One pipe can fill a pool 1.25 times faster than a second

Question

One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

I would appreciate if someone can explain me this using the Rate Time Work chart (as used by Manhattan). That will help me understand better and easier.

Thank you!

m3equals333 · Answer

Not sure about the chart, but this is how I would approach it...I'm sure there are shortcuts/more efficient methods.

Slower Pipe: 1/x (1 Pool in x hours)
Faster Pipe: 1.25/x (1.25 Pools in x hours)

Combined Rate: 2.25x/x^2
Combined Rate to Fill 1 Pool: x^2/2.25x = 5hours

x=0 or 11.25

Slower Pipe Rate: 1/11.25 (1 pool in 11.25hours)

Bunuel · Answer

Say the rate of the slower pipe is R pool/hour, then the rate of the faster pipe would be 1.25R=5R/4. Since when both pipes are opened, they fill the pool in five hours, then their combined rate is 1/5 pool/hour.

Thus we have that R + 5R/4 = 1/5 --> R = 4/45 pool/hour --> time is reciprocal of rate thus it's 45/4 =11.25 hours.

Answer: A.

Hope it's clear.

prasannajeet · Answer

Hi
Trying to solve the way you want
From the table you can find the rate,R, of smaller pipe. Then time is inverse of rate so answer would be A...

Rgds
Prasannajeet

dkumar50 · Answer

Let's say slower pipes speed = v lph (litter per hr)
then faster = 1.25v lph

Together = 2.25v lph
Time = 5 hrs

Volume of Pool = 2.25v * 5
Time taken by slower pipe alone = 2.25v*5/v = 11.25 hrs

ravishankar1788 · Answer

Rate of pipe 1 - 1/a
pipe2 - 1/b
given rate pipe1=1.125pipe2
1/a=1.25(1/b)
also given 1/a+1/b=1/5
hence b=2.25*5=11.25

MyRedemption · Answer

entire job is done in 5 hours, hence 20% of the job is done per hour. slower one does 8.88% and faster one does 11.12% of it. so the slower pipe completes the task in 100/8.88 = 11.25 hours

iaratul · Answer

Let me give you an elaborate explanation on how you can use this (I have also used MGMAT's way of approaching the problems, but adopted according to the time needs) Let's assume the rate of the slower pipe is: x So, the rate of the faster pipe would be: 1.25x Working together, their rate will be x + 1.25x = 2.25x As the question says, they take 5 hours to fill up the tank working together, so: Rate x Time = Work 2.25x x 5 = 1 2.25x = 1/5 x = 1/5 / 2.25 = 4/45 So, the rate of the slower pipe, x = 4/45 Now, let's see how long it takes for the slower pipe to complete the task: Rate x Time = Work 4/45 x Time = 1 Time = 1 / 4/45 = 45/4 = 11.25 Hope it's clear

iaratul · Answer

Let me give you an elaborate explanation on how you can use this (I have also used MGMAT's way of approaching the problems, but adopted according to the time needs) Let's assume the rate of the slower pipe is: x So, the rate of the faster pipe would be: 1.25x Working together, their rate will be x + 1.25x = 2.25x As the question says, they take 5 hours to fill up the tank working together, so: Rate x Time = Work 2.25x x 5 = 1 2.25x = 1/5 x = 1/5 / 2.25 = 4/45 So, the rate of the slower pipe, x = 4/45 Now, let's see how long it takes for the slower pipe to complete the task: Rate x Time = Work 4/45 x Time = 1 Time = 1 / 4/45 = 45/4 = 11.25 Hope it's clear

bankerboy30 · Answer

I set it up as 1/r + 1/(5r/4) and derived 1/r + 4/5r and got r=9. What's wrong with that?

Bunuel · Answer

If r is the rate, then when you reciprocate it you'll get the time, so 1/r + 1/(5r/4) is the sum of the times and cannot equal to the rate (1/5).

Does this make sense?

gracie · Answer

One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

let s=slower pipe rate
5(s+1.25s)=1
s=4/45
inverting, 45/4=11.25 hours slower time
A

GMATGuruNY · Answer

Let the rate for the slower pipe = 4 liters per hour, implying that the rate for the faster pipe = (1.25)(4) = 5 liters per hour.
Combined rate for both pipes = 4+5 = 9 liters per hour.
Since it takes 5 hours for the two pipes to fill the pool at their combined rate of 9 liters per hour, the pool = (rate)(time) = 9*5 = 45 liters.
Since the rate for the slower pipe = 4 liters per hour, the time for the slower pipe to fill the 45-liter pool =  \frac{work}{rate} = \frac{45}{4} = 11.25  hours.

A

pappal · Answer

rate 
 time 
 work=rate*time 
 
  one 
 1.25x 
 1/1.25x 
 1 
 
  two(slower) 
 x let 
 1/x 
 1 
 
  combined 
 2.25x 
 1/2.25x =5 hrs given 
 1

so time to complete the required work by the slower=1/x=5*2.25=11.25

One pipe can fill a pool 1.25 times faster than a second

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	rate	time	work=rate*time
one	1.25x	1/1.25x	1
two(slower)	x let	1/x	1
combined	2.25x	1/2.25x =5 hrs given	1

GMAT Problem Solving (PS) Questions

One pipe can fill a pool 1.25 times faster than a second

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards