The sum of a two digit number and the number obtained by reversing its digits is 121. Find the number if
its units place digit is greater than the tens place digit by 7
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GIVEN :-
- Sum of two digit number and the number obtained by reversing its digits is 121.
- Unit place digit is greater than tens place digit by 7.
TO FIND :-
- The Original Number.
SOLUTION :-
Let the tens place digit be 'x'.
Unit place digit will be 'x+7'.
Original Number → 10x + (x+7)
When we reverse the digits,
Tens place digit will be 'x+7'.
Unit place digit will be 'x'.
Reversed Number → 10(x+7) + x
According to the question, sum of Original number and Reversed Number is 121.
→ 10x + (x+7) + 10(x+7) + x = 121
→ 10x + x + 7 + 10x + 70 + x = 121
→ 22x + 77 = 121
→ 22x = 121 - 77
→ 22x = 44
→ x = 44/22
→ x = 2
____________________
We know ,
Original number → 10x + (x+7)
Putting x = 2,
Original number → 10(2) + 2 + 7
Original number → 20 + 9
Original number → 29
Hence , Original number is 29.
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