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.

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

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.

Thanks Xilef. What a relief I got it

~R

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!!

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.

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.

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

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!

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.

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.

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.

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!