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Two taps A and B can fill a vessel in 12 and 15 minutes, respectively, but a third pipe C can empty the full tank in 60 minutes. A and B is kept open for 5 minutes in the beginning and then C is also opened. In what time will the vessel be emptied?

Answer:

45 minutes


Attempted solution:

Portion filled in 5 min by A and B = (1/12 + 1/15)*5 = 9/12.

When all the three pipes are open, the vessel has to be emptied. So the rate of emptying the vessel should be higher than the rate of filling the vessel. But the rate of emptying the vessel (1/60 portion per minute) is slower than the rate of filling the vessel by taps A and B individually (1/12 th portion per minute and 1/15th portion per minute ). So the tap should fill up instead of getting emptied!

Nevertheless,

Since the tank has to be emptied in lets say time t, we have the following equation:

(1/60 - 1/12 -1/15)*t = 9/12

=>(1-5-4)t/60 = 9/12

which results in a negative t = -45/8 minutes.

Please help me with this. Please correct me if my concept is wrong.

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  • $\begingroup$ You are right, if all three pipes A,B,C are operating then the rate going in is larger than the rate out. But, if you assume you shut off the A and B taps after 5 minutes, so that only C drains out from then on, the answer is 45 minutes. So likely that is the intended scenario. It is just the time for C to drain 9/12 of the tank. $\endgroup$
    – Michael
    Dec 20, 2016 at 17:01
  • $\begingroup$ 45/8 is a correct answer. $\endgroup$ Dec 20, 2016 at 17:19
  • $\begingroup$ @KanwaljitSingh But time is coming negative! $\endgroup$
    – Soumee
    Dec 22, 2016 at 10:18
  • $\begingroup$ No you can see in my answer time is positive. $\endgroup$ Dec 22, 2016 at 10:32
  • $\begingroup$ You are doing it wrong. When something is empty by a pipe we take its value negative and pipes filling we take their values positive. $\endgroup$ Dec 22, 2016 at 10:33

2 Answers 2

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Portion filled in 5 min by A and B = $\left(\frac{1}{12} + \frac{1}{15}\right) * 5 = \frac{9}{12} = \frac{3}{4}$

After 5 minute,

C - A - B = $\left(\frac{1}{60} - \frac{1}{12} - \frac{1}{15}\right)$

= $\frac{-8}{60}$ or $\frac{-2}{15}$

Answer is negative because speed of empty pipe is so slower.

So tank will be not emptying it is filling.

Edit - The question asked by OP has an answer 45 minutes but he is getting answer -45/8 minutes. It means vessel is not emptying. But the above question has some mistake C empty in 6 minutes not 60. So if we do above question using 6 minutes we have correct result 45 minutes (Given below).

Original question -

page 1

Solution -

page 2

page 3

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  • $\begingroup$ You are amazing bro. Thank you for helping me with this. May I ask which book you are following. I have taken this question from Pearson Guide to Quantitative Aptitude and Data Interpretation for the CAT. Pg-1.181 . Q.26. Seems like the question is indeed wrong there. Thanks for helping me with this... $\endgroup$
    – Soumee
    Dec 22, 2016 at 12:42
  • $\begingroup$ I have upvoted your answer. Would you mind changing A + B - C to C-A-B in your answer so that those viewing this question wont get the same doubt as I did. $\endgroup$
    – Soumee
    Dec 22, 2016 at 12:47
  • $\begingroup$ I have taken this pics from R.S Aggarwal (quantative aptitude book). One of the best book for aptitude. $\endgroup$ Dec 22, 2016 at 12:56
  • $\begingroup$ According to your question it should be A+B-C. Because empty speed is less than filling. And according to book it is C-A-B. As emptying speed is more than filling. $\endgroup$ Dec 22, 2016 at 13:01
  • $\begingroup$ Now I got the reason why your answer is negative. As empty speed is so slow. $\endgroup$ Dec 22, 2016 at 13:02
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The flow rates are

$$\varphi_A = \frac{1}{12} \,\rm{vpm} \qquad \qquad \qquad \varphi_B = \frac{1}{15} \,\rm{vpm} \qquad \qquad \qquad \varphi_C = \frac{1}{6} \,\rm{vpm}$$

where $\rm{vpm}$ stands for "vessels per minute". Since $A$ and $B$ are open for $5$ minutes, the following volume of water flows in

$$\left(\frac{1}{12} + \frac{1}{15}\right) \rm{vpm} \cdot 5 \, \rm{minutes} = \frac 34 \,\rm{vessels}$$

Once valve $C$ is open, the flow rate is

$$\varphi_A + \varphi_B - \varphi_C = - \frac{1}{60} \,\rm{vpm}$$

A negative flow rate tells us that the vessel is emptying. It takes

$$\dfrac{\frac 34 \,\rm{vessels}}{\frac{1}{60} \,\rm{vpm}} = 45 \, \rm{minutes}$$

for the vessel to be emptied.

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  • $\begingroup$ If we use value of C as 6 we got answer 45 minutes. But OP uses C = 60. $\endgroup$ Dec 22, 2016 at 15:50
  • $\begingroup$ @KanwaljitSingh The OP must be wrong. If $\varphi_C = \frac{1}{60}$, then $$\varphi_A + \varphi_B - \varphi_C = \frac{2}{15} > 0$$ which means that the tank will never empty. $\endgroup$ Dec 22, 2016 at 15:52
  • $\begingroup$ Yes you are right. Thank you for answer. $\endgroup$ Dec 22, 2016 at 16:09

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