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ATQ,
Let ones and tens digit of the number be 'a' and 'b' respectively.
So, original number = 10b + a
Reverse number = 10a + b
So,
a + b = 9 --------- (I)
And, 10b + a - 27 = 10a + b
Or, 9b - 9a = 27
Or, b - a = 3 ---------- (II)
On adding equation I and II,
We get, a + b + b - a = 9 + 3
Or, 2b = 12
Or, 'b' = 6
On putting value of 'b' in equation I,
We get, 6 + a = 9
Or, 'a' = 3
Required number = 10 × 6 + 3 = 63
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