Quadratic equation application

[paulmdrdo1]

please check my work.

Pipe A can fill a given tank in 4 hr. If pipe B works alone, it takes 3 hr longer to fill the tank than if pipes A and B act together. How long will it take pipe B working alone?

let $x=$ required time for B working together with A
$x+3=$ required time for B working alone

$\frac{1}{x}-\frac{1}{4}=\frac{1}{x+3}$

solving for x I get 5.3 hr (approximately)

can you suggest another way of solving it? thanks!

 

[MarkFL]

Here is another way to work the problem. Let $B$ be the time (in hours) it takes pipe $B$ working alone to fill the tank. In $4B$ hours, pipes $A$ and $B$ working together can fill $4+B$ tanks, which means it takes:

\(\displaystyle \frac{4B}{4+B}\)

hours for them to fill one tank. Thus, given it takes pipe $B$ 3 hours more than this to fill the tank alone, we may write:

\(\displaystyle B=\frac{4B}{4+B}+3\)

which may be written as:

\(\displaystyle B^2-3B-12=0\)

Applying the quadratic formula and discarding the negative root, we obtain:

\(\displaystyle B=\frac{3+\sqrt{57}}{2}\approx5.3\)

 

[paulmdrdo1]

Here is another way to work the problem. Let $B$ be the time (in hours) it takes pipe $B$ working alone to fill the tank. In $4B$ hours, pipes $A$ and $B$ working together can fill $4+B$ tanks, which means it takes:

\(\displaystyle \frac{4B}{4+B}\)

hours for them to fill one tank. Thus, given it takes pipe $B$ 3 hours more than this to fill the tank alone, we may write:

\(\displaystyle B=\frac{4B}{4+B}+3\)

which may be written as:

\(\displaystyle B^2-3B-12=0\)

Applying the quadratic formula and discarding the negative root, we obtain:

\(\displaystyle B=\frac{3+\sqrt{57}}{2}\approx5.3\)

thanks for that. Is my solution above correct?

 

[MarkFL]

thanks for that. Is my solution above correct?

Yes, adding 3 to the positive root of your equation is the same solution. :D

 

[paulmdrdo1]

I have another question, where did you get 4B? why did you multiply them? If they are working together aren't they suppose to be added?

 

[MarkFL]

I have another question, where did you get 4B? why did you multiply them? If they are working together aren't they suppose to be added?

$4B$ is the LCM of $4$ and $B$, and so in that amount of time, we know the two pipes can fill $4+B$ tanks (this is where they are added).

 

[paulmdrdo1]

Is this how you get 4B by adding their work rate?

$\frac{1}{b}+\frac{1}{4}=\frac{4+b}{4b}$

 

[MarkFL]

Is this how you get 4B by adding their work rate?

$\frac{1}{b}+\frac{1}{4}=\frac{4+b}{4b}$

While that equation is true, that's not how I think of it. I simply use the LCM of the two times.

 

[paulmdrdo1]

While that equation is true, that's not how I think of it. I simply use the LCM of the two times.

I'm still confused. 4B here means the time that they work together right? In that span of time they can fill 4+B tanks. which means their work rate together is $\frac{4+B}{4B}$

if I do this $\frac{4+B}{4B}-\frac{1}{4}=\frac{1}{B+3}$

please tell me if this is correct.

 

[MarkFL]

This means it takes \(\displaystyle \frac{4B}{4+B}\) hours to fill one tank. Then we add 3 to that to equal $B$. That's what I did.

 

[paulmdrdo1]

yes I kind of understand your solution. What I'm thinking now is

since $\frac{4+B}{4B}$ is their work rate working together if I subtract the work rate of pipe A working alone from their work rate working together I get this,

$\frac{4+B}{4B}-\frac{1}{4}=\frac{1}{B+3}$

I did this because when B is working alone his required time will be 3 hours more.