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Bunuel
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I have checked and rechecked but I am getting 49/79 as the ratio of Milk:Water in the resultant solution.

Initial qty of milk in the 80 L solution is (80/7)*4=320/7. Qty of milk remaining after 10 L of solution is removed and replaced with 10 L of water three times is (320/7)*(7/8)^3=(5*49)/8. Since the qty of solution remains unchanged, the qty of water becomes 80-(5*49)/8. Therefore, the ratio of Milk to Water in the final solution is 49/79. IMO, had the question asked for the FRACTION of Milk in the final solution then the answer would be 49/(79+49)=49/128.
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Final qty of milk =
4/7×(1-10/80)^3
= 49/128
Final ratio of milk and water
= 49/79
Final qty=
initial qty(1- qty repl/tot qty)^number of times repl

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Hi,

Can someone please confirm this.
The question asks for ratio of milk to water which is 49:79
ratio of milk to solution = 49:128
(this is also confirmed in the twin question as mentioned by Bunnel above which asks quantity of milk in final solution = 49/128 * 80 = 245/8 L)
ratio of water to solution = 79 : 128

Similar question:
https://gmatclub.com/forum/12-5-of-a-so ... 89617.html
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Bunuel
10 litres are removed from 80 litres of solution of milk and water having ratio of milk and water as 4 : 3 and replaced with water. This operation is done a total of three times. What is the ratio of milk and water in the solution now ?


A. 47/128
B. 48/128
C. 49/128
D. 50/128
E. 51/128

Hi Bunuel,

I am getting ratio of milk and water as 49: 79. Could you please check the options.

Regards,
Abhee
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There is an error in the options. The denominator has to be 79 and not 128 (which is the denominator if you're comparing the ratio of milk to new solution).
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Yeah i think the question is phrased incorrectly.
The solutions are for the final result of water in the mixture (49/128), not for the ratio of milk to water.

4/7*(1-10/80)^3
4/7(70/80)^3
4*7*(7/8)^3
1/2*7/8*7/8 = 49/128
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