CAT 2006 — QA Question 10

The four consecutive two-digit odd numbers will have (1, 3, 5, 7) or (3, 5, 7, 9) or (5, 7, 9, 1) or (7, 9, 1, 3) or (9, 1, 3, 5) as units digits.

As the sum divided by 10 yields a perfect square, the sum is a multiple of 10.

∴ The units digits have to be (7, 9, 1, 3).

Thus the four numbers will be (10x + 7), (10x + 9), (10x + 11) and (10x + 13),

where 0 < x < 9 (as each of these numbers is a two digit number)

Sum of these numbers = 40x + 40 = 40(x + 1)

Now, 40(x + 1)/10 = 4(x + 1) is a perfect square

As 4 is a perfect square, (x + 1) is some perfect square < 10

If x + 1 = 4, x = 3, and the four numbers are 37, 39, 41 and 43

If x + 1 = 9, x = 8, and the four numbers are 87, 89, 91 and 93

Hence, option (c).