royQV wrote:
Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?
METHOD ONE: Algebraic approach
This approach to this question involves some tricky algebra.
Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.
In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of
The time it will take the two pumps, working at the combined rate, to drain this, is:
That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)
Answer = B
METHOD TWO: Numerical approach
Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.
Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.
Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?
Add the first minute for total time.
If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?
Only answer (B) works, so that the correct answer.
Can you please post the OA?
Rate of Pump A \(= \frac{1}{a}\)
Rate of Pump B \(= \frac{1}{b}\)
Work done by Pump A in 1 min \(= \frac{1}{a} * 1 = \frac{1}{a}\)
Work remaining (To be done by both pumps) \(= 1 - \frac{1}{a} = \frac{a-1}{a}\)
Combined rate of Pump A & B\(= \frac{1}{a} + \frac{1}{b}\)
Time required to complete the remaining work = t
\(\frac{a+b}{ab} * t = \frac{a-1}{a}\)
\(t = \frac{b(a-1)}{a+b}\)
Total time required
= \(\frac{b(a-1)}{a+b} + 1\)
\(= \frac{a(b-1)}{a+b}\)