A can contains a mixture of two liquids A and B in proportion 3:5. Whe

Question

A can contains a mixture of two liquids A and B in proportion 3:5. When 16 L of mixture is drawn off and the same quantity of B is added, then the proportion of A and B becomes 1 : 2. How many litres of liquid A was contained by the can initially?

(A) 25 L
(B) 38 L
(C) 45 L
(D) 54 L
(E) 108 L

(A) 25 L 
(B) 38 L 
(C) 45 L 
(D) 54 L
(E) 108 L

HarshavardhanR · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=83009 Harsha GMAT Tutor | gmatanchor.com GMAT-Club-Forum-0f8egtbu.png

imeanup · Answer

Ans D

Refer to the attachment

Harsh9676 · Answer

IMO D

So the ratio is

==> (3X - 16(3/(3+5)) ) / (5x - 16 (5/(3+5)) + 16) = 1/2

==> 6X - 12 = 5X + 6

==> X = 18

Amount of A = 3X = 3 * 18 = 54 L

wishmasterdj · Answer

Amount of A in solution = X*3/8
Amount of A taken out = 16*3/8
Resultant Amount of X in solution = X * 1/3

3X/8 - 16*3/8 = X/3
X=144

Calculate X*3/8 for the answer -> 144 * 3/8 = 54

rsrighosh · Answer

I used the replacement theory described here
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/01 ... -mixtures/

Note that in the entire replacement process,
a) Mixture is homogenous (as its all liquid).Old Mixture:- \(\frac{A}{B}\) = \(\frac{3}{5}\). New mixture = \(\frac{A}{B}\) = \(\frac{1}{2}\)
Therefore old concentration \(A =\frac{ 3}{8}, B = \frac{5}{8}\).... Volume = \(8x\)
b) Quantity of A is untouched (constant) => \(\frac{3}{8}\) in 8x will be changed to \(\frac{1}{3}\) after \(16\)L of B is replaced by pure melted chocolate
c) quantity of B is varying (B is being added to change the concentration)

Hence going by the replacement formula,
we have to go by A (constant)

Ci*Vi = Cf*Vf
\(\frac{3}{8}(8x-16) = \frac{1}{3 }8x\)
x=\(\frac{144}{8}\) = \(18\)
AInitial Volume = \(18*3\) =\( 54\)L

Answer: D

sujoykrdatta · Answer

There are many ways to do this. Let's discuss 2 such methods: Method 1 (More logic based): After removal of the mixture, the ratio of A and B does NOT change. Thus, we still have A : B = 3 : 5 We now add B and the ratio changes to 1 : 2 However, we must keep the value for A unchanged since we did not change A while adding B. Thus, we rewrite the new ratio of A : B as 3 : 6 Hence, we see that B has increased by 1. But, we know that this '1' actually represents 16L Thus, A : B = 3 : 5 implies that A + B was '8', which actually equals 16 x 8 = 128L This is AFTER 16L of the mix was removed. Thus, the initial volume was 128 + 16 = 144L Hence, volume of A = 3/(3+5) * 144 = 3 * 18 = 54L Answer D Method 2 (Alligation - anyone?): After removal of the mixture, the ratio of A and B does NOT change. Thus, we still have A : B = 3 : 5 Initial proportion of A = 3/8 Proportion of A added = 0 Final proportion of A = 1/3 Thus, we have: 1.png Answer D

anniaustin · Answer

Let initial total = x lts
We need to find out 3x/8

After the substitution of 16 lts,
remaining A = 3(x-16)/8
remaining B = 5(x-16)/8 + 16
ratio of remaining A and remaining B = 1/2
Solving for x, x=144
So 3x/8 = 54

Answer is D

Mehta17 · Answer

3X : 5X 
3x-6 : 5x-10 (removed in proportion of 3:5)
3x-6 + 0 : 5x-10+16
3x-6: 5x+6 = 1:2 
x = 18 
3x = 54

A can contains a mixture of two liquids A and B in proportion 3:5. Whe

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A can contains a mixture of two liquids A and B in proportion 3:5. Whe

Prep Toolkit

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

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My Rewards