NoHalfMeasures wrote:
Yves can paint a certain fence in 1/2 the time it takes Marcel to paint the same fence. If they work together, each at his own constant rate, how many hours will it take them to paint the fence?
(1) Yves can paint the fence by himself in 3 hours.
(2) Working together, each at his own constant rate, they can paint the fence in 1/2 the time it would take Marcel, working alone, to paint the fence.
Key to solving such problem is to realize that rates are additive.
Statement (1) Yves can paint the fence in 3 hours. Therefore in one hour Y can paint (1/3) fence
Now it is given that Yves can paint a certain fence in 1/2 the time it takes Marcel to paint the same fence.
So Marcel can paint the fence in 6 hours.
In other words, M can paint (1/6) fence in one hour.
in one hour both can paint (\(\frac{1}{3}\)+\(\frac{1}{6}\)) fence in one hour
This is (\(\frac{2}{6}\)+\(\frac{1}{6}\)) = \(\frac{1}{2}\))
Together in an hour they paint (1/2) fence. they can paint the fence in two hours. Sufficient.
Statement (2) Working together, each at his own constant rate, they can paint the fence in 1/2 the time it would take Marcel, working alone, to paint the fence.
Suppose M completes fence in x hours.
So Y competes fence in x/2 hours (this is given)
Also given when they work together they complete fence in x/2 hours. --> This appears poblematic. Has to be less than x/2.
In one hour say M paints (1/x) part of fence.
In one hour say Y paints (2/x) part of fence.
So in one hour both will paint (3/x) part of fence.
In other words they will complete fence in 1/ (3/x) = (x/3) hours.
(x/3) = (x/2)
This gives no solution for x.
Not sufficient.
Hence Answer is A