CAT 2003 Slot 2 — QA Question 44

Any number that gives a remainder of 3 when divided by 7 will be of the form 7k + 3.

Since we only need two-digit numbers, k will range from 1 to 13 {where 7(1) + 3 = 10 and 7(13) + 3 = 94}

Sum of all these numbers = $\sum _{k=1}^{13}7k+3$

= 13 × 3 + 7(1 + 2 + ... + 13)

= 39 + $\frac{7\times 13\times 14}{2}$

= 39 + 637 = 676

Hence, option (b).