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Two water pumps working together at respective constant

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Two water pumps working together at respective constant [#permalink]

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New post 23 Jul 2006, 15:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Two water pumps working together at respective constant rates took 4 hrs to fill a pool. If constant rate of one pump was 1.5 times the constant rate of the other, then how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

5
16/3
11/2
6
20/3

I got B, but the answer is E....please explain.
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New post 23 Jul 2006, 16:19
Rate of First Pump = R
Rate of Second Pump = (3/2) R

Work Done = Rate * time

(R + 3/2 R)*4 = 1

R = 1/10

Rate of Fater Pump = (3/2)R = 3/20
Time = W/Rate = 1/(3/20) = 20/3
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New post 23 Jul 2006, 16:31
HI....

Remember that in this case

1/time machine A + 1/time amchine B = 1/Total time


..
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New post 24 Jul 2006, 00:34
Let rates per hour be a,b.
Then, (a+b)4 = full (f)

now let a=1.5b

2.5b. 10 =f

b=f/10.

(a+f/10)4 =f
4a +2f/5 = f
4a = 3f/5

a=3f/20

comparing a and b, we can see a is the faster rated pump.

in 1hr a completes 3/20 of tank
time taken by to complete tank =1/(3/20) = 20/3
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New post 24 Jul 2006, 08:32
I would have used numbers for this one.

let the smaller pump have a capacity of 200 gallons per hours
Then the faster one is 300 gallons per hour, since it is 1.5 times bigger

Both pump operating for 4 hours will fill up 200*4 + 300*4 = 2000 gallons

You can see the answer the moment you get 2000. The answer is 2000/300 = 20/3
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New post 24 Jul 2006, 12:25
This is how i calculated. Pls let me know if this is wrong way of doing it.

Formula to find the combined effort of A & B is

AB/A+B
we are given 2 equations.
1. Combined work is 4. so AB/A+B = 4
2. A=1.5B ( Pump A is 1.5 times faster than B)

substituting value of A from equation 2 in equation 1 we get B=20/3
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New post 24 Jul 2006, 13:26
I did it like this:

Faster pump is 1.5 times faster than the slower.
So slower pump will fill in 4 * 5/2 = 10 hours

So faster pump will fill in 10 * 2/3 = 20/3
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answering thru proportion [#permalink]

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New post 24 Jul 2006, 19:47
higher rate lesser time
let speed be x , time = y

applying inverse proportion

speed-------------time
| x + 1.5x---------- 4 ^
V 1.5 x--- ---------(y) |

Therefore

x(1 + 1.5)
------------ = y/ 4
1.5x

y = 20/3

if practised upon, it takes hardly 1 min to get the answer
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New post 24 Jul 2006, 21:50
E.

Let the slower one can complete 1 work in x hrs
Hence the rate is 1/x

Rate of the faster one is 1.5/x

Given they both working together take 4 hrs

Hence
1/x +1.5/x = 1/4
2.5/x = 1/4
x = 10
Hence faster one can finish in x/1.5 = 10/1.5 = 20/3
  [#permalink] 24 Jul 2006, 21:50
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Two water pumps working together at respective constant

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