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

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New post 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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New post 10 May 2016, 17:21
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Attached is a visual that should help.
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Screen Shot 2016-05-10 at 5.50.37 PM.png
Screen Shot 2016-05-10 at 5.50.37 PM.png [ 100.66 KiB | Viewed 24130 times ]

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

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New post 16 Oct 2015, 07:06
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Tank A does 1/2 in 3 hrs so we can fine for full capacity it would take 6 hrs.
Similarly , Tank B takes 2/3rd of tank to be filled in 6 hrs so it would fill to full capacity in 6/(2/3) =>18/2 => 9

Now simple work problem states that it would fill the whole tank in :

1/9 + 1/6 = 5/18 unit =>18/5 hrs = 3.6 hrs

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

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New post 16 Oct 2015, 06:48
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pipe 1 = 1/6 an hour
pipe 2 = 1/9 an hour

Total time taken =1/6 + 1/9 =3.6 Hours
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Re: Pumping alone at their respective constant rates, one inlet pipe fills  [#permalink]

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New post 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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New post 30 Nov 2010, 06:30
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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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New post 01 Dec 2015, 01:51
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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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New post 17 Feb 2016, 15:47
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Time taken by First tank Fill 1/2 the tank = 3 hours
i.e. Time taken by First tank to Fill 1 complete the tank = 6 hours

Time taken by Second tank Fill 2/3 the tank = 6 hours
i.e. Time taken by First tank to Fill 1 complete the tank = (3/2)*6 = 9 hours

in 1 Hour, Both tanks together Fill the tank = (1/6)+(1/9) = 5/18 tank
i.e. Time taken by Both tank to Fill 1 complete the tank = 18/5 hours = 3.6 hours

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

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New post 04 Jun 2017, 14:59
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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

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

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New post 08 Sep 2017, 00:03
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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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New post 17 Oct 2015, 03:48
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Bunuel 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

Kudos for a correct solution.


Total capacity = C

Rate of 1 = C/6
Rate of 2 = C/9
Combined rate = 5C/18
Time = 18/5 = 3.6 (B)
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Re: Pumping alone at their respective constant rates, one inlet pipe fills  [#permalink]

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New post 11 Dec 2015, 09:17
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let c be tank capacity.
Time taken by First pipe to fill 1/2 the tank = 3 hours. So work done in 1 hr= c*{1/(2*3)}= c/6

Time taken by Second pipe to Fill 2/3 the tank = 6 hours. So work done in 1hr= c*{2/(3*6)}=c/9

in 1 Hour, Both tanks together Fill the tank = (1c/6)+(c1/9) = 5c/18
i.e. Time taken by Both tank to Fill 1 complete the tank = 18/5 hours = 3.6 hours

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

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New post 11 Sep 2017, 15:02
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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

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

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New post 30 Aug 2018, 06:38
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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


Let's assign a nice value to the volume of the tank. We want a volume that works well with the given information (1/2, 2/3, 3 hours and 6 hours).
So, let's say the tank has a total volume of 18 gallons

One inlet pipe fills an empty tank to 1/2 of capacity in 3 hours
1/2 the tank is 9 gallons.
So, this pipe fills 9 gallons in 3 hours.
So, the RATE of this pipe = 3 gallons per hour

A second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours
2/3 the tank is 12 gallons.
So, this pipe fills 12 gallons in 6 hours.
So, the RATE of this pipe = 2 gallons per hour

So, the COMBINED rate of BOTH pumps = 3 gallons per hour + 2 gallons per hour = 5 gallons per hour

How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
We need to pump 18 gallons of water, and the combined rate is 5 gallons per hour
Time = output/rate
= 18/5
= 3.6 hours

Answer: B

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

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New post 17 Oct 2015, 06:42
Bunuel 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

Kudos for a correct solution.


Time taken by First tank Fill 1/2 the tank = 3 hours
i.e. Time taken by First tank to Fill 1 complete the tank = 6 hours

Time taken by Second tank Fill 2/3 the tank = 6 hours
i.e. Time taken by First tank to Fill 1 complete the tank = (3/2)*6 = 9 hours

in 1 Hour, Both tanks together Fill the tank = (1/6)+(1/9) = 5/18 tank
i.e. Time taken by Both tank to Fill 1 complete the tank = 18/5 hours = 3.6 hours

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

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

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New post 18 Aug 2018, 19:12
Bunuel 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


We are given that the first inlet pipe fills an empty tank to 1/2 capacity in 3 hours. Since rate = work/time, the rate of the first inlet pipe is (1/2)/3 = 1/6.

We are also given that the second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate of the second inlet pipe is (2/3)/6 = 1/9.

We need to determine how many hours it will take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity.

If we let t = the time in hours the two inlet pipes are working together, then the work of the first inlet pipe = (1/6)t and the work of the second inlet pipe = (1/9)t.

Since the tank is filled, we can set total work to 1 and create the following equation:

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

Multiplying the entire equation by 18, we obtain:

3t + 2t = 18

5t = 18

t = 18/5 = 3.6

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

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New post 14 May 2020, 11:17
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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New post 15 May 2020, 23:38
Again a longer solution

What we know
> Tank size is the same hence Tank Size is = 1

Part 1.
>Inflow one takes 3 hrs to fill half
> hence it takes the 6 hrs to fill the tank
> hence it fills @ 1/6 per hour

Part 2
> inflow takes 6 hrs to fill 2/3
> hence the remaining 1/3 will take 3 hrs
> hence 6+3= 9 hrs to fill 100%
> hence it fils @ 1/9 per hour

We solve
(1/9)x + (1/6)x = 1
18*(1/9)x + 18(1/6)x = 1*18
2x+3x = 18
5x = 18
x =18/5
x = 3.6
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Re: Pumping alone at their respective constant rates, one inlet pipe fills   [#permalink] 15 May 2020, 23:38

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

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