# Ncert Solutions For Class 10 Maths Ex 1.1

## Ncert Solutions For Class 10 Maths Chapter 1 Ex 1.1

QUESTION-1

Use Euclid’s division algorithm to find the HCF of:

i)135 and 225

ii)196 and 38220

iii)867 and 225

Solution:

i) We start with the larger number i.e 225

By Euclid’s division algorithm,we have

225=1×$\times$135+90

135=1×$\times$90+45

90=2×$\times$45+0

Hence, HCF(225,135)=HCF(135,90)=HCF(90,45)=45

Therefore,the HCF of 135 and 225 is 45

ii) We start with the larger number i.e 38220

By Euclid’s division algorithm,we have

38220=196×$\times$195+0

196=196×$\times$1+0

Hence , HCF(196,38220)=196

Therefore,the HCF of 196 and 38220 is 196

iii) We start with the larger number i.e 867

By Euclid’s division algorithm,we have

867=225×$\times$3+102

225=102×$\times$2+51

102=51×$\times$2+0

Hence, HCF(867,225)=HCF(225,102)=HCF(102,51)=51

Therefore,the HCF of   867 and 225 is 51

Question 2:

Show that any positive odd integer is of the form 6q+1,6q+3 & 6q+5

where q is some integer.

Solution :

Using Euclid’s division algorithm,we have

x=bq+r {0r<b$0\leq r< b$}…..(1)

Substituting b=6in equation(1)

So, x=6q+r,where r=0,1,2,3,4,5

If r=0, x=6q+0(divisible by 2)…..even

r=1, x=6q+1(not divisible by 2)…..odd

r=2, x=6q+2(divisible by 2)…..even

r=3, x=6q+3(not divisible by 2)…..odd

r=4, x=6q+4(divisible by 2)…..even

r=5, x=6q+5(not divisible by 2)…..odd

Therefore,the number 6q,6q+1,6q+2,6q+3,6q+4,6q+5are either even or odd.Hence ,any positive odd integer is of the form 6q+1,6q+3 & 6q+5Where q is some integer.

Question 3:

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution :

The maximum number of columns would be the HCF of (616,32)

We can find the HCF of 616 and 32 by using Euclid Division Algorithm.

Therefore,

616=19×$\times$32+8

32=4×$\times$8+0

8=8×$\times$1+0

So , HCF of 6161 and 32 is 8

Hence ,the maximum number of columns in which they can march is 8.

Question 4:

Use Euclid’s Division lemma to show that the square of any positive integer is either of the form 3m, 3m+1 for some integer m.

Solution :

According to Euclid Algorithm ,

We have x=bq+r …..(1)

Substituting , b=3

We get,x=3q+r (where,0r<3$0\leq r< 3$

i.e r=0,1,2 )

When ,r=0 ,x=3q…….(A)

r=1 ,x=3q+1…(B)

r=2 ,x=3q+2…(C)

So, Squaring eqn$eq^{n}$(A),(B) and (C)

We get,

From eqn$eq^{n}$ (A),

x2=9q2$x^{2}=9q^{2}$ x2=3×3q2$x^{2}=3\times 3q^{2}$

x2=3×m$x^{2}=3\times m$ (where,m=3q2$m=3q^{2}$)

From eqn$eq^{n}$ (B),

x2=(3q+1)2$x^{2}=(3q+1)^{2}$

=9q2+1+6q$=9q^{2}+1+6q$

=9q2+6q+1$=9q^{2}+6q+1$

=3(3q2+2q)+1$=3(3q^{2}+2q)+1$

=3m+1$=3m+1$ (where,m=3q2+2q$3q^{2}+2q$)

From eqn$eq^{n}$ (C),

x2=(3q+2)2$x^{2}=(3q+2)^{2}$

=9q2+4+12q$=9q^{2}+4+12q$

=9q2+12q+3+1$=9q^{2}+12q+3+1$

=3(3q2+4q+1)+1$=3(3q^{2}+4q+1)+1$

=3m+1$=3m+1$ (where,m=3q2+4q+1$3q^{2}+4q+1$)

Hence, any positive integer is either of the form 3m,3m+1 for some integer m.

