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27 Oct 2021, 04:30
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B
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Difficulty:
35%
(medium)
Question Stats:
68% (01:35) correct
32%
(01:57)
wrong
based on 413
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A mixture consists of liquids A, B, and C in the ratio of 1:4:6, respectively. What is the total volume of the mixture in liters?
(1) If the amount of liquid A were doubled, the ratio would be 1:2:3.
(2) If 5 liters of liquid A were added, the ratio would be 1:2:3.
(1) If the amount of liquid A were doubled, the ratio would be 1:2:3.
(2) If 5 liters of liquid A were added, the ratio would be 1:2:3.
27 Oct 2021, 04:48
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Bunuel
Solution:
Let us assume the value of A, B and C in the mixture to be \(x, 4x\) and \(6x\) respectively.
Total volume \(= x+4x+6x=11x\). To get the volume we will need the value of \(x\).
Statement 1:
When we double the amount of liquid A, we get the volume of A, B and C in the mixture \(= 2x, 4x\) and \(6x\) respectively.
Now \(2x, 4x\) and \(6x\) is already in the ratio \(1:2:3\).
Thus statement 1 alone is not only insufficient but also redundant. We can eliminate options A and D.
Statement 2:
When we add 5 liters of liquid A, volume of A, B and C in the mixture \(= x+5, 4x\) and \(6x\).
The ratio of these volumes are given to us as \(1:2:3\). So, let us assume the volumes to be \(y, 2y\) and \(3y\) respectively.
So, we can write \(6x=3y\)
\(⇒ x=\frac{y}{2}\).......(i)
Upon plugging this value in \(x+5=y\), we get:
\(⇒ \frac{y}{2}+5=y\)
\(⇒ y-\frac{y}{2}=5\)
\(⇒ \frac{y}{2}=5\)
\(⇒ y=10\)
PLuggin this in eq i, we get
\(x=\frac{10}{2}=5\)
This is what we needed to get our answer.
Statement 2 alone is sufficient.
Hence the right answer is Option B.
13 Nov 2021, 14:57
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GMATWhizTeam
Can you elaborate on why statement 1 is insufficient? I figured it out too, but couldn't do it the way you did - and your way is more effective.
Thank you!
