Manhattan Prep
Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joshep, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?
$$A.\ \frac{1}{3x}$$
$$B.\ \frac{x}{\left(x-3\right)}$$
$$C.\ \frac{\left(x-1\right)}{3x}$$
$$D.\ \frac{x}{\left(x-1\right)}$$
$$E.\ \frac{\left(x-1\right)}{x}$$
OA C.
Lindsay can paint 1/x of a certain room in one hour. If
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Let the room = 20 units.AAPL wrote:Manhattan Prep
Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joshep, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?
$$A.\ \frac{1}{3x}$$
$$B.\ \frac{x}{\left(x-3\right)}$$
$$C.\ \frac{\left(x-1\right)}{3x}$$
$$D.\ \frac{x}{\left(x-1\right)}$$
$$E.\ \frac{\left(x-1\right)}{x}$$
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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This appears to be a slightly different version of this official question: https://www.beatthegmat.com/work-rate-p ... 92679.htmlAAPL wrote:Manhattan Prep
Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joshep, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?
$$A.\ \frac{1}{3x}$$
$$B.\ \frac{x}{\left(x-3\right)}$$
$$C.\ \frac{\left(x-1\right)}{3x}$$
$$D.\ \frac{x}{\left(x-1\right)}$$
$$E.\ \frac{\left(x-1\right)}{x}$$
OA C.
HOWEVER, unless I'm missing something, the correct answer is missing.
Cheers,
Brent