A hotel began draining its swimming pool at a constant rate at 7am. Beginning at 12pm, the rate tripled because of evaporation. If the pool began completely full and was 3/4 full at 9:30am, at what time was the pool completely empty?

A. 12:45pm
B. 1:20pm
C. 1:40pm
D. 2:10pm
E. 2:15pm
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Answer : [C]

1/4 emptied in 2.5 hrs. 1/2 emptied in 5 hrs i.e. till 12PM.

draining rate = (1/4) / 2.5 part/hr = 1/10 part/hr
increased draining rate = 3/10 part /hr

time taken to drain 1/2 at increased draining rate = (1/2) / (3/10) hrs = 5/3 hrs = 1 hr 40 mins (after 12 PM)

Official solution from Veritas Prep.

In this Work/Rate problem, you can start by calculating the constant rate. If between 7am and 9:30am the pool drained by \(\frac{1}{4}\) of its capacity, that means that the pool drained at a rate of \(\frac{1}{4}\) pool per 2.5 hours. That then means that the pool drained by \(\frac{1}{10}\) pool per hour during normal hours, and \(\frac{3}{10}\) pool per hour after noon.

Since there are 5 hours between 7am and noon, the pool had drained by \(5(\frac{1}{10})\) of its capacity by noon, so it was half full. In the next hour (12pm-1pm) it drained by \(\frac{3}{10}\) of its capacity, so by 1pm it had \(\frac{2}{10}\) left to drain. If it drained by \(\frac{3}{10}\) every hour, that means that it only needed \(\frac{2}{3}\) of an hour to finish draining, meaning that it would be done by 1:40pm. Answer choice C is correct.
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The hotel began draining the swimming pool(which was completely full) at 7 AM.

At 9:30 AM(2 hours 30 minutes later), the pool was 3/4th full (or) 1/4th of the pool was drained.
At 12:00 PM(2 hours 30 minutes later), working at the same rate, another 1/4th of the pool was drained.
Now, the pool was half full at 12 AM.

It has been given that the draining rate triples because of evaporation after 12 PM.
Initially, 5 hours(300 minutes) was needed to drain the pool to half its capacity,
at 3 times the rate, we would be needing \(\frac{300}{3} = 100\) minutes to drain the pool now.

Therefore, the pool was completely empty at 1:40 PM(Option C) - 100 minutes after 12 PM
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Solution:

Without evaporation, we see that the pool was drained by 1/4 of its capacity in 2.5 = 5/2 hours. Therefore, the drain rate is (1/4)/(5/2) = 2/20 = 1/10. Furthermore, by 12 pm (i.e., in 5 hours), the pool was drained to 1/10 x 5 = 5/10 = 1/2 of its capacity. In other words, the pool was 1/2 full at noon. With evaporation, the new rate is 3/10. So it will take (1/2) / (3/10) = 10/6 = 5/3 = 1 ⅔ = 1 hour 40 minutes more to empty the pool. In other words, the pool will be empty at 1:40 pm.

Answer: C
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let total capacity of tank be 4
so from 7 am to 9:30 am ; i.e in 150 mins tank emptied is 1/4 * 4 ; 1
i.e the rate must have been ; 1= rate * 150 ; rate = 1/150
so from 7 am to 12 pm ; i.e 5*60 ; 300 mins tank emptied must be
=> 1/150*300 ; 2 units
so now left with 2 units which would be emptied at thrice rate of 1/150 ; i.e 3 * 1/150 ; 1/50
time taken to empty 2 units at 1/50 rate ;
100 mins i.e 1 hour 40 mins ; so from 12 pm + 1 hour 40 mins ; 1:40 PM
OPTION C