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 solve this step by step:

1. Initial Condition:
– A has x coins
– A gives half of coins plus 2 to B

2. After first transfer (A to B):
– A has: x/2 – 2 coins
– B has: x/2 + 2 coins

3. After second transfer (B to C):
– B gives half plus 2 to C
– B has: (x/2 + 2)/2 – 2 = x/4 – 1 coins
– C has: (x/2 + 2)/2 + 2 = x/4 + 3 coins

4. After third transfer (C to D):
– C gives half plus 2 to D
– C has: (x/4 + 3)/2 – 2 = x/8 – 1/2 coins
– D has: (x/4 + 3)/2 + 2 = x/8 + 7/2 coins

5. Given condition:
– D has smallest two-digit number (10)
– Therefore: x/8 + 7/2 = 10

6. Solve for x:
– x/8 + 7/2 = 10
– x/8 = 10 – 7/2
– x/8 = 6.5
– x = 52

Verification:
– A starts with 52 coins
– B gets 28 coins (26 + 2), A keeps 24
– C gets 16 coins (14 + 2), B keeps 12
– D gets 10 coins (8 + 2), C keeps 6

Answer: A had 52 coins initially