A rectangular reservoir is filled with water till one-fifth of the hei

To make things easier on ourselves, let's assign a "nice" value to the total volume of the reservoir.

Let's say the reservoir holds a TOTAL of 20 gallons

A rectangular reservoir is filled with water till one-fifth of the height of the reservoir.
1/5 of 20 = 4
So, the reservoir currently holds 4 gallons of water

If an outlet at the bottom of the reservoir is unplugged, the water in the reservoir will drain completely in 1 hour.
In other words, 4 gallons of water will drain out of the reservoir in one hour.
Rate = amount/time = (4 gallons)/(1 hour) = 4 gallons per hour
So the RATE at which water drains out of the outlet = 4 gallons per hour

If the outlet remains plugged and the inlet tap at the head of the reservoir is opened, the reservoir will fill to the brim in 2 hours
In other words, it will take 2 hours for the inlet tap to add the additional 16 gallons of water it will take to fill the reservoir.
Rate = amount/time = (16 gallons)/(2 hours) = 8 gallons per hour
So the RATE at which water flows into the reservoir = 8 gallons per hour


In how much time will the reservoir fill to the brim if the outlet is unplugged 30 minutes after the inlet tap is opened ?
If the inlet tap is opened for 30 minutes (aka 0.5 HOURS), the amount of water flowing into the reservoir = (0.5 hours)(8 gallons per hour) = 4 gallons.
So, after 30 minutes, the volume in the reservoir = 4 + 4 = 8 gallons.
So in order to FILL the reservoir to the top, we must add an additional 12 gallons of water

Once the outlet is unplugged, we have water flowing IN at a rate of 8 gallons per hour, and we have water flowing OUT at a rate of 4 gallons per hour
So the NET EFFECT = 8 gallons per hour - 4 gallons per hour = 4 gallons per hour

We need to add an additional 12 gallons of water, and we are doing so at a rate of 4 gallons per hour
Time = amount/rate
= (12 gallons)/(4 gallons per hour)
= 3 hours

IMPORTANT: Now we must add the 30 minutes that elapsed before the outlet was unplugged.
We get 3 hours + 30 minutes

Answer: B

Cheers,
Brent

Reservoir: ( \(\frac{1}{5}\)) — filled with water
—> inlet tap can fill \(\frac{4}{5}\) of the reservoir in 2 hours ( \(\frac{2}{5}\) —in one hour, \(\frac{1}{5}\) — in 30 minutes)

—> \(\frac{4}{5} —\frac{1}{5} = (\frac{2}{5} —\frac{1}{5})*t\)

\(\frac{3}{5} = \frac{1}{5}*t\)
—> t = 3 hours

The answer is A

Posted from my mobile device

Ughh, silly mistake (it's like Bunuel predicted my error and made a corresponding answer choice specifically for me!)
Good catch!! I've edited my solution.
KUDOS FOR YOU!!

Cheers,
Brent

Good question!
The question is a little bit ambiguous: " In how much time will the reservoir fill to the brim if the outlet is unplugged 30 minutes after the inlet tap is opened ?"
So, are we starting the clock AFTER we unplug the outlet valve, or when we first open the INLET TAP?

If we start the clock when we unplug the outlet valve, then the answer is 3 hours
If we start the clock when we first open the INLET TAP, then the answer is 3.5 hours

Cheers,
Brent

The problem is, in an actual test, we are not allowed to make two choices so we have to analyze the language of the question and make the right call. In this context, this question is more of an exercise in reading comprehension than a mathematical problem since the actual calculation is fairly simple. So let's see what we have:

A reservoir with an inlet tap at the head and an outlet at the bottom is filled up to one-fifth of its capacity. I think that we can safely assume that both the inlet and outlet are closed at this point. From this point on, we are presented with three different scenarios each involving a certain course of action and the result thereof:
(a) The outlet is unplugged and consequently the reservoir is completely drained in 1 hour which gives us the information required to calculate the rate/efficiency of the outlet valve.
(b) The inlet tap is opened and the reservoir fills in 2 hours which gives us the information needed to calculate the efficiency of the inlet tap.
(c) This is the tricky part. The question stem asks: "In how much time will the reservoir fill to the brim if the outlet is unplugged 30 minutes after the inlet tap is opened ?" In my opinion, the answer should be the TOTAL time (the 30 minutes when only the inlet tap is in play plus the time until the reservoir is full during which period both the inlet tap and the outlet valve are in play). There are two reasons why I have made this assumption:
(1) In all three scenarios, the actions start with the reservoir one-fifth full and both the inlet and outlet closed. So the time for the result of each action to culminate (draining in the first scenario and filling to the brim in the second and third scenarios) should be calculated from this starting point. In fact, in the given context the above question can be rephrased thus: " As in Scenario (b), the inlet tap is opened but this time the outlet valve is unplugged after 30 minutes. In this case, how much time will it take to fill up the reservoir?"
(2) Option B would have been the correct choice if the above question would have used the word "after" instead of "if": "In how much time will the the reservoir fill to the brim AFTER the outlet is unplugged 30 minutes from the time the inlet tap is opened?"

