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Pumping alone at their respective constant rates, one inlet pipe fills

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Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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30 Nov 2010, 06:03
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Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5
[Reveal] Spoiler: OA
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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30 Nov 2010, 06:19
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1/2 of capacity in 3 hours

full capacity in 6 hours

---> pump A.

2/3 capacity in 6 hours

(2/3 +1/3) in 6+3 = 9 hours

---> pump B

in 1 hour filled 1/6 +1/9 = 5/18 parts
full in 18/5 hours 3.6 hrs.

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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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30 Nov 2010, 06:30
1
KUDOS
Inlet 1 fills x/6 in 1 hour
inlet 2 fills (2x/3)/6 = x/9 in 1 hour

so together in 1 hour they will fill x/6 + x/9 = 5x/18
which means to fill 5x/18 it takes 1 hour => to fill x it will take 18/5=3.6 hrs
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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01 Dec 2015, 01:51
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Expert's post
rite2deepti wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5

Assume the volume of the container to be 6 Lts
We have assumed 6 because 6 is the LCM of 2 and 3, the denominators of the capacity filled.

Inlet 1: 1/2 of capacity in 3 hours i.e. 3 lts in 3 hours
Hence 1 lts/hour

Inlet 2: 2/3 of capacity in 6 hours i.e. 4 lts in 6 hours
Hence 2/3 lts/hour

Inlet 1 + Inlet 2 combined rate = 6/(1 + 2/3) hours= 6/ (5/3) hours = 18/5 = 3.6 hours
Option B
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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01 Dec 2015, 12:04
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rite2deepti wrote:
one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours

Time required to fill the full tank is 6 hours
rite2deepti wrote:
a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours.

Time required to fill the full tank is 9 hours
rite2deepti wrote:
How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

Use the formula - $$\frac{AB}{(A + B)}$$

Or, $$\frac{6*9}{(6 + 9)}$$

Or, $$\frac{54}{15}$$

Or, $$\frac{18}{5}$$

Or, $$3.6$$

Hence answer is (B) 3.6
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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10 May 2016, 17:21
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KUDOS
Attached is a visual that should help.
Attachments

Screen Shot 2016-05-10 at 5.50.37 PM.png [ 100.66 KiB | Viewed 7167 times ]

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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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03 Jun 2017, 00:08
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KUDOS
The first pipe fills the empty pool to half in 3 hours --> meaning it will take the first pipe double the time, 6h to fill the whole pool.
Now we have the rate of the first pipe: 1/6
The second pipe fills the pool to 2/3 in 6 hours. Multiply capacity and and time by 3/2 to obtain the rate it will take pipe 2 to fill to full capacity: 1/9
Now as we have two machines, two entities working together we can use the work formula: A*B/(A+B)= 6*9/(6+9) =54/15= 3.6
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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04 Jun 2017, 14:59
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Expert's post
Hi All,

This question can be solved in a couple of different ways, but it's essentially just a Work Formula question.

Work = (A)(B)/(A+B) where A and B are the respective times it takes for two entities to individually complete a task.

To start, we have to figure out how long it takes each pipe to fill the FULL tank....

First pipe = fills 1/2 the tank in 3 hours... so it fills the FULL tank in 6 hours
Second pipe = fills 2/3 the tank in 6 hours... so it fills the FULL tank in 9 hours

Working together, the two pipes will fill the tank in (6)(9)/(6+9) = 54/15 = 3 9/15 hours = 3.6 hours

[Reveal] Spoiler:
B

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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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14 Jul 2017, 22:18
Hi Sir,
Could you please explain me your reasoning why you multiplied 2/3*6 hrs = 9 hrs. { Also why you did the reciprocal of 3/2*2/3 = 1}
mcelroytutoring wrote:
Attached is a visual that should help.

Thank you
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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08 Sep 2017, 00:03
Rate= Work/Time

Let total work = x
Inlet pipe 1

Time= 3 hours
Work done= 1/2 *x

Rate= x/2/3= x/6

Inlet pipe 2

Time= 6 hours
Work done= 2/3 of x

Rate= 2x/3/6 = x/9

Combined rate=x/6 + x/9= 5x/18

Again the same formula:

Rate= Work/Time
5x/18= x/Time
Time= x/5x/18= 18/5= 3.6

Another method is to consider the total work as the LCM (2,3) (Denominators)

Total work= 6

Inlet pipe 1
Time= 3
Work done= 1/2 *6= 3

Rate= 3/3= 1

Inlet pipe 2
Time= 6
Work done= 2/3 * 6= 4
Rate= 4/6= 2/3

Combined rate= 1+2/3= 5/3

Work= 6

Time= 6/5/3= 18/5= 3.6
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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10 Sep 2017, 19:54
rite2deepti wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5

1st pipe fills 1/6 cap in 1 hr.

2nd pipe fills 1/9 cap in 1hr

Together 1/6+1/9 = 5/18 in 1 hr.

Full capacity in 18/5 hrs i.e 3.6 hrs.

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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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11 Sep 2017, 15:02
1
KUDOS
Expert's post
rite2deepti wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5

The rate for the first inlet pipe is (1/2)/3 = 1/6 and the rate for the second pipe is (2/3)/6 = 2/18 = 1/9.

If we let t = the number of hours they work together to fill the empty tank, we have:

(1/6)t + (1/9)t = 1

t/6 + t/9 = 1

Multiplying the entire equation by 18, we have:

3t + 2t = 18

5t = 18

t = 18/5 = 3 ⅗ = 3.6

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Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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23 Jan 2018, 23:23
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I have a rather random question for dabral and Bunuel: why can't we use the harmonic mean (average of rates) formula here? Is it sheer coincidence that the answer I am getting (3.75) with this equation is so close to the actual answer (3.6)?

1st pipe rate = 1/6 tank per hour

2nd pipe rate = 1/9 tank per hour

Harmonic mean (average of rates) = #/sum of inverse of rates = 2 / (6 + 9) = (2/15) x 2 = 4/15 per hour. (4/15)x = 1, x = 15/4 = 3.75

Yet the answer is 3.6 with the more traditional (and admittedly easy and straightforward) method. Still, what am I doing wrong? Thanks!
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Pumping alone at their respective constant rates, one inlet pipe fills   [#permalink] 23 Jan 2018, 23:23
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