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PREFACE

“The Objective of Education is to prepare the young to Educatethemselves throughout their Lives”

This philosophy has always been followed by Gujarat Secondary Education Board (GSEB), whether through their education system framework or regular enhancement in curriculum. GSEB ensures better access, equality and quality in elementary education for school students. In order to achieve aforesaid objectives, Gujarat State Board of School Textbooks (GSSTB) has proposed a new syllabus for school textbooks, which will be aligned with NCERT.

We at Oswaal Books, welcome the above decision of GSEB and have ensured our offerings include updated content, aligned with the latest syllabus as directed by the Board. Oswaal GSEB NCERT Problems-Solutions are designed as per the latest curriculum of Gujarat Board and emphasize on nurturing individuality thus enhancing one's innate potentials which help in increasing self confidence.

We believe that OSWAAL GSEB NCERT PROBLEMS-SOLUTIONS will help the students in school and after school in practicing and preparing extensively for both, Final Examinations as well as Competitive Examinations with utmost confidence!

Some of the Key Highlights of Oswaal GSEB NCERT Problems-Solutions are :

l Latest content : Strictly based on the latest GSEB Curriculuml GSSTB (NCERT) Textbook Questions : Fully Solvedl Some Important Questions developed by ‘Oswaal Editorial Board’.l Chapter-wise & Topic-wise presentationl Chapter Objectives : A sneak peek into the chapterl Mind Map : A single page snapshot of the entire chapter l Quick Review : Concept-based study materiall Tips & Tricks : Useful guidelines for attempting each question perfectly l Some Commonly Made Errors : Most common and unidentified errors made by students discussed l Expert Advice : Oswaal Expert Advice on how to score more!l Oswaal QR Codes : For a Digital Learning Experience

We hope that OSWAAL GSEB NCERT PROBLEMS-SOLUTIONS will help you at every step as you move closer to your educational goals. We wish you all great success ahead!!

– Team Oswaal

( 6 )

CONTENTS

l Mind Maps 9 - 23

1. REAL NUMBERS 1 - 13 l Topic 1. Euclid's Division Lemma and Fundamental Theorem of Arithmetic

l Topic 2. Rational and Irrational Numbers

2. POLYNOMIALS 14 - 27

3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 28 - 66

4. QUADRATIC EQUATIONS 67 - 88

5. ARITHMETIC PROGRESSION 89 - 120

6. TRIANGLES 121 - 153

l Topic 1. BPT and Similarity

l Topic 2. Pythagoras Theorem

7. COORDINATE GEOMETRY 154 - 190

8. INTRODUCTION TO TRIGONOMETRY 191 - 209

9. SOME APPLICATIONS OF TRIGONOMETRY 210 - 226

10. CIRCLES 227 - 249

11. CONSTRUCTIONS 250 - 264

12. AREAS RELATED TO CIRCLES 265 - 292

13. SURFACE AREAS AND VOLUMES 293 - 321

14. STATISTICS 322 - 350

15. PROBABILITY 351 - 368

qq

( 7 )

( 8 )

MIND MAP | 9

Mind

map

: lea

rning m

ade

simpl

e

Theo

rem

s

Rea

lN

umbe

rs

Eucl

id’s

Divisi

on L

emm

a

Divisio

n Alg

orith

m

Fundamental Theorem of

S.N

o.St

atem

ent

1.Le

t p b

e a

prim

e nu

mbe

r.If

p d

ivid

es a

2 , th

en p

div

ides

a,

whe

re a

is a

pos

itive

inte

ger

2. 3.Le

t x b

e a

ratio

nal n

umbe

rw

hose

dec

imal

exp

ansi

onte

rmin

ates

. The

n x

can

beex

pres

sed

in th

e fo

rm

,

w

here

p &

q a

re c

opri

me,

the

prim

e fa

ctor

isat

ion

of q

is o

f the

form

2n ,

5m

whe

re n

, mar

e no

n-ne

gativ

e in

tege

rs

√2 is

irra

tiona

l

4.Le

t x =

b

e a

ratio

nal n

umbe

rsu

ch th

at th

e pr

ime

fact

oris

atio

nof

q is

of t

he fo

rm 2

n , 5

m w

here

n, m

are

non

-neg

ativ

e in

tege

rs.

