई नहीं
The sum of two even numbers is always:
A. Even number B. odd number
C. Even and odd number D. none of these

रिक्त स्थानों की पणू िि कीणजए (Fill in the blanks): 1 अांक (1 Mark)
1 दो सख्ं याओ ं का जोड _____________सख्ं या होता है।
The sum of two numbers is _____________ number.
2 त्रवषम सख्ं या का वगु _____________ सख्ं या होता है।
The square of an odd number is _____________ number.
3 त्रकसी त्रिभजु के अतं ः कोणों का योग _____________ होता है।
The sum of the interior angles of a triangle is _____________.
52

सत्य/असत्य णिणिए (Write true/false:): 1 अांक (1 Mark)
1
त्रकसी वास्तत्रवक संख्या x के त्रलए x2 ≥ 0.
For any real number x, x2 ≥ 0.
2 सभी अभाज्य संख्या त्रवषम होती है।
All prime numbers are odd.
3 सभी बहुभजु पचं भजु होते हैं।
All polygons have five sides.
4 सभी वास्तत्रवक संख्याएं अपररमेय होती है।
All real numbers are irrational.
5 सभी सम सख्ं याएं 2 से भाज्य नहीं होती है।
All even numbers are not divisible by 2.
िघु उत्तिीय प्रश्न-3 अांक (Short Answers Questions-3 Marks):
1 त्रसद्ध कीत्रजए त्रक 2K+7 एक त्रवषम पणू ाांक है। जहां K एक पणू ाांक है।
Prove that 2K+7 is an odd integer, where K is an integer.
2 त्रसद्ध कीत्रजए त्रक 4m +9 एक त्रवषम पणू ाांक है जहां m एक पणू ाांक है।
Prove that 4m + 9 is an odd integer, where m is an integer.
3 त्रसद्ध कीत्रजए त्रक त्रवषम संख्याओ ं का वगु त्रवषम संख्या होती है।
Prove that the square of an odd number is an odd number.
4 त्रसद्ध कीत्रजए त्रक दो संख्या का जोड हमेशा त्रवषम संख्या होती है।
Prove that the sum of two numbers is always an odd number.
5 त्रसद्ध कीत्रजए त्रक दो संख्याओ ं का जोड सदैव सम संख्या होती है।
Prove that the sum of two numbers is always an even number.
6 त्रसद्ध कीत्रजए त्रक त्रकसी भी तीन क्रमागत सम संख्याओ ं का योग हमेशा 6 का गणु ज होता है।
Prove that the sum of any three consecutive even numbers is always a multiple
of 6.

53

इकाई 7 – क्षेत्रणमणि
UNIT 7 - MENSURATION

सही णिकल्प का चयन कीणजए (Select the correct option:): 1 अांक (1 Mark)
1. r त्रिज्या वाले अर्ुगोले का सपं णू ु पृष्ठ होता है:

अ. 2𝜋r2 ब. 3𝜋r2

स. 4𝜋r2 द. 𝜋r2
The whole surface of a hemisphere of radius r is:

A. 2𝜋r2 B. 3𝜋r2

C. 4𝜋r2 D. 𝜋r2
2. यत्रद घनाभ की भजु ाएं a इकाई, b इकाई, c इकाई हो तो घनाभ का आकाशीय त्रवकणु होगा:

अ. √𝑎2 + 𝑏2 ब. √𝑏2 + 𝑐 2

स. √𝑎2 + 𝑐 2 द. √𝑎2 + 𝑏2 + 𝑐 2
If the edges of a cuboid are of lengths a units, b units, c units, then the space
diagonal of the cuboid will be:

A. √𝑎2 + 𝑏2 B. √𝑏2 + 𝑐 2

C. √𝑎2 + 𝑐 2 D. √𝑎2 + 𝑏2 + 𝑐 2
3. एक शंकु का व्यास 6 सेंटीमीटर और ऊंचाई 4 सेंटीमीटर है तो शंकु की त्रतयुक ऊंचाई होगी:
अ. 3 सेंटीमीटर ब. 4 सेंटीमीटर
स. 5 सेंटीमीटर द. 6 सेंटीमीटर
54

If the diameter of a cone is 6 cm and height is 4 cm, then the slant height of
the cone will be:
A. 3 cm B. 4 cm
C. 5 cm D. 6 cm

4. बेलन के आर्ार का क्षेिफल 16𝜋 वगु सेंटीमीटर है, तो उसकी त्रिज्या होगी:
अ. 2 सेंटीमीटर ब. 4 सेंटीमीटर

