GMAT PREP WORK PROB?

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dferm: Master | Next Rank: 500 Posts; Posts: 446; Joined: Thu Jul 26, 2007 1:07 pm; Thanked: 6 times

by dferm » Wed Mar 26, 2008 8:31 am

Two water pumps, working simultaneously at their respective constant 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 it's constant rate?

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

Please explain....

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Source: Beat The GMAT — Problem Solving | Wed Mar 26, 2008 8:31 am

xilef: Master | Next Rank: 500 Posts; Posts: 111; Joined: Thu Jan 31, 2008 4:05 pmFollowed by: 1 member; Thanked: 18 times

by xilef » Wed Mar 26, 2008 9:15 am

x- faster pump
y - slower pump

Work formula:

xy/(x+y)=4

second Eq:

3x/2=y

Solving for x:

[x(3x/2)]/(x+3x/2)=4
3x^2=20x
x^2=20x/3
x(x-20/3)=0

x=0 or x=20/3

Answer E.

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rajmirra: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Tue Jan 08, 2008 10:40 am; Location: Canada; Thanked: 4 times

by rajmirra » Wed Mar 26, 2008 9:50 am

I am loosing something here. Could you please detail Xilef. You solved for Y which is the slowest pump and the question asks for the time taken by the fastest pump. Please explain.

Thanks
~R

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xilef: Master | Next Rank: 500 Posts; Posts: 111; Joined: Thu Jan 31, 2008 4:05 pmFollowed by: 1 member; Thanked: 18 times

by xilef » Wed Mar 26, 2008 10:00 am

rajmirra wrote:I am loosing something here. Could you please detail Xilef. You solved for Y which is the slowest pump and the question asks for the time taken by the fastest pump. Please explain.

Thanks
~R

Made a typo. Solved for x. See above.

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rajmirra: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Tue Jan 08, 2008 10:40 am; Location: Canada; Thanked: 4 times

by rajmirra » Wed Mar 26, 2008 10:08 am

Thanks Xilef. What a relief I got it

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freedsl: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Wed Mar 26, 2008 10:30 am; Location: so cal

confused on the rates formula

by freedsl » Wed Mar 26, 2008 10:38 am

if the faster pump is 1.5 times the speed of the slower pump, then

x=1.5 y or x = 3y/2

then 2x/3 = y

how do you get:

second Eq:
3x/2=y

Thank, I'm all confused!!

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xilef: Master | Next Rank: 500 Posts; Posts: 111; Joined: Thu Jan 31, 2008 4:05 pmFollowed by: 1 member; Thanked: 18 times

by xilef » Wed Mar 26, 2008 1:36 pm

You are confusing the fact that x and y are the number of hours it takes for the pump to fill the pool - faster means that the number of hours is smaller not larger.

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moneyman: Master | Next Rank: 500 Posts; Posts: 468; Joined: Sat Mar 03, 2007 10:17 pm; Thanked: 5 times

by moneyman » Wed Mar 26, 2008 9:16 pm

Let A and B be the two pumps. From the question we understand that,
1/A+1/B=1/4

We also know that 1/A=1/1.5B

Substitute and we get

1/B+1/1.5B=1/4

which is equal to 2.5/1.5B=1/4 (Using LCM)

and so B(which is the faster one) would take 20/3 hours. So the Answer is E.

Maxx

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resilient: Legendary Member; Posts: 789; Joined: Sun May 06, 2007 1:25 amFollowed by: 6 members; Location: Southern California, USA; Thanked: 15 times

hey folks

by resilient » Mon Mar 31, 2008 8:42 pm

If we stick to just the basics; ( which gets me out of a lot of trouble)

w=r t

pump 1 slower = 1 job per hour
pump 2 faster = 1.5 jobs per hour

combined rate = 2.5 jobs per hour multiplied by the 4 hour work time.

This yields the W work, which is 10. so simply plug plug back into the w = r t formula. but this time we manipulate the w = rt to t=w/r

w= 10
r= (faster pump) 1.5

so 10/1.5 = 20/3. qa is e

Appetite for 700 and I scraped my plate!

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cat er ina: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sat Feb 13, 2010 1:36 pm

by cat er ina » Wed Apr 28, 2010 1:21 am

resilient wrote:If we stick to just the basics; ( which gets me out of a lot of trouble)

w=r t

pump 1 slower = 1 job per hour
pump 2 faster = 1.5 jobs per hour

combined rate = 2.5 jobs per hour multiplied by the 4 hour work time.

This yields the W work, which is 10. so simply plug plug back into the w = r t formula. but this time we manipulate the w = rt to t=w/r

w= 10
r= (faster pump) 1.5

so 10/1.5 = 20/3. qa is e

Your are a genius..I hadn t thought to the plug in strategy for this!

