Bunuel wrote:
Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?
(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)^2=400
Question Stem : In one hour Jane can paint \(\frac{1}{J}\) of the wall and Bill can paint \(\frac{1}{B}\) of the wall.
Together, in one hour, they can paint \(\frac{1}{J} + \frac{1}{B}\) of the wall.
Starting time : 12:00 p.m.
If J and B are even, can we tell if J = B ?
St. (1) : Finish time : 4:48 p.m.
This tells us that when they worked together, they finished painting the wall in 288 minutes or \(\frac{288}{60}\) hours.
Therefore, in one hour, working together, they finished painting \(\frac{60}{288}\) of the wall.
Therefore we get the equation : \(\frac{1}{J} + \frac{1}{B}\) = \(\frac{60}{288}\)
This can be reduced to \(\frac{J+B}{J*B} = \frac{5}{24}\)
Now let us assume that A and B are equal.
Therefore, \(\frac{J+J}{J*J} = \frac{5}{24}\) ---> \(\frac{2*J}{J^2} = \frac{5}{24}\) ---> \(J = \frac{48}{5}\)
Thus it is obvious that if J is equal to B, in order to satisfy statement 2, J and B cannot be even. Since we are given that J and B have to be even, it is possible to conclude that J is not equal to B.
Hence this statement is Sufficient.
St. (2) : \((J + B)^2 = 400\)
This one is easy to discard by just plugging in some values.
The statement holds true for J = B = 10 as well as for J = 12 and B = 8.
Since it gives contradicting solutions for the question stem (J=B in case and J#B in the other), this statement is Insufficient.
Answer : A _________________
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