Official Solution:A swimming pool has two water pumps, A and B, and one drain, C. Pump A can fill the empty pool by itself in \(x\) hours, while pump B can do so in \(y\) hours. The drain C can empty the entire pool in \(z\) hours, with \(z > x\). When both pumps A and B are operating simultaneously and drain C is open until the pool is filled, which of the following represents the fraction of the total pool volume added by pump A until the pool is filled? A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xz-xy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xz-xy}\)
E. \(\frac{yz+xz-xy}{yz}\)
When both pumps A and B are running, and the drain is open, the pool fills at a rate of \(\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac{yz + xz - xy}{xyz}\) pools per hour. Therefore, it takes \(\frac{xyz}{yz + xz - xy}\) hours to fill the pool (since time is the reciprocal of rate).
In \(\frac{xyz}{yz+xz-xy}\) hours, pump A will contribute \(rate * time = \frac{1}{x} * \frac{xyz}{yz+xz-xy} = \frac{yz}{yz+xz-xy}\) of the total pool volume.
For instance, suppose the pool will be filled in 10 hours, and pump A alone can fill the entire pool in 20 hours. In 10 hours, pump A will pump \(\frac{1}{20} * 10 = \frac{1}{2}\) of the total pool volume.
Answer: B