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.
Solution A is 40% chlorine by volume, and Solution B is 60%
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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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: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.
(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
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