A hotel began draining its swimming pool at a constant rate at 7am. Be

let the quantity of water in the swimming pool be x
so at 7 am it is x and at 9:30 am it is 3x/4 Therefore constant draining rate = 0.25x/150mins

At 12 the pool will be half full i.e. x/2 (since x/4 will be drained from 9:30 to 12pm)
Also at 12pm the rate gets tripled, new draining rate = (3*0.25x)/150 = 0.75x/150

if 3/4 of the pool is emptied in 150 mins then to empty the remaining half of the pool i.e. x/2 will take t mins. So

0.75x/100 = 0.5x/ t
t= (0.5x*150)/0.75x = 100 mins
adding 100 mins to 12pm will result in 1:40 pm

Hence Option C is correct.
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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.

Let the capacity be 1000. According to the question, 1/4th of the water i.e. 250 got drained in 150 minutes
rate =250/150 = 5/3

total time the swimming pool got drained with this rate from 7 am to 12 pm = 5*60 = 300 minutes

total water drained = 5/3*300 = 500
left water = 1000 - 500 = 500
as the rate tripled after 12pm, new rate will be 5/3 *3 = 5

time taken to drain rest of the water with this rate = 500/5 = 100 minutes = 1 hr 40 minutes.

hence the swimming pool will get evacuated by 1:40 pm

Option C

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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

Info that we have
> they start at 7
> at 9:30 3/4 is left
>> hence 1/4 is drained
>> 2 hours 3 mins or 2.5 of an hour = 1/4
>> hence in 5 hours it will drain 1/2
>> hence in 10 hours it will drain 1
>> hence the rate of drain = 1/10


Solve
7 to 12 - First 5 hrs @ 1/10 half the pool or 1/2 is drained
12 onwards - The rate triples hence 3*(1/10)

Total drain is
> 5 hrs * @ (1/10) + x hrs * @ (1*(3 increased speed)/10) = 1
> (5/10) + (3x/10) = 1
> 10* (5/10) + 10*(3x/10) = 10*1
> 5+ 3x = 10
> 3x = 5
> x = 5/3 *60
> x = 5*20
> x = 100
> x = 1 hour 40 mins

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

Given: A hotel began draining its swimming pool at a constant rate at 7am. Beginning at 12pm, the rate tripled because of evaporation.

Asked: If the pool began completely full and was 3/4 full at 9:30am, at what time was the pool completely empty?

1/4 of pool was empty in 2.5 hrs
Normal rate of draining = x/4*2.5 = x/10 litres/hour; where x is capacity of the pool in litres.
In 5 hours, pool drained = (x/10) * 5 = x/2 litres

Remaining x/2 pool will be drained in = (x/2) /(3x/10) = 5/3 = 1 2/3 hours = 1 hour 40 mins

Pool will be empty at = 12pm + 1 hr 40 min = 1:40pm

IMO C

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

I solved the problem without calculating a specific rate:

7:00 - 9:30 (2.5h): 1/4 drained
9:30 - 12 (2.5h): 1/4 drained (constant rate)
Therefore: 1/2 drained in 5h

at 12: triple the rate: means that:
1/2 drained in 5/3 hours
5/3 hours equals 1h40

add the time to 12pm: drained at 1:40pm

Answer C
solved in less than 1min with this approach