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

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Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:07
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Question Stats:

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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

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Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:18
adkikani wrote:
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


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
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Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:38
niks18


Quote:
Let Large pump take \(x\) hours to fill the tank. Hence small pump will take \(2x\) hours


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

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


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?
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:41
1
adkikani

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

Hope it helps.
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:42
adkikani wrote:
niks18


Quote:
Let Large pump take \(x\) hours to fill the tank. Hence small pump will take \(2x\) hours


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


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


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?


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.
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:45
niks18

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


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?
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 02:52
adkikani wrote:
niks18

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


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?


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.
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 04:35
adkikani wrote:
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


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.
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Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 06:02
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pushpitkc

Quote:
I prefer numbers to solve such problems. Assume that the tank has 180 units of water.


Is this 'random' number LCM of answer choices?

Quote:
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)


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.
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 06:03
adkikani wrote:
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
Attachment:
TIME AND EFFICIENCY.PNG
TIME AND EFFICIENCY.PNG [ 1.82 KiB | Viewed 657 times ]
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.
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Re: Working alone, a small pump takes twice as long as large pump takes to  [#permalink]

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New post 24 Mar 2018, 11:49
adkikani wrote:
pushpitkc

Quote:
I prefer numbers to solve such problems. Assume that the tank has 180 units of water.


Is this 'random' number LCM of answer choices?

Quote:
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)


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.


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!
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Re: Working alone, a small pump takes twice as long as large pump takes to &nbs [#permalink] 24 Mar 2018, 11:49
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