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Solution A is 40% chlorine by volume, and Solution B is 60%

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by BTGmoderatorLU » Fri Aug 10, 2018 2:01 pm

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Source: Veritas Prep

Solution A is 40% chlorine by volume, and Solution B is 60% chlorine by volume. If a tank currently holds 40 gallons of Solution A, how many gallons of Solution B must be added so that the liquid in the tank is 50% chlorine?

A. 40 gallons.
B. 50 gallons.
C. 60 gallons.
D. 80 gallons.
E. 100 gallons.

The OA is A.
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by [email protected] » Sun Aug 12, 2018 2:14 pm
Hi All,

We're told that Solution A is 40% chlorine by volume, and Solution B is 60% chlorine by volume and that a tank currently holds 40 gallons of Solution A. We're asked for the number of gallons of Solution B must be added so that the liquid in the tank becomes 50% chlorine. This is a 'Weighted Average' question and can be solved in a number of different ways. Based on the percentages involved, just a bit of logic and Arithmetic is all that's needed to get the correct answer.

When averaging two percentages - as we are in this question - if we have an EQUAL amount of each item, then the average will be the average of the two percentages. For example, if we have the SAME amount of a 10% solution and a 20% solution, then mixing the two solutions together will gives us a total that is (10+20)/2 = a 15% solution.

In this prompt, we're mixing a 40% solution and a 60% solution. Since the result will be a 50% solution - which is the EXACT average of 40 and 60, we know that we have the SAME amount of Solution A and Solution B. The prompt tells us that we have 40 gallons of A, so we will need 40 gallons of B to hit that 50% average.

Final Answer: A

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by Jeff@TargetTestPrep » Sat Aug 18, 2018 6:47 pm
BTGmoderatorLU wrote:Source: Veritas Prep

Solution A is 40% chlorine by volume, and Solution B is 60% chlorine by volume. If a tank currently holds 40 gallons of Solution A, how many gallons of Solution B must be added so that the liquid in the tank is 50% chlorine?

A. 40 gallons.
B. 50 gallons.
C. 60 gallons.
D. 80 gallons.
E. 100 gallons.
Current there are 40 x 0.4 = 16 gallons of chlorine from solution A. We can let the amount of solution B = n and create the equation:

(16 + 0.6n)/(40 + n) = 1/2

2(16 + 0.6n) = 40 + n

32 + 1.2n = 40 + n

0.2n = 8

n = 40

Alternate Solution:

We start with 40 gallons of 40% chlorine, and we add to it n gallons of 60% chlorine, obtaining (40 + n) gallons of 50% chlorine. Putting this into an equation, we have:

40(0.4) + n(0.6) = (40 + n)(0.5)

16 + 0.6n = 20 + 0.5n

0.1n = 4

n = 40

Answer: A

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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