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Asked by Sandeep Last Modified
Answer 
assume the number as "xy" where x and y are first and second digits respectively.
x+y=9 ...............(1)
when interchanging the digits number becomes yx.
yx=xy+27
that is
y*10+x=x*10+y+27
10y+x=10x+y+27
10y-y=10x-x+27
9y=9x+27.................(2)
dividing (2) by 9
y=x+3....................(3)
substituting (3) in (1)
x+y=9
x+x+3=9
2x=9-3
2x=6
x=3
from (3)
y=x+3=3+3=6
number=xy=36
read lessLet 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
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