Then

, x h

as a

dec

imal

exp

ansi

onw

hich

term

inat

es.

p q

5.Le

t x =

b

e a

ratio

nal n

umbe

r,su

ch th

at th

e pr

ime

fact

oris

atio

nof

q is

not

of t

he fo

rm o

f 2n 5

m

whe

re n

, m a

re n

on-n

egat

ive

inte

gers

. The

n, x

has

a d

ecim

alex

pans

ion

whi

ch is

non-

term

inat

ing

repe

atin

g

p q

p q

For

any

two

posi

tive

inte

gers

, a a

nd b

HC

F (a

, b) ×

LC

M (a

, b)

= a

× b

Fo

r Ex

ampl

ef(x

) = 3

x2 yg(

x) =

6xy

2

HC

F =

3xy

LCM

= 6

x2 y2

Giv

en p

ositi

ve in

tege

rs a

and

b,

ther

e ex

ist u

niqu

e in

tege

rs q

and

rsa

tisfy

ing

a =

bq

+ r

; 0<

r<b

Step

s to

obt

ain

the

HC

F of

two

posi

tive

inte

gers

, say

c a

nd d

,w

ith c

>d

Step

1: A

pply

Euc

lid’s

Div

isio

n

L

emm

a, to

c &

d. c

= d

q +

r

Step

3: C

ontin

ue th

e pr

oces

s til

l

th

e re

mai

nder

is z

ero

Step

2: I

f r =

zer

o, d

is th

e H

CF

of

c

and

d

If

r ≠

0, a

pply

Euc

lid’s

Div

isio

n to

d a

nd r

Ever

y co

mpo

site

num

ber

can

beex

pres

sed

as a

pro

duct

of p

rim

es,

and

this

fact

oris

atio

n is

uni

que,

apar

t fro

m th

e or

der

in w

hich

the

prim

e fa

ctor

s oc

cur

Com

posi

te N

umbe

r x

= P

1P2

... P

n,w

here

P1P

2 ...

Pn

are

prim

e nu

mbe

rs

Prime Factorization Method

Arithmetic

10 | Oswaal GSEB NCERT Solutions – Textbook + Exemplar, MATHEMATICS, Class-X

Mind

map

: lea

rning m

ade

simpl

e

Div

isio

n Al

gorit

hm

Pol

ynom

ials

Para

bola

Types

Graph

ical R

epre

sent

atio

n-

Deg

ree

of P

olyn

omial

Qua

drat

ic P

olyn

omia

l

Rel

atio

nshi

p-Z

eroe

s an

d

Coe

ffic

ient

of P

olyn

omia

ls

Qua

drat

ic

Cubic

If p

(x) a

nd g

(x) a

re tw

opo

lyno

mia

ls w

ithg(

x) ≠

0, t

hen

–p(

x)=

g(x

) × q

(x) +

r(x

)w

here

, r(x

) = 0

or

degr

ee o

f r(x

) < d

egre

eof

g(x

)

α a

nd β

are

zer

oes

ofQ

uadr

atic

Pol

ynom

ial

Then

,Su

m o

f zer

oes,

ax2 +

bx

+ c

Hig

hest

pow

er o

f x in

Poly

nom

ial,

p(x)

