A, B and C have amounts of money in the ratio 3 : 4 : 5. First B gives one fourth of his money to A and one fourth to C. Then C gives one sixth of his resulting amount to A. What is the final ratio of the amounts of A, B and C respectively?

Introduction / Context:
This question involves successive transfers of money between three people whose initial holdings are in a given ratio. It tests the ability to handle ratios and fractional transfers carefully. After two stages of giving and receiving, we need to determine the final ratio of their amounts. Such problems appear frequently in aptitude tests to examine systematic reasoning with ratios.

Given Data / Assumptions:

• Initial ratio of A : B : C = 3 : 4 : 5.
• B gives one fourth of his amount to A.
• B then gives one fourth of his original amount to C as well (so he gives a total of half his starting amount).
• After this, C gives one sixth of his new amount to A.
• No money enters or leaves the group from outside.
• We must find the final ratio A : B : C.

Concept / Approach:
We can assume a common multiplier for their starting amounts, convert the ratio into algebraic values and then apply the described transfers step by step. Because only proportions matter, choosing a convenient multiplier simplifies arithmetic. After both transactions are applied, we express the final amounts as a ratio and simplify it to its lowest terms. This approach keeps the reasoning clear and avoids confusion caused by fractions.

Step-by-Step Solution:
Let the initial amounts be A = 3k, B = 4k and C = 5k. First, B gives one fourth of his amount to A: one fourth of 4k is k. So A becomes 3k + k = 4k. B loses k and becomes 4k − k = 3k at this moment. Next, B gives one fourth of his original amount (that is again k) to C. So B finally becomes 3k − k = 2k. C receives k and becomes 5k + k = 6k. Now C gives one sixth of his amount to A: one sixth of 6k is k. So C becomes 6k − k = 5k. A receives k and becomes 4k + k = 5k. Final amounts are A = 5k, B = 2k and C = 5k. Thus the final ratio A : B : C = 5 : 2 : 5.

Verification / Alternative check:
We can test with a concrete value, for example k = 100. Then initial amounts are 300, 400 and 500. B gives 100 to A and 100 to C, so intermediate amounts are A = 400, B = 200, C = 600. Now C gives one sixth of 600, which is 100, to A. After this, A = 500, B = 200 and C = 500. The ratio 500 : 200 : 500 simplifies exactly to 5 : 2 : 5, confirming the algebraic solution.

Why Other Options Are Wrong:
Ratios like 4 : 3 : 5 or 3 : 4 : 5 mistakenly reflect initial or intermediate states instead of the final distribution. The ratio 5 : 4 : 3 misorders the values, and 6 : 3 : 5 does not correspond to any step in the process. Only 5 : 2 : 5 matches the final amounts after all transfers are complete.

Common Pitfalls:
A frequent source of error is using B's reduced amount, rather than his original amount, to compute the second transfer. The problem states that B gives one fourth to A and one fourth to C, which both refer to the original 4k. Another common mistake is to forget that C's one sixth transfer is based on C's updated amount, not the initial 5k.

Final Answer:
The final ratio of the amounts of A, B and C is 5 : 2 : 5.