## Real Numbers – NCERT Maths Solutions Class 10

In this article, we will discuss the *Class 10 Maths* Chapter 1 **Real Numbers**

*Maths Class 10 **live classes are available.*

In Class 9th Maths, we studied the real numbers, especially about irrational numbers. Real numbers are the numbers that can be easily plotted on a number line.

__Euclid’s Division Algorithm__

For any two given positive integers **a** and **b** there exist unique whole numbers **q** and **r** such that

A=bq +r, where 0<r<b

Here **a** is called dividend is called a divisor, **q** is called a quotient and **r** as a remainder.

Dividend = (divisor*quotient) +remainder

Eg: – If we divide 117 by 14 then we will get 8 as quotient and 5 as remainder.

Dividend =117, divisor=14, quotient=8 and remainder=5

117= (14*8) +5

117=112+5

*Few Solved examples for Maths Class 10th students*

*NCERT Maths Solutions Class 10 *

**Eg 1**-A number, when divided by 53, gives 34 as quotient and 21 as remainder. Find the number.

**Soln**-Using Euclid’s division algorithm we have

Dividend= (divisor*quotient) +remainder

= (53*34) +21

= (1802+21)

=1823

*NCERT Maths Solutions Class 10 *

**Eg 2 – **Show that one and only one out of n, (n+1) and (n+2) is divisible by 3, where n is any positive integer?

Sol – let’s assume that on dividing **n** by 3, **q** will be the quotient and **r** will be the remainder.

N=3q+r where r=0, 1 or 2

N=3q or n =3q+1 or n 3q+2

Case 1 – if n=3q, then n is clearly divisible by 3.

Case 2 – if n= (3q+1), then (n+2)=3q+3=3(q+1),which is clearly divisible by 3,hence (n+2) is divisible by 3

Case 3 – if n= (3q+2), then (n+1)=(3q+3)=3(q+1),which is clearly divisible by 3,hence (n+1) is divisible by 3

Hence only one out of 3 is divisible by 3.

**For any two numbers a and b, we always have**

**(A*B)=product of their HCF and LCM**

**The above result is not true in case of three or more numbers.**

Eg:-The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, find the other.

Soln-For two number **a** and **b**,we know that

(a*b)=HCF of a,b* LCM of a,b

A=54,HCF=27 and LCM =162

54*b=27*162

B=27*162/54=81

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