MHB Quadratic equation application

AI Thread Summary
The discussion focuses on solving a problem involving the time it takes for two pipes, A and B, to fill a tank. Pipe A fills the tank in 4 hours, while pipe B takes 3 hours longer than the combined time of both pipes working together. The initial equation derived is $\frac{1}{x} - \frac{1}{4} = \frac{1}{x+3}$, leading to an approximate solution of 5.3 hours for pipe B. Participants clarify the use of the least common multiple (LCM) in their calculations and discuss the relationship between their work rates. The conversation concludes with a better understanding of how to derive the equations based on their combined work rates.
paulmdrdo1
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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!
 
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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:

$$\frac{4B}{4+B}$$

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:

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

which may be written as:

$$B^2-3B-12=0$$

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

$$B=\frac{3+\sqrt{57}}{2}\approx5.3$$
 
MarkFL said:
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:

$$\frac{4B}{4+B}$$

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:

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

which may be written as:

$$B^2-3B-12=0$$

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

$$B=\frac{3+\sqrt{57}}{2}\approx5.3$$
thanks for that. Is my solution above correct?
 
paulmdrdo said:
thanks for that. Is my solution above correct?

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

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

$\frac{1}{b}+\frac{1}{4}=\frac{4+b}{4b}$
 
paulmdrdo said:
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.
 
MarkFL said:
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.
 
  • #10
This means it takes $$\frac{4B}{4+B}$$ hours to fill one tank. Then we add 3 to that to equal $B$. That's what I did.
 
  • #11
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.
 
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