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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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