Working alone, a small pump takes twice as long as large pump takes to

Question

Working alone, a small pump takes twice as long as a large pump takes to fill an empty tank.
Working together at their respective constant rates, the pump can fill the tank in 6 hours.
How many hours would it take for the small pump to fill the tank working alone?

A. 8
B. 9
C. 12
D. 15
E. 18

A. 8
B. 9
C. 12
D. 15
E. 18

niks18 · Answer

Let Large pump take  x  hours to fill the tank. Hence small pump will take  2x  hours

Working together the time taken will be  \frac{2x*x}{(2x+x)}=6 => x=9

Hence small pump alone will fill the tank in  = 2*9=18  hours

option  E

adkikani · Answer

niks18

Can I add Total time for two pumps for emptying the tank will be :  2x  +  x  = 6

I see what you did here:

Working simultaneously, I can add up individual rates:

Small pump rate:  \frac{1}{2x}

Larger pump rate:  \frac{1}{x}

Combined rate: \frac{1}{6}

Or

\frac{1}{2x}  +  \frac{1}{x}  =  \frac{1}{6}

I can take  \frac{1}{x}  common:

\frac{1}{x}  *  \frac{1}{2}  + 1 =  \frac{1}{6}

\frac{1}{x}  *  \frac{3}{2}  =  \frac{1}{6}

\frac{1}{x}  =  \frac{1}{6}  *  \frac{2}{3}

\frac{1}{x}  =  \frac{1}{6}

Large pump will take 6 hours and small pump will take 12 hours.
Where did I falter?

warrior1991 · Answer

adkikani

See calculation mistake 1/6 * 2/3 = 1/9 not 1/6.

Hope it helps.

niks18 · Answer

Hi  adkikani

What do you mean by emptying the tank. I did not get it.?

Also re-visit your calculation for the highlighted part. You made a mistake there, hence you are getting a different answer.

adkikani · Answer

niks18

Just as we add two rates for two pumps for getting combined rates, can we add individual time to get TOTAL time 
taken by both pumps together to empty the tank?

niks18 · Answer

Hi  adkikani

Ok I got your question. in this scenarios, you cannot simply add the individual times to get total time. Because the rate at which both work is different. Hence Larger pump will empty more water at a faster rate than the smaller pump. Hence smaller pump will have to make "LESS EFFORT" i.e. it will have to work for "LESS TIME". So adding individual times will give you "HIGHER" working hours.

That's why we first calculate the rates and then add. same has been done here. As you can see the individual times are 9 & 18 but when they work together they complete the job in 6 hours only because larger pump works at a faster rate.

pushpitkc · Answer

I prefer numbers to solve such problems. Assume that the tank has 180 units of water.
Since working together they can fill the tank in 6 hours, their combined rate - 30 units.

Working alone, the smaller pump takes twice as long as the large pump to fill the empty tank. 
So if the larger pump fill(s) 2x units, the smaller pump must be filling x units.

Now, 2x+x = 3x = 30 -> x = 10(rate at which smaller pump fills the tank)

Therefore, the smaller tank must take  \frac{180}{10}  or  18 hours(Option E)  to fill the tank.

adkikani · Answer

pushpitkc

Is this 'random' number LCM of answer choices?

Why do we bring an additional variable x? If we know that Small pump takes twice the time to 
empty the tank than large pump, then can I infer RATE of large pump must be double than that of small
pump, summing total rate as 1/6 (total time is 6 hours)

Let 2x: rate of larger pump; x: rate of smaller pump then 3x= 1/6 or x =18

I assume that x in your solution can not be rate since I do not have a time unit (min or hr) in denominator.
Let me know if my approach is correct.

Abhishek009 · Answer

TIME AND EFFICIENCY.PNG Time is Inversely related to efficiency (The more the efficiency the lesser the time required) So, 3e = 18 Or, e = 6 and total work is 18 units So, Time required for the small pump will be (E) 18 hours.

pushpitkc · Answer

Hi  adkikani

Your first observation is spot on. It is generally the LCM of the numbers when
the individual times are given. In this case, we are given the detail of the time
taken by both the pumps(6 hours) to fill the pump. So I went with the number
180.

As for the latter part of my solution, I am using the rate at which the pumps fill
the tank. If x units is the rate at which the small pump fills the tank, the larger
pump will fill 2x units respectively.

This is why we equate the sum to 30(which is calculated from the rate at which
both the pumps fill in an hour). Once we calculate the individual rate of the 
smaller tank to be 10 units, the time taken is  \frac{180}{10}  or 10.

Hope this helps you!

Kinshook · Answer

1/x + 1/2x = 1/6
3/2x = 1/6
x = 18/2 = 9

2x = 18

IMO E

TBT · Answer

Bumping for forum quiz and test.

Lodz697 · Answer

Let's do this:

If the big one works at a 2x speed than the small one, the 6 hours must be divided as follows:
- 4 hours for the bigger
- 2 hours for the smaller

And so:

Rate bigger = \frac{ 4 }{ 6 }=\frac{ 2 }{ 3 } 
Rate smaller = \frac{ 2 }{ 6 }=\frac{ 1 }{ 3 }
 
Needed hours for filling the tank using the smaller:

3 * 6 = 18

Kunal1801 · Answer

Um, that's there 1/x=1/6*2/3=1/6 where you made an error.
that should have been 1/9 as per your calculations

jaideeppal7 · Answer

Given>
P1=1/2R
P2=R
So
1/2R+R=1/6
R=1/9

T2=W/R=1/R=9
T1=W/(1/2R)=18

Aboyhasnoname · Answer

Basically Small machine is doing x part of the work and Large marchine is doing 2x part of the work, if both work the same time... 
Now if large machine is removed, the smaller machine would have to do all the 3x part of the work along......
Now...that means ......first it worked for x part for 6 hours...now it has to complete additional 2x parts ... it will take 12 hours...total time alone it will take is 18 hours E

Working alone, a small pump takes twice as long as large pump takes to

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