Question

The sum of the digits of a two-digit number is 11. When 5 is added to the twice of its tens digits
and 2 is added to its unit place, the new number becomes 139. Find the number.
[Hint : Suppose digit at ten's place = X, .. digit at unit place = 11-x
Now 10 (2x + 5) + (11- x + 2) = 139; on solving x = 4; .. number = 47]​

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Asked on December 20, 2019 by

Krishna Pawar

The sum of digits of a two digit number is 11. If the digit at ten's place is increased by 5 and the digit at unit place is decreased by 5, the digits of the number are found to be reversed. Find the original number.

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ANSWER

Let the units digit be x.

The tens digit =11−x.

Original no. =10(11−x)+x=110−9x

If 5 is added to tens place and 5 subtracted for ones place, number is reversed.

So, 10(11−x+5)+x−5=10x+11−x

110−10x+50+x−5=11−9x

155−11=−18x

144=18x

x=8

Tens place =11−8=3

Therefore, the original number is 38.

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