Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 May 2015, 22:11

### 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 pumps working together at their respective constant

Author Message
TAGS:
Intern
Joined: 17 Dec 2007
Posts: 8
Followers: 0

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

Two pumps working together at their respective constant [#permalink]  18 Dec 2007, 12:35
Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?
Senior Manager
Joined: 16 Apr 2006
Posts: 280
Followers: 2

Kudos [?]: 46 [0], given: 2

Suppose slower pump rate is x gallons/Hr
Faster will be 1.5x gallons/Hr

In 4 Hr both combined will pump: 4x + 1.5x*4 = 10x

Time taken by faster pump to fill: 1.5x*T = 10x

Hence Time Taken, T = 6.667 Hrs.
_________________

Trying hard to achieve something unachievable now....

Manager
Joined: 02 Feb 2007
Posts: 120
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: Work rate [#permalink]  18 Dec 2007, 12:46
hi mbsnyc,

i think it would have helped if you put the answer choices, but regardless, i got 1.48 hrs.
is that correct?

mbsnyc wrote:
Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2149 [0], given: 359

Re: Work rate [#permalink]  18 Dec 2007, 12:49
Expert's post
michaelny2001 wrote:
i got 1.48 hrs. is that correct?

No.

Intern
Joined: 17 Dec 2007
Posts: 8
Followers: 0

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

the answer was (16/3) - sorry I didn't bring all the answer choices with me to work - just had written down the question.

Was anyone able to come up with the solution 16/3?
Senior Manager
Joined: 06 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 23 [0], given: 0

dkverma wrote:
Suppose slower pump rate is x gallons/Hr
Faster will be 1.5x gallons/Hr

In 4 Hr both combined will pump: 4x + 1.5x*4 = 10x

Time taken by faster pump to fill: 1.5x*T = 10x

Hence Time Taken, T = 6.667 Hrs.

This is what I got too:

Let x be the number of gallons the slower pipe pumps per hour.

so in 4 hours it pumps = 4x gallons

The faster pipe pumps = 4 (1.5X) = 6x gallons

So the capacity of the tank is = 10x gallons.

The faster pipe pumped 6x in 4 hours
so to fill 10x it would take

= 10x * 4 / 6x = 20/3 = 6.66 hours
Manager
Joined: 03 Sep 2006
Posts: 233
Followers: 1

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

Here is my way:

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

2.5 / 1.5x = 1/4
1.5x = 10
x = 100/15

Hence,
1.5x = 100 * 3 / 15 * 2 = 10

Checking myself:

1/10 + 1/ (100/15) = 1/T
T = 4
Director
Joined: 08 Jun 2007
Posts: 584
Followers: 2

Kudos [?]: 79 [0], given: 0

Re: Work rate [#permalink]  18 Dec 2007, 19:05
mbsnyc wrote:
Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

x and 1.5x are rates

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

2.5/1.5x = 1/4 => 1.5x = 10 ( We need to calculate 1.5x here)
So 10 hrs.
Senior Manager
Joined: 20 Dec 2004
Posts: 257
Followers: 6

Kudos [?]: 84 [0], given: 0

Work Rate [#permalink]  18 Dec 2007, 19:45
Guys

What is the OA ?

I am getting 20/3 similar to dkverma.

Total Size of tank = 10N gallons. (assuming pump rates to be N & 1.5N/hr)

So to fill the tank with the larger pump it would take

10/1.5 = 20/3 Hrs = 6.667

CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2149 [0], given: 359

Re: Work Rate [#permalink]  18 Dec 2007, 22:23
Expert's post
4*V+4*1.5V=1
V=1/10

t*1.5/10=1
t=20/6=6h40m
Director
Joined: 09 Aug 2006
Posts: 766
Followers: 1

Kudos [?]: 69 [0], given: 0

Re: Work rate [#permalink]  18 Dec 2007, 23:05
mbsnyc wrote:
Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

x + 1.5x = 4 hrs
x = 8/5 & 1.5x = 24/10

Let Pool = 16
Time = 4 hrs
Speed = 4 pool/hour

Speed of slower = x = 8/5 pool/hr
Speed of faster = 1.5x = 24/10 pool/hr

A = 16
S = 24/10 pool/hr
T = 16/24 * 10 = 20/3 = 6 hrs 40 mins
Manager
Joined: 01 Nov 2007
Posts: 83
Followers: 0

Kudos [?]: 4 [0], given: 0

Whatever wrote:
Here is my way:

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

2.5 / 1.5x = 1/4
1.5x = 10
x = 100/15

Hence,
1.5x = 100 * 3 / 15 * 2 = 10

Checking myself:

1/10 + 1/ (100/15) = 1/T
T = 4

I had the same approach - I don't understand why you can't apply it in this case?!
Manager
Joined: 04 Sep 2007
Posts: 215
Followers: 1

Kudos [?]: 9 [0], given: 0

Whatever wrote:
Here is my way:

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

2.5 / 1.5x = 1/4
1.5x = 10
x = 100/15

Hence,
1.5x = 100 * 3 / 15 * 2 = 10

Checking myself:

1/10 + 1/ (100/15) = 1/T
T = 4

This seems to be the correct method.
CEO
Joined: 29 Mar 2007
Posts: 2591
Followers: 16

Kudos [?]: 234 [0], given: 0

Re: Work rate [#permalink]  19 Dec 2007, 13:43
mbsnyc wrote:
Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

we have x and y. x= faster pump.

1.5y+y=2.5y total rate--> 1pool/2.5y ---> 1pool/2.5y=4hrs

y=1/10 3/2*1/10= 3/20 --> 20/3 =6.666 or 6hrs and 40min.
Re: Work rate   [#permalink] 19 Dec 2007, 13:43
Similar topics Replies Last post
Similar
Topics:
6 Two water pumps, working simultaneously at their respective 11 12 Jul 2013, 06:26
Working at a constant rate respectively, pumps X and Y took 4 21 Jan 2010, 16:24
11 Two water pumps, working simultaneously at their respective 7 14 Nov 2006, 12:56
Two water pumps, working simlutaneosly at their respective 5 12 Nov 2006, 04:19
Two water pumps working together at respective constant 8 23 Jul 2006, 15:57
Display posts from previous: Sort by