Hard to understand the question in itself. Wordings are vague. If someone can give me a clear explanation it would be useful

Thank you.

OE


As percentage of milk in solution X is twice the percentage of milk in solution Y. Let us assume percentage of milk in solution Y = a% Then, percentage of milk in solution X = 2a%

Since, Solution X and solution Y are mixed in the ratio 3:2 i.e. 3 portions of solution X and 2 portions of solution Y.

300ml of solution X and 200 ml of solution Y. → final mixture is 500ml

Also, the quantity of milk in the final mixture is 40%.


So, we get: 2a% of 300ml + a% of 200ml = 40% of (500 ml)

2a × 300ml + a × 200ml = 40 × (500 ml)

6a + 2a = 200

8 a = 200

a = 25, i.e. 25% milk in solution Y

Ans.(A)

I would add that the wording of the question is FAR from being vague

You have two solutions that mixed together give you the total of the mixing 500 liters or whatever it might be and the percentage of milk inside which is 40%

At the beginning the two portions are separated and each part has its own percentage and in X the % is double than that in Y.

This is SIMPLE

Moreover, the question can be solved also by logic and a bit of common sense

Suppose in x the percentage of milk is 50%. regardless the amount of X as mix

In Y the mix must be 25% because is half of that in X

When we combine the mixture the total milk is 40%


25%----------------------------------40%----------50%

In the weighted average of the % 25 has more weight of 50 because 40-25=15 but 50-40=10.

So in the answer choices B is impossible because is 40. The others are too big. The only valuable option is A

Perhaps I might have to improve my English to math conversion skill

Thank you Anyways!!

Taake a look here sir https://gre.myprepclub.com/forum/gre-ma ... 24952.html

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