EXERCISE 1.2

Question 6:

Express each number as a product of its prime factors:

i)140

ii)156

iii)3825

iv)5005

v)7429

Solution :

i) 140

Take  LCM of 140 i.e  2×2×5×7×1$2\times 2\times 5\times 7\times 1$

Hence,  140=2×2×5×7×1$140=2\times 2\times 5\times 7\times 1$

ii) 156

Take LCM of 156 i.e 2×2×13×3×1$2\times 2\times 13\times 3\times 1$

Hence, 156=2×2×13×3×1$156=2\times 2\times 13\times 3\times 1$

iii)3825

Take  LCM of 3825 i.e 3×3×5×5×17×1$3\times 3\times 5\times 5\times 17\times 1$

Hence, 3825=3×3×5×5×17×1$3825=3\times 3\times 5\times 5\times 17\times 1$

iv)5005

Take LCM of 5005 i.e 5×7×11×13×1$5\times 7\times 11\times 13\times 1$

Hence, 5005=5×7×11×13×1$5005=5\times 7\times 11\times 13\times 1$

v)7429

Take LCM of 7429 i.e 17×19×23×1$17\times 19\times 23\times 1$

Hence, 7429=17×19×23×1$7429=17\times 19\times 23\times 1$

Question 7:

Find the LCM and HCF of the following pairs of integer and verify that LCM×$\times$HCF=Product of the two numbers.

i) 26 and 91

ii) 510 and 92

iii) 336 and 54

Solution :

i) 26 and 91

26=2×$\times$13×$\times$1(expressing as product of it’s prime factors)

91=7×$\times$13×$\times$1(expressing as product of it’s prime factors)

So, LCM(26,91)=2×$\times$7×$\times$13×$\times$1=182

HCF(26,91)=13×$\times$1=13

Verification:

LCM×$\times$HCF=13×$\times$182=2366

Product of 26 and 91 =2366

Therefore,LCM×$\times$HCF=Product of the two numbers .

i) 510 and 92

510=2×$\times$3×$\times$17×$\times$5×$\times$1(expressing as product of it’s prime factors)

92=2×$\times$2×$\times$23×$\times$1(expressing as product of it’s prime factors)

So,

LCM(510,92)=2×2×3×5×17×23=23,460$LCM(510,92)=2\times 2\times 3\times 5\times 17\times 23=23,460$

HCF(510,92)=2

Verification:

LCM×$\times$HCF=23,460×$\times$2=46,920

Product of 510 and 92 =46,920

Therefore,LCM×$\times$HCF=Product of the two numbers .

iii) 336 and 54

336=2×2×2×2×7×3×1$336=2\times 2\times 2\times 2\times 7\times 3\times 1$(expressing as product of it’s prime factors)

54=2×$\times$3×$\times$3×$\times$3×$\times$1(expressing as product of it’s prime factors)

So,

LCM(336,54)=24×33×7$LCM(336,54)= 2^{4}\times 3^{3}\times 7$=3024

HCF(336,54)=2×$\times$3=6

Verification:

LCM×$\times$HCF=3024×$\times$6=18,144

Product of 336 and 54=18,144

Therefore,LCM×$\times$HCF=Product of the two numbers .

Question 8:

Find the LCM and HCF of the following integers by applying the prime factorization method.

i) 12,15 and 21

ii) 17,23 and 29

iii) 8,9 and 25

Solution :

i) 12,15 and 21

12=2×$\times$2×$\times$3

15=5×$\times$3

21=7×$\times$3

From the above ,HCF(12,15,21)=3and LCM(12,15,21)=420

ii)17,23,and 29

17=17×$\times$1

23=23×$\times$1

29=29×$\times$1

From the above ,HCF(17,23,29)=1and LCM(17,23,29)=11339

iii)8,9 and 25

8=2×$\times$2×$\times$2

9=3×$\times$3

25=5×$\times$5

From the above ,HCF(8,9,25)=1and LCM(8,9,25)=1800

Question 9:

Given that HCF(306,657)=9 .Find LCM(306,657)?

