TS ICET 23 May 2018 Shift 2 Paper

Question : 126 of 200

Marks: +1, -0

The sum of a two digit number and the number obtained by reversing the order of its digits is 121 and the digits differ by 3. The largest of such numbers is

Validate

Solution:

Let the two digit number be ab ie., 10a + b.
Given

10a + b + 10b + a = 121

11(a + b) = 121

a + b = 11

→ (1)
Given

a-b = 3

b-a = 3

→(2)
Solving (1) and (2), a = 7; b = 4 or a = 4; b = 7
Larger of the number = 74.

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