स. 16 सेंटीमीटर द. 𝜋 सेंटीमीटर

If the area of the base of the cylinder is 16 𝜋 square centimeter, then its radius
will be:
A. 2 cm B. 4 cm
C. 16 cm D. 𝜋 centimeter
5. घन के शीषों की सख्ं या होती है:
अ. 6 ब. 8
स. 12 द. 16
The number of vertices of the cube is:
A. 6 B. 8
C. 12 D. 16
रिक्त स्थानों की पूणिि कीणजए (Fill in the blanks):
1 घनाभ में आकाशीय त्रवकणों की कुल संख्या _________ होती है।
The total number of space diagonals in a cuboid is _________.
2 घन के आकाशीय त्रवकणु की लंबाई __________ इकाई होती है ।
The length of the space diagonal of a cube is __________ units.
55

3
बेलन का आयतन = ____________ x शंकु का आयतन ।
The volume of a cylinder = ____________ x volume of a cone.
4 एक इकाई त्रिज्या वाले गोले का आयतन __________ घन इकाई होगा ।
The volume of a sphere with a unit radius will be __________ cubic units.
5 घनाभ के कोरों की संख्या _________ होती है ।
The number of vertices in a cuboid is _________.
6 गोले का पृष्ठीय क्षेिफल, उसी त्रिज्या के वृत्त के क्षेिफल का _______ गनु ा होता है ।
The surface area of a sphere is _______ times the area of a circle with the
same radius.
7 शंकु के आर्ार का क्षेिफल _________ के क्षेिफल के बराबर होता है ।
The base area of a cone is equal to the area of _________.
8 14 सेंटीमीटर व्यास वाले अर्ु गोले की त्रिज्या _________ सेंटीमीटर होगी ।
The radius of a hemisphere with a diameter of 14 cm will be _________ cm.

सत्य/असत्य णिणिए (Write true/false:): 1 अांक (1 Mark)
1 घनाभ की सभी फलके वगाुकार होती है ।
All faces of a cuboid are square-shaped.
2 आयताकार कागज का क्षेिफल बेलन का वक्र पृष्ठ होता है ।
The area of a rectangular paper is the curved surface of a cylinder.
3
बेलन का सपं णू ु पृष्ठ का क्षेिफल = 2πrh.
The total surface area of a cylinder is = 2πrh.
2
4 अर्ु गोले का आयतन 3 πr3 होता है ।
2
The volume of a hemisphere is 3 πr3.
5 घन के कुल त्रवकणों की संख्या 16 होती है।
The total number of diagonals in a cube is 16.

56

िघु उत्तिीय प्रश्न-3 अांक (Short Answers Questions-3 Marks):

1 एक बेलन के आर्ार की त्रिज्या 14 सेंटीमीटर ऊंचाई 10 सेंटीमीटर है बेलन के वक्रपृष्ठ का क्षेिफल
ज्ञात कीत्रजए।
Find the curved surface area of a cylinder with a radius of 14 cm and a height
of 10 cm.
2 दो समान ऊंचाई वाले लंबवृत्तीय बेलनों के आर्ार की त्रिज्या 2:5 के अनपु ात में है, तो इनके
आयतनों का अनपु ात ज्ञात कीत्रजए।
If the bases of two similar cylindrical containers have a ratio of their radii as 2:5
and they have the same height, find the ratio of their volumes.
3 एक शंकु के आर्ार की त्रिज्या 7 सेंटीमीटर और ऊंचाई 15 सेंटीमीटर है। शंकु का आयतन ज्ञात
कीत्रजए।
The base of a cone has a radius of 7 cm and a height of 15 cm. Find the volume
of the cone.
4
एक शक ं ु का वक्रपृष्ठ 35π वगु सेंटीमीटर है। यत्रद इसके आर्ार का व्यास 14 सेंटीमीटर हो तो इसकी
त्रतयुक ऊंचाई ज्ञात कीत्रजए।
The curved surface area of a cone is 35π square cm. If its radius is 14 cm, find
its slant height.
5 एक गोले का पृष्ठीय क्षेिफल 154 वगु सेंटीमीटर है। गोले का व्यास ज्ञात कीत्रजए।
The lateral surface area of a sphere is 154 square cm. Find the diameter of the
sphere.
6 एक गोले का व्यास 28 सेंटीमीटर है। गोले का पृष्ठीय क्षेिफल ज्ञात कीत्रजए।
The diameter of a sphere is 28 cm. Find its surface area.
7 उस बडे से बडे खंभे की लंबाई ज्ञात कीत्रजए, जो 10 मीटर लंबा, 10 मीटर चौडा और 5 मीटर ऊंचे
कमरे में रखा जा सकता है।
Determine the height of the largest pole that can be placed in a room that is 10
meters long, 10 meters wide, and 5 meters high.