α +

β =

–b a

Prod

uct o

f zer

oes

αβ

=c a

α, β

and

ϒ a

re z

eroe

sof

Cub

ic P

olyn

omia

l

Sum

of z

eroe

s,

ax3 +

bx2 +

cx

+ d

α +

β +

ϒ =

–b a

αβ

+ β

ϒ +

ϒα

=c a

αβϒ

= –

d a

Sum

of p

rodu

cts

of th

eze

roes

take

n tw

o at

atim

e

Prod

uct o

f zer

oes

y =

x2 –3

x–4

Poly

nom

ial

Deg

ree

Gen

eral

For

m

Line

ar1

ax+

b

Qua

drat

ic2

ax2 +

bx+

ca

≠ 0

a ≠

0C

ubic

3ax

3 +bx

2 +cx

+d

Cas

eG

raph

Num

ber

of Z

eroe

s

Cas

e1- G

raph

cut

sx-

axis

at 2

poi

nts

2

Cas

e2- G

raph

cut

sx-

axis

at e

xact

lyon

e po

int

1

Cas

e3- G

raph

doe

sno

t cut

x-a

xis

0

Zero

es o

fpo

lyno

mia

lG

raph

ical

ly

MIND MAP | 11

Mind

map

: lea

rning m

ade

simpl

e

Pair

of

Lin

ear

Equa

tions

in

Two

Var

iabl

es

Gen

eral

For

m

By E

limin

atio

n

By Cross-

Multiplication

By Su

bstitution

a 1x+b 1y+

c 1 = 0

a 2x+b 2y+

c 2 = 0

a 1,b1,c

1,a2,b

2,c2, –

Rea

l num

bers

Solu

tion

Gra

phic

ally

Each

sol

utio

n (x

, y),

corr

espo

nds

to a

poi

nt o

n th

e lin

e re

pres

entin

gth

e eq

uatio

n an

d vi

ce-v

ersa

S. N

o.Pa

ir o

f Lin

es

1. 2. 3.

Solv

e: 2

x+3y

–46

= 0

–(

i)

3x

+5y

–74

= 0

–(

ii)

xy

13 5

–46

–74

2 3

3 5

Solu

tion

: By

cros

s-m

ultip

licat

ion

met

hod

Then

,=

x3(

–74)

–5(–

46)

y(–

46)(

3)–(

–74)

(2)

=1

2(5)

–3(3

)

=x

–222

+23

0=

y–1

38+

148

110

–9

=x 8

=1 1

y 10an

d

x =

8 a

nd y

= 1

0

=x 8

1 1=

y 101 1

x–2y

= 0

3x+

4y–2

0 =

0≠

2x+

3y–9

= 0

4x+

6y–1

8 =

0

x+2y

–4 =

02x

+4y

–12

= 0

Inte

rsec

ting

Line

s

Coi

ncid

ent L

ines

Para

llel L

ines

Com

pare

the

Rat

ios

Gra

phic

alR

epre

sent

atio

nA

lgeb

raic

Inte

rpre

tati

on

Exac

tly o

neso

lutio

n –

cons

iste

nt(u

niqu

e)

Infin

itely

man

y so

lutio

ns–

Dep

ende

nt

No

solu

tion

–In

cons

iste

nt

a 1 a 2 1 3 2 4 1 2

b 1 b 2 –2 4 3 6 2 4

c 1 c 2

a 1 a 2b 1 b 2

==

a 1 a 2b 1 b 2

c 1 c 2

=≠

a 1 a 2b 1 b 2

c 1 c 2

0 –20

–9 –18

–4 –12

Solv

e: 7

x–15

y =

2

–(i)

x

+2y

= 3

–(

ii)

Solu

tion

: Fro

m e

quat

ion

(ii),

x =

3–2

y-(ii

i)

sub

stitu

te v

alue

of x

in e

q. (i

)

7

(3–2

y)–1

5y =

2

–29y

= –

19⇔

∴In

eq.

(iii)

x =

3 –

2

=

y =

19 2919 29

49 29

Solv

e: 2

x+3y

= 8

– (i

)

4x+

6y =

7 –

(ii)

Solu

tion

: Fro

m e

q. (i

)x 2

–eq.

(ii)x

1, w

eha

ve(4

x–4x

) + (6

y–6y

) = 1

6–7

0 =

9, w

hich

is a

fals

e st

atem

ent

The

pair

of e

quat

ion

has

no s

olut

ion

⇔

i.e.

Gra

phic

alR

epre

sent

atio

n

Alg

ebra

ic M

etho

ds

12 | Oswaal GSEB NCERT Solutions – Textbook + Exemplar, MATHEMATICS, Class-X

Mind

map

: lea

rning m

ade

simpl

e

Qua

drat

icEq

uatio

nsSo

luti

on o

f a

Qua

drat

ic E

quat

ion

By

Com

ple

tin

g th

e Sq

ua

reMeaning

Equ

atio

n o

f d

egre

e 2,

in o

ne

vari

able

Gen

eral

For

m

Qua

drat

ic F

orm

ula

Nature of Roots

ax2 +

bx+

c =

0a,

b, c

– r

eal n

um

bers

a ≠

0

For

quad

rati

c eq

uat

ion

ax2 +

bx+

c =

0,

b2 –4a

c is

Dis

crim

inan

t (D

)

Roo

ts o

f ax

2 +bx

+c

= 0

are

giv

en b

y

S. N

o.D

iscr

imin

ant

1. 2. 3.

