Question 1197896
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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 = <font color=red>60 minutes</font>


It takes the larger hose 30 min to do the job alone.
It takes the smaller hose <font color=red>60 min</font> 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 = <font color=red>60 minutes</font>


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
<a href = "https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1197891.html">https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1197891.html</a>


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 = <font color=red>60 minutes</font> when working alone.


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Answer: <font color=red>60 minutes</font> (aka 1 hour)
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