# algebra word problems

• Oct 11th 2009, 04:37 AM
zorodzai
algebra word problems
in a factory pf two different machines, A and B, are used to fill bottles. when the machines work together they fill the botlles in 2 hours. machine A works twice as fast as machine B. how long does it take machine B to fill the bottles when it works alone?
• Oct 11th 2009, 05:04 AM
HallsofIvy
Quote:

Originally Posted by zorodzai
in a factory pf two different machines, A and B, are used to fill bottles. when the machines work together they fill the botlles in 2 hours. machine A works twice as fast as machine B. how long does it take machine B to fill the bottles when it works alone?

When two people or machines, etc., work together, their speeds add. Let T be the time, in hours, it takes T hours for B to fill the bottles so its speed is 1/T (jobs/hour). Since A works twice as fast, its speed it 2/T. Working together they have speed 1/T+ 2/T= 3/T. Since it took them 2 two hours to do the job together we know their speed together is really 1/2 job/hour.

3/T= 1/2. Solve for T.

(This time I got in before earboth!)
• Oct 11th 2009, 05:12 AM
reiward
Quote:

Originally Posted by zorodzai
in a factory pf two different machines, A and B, are used to fill bottles. when the machines work together they fill the botlles in 2 hours. machine A works twice as fast as machine B. how long does it take machine B to fill the bottles when it works alone?

Time together = 2 hours

Time alone:
A = B / 2

B = B

You use this equation (time together over time alone is equal to 1 work)

$\displaystyle \frac {2}{A} + \frac{2}{B} = 1$

$\displaystyle \frac {4}{B} + \frac{2}{B} = 1$

$\displaystyle \frac {4}{B} + \frac{2}{B} = 1 ] B$

$\displaystyle 4 + 2 = B$

B = 6 hours

A = 3 hours