Fun question, so we know A/B = 32.16 , which can also be written as 32 + 16/100 = 32 + 4/25. This means that **4 is the remainder** which when divided by 25 gives 0.16 as the latter part of the quotient with which we are concerned. Similarly we can also deduce that any multiple of 4 could be the **remainder** as well for instance, 16/100 where 16 is a **multiple of 4.**

Now, the question is asking sum of all possible two digit remainders of A/B. We know that multiples of 4 will be as 4,8,12,16â€¦infinity, but we only need to concern ourselves with **2 digit remainders**. Hence we are focused on the **range from 12 to 96.**

Finally this becomes an **Arithmetic Progression** problem, where **a = 12, an = 96 & d = 4.**

**an = a +(n-1)d**

96 = 12 + (n-1) 4

**n** = 84/4 +1 = **22.**

**Sum of n = (n/2) * (a + an)** = (22/2) * (12 + 96) = 11 * 108 = **1188 = (c)**

I couldnâ€™t find a shorter method, let me know if anyone knows.

Hope this helps.

Jeet

Great! Thanks.