A certain water tank has two outlet valves, valve A and valve B. Each

"Valve A is opened and valve B is opened an hour later, ..." So, for an hour valve A drained the tank alone. After B joined together they needed (1−a)/(a+b) hours, hence total of 1 + (1−a)/(a+b) hours.

Got it. Thanks Bunuel

Thankyou Bunuel for noticing the mistake and editing this one. I was pretty confused. Heres how I did this:

Let valve A drains 'a' units per hour, let valve B drains 'b' units per hour, and let the water in tank be total 'a*b' units.
So valve A will take = ab/a = b hours, while valve B will take = ab/b = a hours : to drain water on their own each. We need to know whether a > 2 or not?

(1) valve A alone takes 4 hours or b=4. or total water in tank = 4a units. But this doesnt give the answer to our question. Insufficient.


(2) if both valves are opened simultaneously, units of water drained per hour = (a+b), thus time taken = ab/(a+b)
IF instead first A is opened for 1 hour, 'a' units of water already drained in one hour. Units left in tank = (ab-a) ...now both valves are opened and units being drained per hour = a+b, thus further time now required to drain completely = (ab-a)/(a+b). So total time required in this case = 1 + (ab-a)/(a+b) OR (b+ab)/(a+b)

We are given that this latter time is 4/3 or the former time, so we can write this as:
(b+ab)/(a+b) = 4/3 * ab/(a+b) ... Solving we get: a+1 = 4a/3 Or a=3. So we get that a>2 (or valve B will take more than 2 hours to drain the tank). Sufficient.

Hence B answer

From Statement B, you can see that the only different is B opening one hour later and causing delay (or lets say adding) 1/3 more time.
Which means to add delay of 100% it will take 3 hours. In other words it will take 3 hours to empty the tank.

Statement 1:
No info about B.
INSUFFICIENT.

Statement 2:
Let the tank = 3 gallons, A's hourly rate = 1, and B's hourly rate = B, implying that the combined rate for A and B together = 1+B

Since the combined rate for A and B = 1+B, the time for A and B together to empty the 3-gallon tank \(= \frac{work}{combined-rate} = \frac{3}{1+B}\)

If A works on its own for an hour -- emptying 1 gallon of the 3-gallon tank -- the time for A and B together to empty the remaining 2 gallons \(= \frac{remaining-work}{combined-rate} = \frac{2}{1+B}\), with the result that the total number of work-hours to empty the tank = A's time alone + time for A and B together = \(1 + \frac{2}{1+B}\)

Since the second time is equal to 4/3 of the first time, we get:
\(1 + \frac{2}{1+B }= \frac{4}{3}*\frac{3}{1+B}\)
\(\frac{1+B+2}{1+B} = \frac{4}{1+B}\)
\(B+3 = 4\)
\(B = 1\)
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.

GMATGuruNY

I 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.

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