it takes the larger of two hoses 3hrs, working alone to fill a pool. it takes smaller of the two hoses 6hrs, working alone to fill up the same pool. if both of the hoses are used at the same time, how long will it take to fill up the pool.
can you show me how u got the answer?
2007-07-02 17:03:39 · 5 answers · asked by Anonymous in Education & Reference ➔ Homework Help
in 1 hour, the large hose can fill 1/3 of the pool.
in 1 hour, the smaller hose can fill 1/6 of the pool.
1/3t + 1/6t = 1
1/2t = 1
t = 2hr
2007-07-02 17:27:20 · answer #1 · answered by 7 · 0⤊ 0⤋
Time taken for the larger hose alone - 3 hours
Time taken for the smaller hose alone - 6 hours
Time taken if both are employed is say T
1/T = 1/3 + 1/6
= 3/6
t = 6/3 = 2 hours
The speeds are additive and speed is inverse of time. I hope the solution is clear. A similar approach is needed for people working together on one project etc.
2007-07-03 01:19:28 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
This is a little confusing but i hope it works
L= 3hrs
S=6hrs
L=2S
L+S=
2s+s= 3s
it would be like using 3 small ones and instead of multiplying divide.. 6 divided by 3 is 2 so it would be 2 hrs..
2007-07-03 00:14:18 · answer #3 · answered by lilkrazyme3 2 · 0⤊ 0⤋
This is the formula that you need: (l x s)/(l+s)
(3x6)/(3+6)=18/9 = 2hours
2007-07-03 00:50:10 · answer #4 · answered by mathvideosonline.com 2 · 0⤊ 0⤋
Add the 2 together and divide by 2. It would take 4.5 hours. ~
2007-07-03 00:38:12 · answer #5 · answered by Anonymous · 0⤊ 1⤋