Mixtures

[kmittal82]

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?

a) 1/5
b) 1/4
c) 1/2
d) 3/4
e) 4/5

OA is C

[Patrick_GMATFix]

Weighted average problem. No math needed. Reason through.

The starting solution is entirely 50% concentration.

Some solution is removed. We now have a solution that is 50% concentration, but of a lower volume.

This lower volume solution is then mixed with a solution with a 30% concentration.

The final result is a solution of 40% concentration.

To mix a 50% concentrate with a 30% concentrate and end up with something exactly in the middle (40%), it must be true that there are equal volumes of 50% concentrate and 30% concentrate present.

In other words, half of the solution is 50% concentrate. The other half is 30% concentrate. To accomplish this, logically 1/2 of the initial 50% concentrate solution was removed and replaced with a new 1/2 that is 30% concentrate.

Pick C.

Weighted average questions can take you a lot of time or very little time depending on your mastery level. To practice hard weighted avg problems, set topic='Weighted Averages' and difficulty='600-700 & 700+' in the Drill Generator.

-Patrick

[kmittal82]

Weighted average problem. No math needed. Reason through.

The starting solution is entirely 50% concentration.

Some solution is removed. We now have a solution that is 50% concentration, but of a lower volume.

This lower volume solution is then mixed with a solution with a 30% concentration.

The final result is a solution of 40% concentration.

To mix a 50% concentrate with a 30% concentrate and end up with something exactly in the middle (40%), it must be true that there are equal volumes of 50% concentrate and 30% concentrate present.

In other words, half of the solution is 50% concentrate. The other half is 30% concentrate. To accomplish this, logically 1/2 of the initial 50% concentrate solution was removed and replaced with a new 1/2 that is 30% concentrate.

Pick C.

Weighted average questions can take you a lot of time or very little time depending on your mastery level. To practice hard weighted avg problems, set topic='Weighted Averages' and difficulty='600-700 & 700+' in the Drill Generator.

-Patrick

Thanks for the logical explanation there Patrick. However, is there an a more mathematical way of doing this? Say of the final strength of the mixture was 46% ?

[Patrick_GMATFix]

To find out how much of the original solution was removed, we need to know the volume ratio of initial solution to new solution.

The formula for weighted avg is [(quantity 1)(datapoint 1)+(quantity 2)(datapoint 2)]/(quantity 1 + quantity 2)

If solution A is 50% concentrate; solution B is 30% concentrate; mixture is 46% concentrate. We can write: (50a + 30b)/(a+b) = 46. From this we can solve for the ratio: a/b = 4/1. Thus for every 5 ounces in the final mix, there are 4 oz of solution A and 1 oz of solution B. In other words the initial solution had 5x ounces of 50% concentrate. 1x was removed and replaced with the lower concentrate, resulting in the ratio of 4:1.

The solution would have been 1/5 of the solution was removed and replaced.

Notes:
The ratio a/b = 4/1 cn be arrived at very quickly. In weighted avgs, the distance between each data point and the weighted average is always the inverse of the ratio of quantities. In the example above, the data points are 50% concentrate (A) and 30% concentrate (B) and the weighted avg is 46. So A is 4 away from the avg while B is 16 away from the avg. So the ratio of quantities A/B is the inverse of 4:16, or 16/4=4/1

-Patrick