There are four persons A, B, C, D; and A has some coins. A gave half of the coins to B and 4 more besides. B gave half of the coins to C and 4 more besides. C gave half of the coins to D and 4 more besides. Both B and D end up with same number of coins. How many coins did A have originally?

To solve this problem, we work backward or use the options to verify. Let the original number of coins with A be x. Step 1: A gives half and 4 more to B. A gives x/2 + 4 to B. A is left with x/2 - 4. Step 2: B gives half of what he received and 4 more to C. B received x/2 + 4. B gives away (x/2 + 4)/2 + 4 to C. B is left with (x/2 + 4)/2 - 4. Simplifying B's remaining coins: x/4 + 2 - 4 = x/4 - 2. Step 3: C gives half of what he received and 4 more to D. C received (x/4 + 2 + 4) = x/4 + 6. D receives (x/4 + 6)/2 + 4. Simplifying D's coins: x/8 + 3 + 4 = x/8 + 7. Step 4: Set B's remaining coins equal to D's coins. x/4 - 2 = x/8 + 7. Multiply the entire equation by 8 to clear denominators: 2x - 16 = x + 56. x = 72. If we check with 72: A has 72. A gives 36 + 4 = 40 to B. B has 40. B gives 20 + 4 = 24 to C. B is left with 40 - 24 = 16. C has 24. C gives 12 + 4 = 16 to D. D ends up with 16. Since B and D both have 16, the answer is 72.