A woman distributed her savings between her daughters A, B, C and D in the…

Concept

In a ratio-transfer problem, when one party gives money to another, only those two shares change; every other share stays exactly the same. So any equation must be written between the post-transfer amounts and the post-transfer ratio. Here the new four-term ratio is over-stated (the untouched B and D shares cannot fit it), so the standard exam method uses the consistent pair that actually changed — A and C — to fix the scale, then answers the question asked.

Setting up

1. Let the initial shares be A = 6x, B = 7x, C = 9x and D = 16x (the 6 : 7 : 9 : 16 ratio).

2. A gives 600 to C, so after the transfer A = 6x − 600 and C = 9x + 600 (B = 7x and D = 16x are unchanged).

3. In the new ratio 1 : 4 : 2 : 5 the A : C part is 1 : 2, so write (6x − 600)/(9x + 600) = 1/2.

4. Cross-multiply: 2(6x − 600) = 1(9x + 600) -> 12x − 1200 = 9x + 600.

5. Collect terms: 12x − 9x = 600 + 1200 -> 3x = 1800 -> x = 600.

Answer the question (initial sum of A, B and C)

1. Initial A = 6x = 3600, B = 7x = 4200, C = 9x = 5400.

2. Required sum = 6x + 7x + 9x = 22x = 22 × 600 = 13200.

Cross-check

Put x = 600 back into the transfer: A becomes 3600 − 600 = 3000 and C becomes 5400 + 600 = 6000, giving A : C = 3000 : 6000 = 1 : 2, which matches the 1 : … : 2 : … part of the new ratio. Hence the initial sum of A, B and C is 13200.

Note: the four-term new ratio 1 : 4 : 2 : 5 is internally over-determined — B : D would have to be 7 : 16 (unchanged) yet is stated as 4 : 5 — so only the A : C pair is used, which is the accepted method for this previous-year item.