Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 May 2017, 11:04

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Two water pumps, working simlutaneosly at their respective

Author Message
Director
Joined: 06 Feb 2006
Posts: 898
Followers: 3

Kudos [?]: 112 [0], given: 0

Two water pumps, working simlutaneosly at their respective [#permalink]

### Show Tags

12 Nov 2006, 05:19
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Two water pumps, working simlutaneosly at their respective rates, took exactly 4 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 its constant rate?

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

Finding x is easy, but i found difficulty in finding the total work done....
Current Student
Joined: 29 Jan 2005
Posts: 5221
Followers: 26

Kudos [?]: 404 [0], given: 0

### Show Tags

12 Nov 2006, 06:40
Let Y be the rate of the faster pump in terms of the the slower pump:

3/2Y(Y)/3/2Y+Y=4 ---> 3/2Y^2=10Y ---> 3/2Y=10 ---> Y=20/3
Director
Joined: 06 Feb 2006
Posts: 898
Followers: 3

Kudos [?]: 112 [0], given: 0

### Show Tags

12 Nov 2006, 10:26
GMATT73 wrote:
Let Y be the rate of the faster pump in terms of the the slower pump:

3/2Y(Y)/3/2Y+Y=4 ---> 3/2Y^2=10Y ---> 3/2Y=10 ---> Y=20/3

A very unclear equation...

(3/2y)*y / (3/2y)+y=4?
Manager
Joined: 13 Sep 2006
Posts: 212
Followers: 2

Kudos [?]: 2 [0], given: 0

### Show Tags

12 Nov 2006, 17:02
I set up two equations and used the substitution method for this problem:

equation 1

1/x + 1/y = 1/4

equation 2

x = 1.5y

Solve for Y and you get 20/3..
Manager
Joined: 10 Jul 2006
Posts: 74
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

12 Nov 2006, 18:24
Let x be the faster one, y is the slower machine.

Since the rate of the faster one is 3/2 of the slower machine and the rate is expressed as 1/x and 1/y for the 2 machine.
I have: 1/x = (3/2) (1/y) <=> y = (3x)/2 <1>
another equation is 1/x +1/y = 1/4 <2>

Substitute the value of y into the second equation and solve for x, that would be the number of hrs x will take to work alone (20/3)
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5045
Location: Singapore
Followers: 31

Kudos [?]: 376 [0], given: 0

### Show Tags

12 Nov 2006, 20:21
Pump A - A hours to fill 1 pool --> 1/A pool in 1 hour
Pump B - B hours to fill 1 pool --> 1/B pool in 1 hour

Together, they can fill (A+B)/AB pool in 1 hour. They can fill AB/A+B hours, which is 4 hours.

AB/A+B = 4

Assuming B is faster, then A = 1.5B
1.5B^2/2.5B = 4
1.5B = 10
B = 20/3 hours
12 Nov 2006, 20:21
Display posts from previous: Sort by