Question
The sum of the digits of a two-digit number is 7. If 45 is
subtracted from this number, the digits reverse their positions. What is the original number?Solution
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 = 7 --------- (I)
And, 10b + a - 45 = 10a + b
Or, 9b - 9a = 45
Or, b - a = 5 ---------- (II)
On adding equation I and II,
We get, a + b + b - a = 7 + 5
Or, 2b = 12
Or, 'b' = 6
On putting value of 'b' in equation I,
We get, 6 + a = 7
Or, 'a' = 1
Required number = 10 x 6 + 1 = 61
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