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# Two water pumps, working simultaneously at their respective constant

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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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kmasonbx wrote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

sol:

Rates be A and B

(A+B) * Time = Work
(A+B) * 4 = 1 ---->eq 1

A= 3B/2 ----->eq 2

substituting eq 2 in eq 1

B = 1/10 --->eq 3

substituting eq 3 in eq 2

A= 3/20

Time= work/rate
= 1/(3/20) =>20/3
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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Video solution from Quant Reasoning starts at 7:30
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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kmasonbx wrote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

Drawing a chart:

Since they are working together we add rates:
One pump has a rate of 1 and the other 3/2, together their t=4
so combined 2/2+3/2 * 4 = 20/2 = 10
together they do 10 units of work

To find the individual time we divide total work by individual rate
so 10/(3/2) = 20/3, E

Or...

1/1r + 1/(3/2r) = 1/4
1r+(3/2r) / (3/2)r^2 = 1/4
(3/2)r^2 / (2/2)r+(3/2r) = 4
(3/2)r / (5/2) = 4
(3/5)r = 4
r = 4*(5/3) = 20/3
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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AMITAGARWAL2 wrote:
i am little confused. If i take slower pump as X then 1/X + 1/1.5X = 1/4 which would result X = 20/3 and so faster pump as 10..... where i am going wrong...

In my solution x is the rate in your solution x is the time.

In your solution x is the time of the faster pump and 1.5x is the time of the slower pump (faster pump needs less time).

Hope it's clear.
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Skag55 wrote:
I did,
A+B = 4
A = 1.5B

4-B = 1.5B => B = 8/5

Why is this wrong?

We are told that "two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool".

If one pump needs A hours to fill the pool (rate=1/A) and another need B hours to fill the same pool (rate=1/B), then 1/A + 1/B = 1/4.

Solving 1/A + 1/B = 1/4 and A = 1.5B gives A=10 and B=20/3.

Check the links provided here: two-water-pumps-working-simultaneously-at-their-respective-155865.html#p1245761 for more.
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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xLUCAJx wrote:
How would you solve this problem by saying the faster pump 1.5x and slower x. I cant seem to figure that out.

Plug in some values-
Attachment:

Plug in.PNG [ 2 KiB | Viewed 96780 times ]

Quote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool.

Capacity of the swimming pool is -

5*4 = 20

Quote:
how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

The faster pump is pump B, so time required by fill the swimming pool alone will be = Total Capacity of the pool/Efficiency of Pipe B

20/3

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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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i am little confused. If i take slower pump as X then 1/X + 1/1.5X = 1/4 which would result X = 20/3 and so faster pump as 10..... where i am going wrong...
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kmasonbx wrote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

fast pump takes x hour
Slow pump takes 1.5x hour

so

1/x+1/1.5x = 1/4

> (1.5+1)/1.5x = 1/4
> 2.5/1.5x = 1/4
> 1.5 x = 10
>x = 10/1.5
>x = 20/3

Somebody confirm whether this is a right approach to do this type of problem or not. Thanks
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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kmasonbx wrote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

We are given that the rate of 1 pump is 1.5 times faster than the rate of the other pump. Since 1 pool is being filled and rate = work/time, the rate of the faster pump is 1/x, in which x = the time it takes for the faster pump to fill the pool, and the rate of the slower pump = 1/(1.5x) = 1/(3x/2) = 2/3x.

Since when the pumps work together they take 4 hours to fill 1 pool, we can create the following equation:

work of faster pump + work of slower pump = 1

(1/x)4 + (2/3x)4 = 1

4/x + 8/3x = 1

Multiplying the entire equation by 3x, we have:

12 + 8 = 3x

20 = 3x

20/3 = x

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Irising wrote:
Bunuel wrote:
Irising wrote:
I read quite a few solutions and they all have the following equation set up: 1/x+1/1.5x=1/4. But I have 1/x+1/1.5x=4. I don't understand why "x" is time. "x" to me is rate because the question states "constant rate of one pump was 1.5 times the constant rate of the other".

x is an unknown and you can denote by x either time or rate, it's your choice. Doesn't really matter.