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sk818020: Master | Next Rank: 500 Posts; Posts: 214; Joined: Mon Mar 29, 2010 1:46 pm; Location: Houston, TX; Thanked: 37 times; GMAT Score:700

by sk818020 » Wed Apr 28, 2010 10:43 am

I'm just going to throw in another way of thinking about the problem. Maybe it will help you.

The first sentence states, "Two water pumps, working simultaneously at their respective constant rates, took exactly 4 hours to fill a certain swimming pool."

Mathematically this can be expressed as:

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

Logically we can think of this as Pump X completes 1 unit per every x hours (1/x). Pump Y completes 1 unit every y hours. Together the pumps complete 1 unit every 4 hours. If we can figure out what y is, then we know how long it takes Pump Y to complete 1 job.

Next, "the constant rate of one pump was 1.5 times the constant rate of the other."

In this problem we will say Pump Y is the faster pump, so the amount of time it takes Pump X to complete the job (x) is 1.5 times greater than Pump Y. Or mathematically;

x=1.5y

With this information we can solve the problem by substitution. We want to solve for y so we will substitute 1.5y for any x we see. Thus,

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

To get rid of all the fractions multiply the entire equation by (1.5y)(y)(4), resulting in:

4y + 6y = (1.5y)y ***see note at bottom.

Divide all terms by y to get;

4 + 6 = 1.5y

Simplify;

10=1.5y

10/1.5 = y

y = 20/3

Therefore, Pump y completes 1 job every 20/3 hours (ratio expressed mathematically as 1/y or 1/(20/3).

Hope that helps some.

Thanks,

Jared

*** You want to be careful not to devide variables because they may be 0, but in this problem we know that one pump is working at 1.5 times the rate of another pump, which wouldn't make sense if either number was 0. Thus, i felt it was safe to devide the variable y here.

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this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

by this_time_i_will » Sat May 01, 2010 10:59 am

dferm wrote:Two water pumps, working simultaneously at their respective constant 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 it's constant rate?

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

Please explain....

Faster takes x hrs. slower takes = 1.5x hours.

4 = x*1.5x/(x+1.5x)
=>x = 20/3.

My Blog:
https://www.iwantutopia.blocked/

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Sat May 01, 2010 11:16 am

sk818020 wrote:I'm just going to throw in another way of thinking about the problem. Maybe it will help you.

The first sentence states, "Two water pumps, working simultaneously at their respective constant rates, took exactly 4 hours to fill a certain swimming pool."

Mathematically this can be expressed as:

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

Logically we can think of this as Pump X completes 1 unit per every x hours (1/x). Pump Y completes 1 unit every y hours. Together the pumps complete 1 unit every 4 hours. If we can figure out what y is, then we know how long it takes Pump Y to complete 1 job.

Next, "the constant rate of one pump was 1.5 times the constant rate of the other."

In this problem we will say Pump Y is the faster pump, so the amount of time it takes Pump X to complete the job (x) is 1.5 times greater than Pump Y. Or mathematically;

x=1.5y

With this information we can solve the problem by substitution. We want to solve for y so we will substitute 1.5y for any x we see. Thus,

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

To get rid of all the fractions multiply the entire equation by (1.5y)(y)(4), resulting in:

4y + 6y = (1.5y)y ***see note at bottom.

Divide all terms by y to get;

4 + 6 = 1.5y

Simplify;

10=1.5y

10/1.5 = y

y = 20/3

Therefore, Pump y completes 1 job every 20/3 hours (ratio expressed mathematically as 1/y or 1/(20/3).

Hope that helps some.

Thanks,

Jared

*** You want to be careful not to devide variables because they may be 0, but in this problem we know that one pump is working at 1.5 times the rate of another pump, which wouldn't make sense if either number was 0. Thus, i felt it was safe to devide the variable y here.

Jared,

a good one!

speaking of different approaches, here's one from me. Hope you like it.

I think there is a reason to give the faster rate as 1.5 times the other one. Why?

let the slower rate be x , so the other one becomes 1.5x. combined rate 2.5x which is given to be 1/4 or .25 . See?

therefore , x is .1 and the faster is .15 or 15/100 .

The times is the inverse of this figure or 100/15.

Regards,
Harsha

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jcbruin: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Mon Jan 18, 2010 9:36 pm; Thanked: 1 times

by jcbruin » Tue Jun 01, 2010 7:09 pm

resilient wrote:If we stick to just the basics; ( which gets me out of a lot of trouble)

w=r t

pump 1 slower = 1 job per hour
pump 2 faster = 1.5 jobs per hour

combined rate = 2.5 jobs per hour multiplied by the 4 hour work time.

This yields the W work, which is 10. so simply plug plug back into the w = r t formula. but this time we manipulate the w = rt to t=w/r

w= 10
r= (faster pump) 1.5

so 10/1.5 = 20/3. qa is e

Thanks for keeping it simple! Kudos!

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