There are 12 pipes attached to a tank. Some of them are fill pipes and

Question

There are 12 pipes attached to a tank. Some of them are fill pipes and some are drain pipes. Each of the fill pipes can fill the tank in 12 hours, while each of the drain pipes will take 24 hours to drain a full tank completely. If all the pipes are kept open when the tank was empty, it takes 2 hours for the tank to overflow. How many of these pipes are drain pipes?

A. 4
B. 6
C. 7
D. 8
E. 11

Are You Up For the Challenge: 700 Level Questions

A. 4
B. 6
C. 7
D. 8
E. 11

Are You Up For the Challenge: 700 Level Questions

CrackverbalGMAT · Accepted Answer

CONCEPT: Pipes and Cisterns
Filling capacity = 1/time taken to completely fill a tank
Emptying capacity =1/time taken to completely empty a tank
Filling -Emptying Capacity=Portion of the tank filled in an hour
Emptying-filing capacity =Portion of the tank emptied in an hour
SOLUTION: Let there be x filling pipes and y drain pipes.
Then x +y=12=>y=(12-x)....(i)
Also,
(1/12)*x -(1/24)*y =1/2.....(ii)
Solving (i) and (ii) we get
x/12 -(12-x)/24=1/2
=>x= 8 and hence y-12-8=4
Thus there are 4 drain pipes. (A)

Hope this helps

Devmitra Sen[/b]

Devmitra Sen

UsamaGuddar · Answer

I think the answer A.. but I am a little uncomfortable around we are to construct the equation for this q. Does the fact that the 2 pipes are working against each other mean that there needs to be a subtraction sign between them to achieve one full work? Experts help!

Prowess · Answer

let fill pipe be "n" and drain pipe be "m"

n+m=12------- equ.1

n/12-m/24 = 1/2---- Equ. 2.
on solving equ.2
2n-m=12---- equ. 3

on solving equ. 1 and 3.
n=8, m=4

ananya3 · Answer

IMO A

consider total work = 24units
=> rate of work of fill pipes = work/time = 24/12 = 2units/hour
and rate of drain pipes = work/time = 24/24 = 1u/hr

when all the pipes are open, it takes 2 hours to overflow
=> effective rate of working for the two types of pipes = 24/2 = 12units/hour
=> effective rate of work = combined rate of all fill pipes - combined rate of all drain pipes

let number of drain pipes = x
=> number of fill pipes = 12-x

=> effective rate of work = combined rate of all fill pipes - combined rate of all drain pipes
=> effective rate of work = number of fill pipes * rate of fill pipe - number of drainpipes * rate of drainpipe
=> 12 = (12-x) * 2 - x * 1
=> 12 = 24 - 2x -x
=> 3x = 13
=> x = 4

bond001 · Answer

let x be the number of the fill pipe
so drain pipe will be 12-x
in 2 hours it filled the tank so in 1 hour it did 1/2 
1x/12 - 12-x/24=1/2
2x-12+x/24=1/2
3(x-4)/24=1/2
x-4=4
x=8
drain pipe =12-8=4 
Answer A

chetan2u · Answer

Logical There are so many questions that give away the answer because of the options given. Here too, that is the case. There are 12 pipes. If we have half of each, fill and drain, the work will finish in 24 hrs. Thus, there have to be more number of fill pipes and lesser number of drain pipes. => D<\frac{12}{2}...D<6 Only A fits in. Algebraic If there are f or 12-d fill pipes and d drain pipes, one hour work = \frac{12-d}{12}-\frac{d}{24} = 1-\frac{d}{12}-\frac{d}{24}=1-\frac{d}{8} . This has to be equal to \frac{1}{2} => 1-\frac{d}{8}=\frac{1}{2}....\frac{d}{8}=\frac{1}{2}.....d=4 Other easier way would be to use substitution. Now,the equation is => \frac{f}{12}-\frac{d}{24}=\frac{1}{2}

Kaichi · Answer

This question from the forum quiz is pretty simple.

The equation should be:
1/12 (X) - 1/24 (12-X) = 1/2
X/12 - (12-X)/24 = 1/2
2X - 12 + X / 24 = 1/2
6X - 24 = 24
6X = 48
X = 8

Drain pipe = 12-8 = 4

There are 12 pipes attached to a tank. Some of them are fill pipes and

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There are 12 pipes attached to a tank. Some of them are fill pipes and

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