If you say x is the rate, then you get: x + 2x/3 = 1/4. My solution here uses x as the rate: https://gmatclub.com/forum/two-water-pu ... l#p1245761

If you say x is the time, then you get: 1/x + 2/(3x) = 1/4.

In any case notice that we sum the rates.

Thank you for your reply! Please bare with me as I am still trying to understand the problem. When you say "we sum rates", do you also imply that we cannot sum time? In my equation, I set x as rate. Then I thought 1/x must be time (work/rate=time). And based on that, I sum time to solve for x: 1/x+1/(1.5x)=4, time of the slow pump+time of the fast pump=total time. There must be something wrong with my logic here but I can't seem to figure it out.

Yes, we can sum rates but not times.

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$. For example when we are told that a man can do a certain job in 3 hours we can write: $$3*rate=1$$ --> $$rate=\frac{1}{3}$$ job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then $$5*(2*rate)=1$$ --> so rate of 1 printer is $$rate=\frac{1}{10}$$ job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then $$3*(2*rate)=12$$ --> so rate of 1 printer is $$rate=2$$ pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is $$rate_a=\frac{job}{time}=\frac{1}{2}$$ job/hour and B's rate is $$rate_b=\frac{job}{time}=\frac{1}{3}$$ job/hour. Combined rate of A and B working simultaneously would be $$rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$ job/hour, which means that they will complete $$\frac{5}{6}$$ job in one hour working together.

3. For multiple entities: $$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}$$, where $$T$$ is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}$$, where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$t_1$$ and $$t_2$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}$$ hours.

17. Work/Rate Problems

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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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Hi All,

This is an example of a Work Formula question. Any time you have two entities (people, machines, water pumps, etc.) working on a job together, you can use the following formula:

(AB)/(A+B) = Total time to do the job together

Here, we're told that the total time = 4 hours and that one machine's rate is 1.5 times the other machine's rate…

If B = 1.5A then we have…

(A)(1.5A)/(A + 1.5A) = 4

1.5(A^2)/2.5A = 4

(3/2)(A^2) = 10A

A^2 = 20A/3

A = 20/3 hours to fill the pool alone

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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
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Irising wrote:
I read quite a few solutions and they all have the following equation set up: 1/x+1/1.5x=1/4. But I have 1/x+1/1.5x=4. I don't understand why "x" is time. "x" to me is rate because the question states "constant rate of one pump was 1.5 times the constant rate of the other".

x is an unknown and you can denote by x either time or rate, it's your choice. Doesn't really matter.

If you say x is the rate, then you get: x + 2x/3 = 1/4. My solution here uses x as the rate: https://gmatclub.com/forum/two-water-pu ... l#p1245761

If you say x is the time, then you get: 1/x + 2/(3x) = 1/4.

In any case notice that we sum the rates.
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
I did,
A+B = 4
A = 1.5B

4-B = 1.5B => B = 8/5

Why is this wrong?
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
How would you solve this problem by saying the faster pump 1.5x and slower x. I cant seem to figure that out.
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
Bunuel wrote:
kmasonbx wrote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour.

Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 --> x=3/20 pool hour.

The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone.

Theory on work/rate problems: https://gmatclub.com/forum/work-word-pro ... 87357.html

All DS work/rate problems to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=46
All PS work/rate problems to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=66

I'm having trouble seeing how you got x= 3/20 from x+2x = 1/4. Do you mind showing step by step? thank you!!
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
rnz wrote:
Bunuel wrote:
kmasonbx wrote:
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour.

Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 --> x=3/20 pool hour.

The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone.

I'm having trouble seeing how you got x= 3/20 from x+2x = 1/4. Do you mind showing step by step? thank you!!

$$x+\frac{2x}{3}=\frac{1}{4}$$;

$$\frac{3x+2x}{3}=\frac{1}{4}$$;

$$\frac{5x}{3}=\frac{1}{4}$$;

$$20x=3$$;

$$x=\frac{3}{20}$$.
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Re: Two water pumps, working simultaneously at their respective constant [#permalink]
I read quite a few solutions and they all have the following equation set up: 1/x+1/1.5x=1/4. But I have 1/x+1/1.5x=4. I don't understand why "x" is time. "x" to me is rate because the question states "constant rate of one pump was 1.5 times the constant rate of the other".