57

िघु उत्तिीय प्रश्न-4 अांक (Short Answers Questions-4 Marks):

1 एक बेलन के आर्ार की त्रिज्या 14 सेंटीमीटर और ऊंचाई 10 सेंटीमीटर है। बेलन का वक्रपृष्ठ तिा
संपणू ु पृष्ठ का क्षेिफल ज्ञात कीत्रजए।
The base radius of a cylinder is 14 cm and its height is 10 cm. Find the curved
surface area and the total surface area of the cylinder.
2 एक बेलन के वक्र पृष्ठ का क्षेिफल 3696 वगु सेंटीमीटर है। यत्रद बेलन के आर्ार की त्रिज्या 14
सेंटीमीटर है तो बेलन की ऊंचाई ज्ञात कीत्रजए।
The curved surface area of a cylinder is 3696 square cm. If the base radius of
the cylinder is 14 cm, find its height.
3 एक बेलन का आयतन और ऊंचाई क्रमशः 3080 घन सेंटीमीटर और 20 सेंटीमीटर है, तब बेलन
का वक्र पृष्ठ का क्षेिफल ज्ञात कीत्रजए।
The volume and height of a cylinder are 3080 cubic cm and 20 cm respectively.
Find the curved surface area of the cylinder.
4 एक लंब वृत्तीय बेलन के आर्ार की पररत्रर् 44 सेंटीमीटर है, यत्रद बेलन की ऊंचाई 10 सेंटीमीटर है।
तो बेलन का वक्र पृष्ठ क्षेिफल और आयतन ज्ञात कीत्रजए।
The circumference of the base of a cylindrical container is 44 cm, and its height
is 10 cm. Find the curved surface area and volume of the cylinder.
5
शंकु के आकार के तंबू में 65π वगु मीटर कपडा लगा है। तंबू की त्रतयुक ऊंचाई 13 मीटर है। तो
उसकी ऊंचाई तिा त्रिज्या ज्ञात कीत्रजए।

A tent shaped like a cylinder has canvas covering an area of 65π square meters.
The tent's slant height is 13 meters. Find its height and radius.

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6 एक शंकु आकार तंबू की ऊंचाई 5 मीटर तिा आर्ार की त्रिज्या 12 मीटर हो तो उसकी त्रतयुक
ऊंचाई तिा तबं ू को बनाने में लगने वाले त्रतरपाल (कै नवास) का लागत मल्ू य ज्ञात कीत्रजए। यत्रद
उसका मल्ू य ₹70 प्रत्रत वगु मीटर हो।
If a tent-shaped like a cone has a height of 5 meters and a base radius of 12
meters, find its slant height and the cost of canvas used to make the tent if it
costs ₹70 per square meter.
7 यत्रद एक शंकु के आर्ार का व्यास 14 सेंटीमीटर तिा ऊंचाई 24 सेंटीमीटर है, तो शंकु का संपणू ु
पृष्ठ का क्षेिफल तिा आयतन ज्ञात कीत्रजए।
If the base radius of a cone is 14 cm and its height is 24 cm, find the total surface
area and volume of the cone.
8 यत्रद गोले का व्यास 12 सेंटीमीटर है, तो गोले का पृष्ठीय क्षेिफल तिा आयतन ज्ञात कीत्रजए।
If the diameter of a sphere is 12 cm, find its surface area and volume.
9 लोहे की तीन गोत्रलयों को त्रजनकी त्रिज्याएँ 6 सेंटीमीटर, 8 सेंटीमीटर और 10 सेंटीमीटर है, को
त्रपघलाकर एक बडा ठोस गोला बनाया जाता है। बनाए गए नए गोले की त्रिज्या ज्ञात कीत्रजए।
Three solid iron spheres with radii of 6 cm, 8 cm, and 10 cm are melted down
to form a single large solid sphere. Find the radius of the resulting sphere.
10 2 सेंटीमीटर त्रिज्या वाले 64 गोत्रलयों को त्रपघलाकर एक बडा गोला बनाया गया। बडे गोले की
त्रिज्या ज्ञात कीत्रजए।
Sixty-four small iron spheres with radii of 2 cm each are melted down to form
a single large solid sphere. Find the radius of the resulting sphere.
11 दो ठोस गोले के आयतनों का अनपु ात 64:27 है। इनके पृष्ठ क्षेिफलों का अनपु ात ज्ञात कीत्रजए।
The ratio of the volumes of two solid spheres is 64:27. Find the ratio of their
surface areas.

