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Lindsay can paint:

Expert replies
by \'manpreet singh » Sat Aug 17, 2013 10:15 pm
Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

1/3x

x/x - 3

(x - 1)/3x

x/x - 1

(x - 1)/x

Ans c
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Source: — Problem Solving |

by raul_200435 » Sat Aug 17, 2013 10:52 pm
c

If Lindsay & Joseph paint the entire room in 1 hour and lindsay paints 1/x part of room, than Joseph paints (x-1)/x part in one hour. In 20 min or 1/3 (20/60) hour it paints (x-1)/3x part of the room.
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by GMATGuruNY » Sun Aug 18, 2013 2:24 am
'manpreet singh wrote:Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

1/3x

x/x - 3

(x - 1)/3x

x/x - 1

(x - 1)/x

Ans c
Let the room = 20 units.
Let x = 4.

Since Lindsay paints 1/x of the room in 1 hour, Lindsay's rate = (1/4) * 20 = 5 units per hour.
Since Lindsay and Joseph paint the entire room in 1 hour, their combined rate = 20 units per hour.
Thus, Joseph's rate = (combined rate) - (Lindsay's rate) = 20-5 = 15 units per hour.

In 20 minutes -- the equivalent of 1/3 of an hour -- the amount of work produced by Joseph = r*t = 15 * (1/3) = 5 units.
Thus, the fraction painted by Joseph = 5/20 = 1/4. This is our target.

Now we plug x=4 into the answers to see which yields our target of 1/4.
Only C works:
(x-1)/3x = (4-1)/(3*4) = 3/12 = 1/4.

The correct answer is C.
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by Brent@GMATPrepNow » Sun Aug 18, 2013 6:42 am
'manpreet singh wrote:Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

1/3x

x/(x - 3)

(x - 1)/3x

x/(x - 1)

(x - 1)/x
NOTE: I added some brackets to your answer choices to avoid confusion.

This approach is similar to raul_200435's great solution. I've just filled in a few extra details.

Given: Lindsay and Joseph can paint the room in one hour.
During that one hour, Lindsay can paint 1/x of the room.
So, during that one hour, Joseph must paint the rest (whatever Lindsay did not paint)
So, during the one hour, the fraction of the room that Joseph paints = 1 - 1/x
= x/x - 1/x
= (x-1)/x

So, (x-1)/x = the fraction of the room that Joseph paints in one hour.
Since 20 minutes = 1/3 of an hour, Joseph can paint (1/3)[(x-1)/x] of the room in 20 minutes.
(1/3)[(x-1)/x] = [spoiler](x-1)/3x = C[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by \'manpreet singh » Wed Aug 21, 2013 12:08 am
Thanks guy!
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by Jeff@TargetTestPrep » Mon Dec 11, 2017 5:23 pm
'manpreet singh wrote:Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

1/3x

x/x - 3

(x - 1)/3x

x/x - 1

(x - 1)/x
We are given that Lindsay's rate to paint a room is 1/x. We are also given that when she works with Joseph, they can paint the room in 1 hour. If we let total work = 1 and j = the number of hours it takes Joseph to paint the room by himself, then Joseph's rate = 1/j. We can create the following equation and isolate j.

work of Lindsay + work of Joseph = 1

(1/x)(1) + (1/j)(1) = 1

1/x + 1/j = 1

Multiplying the entire equation by xj, we obtain:

j + x = xj

x = xj - j

x = j(x - 1)

x/(x-1) = j

Since j = x/(x-1) and 1/j = Joseph's rate, Joseph's rate in terms of x is (x - 1)/x.

Since 20 minutes = 1/3 of an hour, and since work = rate x time, Joseph can complete:

[(x - 1)/x](1/3) = (x - 1)/(3x) of the job in 20 minutes.

Answer: C

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