Suruchim12
GMATGuruNYI have a question with regards to the approach yo have shared. In that, we have assumptions made at 2 steps ; 1 where we assume the tank capacity to be 3 gallons and next when we assume that Valve A when it works on its own for an hour -- empties 1 gallon of the 3-gallon tank.
While the first assumption is understandable. the 2nd assumption isnt something that I'd think of in the exams if I was faced with this problem. Therefore, Is there an alternative approach to solving this question?
One option is to test a second value for A's rate to confirm that we get the same result as when A=1.
Let the tank = 3 gallons, A's hourly rate = 2, and B's hourly rate = B, implying that the combined rate for A and B together = 2+B
Since the combined rate for A and B = 2+B, the time for A and B together to empty the 3-gallon tank \(= \frac{work}{combined-rate} = \frac{3}{2+B}\)
If A works on its own for an hour -- emptying 2 gallons of the 3-gallon tank-- the time for A and B together to empty the remaining 1 gallon \(= \frac{remaining-work}{combined-rate} = \frac{1}{2+B}\), with the result that the total number of work-hours to empty the tank = A's time alone + time for A and B together = \(1 + \frac{1}{2+B}\)
Since the second time is equal to 4/3 of the first time, we get:
\(1 + \frac{1}{2+B }= \frac{4}{3}*\frac{3}{2+B}\)
\(\frac{2+B+1}{2+B} = \frac{4}{A+B}\)
\(B+3 = 4\)
\(B = 1\)
Since B empties 1 gallon per hour, the time for B on its own to empty the 3-gallon tank = \(\frac{work}{rate} = \frac{3}{1} = 3\) hours
B takes 3 hours whether A=1 or A=2.
Thus, the answer to the question stem is YES.
SUFFICIENT.
Plugging in a value for A's rate allows us to perform the algebra using only one unknown (B).
Alternatively, we can leave A's rate as a variable.
Let the tank = 3 gallons, A's hourly rate = A, and B's hourly rate = B, implying that the combined rate for A and B together = A+B
Since the combined rate for A and B = A+B, the time for A and B together to empty the 3-gallon tank \(= \frac{work}{combined-rate} = \frac{3}{A+B}\)
If A works on its own for an hour -- emptying A gallons of the 3-gallon tank -- the time for A and B together to empty the remaining 3-A gallons \(= \frac{remaining-work}{combined-rate} = \frac{3-A}{A+B}\), with the result that the total number of work-hours to empty the tank = A's time alone + time for A and B together = \(1 + \frac{3-A}{A+B}\)
Since the second time is equal to 4/3 of the first time, we get:
\(1 + \frac{3-A}{A+B }= \frac{4}{3}*\frac{3}{A+B}\)
\(\frac{A+B+3-A}{A+B} = \frac{4}{A+B}\)
\(B+3 = 4\)
\(B = 1\)
Since B empties 1 gallon per hour, the time for B on its own to empty the 3-gallon tank = \(\frac{work}{rate} = \frac{3}{1} = 3\) hours
Thus, the answer to the question stem is YES.
SUFFICIENT.