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C
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Difficulty:
55%
(hard)
Question Stats:
68% (02:31) correct
32%
(02:48)
wrong
based on 708
sessions
History
Date
Time
Result
Not Attempted Yet
Laura 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?
A. 1/(3x)
B. 3x/(x-3)
C. (x – 3)/(3x)
D. x/(x-3)
E. (x-3)/x
A. 1/(3x)
B. 3x/(x-3)
C. (x – 3)/(3x)
D. x/(x-3)
E. (x-3)/x
10 Aug 2006, 10:16
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20 minutes is 1/3 of an hour. In an hour, Laura would paint
3 * 1/x = 3/x;
Joseph, then, would have painted the rest:
1 - 3/x, or (x-3)/x.
So in 20 minutes Joe would do
(x-3)/3x.
3 * 1/x = 3/x;
Joseph, then, would have painted the rest:
1 - 3/x, or (x-3)/x.
So in 20 minutes Joe would do
(x-3)/3x.
23 Apr 2014, 20:53
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Bookmarks
Rate of Laura \(= \frac{1}{20x}\) (Per Min)
Consider rate of Joseph \(= \frac{1}{j}\) (Per Min)
Both combined perform the task in 60 minutes
\(\frac{1}{20x} + \frac{1}{j} = \frac{1}{60}\)
\(\frac{1}{j} = \frac{1}{60} - \frac{1}{20x}\)
\(\frac{1}{j} = \frac{x-3}{60x}\) (Per Min)
So in 20 minutes, work done by Joseph
\(= \frac{x-3}{60x} * 20\)
\(= \frac{x-3}{3x}\)
Answer = C
Consider rate of Joseph \(= \frac{1}{j}\) (Per Min)
Both combined perform the task in 60 minutes
\(\frac{1}{20x} + \frac{1}{j} = \frac{1}{60}\)
\(\frac{1}{j} = \frac{1}{60} - \frac{1}{20x}\)
\(\frac{1}{j} = \frac{x-3}{60x}\) (Per Min)
So in 20 minutes, work done by Joseph
\(= \frac{x-3}{60x} * 20\)
\(= \frac{x-3}{3x}\)
Answer = C
