Q. A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two- digit number. How many coins does A have in the beginning?

Explanation : 

A has some coins

Let’s say A had x coins in the beginning.
After giving half of the coins and 2 more to B, A has (x/2) – 2 coins left and B has (x/2) + 2 coins.
After B gives half of the coins and 2 more to C, B has ((x/2) + 2)/2 – 2 = (x/4) – 1 coins left and C has ((x/2) + 2)/2 + 2 = (x/4) + 3 coins.
After C gives half of the coins and 2 more to D, C has ((x/4) +3)/2 – 2 = (x/8) -1/2 coins left and D has ((x/4)+3)/2+2 = (x/8)+7/2 coins.

Since D has the smallest two-digit number which is 10, we can set up an equation: (x/8)+7/2 =10. Solving for x we get: x=52.

So A had 52 coins in the beginning.