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

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:


Answer D

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

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

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

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