SOLUTION: Working together, it takes two different sized hoses
30 minutes to fill a small swimming pool. If it takes
45 minutes for the larger hose to fill the swimming pool by itself,
Algebra.Com
Question 1135820: Working together, it takes two different sized hoses
30 minutes to fill a small swimming pool. If it takes
45 minutes for the larger hose to fill the swimming pool by itself, how long will it take the smaller hose to fill the pool on its own?
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Working together, it takes two different sized hoses 30 minutes to fill a small swimming pool.
If it takes 45 minutes for the larger hose to fill the swimming pool by itself, how long will it take the smaller hose to fill the pool on its own?
:
let t = time required by the smaller hose to fill the pool
let the completed job = 1 (a full pool)
:
Each will do a fraction of the job, the two fractions add up to 1
+ = 1
Reduce fraction
+ = 1
multiply equation by 3t
2t + 3(30) = 3t
90 = 3t - 2t
t = 90 min the smaller hose to fill the pool
Answer by greenestamps(13208) (Show Source): You can put this solution on YOUR website!
Here is an alternative to the standard algebraic method for solving "working together" problems like this that I find many students prefer.
(1) Find the least common multiple of the given times.
The LCM of 30 and 45 is 90.
(2) Determine how many times each hose could fill the pool in that amount of time.
The two hoses together could fill the pool 90/30 = 3 times.
The larger hose could fill the pool 90/45 = 2 times.
The large hose alone could fill the pool 2 times in 90 minutes; the two together could fill the pool 3 times. That means that in 90 minutes the smaller hose could fill the pool once.
So the time it takes the smaller hose to fill the pool is 90 minutes.
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