Solution :

We know,

HCF×$\times$LCM=Product of two numbers

i.e 9×$\times$LCM=306×$\times$657

LCM=306×6579=22338$\frac{306\times 657}{9}=22338$

Question 10:

Check whether 6n$6^{n}$ can end with the digit 0 for any natural number n.

Solution :

If the number 6n$6^{n}$ ends with the digit zero,then it is divisible by 5.Therefore the prime factorization of 6n$6^{n}$ contains the prime 5.This is not possible because the only prime in the factorization of 6n$6^{n}$ is 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the factorization of 6n$6^{n}$

Hence, it is very clear that there is no value of n in natural number for which 6n$6^{n}$ ends with the digit zero.

Question 11:

Explain why 7×$\times$11×$\times$13+13 and 7×$\times$6×$\times$5×$\times$4×$\times$3×$\times$2×$\times$1+5 are composite number.

Solution :

We have,

7×$\times$11×$\times$13+13

=13(7×$\times$11×$\times$1+1)

=13×$\times$78

=13×$\times$3×$\times$2×$\times$13

Hence, it is a composite number .

We have,               7×$\times$6×$\times$5×$\times$4×$\times$3×$\times$2×$\times$1+5

=5(7×$\times$6×$\times$4×$\times$3×$\times$2×$\times$1+1)

=5(1008+1)

=5×$\times$1009

Hence, it is a composite number .

Question 12:

There is a circular path around a sports field.Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction, after how many minutes will they meet again at the starting point?

Solution :

Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM(18,12) is 2×$\times$3×$\times$3×$\times$2×$\times$1=36

Therefore, Sonia and Ravi will meet again after 36 minutes.

EXERCISE 1.3

Question 13:

Prove that 3$\sqrt{3}$ is irrational .

Solution :

Let us assume ,that 3$\sqrt{3}$ is rational

i.e 3=xy$\sqrt{3}=\frac{x}{y}$ (where,x and y are co-primes)

y3=x$y\sqrt{3}=x$

Squaring both sides

We get,(y3)2=x2$(y\sqrt{3})^{2}=x^{2}$

3y2=x2$3y^{2}=x^{2}$……..(1)

x2$x^{2}$ is divisible by 3

So, x is also divisible by 3

therefore$\ therefore$ we can write x=3k (for some values of k)

Substituting ,x=3k in eqn$eq^{n}$ 1

3y2=(3k)2$3y^{2}=(3k)^{2}$ y2=3k2$y^{2}=3k^{2}$

y2$y^{2}$ is divisible by 3 it means y is divisible by 3

therefore$\ therefore$ x and y are co-primes.

Since ,our assumption about 3$\sqrt{3}$ is rational is incorrect .

Hence, 3$\sqrt{3}$ is irrational number.

Question 14:

Prove that  3+23$3+2\sqrt{3}$ is irrational .

Solution :

Let us assume that  3+23$3+2\sqrt{3}$ is rational .

So, x= 3+23$3+2\sqrt{3}$

x2=(3+23)2$x^{2}=(3+2\sqrt{3})^{2}$ x2=21+123$x^{2}=21+12\sqrt{3}$

3=x22112$\sqrt{3}=\frac{x^{2}-21}{12}$…….(1)

because$\ because$ x is a  rational number

So, the expression  x22112$\frac{x^{2}-21}{12}$is also a rational number.This is a contradiction .Hence, 3+23$3+2\sqrt{3}$ is irrational .

