Lindsay

[vietst]

Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?

CAT 4 .

1 hour to finish the job, then 20 minutes, finish 1/3
call 1/y the fraction of the same room can J paint in 20 minutes.
1/x + 1/y = 1/3
then 1/y = 1/3 - 1/x = (x-3)/3x
C.
I think It is much shorter.

[StaceyKoprince]

Very nice, and yes this is a good technique if you are extremely confident with this type of problem. Many people are more likely to make mistakes with this method, though, than they will when just picking numbers to plug in. But if you are confident you will never make a mistake with this (and remember that the wrong answers will be based upon the common mistakes people make with the algebra), then definitely go ahead and use the algebra!

[as2764]

1 hour to finish the job, then 20 minutes, finish 1/3
call 1/y the fraction of the same room can J paint in 20 minutes.
1/x + 1/y = 1/3
then 1/y = 1/3 - 1/x = (x-3)/3x
C.
I think It is much shorter.

hi -- you've overlooked a minor error.
lindsay and joseph together paint 1 room in 1 hour, so:
1/x + 1/y = 1
1/y = 1-1/x = (x-1)/x --> this is in 1hr=60min
thus, in 20min=60min/3, --> 1/3y = (x-1)/3x <-- right answer

alternatively,
you can say in 20 minutes, L and J together paint a 1/3rd of the room:
1/3*(1/x + 1/y) = 1/3
1/3x + 1/3y = 1/3
1/3y = 1/3 - 1/3x = (x-1)/3x

as Stacey pointed out, unless one is confident, mistakes can happen.

[tim]

i think you may be defining your variables differently than the original poster. there are of course several ways to solve most math problems, and you should try as many of them as you can..

[as2764]

i think there are two variations of the 'Lindsay' problem; here's the other:
lindsay-paints-t7598.html

the one i posted was for the other variation. sorry for the confusion, but it was really hard to see because the original poster did not include the answer choices w/ the question.

[jnelson0612]

Thanks as2764.