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10 May 2018, 10:59
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:08) correct
29%
(02:01)
wrong
based on 158
sessions
History
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Not Attempted Yet
A tank is fitted with an inlet pipe and an outlet pipe. This tank was initially filled with exactly X litres of water. Then the inlet pipe is opened and another Y litres of water is added to the tank, post which the inlet pipe is closed. Immediately then the outlet pipe is opened and kept open till it drains 1/3 of total water present in the tank. After this the outlet pipe is also closed. Y is what percent of X?
(1) X = 90 litres.
(2) After the outlet pipe is closed, the water remaining in the tank is 20% more than X litres.
(1) X = 90 litres.
(2) After the outlet pipe is closed, the water remaining in the tank is 20% more than X litres.
10 May 2018, 14:25
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amanvermagmat
According to the Q-Stem:
Initial water at the tank= X
Water at the tank when the inlet is closed = X+Y
Water at the tank when the outlet is closed =\(1 -\frac{1}{3}*(X+Y)\)= \(\frac{2}{3}*(X+Y)\)
(1) X = 90 litres...................No information about Y.........Insufficient .
(2) After the outlet pipe is closed, the water remaining in the tank is 20% more than X litres.
==>\(\frac{2}{3}*(X+Y)\)=\(1.2*X\)
or \(\frac{2}{3}*(X+Y)\)=\(\frac{6}{5}*X\)
or \(\frac{1}{3}*(X+Y)\)=\(\frac{3}{5}*X\)
or \(\frac{1}{3}*(X+Y)\)=\(\frac{3}{5}*X\)
or \(5X+5Y\)=\(9X\)
or \(5Y\)=\(4X\)
or \(125Y\)=\(100X\)... Y is 125% of X.................Sufficient....................Thus I would like to go with option B.
10 May 2018, 23:24
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It is a DS question. So, if we can determine that Y can be expressed in terms of X, it is enough. It might be a waste of time trying to find the answer as we do in PS.
Initially, tank contains X litres.
After inlet pipe is closed, tank will hold (X + Y) litres
Outlet pipe drains 1/3(X + Y)
After outlet pipe is closed, tank will hold 2/3(X + Y) litres
Statement 1: X = 90. No information about Y. Not sufficient.
Statement 2: After the outlet pipe is closed, the water remaining in the tank is 20% more than X litres. i.e., 6/5(X)
From stem, we know the tank has 2/3(X + Y) litres of water after the outlet pipe is closed.
Equating information from stem and statement 2, 2/3(X + Y) = 6/5(X)
The equation makes it evident that we will be able to express Y in terms of X in a unique manner. Statement 2 ALONE is sufficient. Choice B.
If we solve this equation, we will get Y = 4/5X or 80%X. But certainly an avoidable step.
Initially, tank contains X litres.
After inlet pipe is closed, tank will hold (X + Y) litres
Outlet pipe drains 1/3(X + Y)
After outlet pipe is closed, tank will hold 2/3(X + Y) litres
Statement 1: X = 90. No information about Y. Not sufficient.
Statement 2: After the outlet pipe is closed, the water remaining in the tank is 20% more than X litres. i.e., 6/5(X)
From stem, we know the tank has 2/3(X + Y) litres of water after the outlet pipe is closed.
Equating information from stem and statement 2, 2/3(X + Y) = 6/5(X)
The equation makes it evident that we will be able to express Y in terms of X in a unique manner. Statement 2 ALONE is sufficient. Choice B.
If we solve this equation, we will get Y = 4/5X or 80%X. But certainly an avoidable step.
