Question

Four children A, B, C & D are having some chocolates each.
A gives B as many as he already has, he gives C twice of what C already has and he gives D thrice of what D already has.
Now, D gives (1/8)th of his own chocolates to B.
Then A gives 10% chocolates he now owns to C and 20% to B.
Finally, all of them have 35 chocolates each.
What is the original number of chocolates each had in the beginning?

• A. A - 110, B - 10, C - 10, D - 10
• B. A - 90, B - 20, C - 20, D - 10
• C. A - 70, B - 25, C - 25, D - 20
• D. A - 125, B - 5, C - 5, D - 5

Hint:

Work backwards from the final equal amount of 35, undoing each transaction one at a time, starting with A's last transfer of 10% and 20%.

Answer 1

The Correct Option is A

Instead of forming algebraic equations, we can verify the answer by directly plugging option (1) into the sequence of transactions described in the question and checking whether everyone ends up with 35 chocolates.

1. Start with option (1): A = 110, B = 10, C = 10, D = 10.
2. Round 1: A gives B an amount equal to B's holding (10), so B becomes 20 and A becomes 100. A gives C twice of C's holding (20), so C becomes 30 and A becomes 80. A gives D thrice of D's holding (30), so D becomes 40 and A becomes 50. Now A = 50, B = 20, C = 30, D = 40.
3. Round 2: D gives 1/8th of his 40 chocolates, which is 5, to B. D becomes 35 and B becomes 25. Now A = 50, B = 25, C = 30, D = 35.
4. Round 3: 10% of A's 50 is 5, and 20% of A's 50 is 10. A gives 5 to C and 10 to B, keeping 50 - 15 = 35 for himself. C becomes 30 + 5 = 35, and B becomes 25 + 10 = 35.
5. Final amounts: A = 35, B = 35, C = 35, D = 35, matching the condition given in the question exactly.
6. Trying any other option, such as A = 90, B = 20, C = 20, D = 10, through the same steps does not produce 35 chocolates for all four children, so those options fail.

Therefore, the original number of chocolates was A = 110, B = 10, C = 10 and D = 10, confirming option (1) as the correct answer.