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ATQ,
Let the digits be 'a' (ones) and 'b' (tens).
Original number = 10b + a
Reversed = 10a + b
Given:
a + b = 9 --------- (I)
And, 10b + a - 27 = 10a + b
⇒ 9b - 9a = 27
⇒ b - a = 3 --------- (II)
Add (I) and (II):
a + b + b - a = 9 + 3
⇒ 2b = 12 ⇒ b = 6
Then a = 3
Required number = 10 × 6 + 3 = 63
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