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

Question

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

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

DavidTutorexamPAL · Accepted Answer

As mixture problems can often be solved based on properties of ratios, without explicit calculation, we'll look for such a solution.
This is a Logical approach.

We'd like the final concentration of chlorine in the tank to be 50%, exactly in the middle of the concentrations of solution A (40%) and solution B (60%).
That is, we want both of the solutions to affect the final concentration equally, meaning we want them to have equal volumes.
Since there were 40 gallons fo Solution A, we need to add 40 gallons of Solution B.

(A) is our answer.

*Note: if the conentration weren't exactly in the middle, we would have needed to adjust accordingly. Say we want the final concentration to be 45%. This is 5% away from Solution A and 15% away from solution B for a ratio of 1:3. Then we need 3 times as much solution A as solution B (this ensure is has '3 times the effect' and our answer would be 40/3 gallons.

GMATinsight · Answer

Please find the solution as attached.

Two methods mentioned.

Answer: option A

gracie · Answer

let x=gallons of solution B to be added
.4*40+.6*x=.5(40+x)
x=40 gallons
A

generis · Answer

10-second method :
The final solution's concentration (50%) is the average of A's (40%) and B's (60%) concentrations.

That can happen only if A and B have identical volumes. B's volume = A's = 40 gallons

Answer A

Weighted average :

One version of weighted average formula:
 (% A)(Vol A) + (% B)(Vol B) = (% of A+B)(Vol A+B)

A = 40% chlorine = .40
B = .60 chlorine
A's volume = 40
B's volume = x

.40(40) +.60(x) = .50(40+x) 
 16 + .60x = 20 + .50x 
 .10x = 4 
 x=\frac{4}{.1}=\frac{40}{1}=40   gallons

Answer A

Funsho84 · Answer

40A + 60B = 50A + 50B

10B = 10A

A/B = 1/1

For 1A added, there's 1B added since the ratio is 1 to 1. So if 40 gallons of A is in the tank, 40 gallons of B will need to be added to maintain the ratio. 
Answer A

ScottTargetTestPrep · Answer

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

Answer: A

Mizar18 · Answer

we have that(solution type - gallons - concentration) :

A - 40 - 40%
B - X - 60%

then, if we get a resultant solution of 50%, the formula is:

40(40%) + (X)60% = (40+x)50%

40*(40/100) + X*(60/100) = (40+x)*(50/100)

X= 40, (A) is the answer

bumpbot · Answer

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

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GMAT Problem Solving (PS) Questions

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

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