So, IMO, the answer is 3 hours 30 minutes (B)

let total reservoir be 10 ltrs
1/5 th filled = 2 ltrs
and outlet rate ; 2 ltr per hr
and intlet rate ; 10/2 ; 5 ltr per hr
net rate ; 5-2 ; 3 ltr per hr
for first 30 mins the reservoir would be filled ; 2*30/60 ; 1 ltr
now left with 9 ltrs would be filled in 9/3 ; 3 hrs
total time taken ; 3hrs and 30 mins
IMO B

rate*time=work,distance,fill

currently: 1/5 tank filled

rate_out * 1hr = 1/5, rate_out = 1/5
rate_in * 2hr = 4/5, rate_in = 2/5
rate_in * 0.5hr = 2/5*1/2 = 1/5
rate_in-rate_out * time = 1-(1/5+1/5) = 3/5
(2/5-1/5)*t=3/5
t=3/5*5/1=3 hours to fill after unplugging
3:30 hours to fill since beginning

Ans (B)

It takes 1 hour to drain 1/5 of the reservoir by the outlet tap and in 1 hour the inlet tap can fill 2/5 of the reservoir.
So when the outlet tap was opened, the reservoir already fiiled upto 2/5 . In the next one hour the filled part will be 1/5. So to fill the remaining 3/5, the required hours are 3. So, after the initial 1/5, to fill the 4/5 part it needs 3 and 30 minutes.

According to the question stem, we have the following data
Rate of Inlet/filling = 2/5 per hour (Also note, 1/5 every half hour)
Rate of Outlet/draining = 1/5 per hour
So, from the stem we see that for the 1st half hour only inlet pipe was working, So 2/5th of the total was filled, Now the outlet tank was opened,
Effective rate when Inlet and outlet Pipes are working together = 2/5-1/5 = 1/5
So, when both pipes are working together we are filling 2/5th of the tank evert hour.
We are already at 2/5th capacity and we derived that from now onwards wee gain 1/5 every hour which will take 3 hours to fill to the brim (max capacity)
So in all half hour earlier and these 3 hours together make 3:30 hours.
So to fill to the brim it takes 3 and half hours
Bunuel BrentGMATPrepNow
Please provide your inputs if you think the logic is flawed!

Filling efficiency= 50%
Emptying efficiency= 20%
So, net efficiency= 30%
So, Required time= 100/30= 3.5hrs (Approx) [Option B]
Bunuel, correct me if i am wrong..

Thanks

rajchotu02

I think you've got some fuzzy math that happened to work out to be close enough to the correct answer that you selected the right one by luck!!

The filling efficiency is not 50%. Yes, it will fill in 2 hours, but only because it starts out 20% full. If it started out 90% full and took 2 hours to fill, would you still call that 50% filling efficiency?

Let's say the thing can hold 5 gallons.
Outlet: It currently has 1 gallon and would take 1 hour to drain, so the outlet drains at a rate of 1 gallon per hour.
Inlet: It currently has 1 gallon and would take 2 hours to fill, so the inlet fills at a rate of 4 gallons per 2 hours, or 2 gallons per hour.
Net: +2 gallons per hour - 1 gallon per hour = 1 gallon per hour.

We turn on the inlet for 30 minutes. At a rate of 2 gallons per hour, 1 gallon comes in. We now have 2 gallons in and need 3 more. 30 minutes elapsed.
At a net rate of +1 gallon per hour, we can fill the remaining 3 gallons in 3 hours. 3 hours elapsed.

30 minutes + 3 hours = 3.5 hours.

Answer choice B.