D >

0

Roo

ts

Tw

o d

isti

nct

real

roo

ts

Tw

o eq

ual

real

roo

ts

No

real

roo

ts(i

mag

inar

y)

D =

0

D <

0

2a–b

± √

b2 –4a

c

Sol

ve: 2

x2 –5x

+3

= 0

Sol

uti

on:

2

x2 –5x

+3

= 0

x2 –2

2–

+

x+ =

0

= 0

⇔

⇔

5 23 2

x –

5 4

x –

5 4

5 43 2

=1 4

±

=x

or5 4

1 4+

=x

5 41 4

–

=x

or3 2

x =

1

By Factorization

Fin

d r

oots

of

6x2 –

x–2=

0

Sol

uti

on:

6

x2 +3x

–4x–

2 =

03x

(2x+

1)–2

(2x+

1) =

0(3

x–2)

(2x+

1) =

0

Th

e ro

ots

of 6

x2 –x–

2=0

Roo

ts a

re

(3x–

2) =

0 o

r (2

x+1)

= 0

2 3x

=or

–1 2x

=

2 3,

–1 2

2 –=

0x

5 41 16

2=

1 16

–

x5 4

–

MIND MAP | 13

Mind

map

: lea

rning m

ade

simpl

e

s n =

n(a+

l)2 Ari

thm

etic

Pro

gres

sion

s

Defin

ition

Arit

hmet

ic m

ean

Sum (s)

List

of n

umbe

rs in

whi

ch e

ach

term

is o

btai

ned

by a

ddin

g a

fixed

num

ber

to th

e pr

eced

ing

term

exc

ept t

he fi

rst t

erm

. Fi

xed

num

ber

is c

alle

d co

mm

on d

iffer

ence

.

Gen

eral

form

Comm

on D

iffer

ence

nth Term

a, a

+d,

a+

2d, a

+3d

, ...

a+(n

–1)

d

From

beg

inni

nga n

= a

+(n

–1)d

Her

ea

– fir

st te

rmd

– co

mm

on d

iffer

ence

a n –

nth

term

From

the

end

a n =

l –

(n–1

)d

Her

el –

last

term

d –

com

mon

diff

eren

cea n

– n

th te

rm

Fixe

d nu

mbe

r in

arith

met

icpr

ogre

ssio

n w

hich

pro

vide

s

the

to a

nd fr

o te

rms

by a

ddin

g/su

btra

ctin

g fr

om th

e pr

esen

t nu

mbe

r.C

an b

e po

sitiv

e or

neg

ativ

e.

Examples

s =

(a+

a n)

a –

first

term

n –

tota

l ter

ms

a n –

nth

term

l – la

st te

rm

orn 2

s =

(a+

l)n 2

a –

first

term

d –

com

mon

diff

eren

cen

– to

tal t

erm

s

s =

(2a+

(n–1

)d)

n 2

If a

, b, c

, are

in A

P,

b is

ari

thm

etic

mea

n

b =

a +

c2

How

man

y 2-

digi

t num

bers

are

divi

sibl

e by

3?

2-di

git n

umbe

rs d

ivis

ible

by

312

, 15,

18,

... 9

9a

= 1

2, d

= 3

, an =

99

a n =

a +

(n–1

)d99

= 1

2 +

(n–1

)3

n =

30

n–1

=i.e

.,=

29

87 3

Sum

of f

irst

n p

ositi

ve in

tege

rs

Let s

n =

1 +

2 +

3 +

... n

a =

1, l

ast t

erm

l =

n

s n =

n(a+

l)2

=

or

n(1+

n)2

s n =

n(n+

l)2

Whe

n fir

st te

rm o

fco

mm

on d

iffer

nce

isgi

ven

: W

hen

first

& la

stte

rms

are

give

n :

14 | Oswaal GSEB NCERT Solutions – Textbook + Exemplar, MATHEMATICS, Class-X

Mind

map

: lea

rning m

ade

simpl

e

Tri

angl

esSimilarit

y

Theo

rem

s

i) C

orre

spon

ding

ang

les

are

equa

l

Stat

emen

tFi

gure

1. If

a p

erpe

ndic

ular

is d

raw

n fr

om th

e ve

rtex

of t

he

rig

ht a

ngle

of a

rig

ht tr

iang

le to

the

hypo

tenu

se

then

tria

ngle

s on

bot

h si

des

of th

e pe

rpen

dicu

lar

a

re s

imila

r to

the

who

le tr

iang

le a

nd to

eac

h ot

her.

