A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Eac

Question

A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 3/7
(E) 4/7

Different ways of doing this question will be highly appreciated.

(A) 1/2
(B) 2/3
(C) 3/4
(D) 3/7
(E) 4/7

EgmatQuantExpert · Accepted Answer

The question plays on the concept that time taken to do a work is inversely proportional to the rate of doing the work. We can deduce this relation from the expression

Work = Rate * Time i.e. Time = Work/Rate  .

If the work done by two entities are the same, the ratio of time taken by them would be the inverse ratio of their rate of doing work.

Given:  
The question gives us the relation between the rates of the wide and narrow pipes and asks us to find the ratio of time taken by all pipes working together to that of only wide pipes working together for filling up the tank.

Approach  
The work done by all pipes working together and only wide pipes working together is the same i.e. filling the tank. As the work to be done is same, the ratio of time taken will be inversely proportional to the rate.So, inverting the ratio of rates would gives us the ratio of time. Let's see what information has been given to us about the rates of the pipes.

Working Out:  
We are told that rate of work done by wide pipe is double the rate of work done by narrow pipe. If we assume the rate of narrow pipe to be r, the rate of wider pipe would be 2r.

We are asked to find (Time taken by all the pipes)/(Time taken by two wide pipes) = (Rate of two wide pipes)/ (Rate of all the pipes)= \frac{( 2r + 2r)}{( r +r + r +2r +2r)} = \frac{4}{7}  which is our answer.

Takeaway   
For time and work question, remember the relation between the work, time and rate for a set of different entities:

a. Greater the rate of doing work, lesser the amount of time taken, assuming the work done to be constant
b. Greater the amount of time taken, lesser the rate of doing work, assuming the work done to be constant
c. Greater the amount of time taken, greater the amount of work done, assuming the rate to be constant
d. Greater the rate of doing work, greater the amount of work done, assuming time taken to be constant

Regards
Harsh

sudhir18n · Answer

I always plug in for these questions.. I find them easier

Let 2 large pipes each work fill 10lts/hr
small ones = 1/2*10 = 5
total 5 = 10+10+5+5+5 = 35 lts/hrs

now assume the total capacity = 140 lts ( LCM of 20,35)

Qn :All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank

all working together will take 140/35 = 4 hrs
two large pipes working will take 140/20 = 7 hrs

hence ratio = 4/7 = E

GMATPASSION · Answer

@ Sudhir Gr8 solution. That LCM part was really awesome.Looking forward for some gr8 solution from other members.

Anonymous · Answer

I answered this in a similar way with a different twist...

Narrow = 3; Wide = 2;

Assume each Wide = 10 liters per hour (10 l/hr)
Assume each narrow = 10*0.5 = 5 liters per hour (5 l/hr)

Wide total work = 2*10 l/hr = 20 liters per hour
Total work = 3*5 + 2*10 = 35 liters per hour

ratio of work = 35/20 = 7/4
W = R*t ---> Time is just the inverse of Work rate ---> 4/7...

chawlavinu · Answer

Another approch  without plugging  in:

Let the wide pipe's rate be : x
Let the thin pipe's rate be : x/2

Formula: Speed * time= work
Eq1 : speed of all pipes together * T1 = W (Work is going to be constant)
Total speed: 3(x/2) + 2x = 7x/2
7x/2 * T1 = W

Eq2: Speed of 2 wide pipes only * T2 = W 
Total speed: 2x
2x * T2 = W

The expectation is to find out T1/T2= 4/7 
Cheers!!!!

mejia401 · Answer

I'm sure the long responses are meant to help explain the solution. I want to stress that the answer should be a simplistic as possible, without losing quality, of course. The approach I went with is below:

\frac{w}{2}  = small pipe
w = wide pipe

Total flow:  \frac{w}{2} * 3 +2w = \frac{7}{2}w 
Ratio of flows:  \frac{7w}{2} : 2w = 7:4

deepthit · Answer

Let wide pipe rate = 6 
smaller pipe rate = 3

total rate of all pipes working together = 3  + 2  = 7/6 
rate of wide pipes = 2/3

/   = 4/7

AR15J · Answer

Hi  Sudhir ,

I have a small doubt.

In the solution, till the below line I understood

time taken by all pipes together : time taken by 2 wide pipes=4:7

but in question we are asked for the fraction, why did we not make it 4/11?

I guess there is a gap in my understanding. Please help !

RMD007 · Answer

Because the time taken by two large pipes is already accounted in 7. 
7 actually consists of 4(time taken by large pipes) and 3(time taken by small pipes).

Hope its clear..

Chemerical71 · Answer

thanks for the explanation :-D

KarishmaB · Answer

We know rates are additive so let's work with rates. We will convert it to time later. 
Say rate of work of each narrow pipe is R. So rate of work of each wide pipe is 2R.

