SOLUTION: It takes a smaller hose twice as long to fill a small swimming pool as it does a larger hose. It takes both hoses working together 20 minutes to fill the swimming pool. How long wi

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Question 1197896: It takes a smaller hose twice as long to fill a small swimming pool as it does a larger hose. It takes both hoses working together 20 minutes to fill the swimming pool. How long will it take the smaller hose to fill the pool by itself?
Do not do any rounding.

Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Method 1

x = time it takes the larger hose to fill the pool if working alone
2x = time it takes the smaller hose to fill the pool if working alone

The jump from x to 2x is twice as much to reflect it takes twice as long.

1/x and 1/(2x) are the unit rates for each hose

They add to...
1/x + 1/(2x)
2/(2x) + 1/(2x)
(2+1)/(2x)
3/(2x)
This is the combined unit rate for both hoses.

(unit rate)*(time) = amount done
( 3/(2x) )*( 20 min ) = 1 job
60/(2x) = 1
30/x = 1
x = 30

If x = 30, then 2x = 2*30 = 60 minutes

It takes the larger hose 30 min to do the job alone.
It takes the smaller hose 60 min to do the job alone.
1/30 + 1/60 = 2/60 + 1/60 = 3/60 = 1/20 is the combined unit rate.
(unit rate)*(time) = amount done
(1/20 job per min)*(20 min) = 1 job
This confirms our answer is correct.

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Method 2

x = time it takes for the larger hose working alone
2x = time it takes for the smaller hose working alone

The two hoses, when working together, need T = 20 minutes
1/x + 1/(2x) = 1/T
1/x + 1/(2x) = 1/20
20x * [ 1/x + 1/(2x) ] = 20x*(1/20)
20x*(1/x) + 20x*( 1/(2x) ) = 20x*(1/20)
20 + 10 = x
30 = x
x = 30
2x = 2*30 = 60 minutes

This method is very similar to the first section, just a slight rephrasing.

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Method 3

Let m and n represent the time values for each hose when they work individually.
Their respective unit rates are 1/m and 1/n
Add them up to find the combined unit rate
1/m + 1/n
n/(mn) + m/(mn)
(n+m)/(mn)
(m+n)/(mn)

Apply the reciprocal because
time = (amount done)/(rate)
time = (1 job)/(rate)
Effectively informally saying
time = 1/rate

The reciprocal of (m+n)/(mn) is (m*n)/(m+n)

Therefore this formula
x = (m*n)/(m+n)
gives the time it takes both hoses to work together given the input time values m and n for each hose working alone.

Refer to method 3 of the link below to see an example
https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1197891.html

Let
m = amount of time for the larger hose
n = 2m = amount of time for the smaller hose (that takes twice as long)

So we can further say
x = (m*n)/(m+n)
x = (m*2m)/(m+2m)
x = (2m^2)/(3m)

From here we replace x with 20 since we know the two hoses work together needing only 20 minutes. Then solve for m.
x = (2m^2)/(3m)
20 = (2m^2)/(3m)
20*3m = 2m^2
60m = 2m^2
60m - 2m^2 = 0
2m(30 - m) = 0
2m = 0 or 30-m = 0
m = 0 or m = 30

Ignore m = 0 as it's a nonsensical answer.
m = 30 fits with the conclusion reached in the earlier sections where the larger hose needs 30 minutes if working alone.
The smaller hose then needs 2m = 2*30 = 60 minutes when working alone.

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Answer: 60 minutes (aka 1 hour)

Answer by greenestamps(13209) (Show Source):
You can put this solution on YOUR website!

For a quick informal solution, if a formal algebraic solution is not required....

The smaller hose takes twice as long as the larger to fill the pool alone; that means it works half as fast.
So when the two hoses together fill the pool in 20 minutes, the smaller hose is doing 1/3 of the job and the larger hose is doing 2/3 of the job.
So the smaller hose fills 1/3 of the pool in 20 minutes; that means it would take it 3*20 = 60 minutes to fill the whole tub alone.

ANSWER: 60 minutes

NOTE: Even if a formal algebraic solution is needed, you can get good mental exercise by solving the problem in a manner similar to the above, using logical reasoning and simple arithmetic.