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Dear Student,


Let the digits at tens place and ones place be x and 9 − x respectively.

Therefore, original number = 10x + (9 − x) = 9x + 9

On interchanging the digits, the digits at ones place and tens place will be x and 9 − x respectively.

Therefore, new number after interchanging the digits = 10(9 − x) + x

= 90 − 10x + x

= 90 − 9x

According to the given question,

New number = Original number + 27

90 − 9x = 9x + 9 + 27

90 − 9x = 9x + 36

Transposing 9x to R.H.S and 36 to L.H.S, we obtain

90 − 36 = 18x

54 = 18x

Dividing both sides by 18, we obtain

3 = x and 9 − x = 6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore, the two-digit number is 9x + 9 = 9 × 3 + 9 = 36


Regards!