Having problems solving this Algebra Word problem. I'm taking a Algebra 2 class and so far its like looking at chinese.

Question

. one hose requires 3 hours to fill the pool. a second hose turned on, the pool be filled in 1.2 hours. How long the second hose to fill the pool if it worked alone?

Answer 1

The first hose can fill 1/3 of the pool in one hour.

If 1.2 hours have passed, then what fraction of the pool has been filled by the first hose?

 

We can set up a proportion:

(1/3)/(1) = (z)/(1.2)

 

z=2/5. The first hose fills 2/5 of the pool.

 

The second hose must have filled the rest of the pool (3/5 of the pool).

 

In 1.2 hours, the second hose filled 3/5 of the pool.

So, in x hours, the second hose would fill the whole pool, on its own.

 

Another proportion:

(3/5)/(1.2) = (1)/(x)

 

x= two hours.

 

The second hose takes two hours to fill the pool if on its own.

Answer 2

One good way to do these problems is this:

You're given "#hours to fill a pool". Turn it around into "fraction of a pool per hour". So you change "3 hours to fill a pool" into "1/3 of a pool per hour". Hose 2 is an unknown fraction per hour at this point.

 

So we have for both hoses: 1/3 + x = 1/1.2    1/1.2 = 1//6/5 = 5/6

So now:  1/3 + x = 5/6 

                         x = 5/6 - 1/3 = 1/2 pool per hour for hose 2

 

This means it takes 2 hours for hose 2 by itself to fill the pool

Answer 3

x = time second hose needs to fill the pool alone.

 

For the first hose, it does 1/3 of the pool in one hour

For the second hose, it does 1/x of the pool in one hour.

 

Working together, it takes 1.2 hours, so

 

For the first hose, it does 1.2/3 of the pool

For the second hose, it does 1.2/x

 

Together they do the 1 job

 

1.2/3 + 1.2/x = 1                      Multiply both sides by LCD

1.2x + 3.6 = 3x

3.6 = 1.8x

2 = x

 

Working alone the second hose would need 2 hours.