Question

Rates question. Please help!?

Two water pipes lead into a tank. Each pipe provides water to the tank at its own constant rate. If both pipe are turned on, the tank fills in 80 minutes. If only one pipe is turned on, one of the pipes takes two hours longer than the other to fill the tank. How long does each pipe take to fill on its own?

Answer 1

2 hours and 4 hours, respectively.

Explanation:

Let the faster of the two pipes take `x` hours to fill the tank on its own. The other one will take `x+2` hours.

In one hour, the two pipes will fill, `1/x` and `1/{x+2}` fractions of the tank, respectively, on their own.

If both the pipes are opened on, the fraction of the tank that will fill up in one hour is `1/x+1/{x+2} = {2x+2}/{x(x+2)}`. Thus the time that will take to fill the tank is `{x(x+2)}/{2x+2}`.

Given
`{x(x+2)}/{2x+2} = 80/60 = 4/3`

Thus
`3x^2+6x = 8x+8 implies 3x^2-2x-8=0`
`3x^2-6x+4x-8 = 0 implies 3x(x-2)+4(x-2)=0`
so that
`(3x+4)(x-2)=0`

Since `x` must be positive, it has to be 2.

Answer 2

Read below. I used hose instead of pipe.

Explanation:

So we know the following:

Hose A and B working together take 80 minutes to fill the tank.

Hose A takes two hours longer than B to fill the tank.

Let `t` represent the amount of time hose B needs to fill the tank.

Since hose A takes two hours longer to fill the tank, it takes `t+2` hours

Remember the formula `Q=rt`
(Quantity equals rate times time)

The quantity is one tank for all cases

For hose A:

`1=r(t+2)` divide both sides by `t+2`

`1/(t+2)=r`

The rate of hose A is therefore `1/(t+2)`.

Similarly, we can find the rate for hose B.

`1=rt`

`1/t=r`

Now when hoses A and B are working together:

`1=r1 1/3`(`80`min.`=1 1/3`
hour)

`1÷1 1/3=r`

`3/4=r`
Now, we use logic here:

When hoses A and B are working together, their rate is added together.

For example, if a worker could build a statue per week and another worker could build two statues per week, then they would build 3 statues per week if they work together.

Therefore,

Rate of hose A plus the rate of hose B equals their total rate.

`1/(t+2)+1/t=3/4`

We try to find the GCF between `t` and `t+2`

It is simply t(t+2)

We now have:

`1/cancel(t+2)*(tcancel(t+2))/(t(t+2))+1/cancelt*(cancelt(t+2))/(t(t+2))=3/4`

We now have:

`t/(t(t+2))+(t+2)/(t(t+2))=3/4`

`(t+(t+2))/(t(t+2))=3/4`

`(2t+2)/(t^2+2t)=3/4` cross multiply

`4(2t+2)=3(t^2+2t)`

`8t+8=3t^2+6t`

`0=3t^2-2t-8` factor

`0=3t^2-6t+4t-8`

`0=3t(t-2)+4(t-2)`

`0=(3t+4)(t-2)`

`-4/3=t=2`

In our normal situations, time is positive.

So it takes hose B 2 hours, hose A 4 hours to fill the tank.