NCERT solutions for class 10 maths chapter 1 Real Numbers can be checked from here. In NCERT class 10 maths book, the first chapter is real numbers. In this chapter, the concepts of real numbers introduced in class 9 are continued. In class 10 chapter 1 real numbers, the concepts like Euclid’s division algorithm, the Fundamental Theorem of Arithmetic, etc. are introduced which are extremely crucial.

NCERT class 10 maths solutions for chapter 1 real numbers are given here in a very simple and easy to understand way to help the students clear their doubts and understand the concepts better. Check the class 10 NCERT maths solutions for chapter 1 below. The NCERT solutions for class 10 maths chapter 1 Real Numbers PDF is also available here.

### NCERT Solutions Class 10 Maths Chapter 1 Exercises

**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

135=1

90=2

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

196=196

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

225=102

102=51

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 {**(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

32=4

8=8

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,

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

We get,

From

From

From

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

Hence,

ii) 156

Take LCM of 156 i.e

Hence,

iii)3825

Take LCM of 3825 i.e

Hence,

iv)5005

Take LCM of 5005 i.e

Hence,

v)7429

Take LCM of 7429 i.e

Hence,

**Question 7:**

Find the LCM and HCF of the following pairs of integer and verify that LCM

i) 26 and 91

ii) 510 and 92

iii) 336 and 54

**Solution :**

i) 26 and 91

26=2

91=7

So, LCM(26,91)=2

HCF(26,91)=13

__Verification:__

LCM

Product of 26 and 91 =2366

Therefore,LCM

i) 510 and 92

510=2

92=2

So,

HCF(510,92)=2

__Verification:__

LCM

Product of 510 and 92 =46,920

Therefore,LCM

iii) 336 and 54

54=2

So,

HCF(336,54)=2

Verification:

LCM

Product of 336 and 54=18,144

Therefore,LCM

**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

15=5

21=7

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

ii)17,23,and 29

17=17

23=23

29=29

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

iii)8,9 and 25

8=2

9=3

25=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

i.e 9

LCM=

**Question 10:**

Check whether

**Solution :**

If the number

Hence, it is very clear that there is no value of n in natural number for which

**Question 11:**

Explain why 7

**Solution :**

We have,

7

=13(7

=13

=13

Hence, it is a composite number .

We have, 7

=5(7

=5(1008+1)

=5

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

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

**EXERCISE 1.3**

**Question 13:**

Prove that

**Solution :**

Let us assume ,that

i.e

Squaring both sides

We get,

So, x is also divisible by 3

Substituting ,x=3k in

Since ,our assumption about

Hence,

**Question 14:**

Prove that

**Solution :**

Let us assume that

So, x=

So, the expression

**Question 15:**

Prove that the following are irrational numbers.

i)

ii)

iii)

**Solution :**

i)

Let, a=

3a is a rational number .Since product of any two rational numbers is a rational number which will imply that

Hence,

ii)

Let, a=

is a rational number .But,it contradicts since

Hence,

iii)

Let, a=

Squaring ,

Since, ais a rational number

So, the expression

This is a contradiction.

Hence,

**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)

ii)

iii)

iv)

v)

vi)

vii)

viii)

ix)

x)

**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)

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

ii)

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

iii)

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

iv)

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

v)

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

vi)

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

vii)

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

viii)

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

ix)

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

x)

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)

ii)

iii)

iv)

v)

vi)

vii)

viii)

ix)

x)

**Solution;**

i)

ii)

iii)

iv)

v)

vi)

vii)

viii)

ix)

x)

**Question 18:**

Decide whether the real numbers are rational or not. If they are rational ,then write its

i)24.1352436789

ii).12346783940564543……

iii)

**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)

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

__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;

**IRRATIONAL NUMBERS- **Number which can’t be represented in

- 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*.