Question

The sum of two digit number and the number obtained by reversing the order of its digit is 121, and the two digits differ by 3 find the number (no spams pls ,pls give full steps)

Answer 1

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Answer 2

Hi there !!
Here's your answer

Given,
the digits of a two digit number differ by 3
So,
Let the digit in the units place be x
Digit in the tens place = x + 3

The original number is
10(x+3) + x = 10x + 30 + x
= 11x + 30 _____(i)

By interchanging the digits,
we have,
digit in units place = x + 3
digit in tens place = x
The new number is 10(x) + x + 3
= 10x + x + 3
= 11x + 3 __________(ii)

Given that,
the sum of original number and new number is 121

So,
the following balanced equation will be formed

11x + 30 + 11x + 3 = 121

22x + 33 = 121

22x = 121 - 33

22x = 88

x = 88/22

x = 4

Therefore,
digit in the units place = x = 4
Digit in tens place = x + 3 = 4 + 3 = 7

Thus,
the number formed is 47


NOTE : The number can be 74 also if you take the digit in tens place as x and digit in units place as x + 3