Rate of work of two wide pipes = 2*2R = 4R
Rate of work of all 5 pipes = 3R + 4R = 7R

Ratio of rate of work of all 5 pipes: 2 wide pipes = 7:4
Rati of time taken by all 5 pipes: 2 wide pipes = 4:7

So all 5 pipes take 4/7 of the time taken by 2 wide pipes alone.

gracie · Answer

let combined rate of 2 wide pipes=2*2=4
combined rate of 3 narrow pipes=3*1=3
time for all 5 pipes=1/(4+3)=1/7 
time for 2 wide pipes=1/4
(1/7)/(1/4)=4/7
E

fskilnik · Answer

5\,\,{	ext{pipes}}\,\,\,\left\{ \begin{gathered}
 \,3\,\,{	ext{narrow}}\,\,\,\, 	o \,\,\,{	ext{each}}\,\,\,1\,\,{	ext{gallons}}/\min \,\,\, \hfill \
 \,2\,\,{	ext{wide}}\,\,\,\,\,\,\,\,\, 	o \,\,\,{	ext{each}}\,\,\,2\,\,{	ext{gallons}}/\min \hfill \ 
\end{gathered} ight.\,\,\,\,\,\left( {{	ext{particular}}\,\,{	ext{case}}!} ight)

{	ext{A}}\,\,\,{	ext{ = }}\,\,\,{	ext{2}}\,\,{	ext{wide}}\,\,{	ext{together}}\,\,\,{	ext{:}}\,\,\,\,2 \cdot 2 = 4\,\,{	ext{gallons/min}}

{	ext{B}}\,\,\,{	ext{ = }}\,\,\,{	ext{all}}\,\,{	ext{5}}\,\,{	ext{together}}\,\,\,{	ext{:}}\,\,\,\,3 \cdot 1 + 2 \cdot 2 = 7\,\,{	ext{gallons/min}}

{	ext{B:A}}\,\,\underline {{	ext{work}}} \,\,{	ext{ratio}}\,\,\left( {{	ext{per}}\,\,{	ext{any}}\,\,{	ext{time}}} ight)\,\,\,{	ext{ = }}\,\,\,\,\frac{7}{4}\,\,\,\,\,

?\,\,\, = \,\,\,B:A\,\,\underline {{	ext{time}}} \,\,{	ext{ratio}}\,\,\,\left( {{	ext{per}}\,\,{	ext{any}}\,\,{	ext{work}}} ight)\,\,\, = \,\,\,{\left( {\frac{7}{4}} ight)^{ - 1}} = \,\,\,\frac{4}{7}

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

dabaobao · Answer

TL;DR

LCM Method:
Speed of Narrow Pipe: 1
Speed of Wide Pipe: 2
Total speed of all pipes = 2*2 + 1*3 = 7
Total speed of 2 wide pipes = 2*2 = 4

Time spent by all pipes = 1/7
Time spent by 2 wide pipes = 1/4

Time for all pipes/Time for 2 wide pipes = (1/7)/(1/4) = 4/7 (E)

Veritas Prep Official Solution

We are given that rate of work of 1 narrow pipe : rate of work of 1 wide pipe = 1:2

If we can find the ratio of rate of work of 2 wide pipes : rate of work of all pipes together, then we can easily get the ratio of time taken by 2 wide pipes : time taken by all pipes together. This is because ratio of time taken will be inverse of the ratio of rate of work since work done in both the cases is the same. (For a further explanation of this concept, check out the previous post)

In ratio terms, rate of work of 3 narrow pipes is 1*3 and rate of work of 2 wide pipes is 2*2

Therefore, rate of work of 3 narrow pipes : rate of work of 2 wide pipes = 3:4

Or we can say rate of work of 2 wide pipes : rate of work of all pipes together = 4 : (3+4) = 4:7

Then, time taken by 2 wide pipes : time taken by all pipes together = 7:4 (i.e. inverse of 4:7)

So all the pipes together will take 4/7 th of the time taken by the two wide pipes.

Answer (E)

kitipriyanka · Answer

A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 3/7
(E) 4/7

Different ways of doing this question will be highly appreciated.

Given:
1. 3 narrow(N) & 2 wide(W) pipes
2. Speed of 1N=1/2 Speed of 1W i.e if Speed of 1W=2x, then Speed of 1N=x 
 Further to this Speed of 2W=4x & Speed of 3N=3x & Speed of 2W+3N=7x

Find:  \frac{time taken by 3N+2W}{time taken by 2W}

Say, Say volume of tank=84L(multiple of 4,3 & 7)
then Time taken by 3N+2W=  \frac{84}{7x} = \frac{12}{x} 
time taken by 2W =  \frac{84}{4x} = \frac{21}{x}

THerefore,  \frac{time taken by 3N+2W}{time taken by 2W} = \frac{21}{x}/\frac{12}{x}  =  \frac{21}{12} = \frac{4}{7}

Hence E

sachinsavailable · Answer

I prefer to work these problems using per minute/ per hour rate. So considering the question:

W1 takes 2 unit/ hr to fill

w2 takes 2 unit/hr to fill

N1 takes 1 unit/hr to fill

N2 Takes 1 unit/ hr to fill

N3 takes 1 unit/hr to fill

All pipes in 1 hr fill 7 units

Wide pipe (W1 and W2 ) fill 4 units in 1 hr.

Answer: 4/7

gradschoolbound · Answer

Why can't you apply the theory of 1 job in x hours?

When I solved this, I did: N = 2x, W = x. Therefore
1/2x + 1/2x + 1/2x + 1/x + 1/x = 1/8x

Width Pipes only:
1/x + 1/x = 1/2x

Therefore proportion would be: (1/8x)/(1/2x) = 1/4

Bunuel  can you explain why this method doesn't work here?

800GMAT2019 · Answer

how many seconds it take you guys to finish this? 
I needed 2 minutes...I think i may be writing out TOO MUCH on my scratch paper. Thank you!

Nr967 · Answer

Think of this in terms of work=Rate x Time x number of pipes.

Narrow pipes: 1= (w/2)*t*3
t= 2/(3w)

Wide pipes: 1= w*t*2
t= 1/(2w)

Now you need to find the rate of the pipes working together (the inverse of time): 3w/2 + 2w = 7w/2

Now you can solve for the question in terms of time: (2/7w)/(1/2w) = 4w/7

A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Eac

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A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Eac

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