Question 15:

Prove that the following are irrational numbers.

i)13$\frac{1}{\sqrt{3}}$

ii)73$7\sqrt{3}$

iii)6+5$6+\sqrt{5}$

Solution :

i)13$\frac{1}{\sqrt{3}}$

13=333$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}\sqrt{3}}$

$\Rightarrow$ 13=33$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

$\Rightarrow$ 13=13×3$\frac{1}{\sqrt{3}}=\frac{1}{3}\times \sqrt{3}$

Let, a=13=13×3$\frac{1}{\sqrt{3}}=\frac{1}{3}\times \sqrt{3}$ be a rational number .

$\Rightarrow$ 3a=3$3a=\sqrt{3}$

3a is a rational number .Since product of any two rational numbers is a rational number which will imply that 3$^{\sqrt{3}}$ is a rational number .But,it contradicts since 3$^{\sqrt{3}}$ is a irrational number .

therefore$\ therefore$ 3ais a irrational or ais irrational.

Hence, 13$\frac{1}{\sqrt{3}}$ is irrational .

ii)73$7\sqrt{3}$

Let, a=73$7\sqrt{3}$ be a rational number .

a7=3$\Rightarrow \frac{a}{7}=\sqrt{3}$

a7$\frac{a}{7}$ is a rational number .Since product of two rational number is a rational number. Which will imply that 7$^{\sqrt{7}}$

is a rational number .But,it contradicts since7$^{\sqrt{7}}$ is a irrational number .

therefore$\ therefore$a7$\frac{a}{7}$is a irrational or ais irrational.

Hence,73$7\sqrt{3}$ is irrational .

iii)6+5$6+\sqrt{5}$

Let, a=6+5$6+\sqrt{5}$ be a rational number .

Squaring , a2=$a^{2}=$(6+3)2$(6+\sqrt{3})^{2}$

a2=$a^{2}=$36+23+3$36+2\sqrt{3}+3$

a2=$a^{2}=$39+23$39+2\sqrt{3}$

3=a23912$\sqrt{3}=\frac{a^{2}-39}{12}$……………..(1)

Since, ais a rational number

So, the expression  a23912$\frac{a^{2}-39}{12}$ is also rational number.

$\Rightarrow$3$\sqrt{3}$ is a rational number .

This is a contradiction.

Hence, 6+5$6+\sqrt{5}$ is irrational.

EXERCISE 1.4

Question 16:

Without actually performing the long division, state whether the following rational numbers have a terminating decimal expansion or a non-terminating repeating decimal expansion.

i)233125$\frac{23}{3125}$

ii)2732$\frac{27}{32}$

iii)3235$\frac{32}{35}$

iv)51600$\frac{5}{1600}$

v)2949$\frac{29}{49}$

vi)2724×53$\frac{27}{2^{4}\times 5^{3}}$

vii)2922×56×73$\frac{29}{2^{2}\times 5^{6}\times 7^{3}}$

viii)315$\frac{3}{15}$

ix)3550$\frac{35}{50}$

x)77210$\frac{77}{210}$

Solution :

Note; If the denominator has only factors of 2 and 5 then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

i)233125$\frac{23}{3125}$=2352$=\frac{23}{5^{2}}$

Since, the denominator has only 5 as its factor, it has a terminating decimal expansion.

ii)2732$\frac{27}{32}$=2725$=\frac{27}{2^{5}}$

Since, the denominator has only 2 as its factor, it has a terminating decimal expansion.

iii)3235$\frac{32}{35}$=323×7$=\frac{32}{3\times 7}$

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

iv)51600$\frac{5}{1600}$=526×52$=\frac{5}{2^{6}\times 5^{2}}$

=126×51$=\frac{1}{2^{6}\times 5^{1}}$

Since the denominator has only  2 and 5 as its factors, it has a terminating decimal expansion.

v)2949$\frac{29}{49}$=2972$=\frac{29}{7^{2}}$

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

vi)2724×53$\frac{27}{2^{4}\times 5^{3}}$

Since the denominator has only  2 and 5 as its factors, it has a terminating decimal expansion.

