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03 Feb 2021, 10:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
86% (01:18) correct
14%
(01:52)
wrong
based on 57
sessions
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Two pipes can fill a tank in 10 h and 20 h respectively, while the third can empty it in 12 h. If all the pipes are opened together, then the tank will be filled in
(A) 7½ h
(B) 8 h
(C) 10 h
(D) 12 h
(E) 15 h
(A) 7½ h
(B) 8 h
(C) 10 h
(D) 12 h
(E) 15 h
03 Feb 2021, 10:14
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I'm too lazy to draw a diagram, but anyway its an easy one
Ans E 15 hours
Ans E 15 hours
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03 Feb 2021, 10:26
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Pipe 1 can fill a tank in 10 hours, in 1 hour it can fill \(\frac{1}{10}\)th of the tank.
Pipe 2 can fill a tank in 20 hours, in 1 hour it can fill \(\frac{1}{20}\)th of the tank.
Pipe 3 can empty the tank in 12 hours, in 1 hour it will empty \(\frac{1}{12}\)th of the tank.
The tank will be filled in \(\frac{1}{10}\)+\(\frac{1}{20}\)-\(\frac{1}{12}\)
Taking common LCM as 60.
\(\frac{1*6}{10*6}\)+\(\frac{1*3}{20*3}\)-\(\frac{1*5}{12*5}\)
\(\frac{6+3-5}{60}\) => \(\frac{4}{60}\),
\(\frac{60}{4} = 15\).
Therefore it takes 15 hours to fill the bucket, and the OA is E.
I personally feel that meanup 's method is easier and simple.
Pipe 2 can fill a tank in 20 hours, in 1 hour it can fill \(\frac{1}{20}\)th of the tank.
Pipe 3 can empty the tank in 12 hours, in 1 hour it will empty \(\frac{1}{12}\)th of the tank.
The tank will be filled in \(\frac{1}{10}\)+\(\frac{1}{20}\)-\(\frac{1}{12}\)
Taking common LCM as 60.
\(\frac{1*6}{10*6}\)+\(\frac{1*3}{20*3}\)-\(\frac{1*5}{12*5}\)
\(\frac{6+3-5}{60}\) => \(\frac{4}{60}\),
\(\frac{60}{4} = 15\).
Therefore it takes 15 hours to fill the bucket, and the OA is E.
I personally feel that meanup 's method is easier and simple.
