15 liters are taken of from a container full of liquid A and replaced

Question

15 liters are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more liters of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16, what is the capacity of the container?

A. 45
E. 42
C. 37.5
D. 36
B. 25

A. 45
E. 42
C. 37.5
D. 36
B. 25

GMATGuruNY · Accepted Answer

After two replacements, the resulting ratio of A to B --  \frac{9}{16}  -- indicates that A constitutes  \frac{9}{25}  of the resulting solution.

Implication:
After each replacement,  \frac{3}{5}  of A remains in the container, with the result that -- after two replacements -- the remaining amount of A =  \frac{3}{5}*\frac{3}{5}=\frac{9}{25} .

Since 3/5 of A remains after the first replacement -- implying the removal of 2/5 of A -- the 15-liter reduction in A must constitute 2/5 of the total volume of A in the full container:
 15 = \frac{2}{5}x 
 75 = 2x 
 x = 37.5

C

RanonBanerjee · Answer

Answer choice C.

Final volume of of A is 9x/25 given x is the capacity of the container, i.e, the volume of A to begin with.

From the equation of multiple replacements,

9x/25= x(1-15/x)^2

9/25= (1-15/x)^2
(3/5)^2 = (1-15/x)^2

3/5 = 1-15/x
X=75/2 =37.5

Posted from my mobile device

Kinshook · Answer

Given:  15 liters are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more liters of the mixture is taken and replaced with liquid B. 
 Asked:  After this process, if the container contains Liquid A and B in the ratio 9:16, what is the capacity of the container?

Let the capacity of the container be V

First Replacement 
Liquid A = V-15
Liquid B = 15
Total = V
Liquid A/Total = (V-15)/V
Liquid B/Total = 15/V

Second replacement 
Liquid A = V-15 - 15(V-15)/V = (V-15)(1-15/V) = (V-15)ˆ2/V
Liquid B = V - (V-15)ˆ2/V = {Vˆ2 - (V-15)ˆ2}/V
Ratio Liquid A and Liquid B = (V-15)ˆ2 / {Vˆ2 - (V-15)ˆ2} = 9/16
(V-15)ˆ2/Vˆ2 = 9/25
(V-15)/V = 3/5
V/15 = 5/2
V = 75/2 = 37.5

IMO C

HarshavardhanR · Answer

-> Inital A:B = x:0
 Operation 1 
-> 15L removed. A:B = x-15:0
-> 15L of B added. A:B = x-15:15
Proportion of A in the mix:  \frac{x-15}{x} 
 
Operation 2
-> 15L of mixture removed. 
Amount of A removed when 15L of mixture is removed =  \frac{x-15}{x}  x 15
Thus,
Amount of A in the mixture after the removal = (x-15) -  \frac{x-15}{x}  x 15 =  (x-15)^{2} /x

-> 15 L of B is added. Final A:B = 9:16.

Final amount of A in the mixture =  (x-15)^{2} /x = 9/25 into final total volume

Final total volume is still x (15 l is removed, 15 l is added in each operation).

So,
 (x-15)^{2} /x = 9/25 (x)
 (x-15)^{2} / x^{2}  = 9/25
=>  \frac{x-15}{x}  = 3/5
=> 5x - 75 = 3x
=> 2x = 75
=> x = 75/2 = 37.5 = capcacity of container. Choice C.

Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

BongBideshini · Answer

Let's take initial capacity C
Initially A=100%,
Final %A = (9/16+9)*100 = 36%
Using the multiple mixture formula, where f = fraction replaced every time = 15/C
100%(1-f)^2 = 36%
(1-f)^2=36/100=9/25
1-f = 3/5
f=2/5
15/C=2/5
C=75/2=37.5

Ans C

15 liters are taken of from a container full of liquid A and replaced

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

15 liters are taken of from a container full of liquid A and replaced

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards