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# Filling a pool math problem?

A well and a spring are filling a swimming pool. Together they can fill the pool in 3 hours. The well, working alone, can fill the pool in 8 hours less time than the spring. How long would the spring take, working alone, to fill the pool?

I don't understand how to do these my math book doesn't explain it at all and I've searched online please help me solve this!!! I need to know how :(

### 2 Answers

- 1 decade agoFavorite Answer
Simple questions that follow the same concept usually goes as follows:

A well can fill up a pool in 3 hours and a spring can fill up the same pool in 4 hours. How long does it take to fill up the pool if both the well and the pool fill it up at the same time.

(This is how I do it but it may be different to other people's working out)

In one hour, the well (lets call W) fills up 1/3 of the pool. In one hour, the spring (call S) fills up 1/4 of the pool. So in one hour altogether, it can fill 1/3 + 1/4 of the pool = 7/12 of the pool. Now we use ratios.

1 hour : 7/12 V (volume of the pool). 12 hr : 7V

So you can work out that it will take 12/7 hours for those two to fill up the pool together.

Now in this instance, you use the same method of approach. Let x = number of hours S would take, so W = 1/(x-8)

In one hour:

1/x of V + 1/(x-8) of V = 1/3 of V ---> Since it would take 3 hours to fill up, one hour/three hours = 1/3, which is where this comes from)

V/x + V/(x-8) = V/3

V(x-8) + V(x) = V(x)(x-8)/3

3 (Vx - 8V + Vx) = Vx^2 - 8Vx

3Vx - 24V + 3Vx = Vx^2 - 8Vx

0 = Vx^2 -14Vx + 24V

0 = x^2 - 14x + 24

0 = (x - 12)(x - 2)

x = 12 or 2

Now that we've got that, we know S takes 12 hours or 2 hours to fill up a pool.

HOWEVER, if S takes 2 hours, W takes x - 8 = 2 - 8 = -6.

This means instead of adding to the pool's volume, it drains it. i.e. W is in face some drain which sucks the water out.

I don't think so.

So the final answer would be 12 hours, working alone, for the spring to fill the pool.

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- 3 years ago
Assuming all aspects are vertical, take the similar old intensity as (a million.a million + 2.a million)/2 - 0.a million = a million.5 m volume of filled pool = 12 x 4 x a million.5 = seventy 2 m^3 a million ml = a million cm^3 a million m^3 = one hundred^3 cm^3 = a million,000,000 cm^3 At 2 hundred cm^3/sec, a million m^3 will fill in a million,000,000/2 hundred = 5,000 sec = a million.38* hr. seventy 2 m^3 will fill in one hundred hr = 4 days and four hours one hundred hr from 9:00 am Monday ends at a million: pm on Friday

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