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Two tanks T1 and T2 have a capacity of 1500 litres and 2000 litres respectively. If a pipe P can fill empty tank T1 in X hours alone and empty tank T2 in Y hours alone while operating at same constant rate in both cases, what is the value of Y?
(1) Y/X = 4/3
(2) X = 6
(1) Y/X = 4/3
(2) X = 6
05 Aug 2018, 06:52
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Let the speed of pipe P = p. We are told that the speed is constant, then
T1/x=T2/y, 1 500/x=2000/y, 3y=4x. So if we find value of X, we will know tha value of Y.
(1) Y/X = 4/3 - Not Suff because it gives us the same 3y=4x as we already know
(2) X = 6[/quote] - Suff, because we can now use the information in the given equation 3y=4x
Answer B.
T1/x=T2/y, 1 500/x=2000/y, 3y=4x. So if we find value of X, we will know tha value of Y.
(1) Y/X = 4/3 - Not Suff because it gives us the same 3y=4x as we already know
(2) X = 6[/quote] - Suff, because we can now use the information in the given equation 3y=4x
Answer B.
17 Sep 2018, 01:26
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Hero8888
Let the speed of pipe P = p. We are told that the speed is constant, then
T1/x=T2/y, 1 500/x=2000/y, 3y=4x. So if we find value of X, we will know tha value of Y.
(1) Y/X = 4/3 - Not Suff because it gives us the same 3y=4x as we already know
(2) X = 6
- Suff, because we can now use the information in the given equation 3y=4xT1/x=T2/y, 1 500/x=2000/y, 3y=4x. So if we find value of X, we will know tha value of Y.
(1) Y/X = 4/3 - Not Suff because it gives us the same 3y=4x as we already know
(2) X = 6
Answer B.[/quote]
Hello, I don't seem to be clear on your explanation. In stat 1 . We are given the ratio of both times and we're told that the pipe is operating at a constant rate in both tanks. Isnt it sufficient to find y?
Hero8888
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