Since they both started working together, they both worked for 1 day. That means, 1/5th + 1/3rd of the pool is filled in 1 day(8/15th). For the rest of the pool(1-8/15), A alone has to fill it.

Since A filles 1/5th of the pool in 1 day, for 7/15th, it will take 7/3 days.

So the total time is 1 + 7/3 = 10/3 days. (we add 1 since both A and B worked for 1 day before B stopped working).

I would asusme there is a easier approach than this.

notgoodinmath wrote:A pipe can fill a pool in 5 days. A 2nd pipe can fill the pool in 3 days. Assuming they start working together at the same time, how long will it take to fill the pool if the second pipe stops working after a day?

Please explain how the answer ends up being 10/3.

Thanks!!!!

This is quite simple if you apply Logitech's method:

x/5 + 1/3 = 1 (b/c p2 worked only 1 day when they start work together)
=> x/5 = 2/3
=> x = 10/3

notgoodinmath wrote:A pipe can fill a pool in 5 days. A 2nd pipe can fill the pool in 3 days. Assuming they start working together at the same time, how long will it take to fill the pool if the second pipe stops working after a day?

Please explain how the answer ends up being 10/3.

Thanks!!!!

since you're not good at math - (you said it, not me ...) - here's how you do it.

First pipe is 1/5 complete after a day and 2nd pipe is 1/3.

After one day they are 8/15 done and 2nd pipe stops working

Day 2 - 1/5 AKA 3/15 more gets finished (total is 11/15)

Day 3 - 1/5 AKA 3/15 more gets finished (total is 14/15)

Day 4 - only need 1/15 left to do, which is only 1/3 of a day or 1/15 of work.

That's pretty clear right?

Words like algebra and using algebra scare people like me so I like to explain things as basic as possible.