NCERT Maths Solutions Class 10 Real Numbers
In this article, we will discuss the Class 10 Maths Chapter 1 Real Numbers
In Class 9 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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