* Mixture problem

[sfbay]

Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

1/5
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1/2
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[RonPurewal]

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[sfbay]

Question is from GMAT club but it is from their MGMAT challenge set.

(C) 2008 GMAT Club

[RonPurewal]

the quickest approach: in this problem, you can notice right away that 40% is the average of 30% and 50%; if you have the proper intuition about averages, you'll know at once that we need equal amounts of the 30% and 50% acid solutions to get this result. so, therefore, one half of the acid was replaced.

if not, two methods:
1) PLUG IN THE ANSWER CHOICES
as is customary with this sort of technique, begin by plugging in the middle choice.
let's say that you start with 100 g of solution total. then, if you replace 1/2 of that with 30% solution, that's 50 g of 30% solution and 50 g of 50% solution.
total amount of acid will be (50)(30%) + (50)(50%) = 15 + 25 = 40 grams of acid, out of a total of 100 g of solution. that's 40% -- which is what we want -- so we're done.
(if the answer were not the middle choice, we would just go up or down according to whether our initial answer gave a result with too much or too little acid.)

2) ALGEBRA
let's say that you have 100 grams of 50% acid to start with, and that you replace X grams with 30% acid.
so, now, you have (100 - x) grams of 50% solution and x grams of 30% solution.
we need a total of 40 g of acid, so
(50%)(100 - x) + (30%)(x) = 40
this gives x = 50, so we are replacing 50 g of acid, which is 1/2 of the total.