vii)2922×56×73$\frac{29}{2^{2}\times 5^{6}\times 7^{3}}$

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

viii)315$\frac{3}{15}$=15$=\frac{1}{5}$

Since, the denominator has  only 5 as its factor, it has a terminating decimal expansion.

ix)3550$\frac{35}{50}$=7×521×52$=\frac{7\times 5}{2^{1}\times 5^{2}}$

=721×51$=\frac{7}{2^{1}\times 5^{1}}$

Since the denominator has only  2 and 5 as its factors, it has a terminating decimal expansion.

x)77210$\frac{77}{210}$=7×1121×51×3×7$=\frac{7\times 11}{2^{1}\times 5^{1}\times 3\times 7}$

=1121×51×3$=\frac{11}{2^{1}\times 5^{1}\times 3}$

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

Question 17:

Write down the decimal expansion of the following rational numbers.

i)233125$\frac{23}{3125}$

ii)2732$\frac{27}{32}$

iii)3235$\frac{32}{35}$

iv)51600$\frac{5}{1600}$

v)2949$\frac{29}{49}$

vi)2724×53$\frac{27}{2^{4}\times 5^{3}}$

vii)2922×56×73$\frac{29}{2^{2}\times 5^{6}\times 7^{3}}$

viii)315$\frac{3}{15}$

ix)3550$\frac{35}{50}$

x)77210$\frac{77}{210}$

Solution;

i)233125$\frac{23}{3125}$=0.00736

ii)2732$\frac{27}{32}$=.84375

iii)3235$\frac{32}{35}$=.91428….

iv)51600$\frac{5}{1600}$=.00312

v)2949$\frac{29}{49}$=.59183…

vi)2724×53$\frac{27}{2^{4}\times 5^{3}}$=.0135

vii)2922×56×73$\frac{29}{2^{2}\times 5^{6}\times 7^{3}}$=.00001…

viii)315$\frac{3}{15}$=.2

ix)3550$\frac{35}{50}$=.7

x)77210$\frac{77}{210}$=.36¯$.3\bar{6}$

Question 18:

Decide whether the real numbers are rational or not. If they are rational ,then write its pq$\frac{p}{q}$ form.What can you say about the prime factors of q?

i)24.1352436789

ii).12346783940564543……

iii)41.2¯$41.\bar{2}$

Solution;

i)24.1352436789

Since it has a terminating decimal expansion, it is a rational number and q has factors of 2 and 5 only.

ii).12346783940564543……

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

iii)41.2¯$41.\bar{2}$

Since, it has non-terminating but repeating decimal expansion, it is a rational number and q has factors other than 2 and 5 .

SUMMARY ;

REAL NUMBER-Numbers that have a finite or infinite sequence of digits when it is represented in decimal form.

It is denoted by “R”

REAL NUMBERS ARE DIVIDED INTO 2 TYPES

RATIONAL NUMBERS- Number which can be represented in pq$\frac{p}{q}$ form.

It is divided into 2 types;

• TERMINATING DECIMAL NUMBER: The terminating decimal number has a finite number of Eg;0.432,653.8523,etc
• NON-TERMINATING RECURRING DECIMAL NUMBER: In a decimal, if a digit or a sequence of digits keeps repeating itself infinitely, then it is known as a non-terminating repeating decimal or recurring decimals.

Note;

It is expressed by putting a bar over the repeating digits.

eg;454.32¯$454.\bar{32}$,0.2¯$0.\bar{2}$,etc

IRRATIONAL NUMBERS- Number which can’t  be represented in pq$\frac{p}{q}$ form.

• NON- RECURRING DECIMAL NUMBER: It has infinite numbers and has the property that no sequence of digits are repeated.

eg-34.0428356393….,0.68384937…, etc.

HENCE, We can conclude that for every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number.