SOLUTION: A chemistry experiment requires that a vacuum be created in a chamber. A small vacuum pump requires 15 seconds to evacuate the chamber. Working together, the chamber can be evacuat

Question 1033132: A chemistry experiment requires that a vacuum be created in a chamber. A small vacuum pump requires 15 seconds to evacuate the chamber. Working together, the chamber can be evacuated in 4 sec. Find the time required for the larger vacuum pump working alone to evacuate the chamber.
Found 3 solutions by fractalier, ikleyn, MathTherapy:
Answer by fractalier(6550)   (Show Source): You can put this solution on YOUR website!
The set up is like this:

where A and B are the times working alone and t is the time working together...thus we have

Multiply by 15B and get

and
hours
Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.
A chemistry experiment requires that a vacuum be created in a chamber. A small vacuum pump requires 15 seconds to evacuate the chamber.
Working together, the chamber can be evacuated in 4 sec.
Find the time required for the larger vacuum pump working alone to evacuate the chamber.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The set up is like this:

where A and B are the times working alone and t is the time working together...thus we have

Multiply by 60 and get

and
= seconds.

Answer. seconds.

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Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

A chemistry experiment requires that a vacuum be created in a chamber. A small vacuum pump requires 15 seconds to evacuate the chamber. Working together, the chamber can be evacuated in 4 sec. Find the time required for the larger vacuum pump working alone to evacuate the chamber.

With larger taking L seconds, we get:

4L + 60 = 15L ------- Multiplying by LCD, 60L
4L - 15L = - 60
- 11L = - 60
L, or time larger takes, alone = , or
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