In r

ight

∆A

BC

, BD

⊥A

C,

then

, ∆A

DB

~ ∆

AB

C

∆BD

C ~

∆A

BC

∆

AD

B ~

∆BD

C

In r

ight

∆A

BC

,B

C2 =

AB2 +

AC

2

If A

C2

= A

B2 +B

C2

then

, ∠B

= 9

0°

ii) C

orre

spon

ding

sid

es a

re in

the

s

ame

ratio

Rig

ht a

ngle

d tr

iang

le th

eore

m

Pyth

agor

as

A

BC

P

QR

2. In

a r

ight

tria

ngle

, the

squ

are

of th

e hy

pote

nuse

is

equ

al to

the

sum

of t

he s

quar

es o

f the

oth

er tw

o

sid

es.

3. In

a tr

iang

le, i

f squ

are

of o

ne s

ide

is e

qual

to th

e

sum

of t

he s

quar

es o

f oth

er tw

o si

des,

then

the

a

ngle

opp

osite

the

first

sid

e is

a r

ight

ang

le.

A

B

CD

A BC

C

BA

Stat

emen

tFi

gure

1. If

a li

ne is

dra

wn

para

llel t

o on

e si

de o

f a tr

iang

le

to in

ters

ect t

he o

ther

two

side

s in

dis

tinct

poi

nts,

th

e ot

her

two

side

s ar

e di

vide

d in

the

sam

e ra

tio.

2. If

a li

ne d

ivid

es a

ny tw

o si

des

of a

tria

ngle

in th

e

sam

e ra

tio, t

hen

the

line

is p

aral

lel t

o th

e th

ird

s

ide.

3. If

in tw

o tr

iang

les,

cor

resp

ondi

ng a

ngle

s ar

e

equ

al, t

hen

thei

r co

rres

pond

ing

side

s ar

e in

the

s

ame

ratio

(or

prop

ortio

n) a

nd h

ence

the

two

tr

iang

les

are

sim

ilar.(

AA

A c

rite

rion

)

4. If

in tw

o tr

iang

les,

sid

es o

f one

tria

ngle

are

p

ropo

rtio

nal t

o (i.

e., i

n th

e sa

me

ratio

of )

the

s

ides

of t

he o

ther

tria

ngle

, the

n th

eir

c

orre

spon

ding

ang

les

are

equa

l and

hen

ce th

e

two

tria

ngle

s ar

e si

mili

ar.(S

SS c

rite

rion

)

5. If

one

ang

le o

f a tr

iang

le is

equ

al to

one

ang

le o

f

the

othe

r tr

iang

le a

nd th

e si

des

incl

udin

g th

ese

a

ngle

s ar

e pr

opor

tiona

l, th

en th

e tw

o tr

iang

les

are

s

imila

r.(SA

S cr

iteri

on)

The

ratio

of t

he a

reas

of tw

o si

mila

r tr

iang

les

is e

qual

to th

e sq

uare

of th

e ra

tio o

f the

irco

rres

pond

ing

side

s

Her

e ∆A

BC

~ ∆

PQR

∆AB

C ~

∆PQ

R

then

, ∆A

BC

~ ∆

DEF

ar(A

BC)

ar(P

QR

)=

B

A

CM

Q

P

RN

Are

a of

Sim

ilar

Tria

ngle

s

AB

PQ2

AB

DE

=BC Q

R2 =

CA

RP

2=

AC

DF

& ∠

A =

∠D

Ifthen

, ∠A

= ∠

D ;

∠B

= ∠

E , ∠

C =

∠F

∆AB

C ≅

∆D

EF

∆AB

C ≅

∆D

EF

then

, DE

|| B

C

If, D

E ||

BC

If ∠

A =

∠D

, ∠B

= ∠

E

∠C

= ∠

Fth

en,

AB

DE

=BC EF

=C

AFD

AB

DE

=BC EF

=A

CD

F

If

AD

DB

=A

EEC

If

AD

DB

=A

EEC

then

Oswaal Gujarat GSEB NCERT Solutions (Textbook +Exemplar) For Class - X Mathematics (For March 2020

Exam)

Publisher : Oswaal Books ISBN : 9789388462655 Author : Panel Of Experts

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