abansal1805 wrote:

A certain water tank has two outlet valves, valve A and valve B. Each valve drains water from the tank at a constant rate. If the tank is full and valve B alone is opened, does it take more than two hours to completely drain the tank?

(1) When the tank is full, if valve A alone is opened, it takes 4 hours to completely drain the tank.

(2) When the tank is full, if valve A is opened and valve B is opened an hour later, it takes 4/3 as long to completely drain the tank as it would if both valves had been opened simultaneously from the start.

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