The Valve that Fills is working at a rate of 1/4 per hour
The Drain is working at a rate of 1/5 per hour
Now The valve is open for 10 hrs ( between 1 Pm to 11 Pm)
Let the drain be open for x hrs.
so work done is (10)/4 (Valve's work)
Drains work is (x)/5
==> (10/4 ) - (x/5) = 1 ( Assuming work is 1)
gives us x= 7.5 hrs ., Drain was open for 7.5 hrs which is 2.5 hrs from 1 Am = 3:30 Am
Hope that helps.
Regards,
Arun
Bunuel wrote:
Official Solution:
A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?
A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm
When only the valve is open, the pool is filling at a rate of \(\frac{1}{4}\) an hour. When both the valve and the drain are open, the pool is filling at a rate of \(\frac{1}{4} - \frac{1}{5} = \frac{1}{20}\) an hour. Let \(x\) denote the time when only the valve was open. Then both the valve and the drain were open for \(11 - 1 - x = 10 - x\)hours. Now we can compose the equation \(\frac{1}{4}x + \frac{1}{20}(10 - x) = 1\) which reduces to \(\frac{x}{5} = 0.5\) from where \(x = 2.5\). Thus, the drain was opened at \(1:00 + 2:30 = 3:30\)pm.
Answer: D