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07 Jun 2024, 14:43
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Pool A and Pool B each have the same volume of water and are drained by Pipe X and Pipe Y, respectively. The draining rate of Pipe X is double that of Pipe Y. If, after \(t\) hours, the remaining volume of water in Pool A is two-thirds that of Pool B, how long, in terms of \(t\), will it take for Pipe X to completely drain Pool A from its full volume?
A. \(\frac{t}{2}\) hours
B. \(\frac{2t}{2}\) hours
C. \(\frac{3t}{2}\) hours
D. \(2t\) hours
E. \(3t\) hours
A. \(\frac{t}{2}\) hours
B. \(\frac{2t}{2}\) hours
C. \(\frac{3t}{2}\) hours
D. \(2t\) hours
E. \(3t\) hours
07 Jun 2024, 14:43
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Official Solution:
Pool A and Pool B each have the same volume of water and are drained by Pipe X and Pipe Y, respectively. The draining rate of Pipe X is double that of Pipe Y. If, after \(t\) hours, the remaining volume of water in Pool A is two-thirds that of Pool B, how long, in terms of \(t\), will it take for Pipe X to completely drain Pool A from its full volume?
A. \(\frac{t}{2}\) hours
B. \(\frac{2t}{2}\) hours
C. \(\frac{3t}{2}\) hours
D. \(2t\) hours
E. \(3t\) hours
Assume Pipe X needs \(x\) hours to completely drain Pool A, then Pipe Y would need \(2x\) hours to drain Pool B.
After \(t\) hours, the volume of water left in Pool A is \(\frac{x - t}{x}\) of the initial volume. For example, if it takes Pipe X 4 hours to empty Pool A, then after 1 hour, \(\frac{4 - 1}{4} = \frac{3}{4}\) of the water remains.
Similarly, after \(t\) hours, the volume of water left in Pool B is \(\frac{2x - t}{2x}\). At this point, the remaining volume in Pool A is two-thirds of the volume in Pool B, which gives us the equation:
\(\frac{x - t}{x}=\frac{2}{3}*\frac{2x - t}{2x}\);
\(x - t=\frac{1}{3}*(2x - t);\)
\(3(x - t)=(2x - t);\)
\(x=2t\).
Answer: D
Pool A and Pool B each have the same volume of water and are drained by Pipe X and Pipe Y, respectively. The draining rate of Pipe X is double that of Pipe Y. If, after \(t\) hours, the remaining volume of water in Pool A is two-thirds that of Pool B, how long, in terms of \(t\), will it take for Pipe X to completely drain Pool A from its full volume?
A. \(\frac{t}{2}\) hours
B. \(\frac{2t}{2}\) hours
C. \(\frac{3t}{2}\) hours
D. \(2t\) hours
E. \(3t\) hours
Assume Pipe X needs \(x\) hours to completely drain Pool A, then Pipe Y would need \(2x\) hours to drain Pool B.
After \(t\) hours, the volume of water left in Pool A is \(\frac{x - t}{x}\) of the initial volume. For example, if it takes Pipe X 4 hours to empty Pool A, then after 1 hour, \(\frac{4 - 1}{4} = \frac{3}{4}\) of the water remains.
Similarly, after \(t\) hours, the volume of water left in Pool B is \(\frac{2x - t}{2x}\). At this point, the remaining volume in Pool A is two-thirds of the volume in Pool B, which gives us the equation:
\(\frac{x - t}{x}=\frac{2}{3}*\frac{2x - t}{2x}\);
\(x - t=\frac{1}{3}*(2x - t);\)
\(3(x - t)=(2x - t);\)
\(x=2t\).
Answer: D
28 Sep 2025, 05:50
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let the capacity be c.
Given that x = 2y;
after t hours, work remaining
c - 2yt = 2/3(c - yt)
1/3c = 4/3yt
c = 4yt
it takes t hours to finish 2yt work, since its 4yt : it takes 2t hours
Given that x = 2y;
after t hours, work remaining
c - 2yt = 2/3(c - yt)
1/3c = 4/3yt
c = 4yt
it takes t hours to finish 2yt work, since its 4yt : it takes 2t hours