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12 त्रमट्टी का एक शंकु त्रजसकी ऊंचाई 24 सेंटीमीटर और आर्ार की त्रिज्या 6 सेंटीमीटर है त्रजसे एक
बच्चा गोले में पररवत्रतुत कर देता है। इस गोले की त्रिज्या ज्ञात कीत्रजए।
A clay cone with a height of 24 cm and a base radius of 6 cm is transformed into
a sphere by a child. Find the radius of the sphere formed.
13 एक गोले का आयतन ज्ञात कीत्रजए, त्रजसका पृष्ठीय क्षेिफल 154 वगु सेंटीमीटर है।
Find the volume of a sphere whose lateral surface area is 154 square cm.
14 एक ठोस शंकु की ऊंचाई 10 सेंटीमीटर और व्यास 20 सेंटीमीटर है इसे गलाकर 2 सेंटीमीटर व्यास
वाले त्रकतने गोले बनाए जा सकते हैं ।
A solid cone has a height of 10 cm and diameter of 20 cm. How many smaller
spheres with a radius of 2 cm can be formed by melting down this larger cone?

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इकाई 8 – साांणययकी
UNIT 8 - STATISTICS
अणि िघु उत्तिीय प्रश्न-2 अांक (Very Short Answers Questions-2 Marks):

1 प्रिम 10 सम संख्याओ ं का औसत ज्ञात कीत्रजए ।
Find the average of the first 10 even numbers.
2 त्रनम्नत्रलत्रखत आँकडों की मात्रययका ज्ञात कीत्रजए ।
22, 20, 24, 16, 18, 26, 21.
Calculate the median of the following numbers:
22, 20, 24, 16, 18, 26, 21.
3 प्रिम दस प्राकृ त सख्ं याओ ं का औसत ज्ञात कीत्रजए।
Determine the mean of the first ten natural numbers.
4 आँकडे 3, 6, 9, 12, 15, 18, 21, 24 का समातं र मायय ज्ञात कीत्रजए।
Find the arithmetic mean of the numbers 3, 6, 9, 12, 15, 18, 21, 24.
5 त्रनम्नत्रलत्रखत आँकडों का बहुलक ज्ञात कीत्रजए।
56, 39, 94, 36, 39, 15, 39, 40.
Calculate the mode of the numbers:
56, 39, 94, 36, 39, 15, 39, 40.
6 त्रनम्नत्रलत्रखत आँकडों की मात्रययका ज्ञात कीत्रजए।
117, 106, 123, 110, 125, 112, 115, 102, 100, 115
Find the median of the following numbers:
117, 106, 123, 110, 125, 112, 115, 102, 100, 115

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िघु उत्तिीय प्रश्न-4 अांक (Short Answers Questions-4 Marks):
1. त्रनम्न सारणी का बहुलक ज्ञात कीत्रजए।
वगाुन्तर 0-10 10-20 20-30 30-40 40-50
आवृत्रत्त 4 10 16 12 8

Calculate the mode of the given table.
Class interval 0-10 10-20 20-30 30-40 40-50
Frequency 4 10 16 12 8

2. त्रनम्न सारणी का मात्रययका ज्ञात कीत्रजए।
वगाुन्तर 0-10 10-20 20-30 30-40 40-50 50-60
आवृत्रत्त 4 6 10 7 3 2

Find the median of the given table.
Class
0-10 10-20 20-30 30-40 40-50 50-60
interval
Frequency 4 6 10 7 3 2

3. त्रनम्न सारणी से समांतर मायय ज्ञात कीत्रजए।
अंक (x) 5 15 25 35 45 55
बारंबारता (f) 5 3 13 18 8 6

Calculate the arithmetic mean of the given table.

Marks (x) 5 15 25 35 45 55
Frequency (f) 5 3 13 18 8 6

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4. एक त्रदवसीय अंतरराष्ट्रीय त्रक्रके ट मैचों में बहुत से गेंदबाजों िारा त्रलए गए कुल त्रवके टों की
सख्ं या के आँकडे तात्रलका में त्रदए गए हैं। इन आँकडों का बहुलक ज्ञात कीत्रजए।
त्रवके टों की
0-50 50-100 100-150 150-200 200-250 250-300
संख्या
गेंदबाजों की
4 5 16 12 3 2
संख्या

Determine the mode of the total wickets taken by various bowlers in a one-day
international cricket match. The numbers are provided in the table.
No. of
0-50 50-100 100-150 150-200 200-250 250-300
wickets
No. of
4 5 16 12 3 2
bowlers

5. उच्चतर माययत्रमक शाला के छोटे-बडे बच्चों (त्रवद्यात्रिुयों) के वजन के आँकडे त्रदए गए हैं।
आँकडों का मायय ज्ञात कीत्रजए।
वजन त्रक.ग्रा.
30-40 40-50 50-60 60-70 70-80
में
त्रवद्यात्रिुयों की
11 29 6 3 1
संख्या

Find the mean of the weights of small and big children (students) of a higher
secondary school.
Weight in
30-40 40-50 50-60 60-70 70-80
kilograms
No. of
11 29 6 3 1
students

ll

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