Two pipes can separately fill a tank in 20 hours and 30 hours

Let the tank = 180 gallons.

Since the first pipe takes 20 hours to fill the 180-gallon tank, the rate for the first pipe\(= \frac{work}{time}= \frac{180}{20} = 9\) gallons per hour.
Since the second pipe takes 30 hours to fill the 180-gallon tank, the rate for the second pipe \(= \frac{work}{time} = \frac{180}{30} = 6\) gallons per hour.
Combined rate for the two pipes = 9+6 = 15 gallons per hour.

\(\frac{1}{3}\) of the 180-gallon tank \(= \frac{1}{3}*180 = 60\) gallons.
Since the combined rate for the two pipes = 15 gallons per hour, the time for the two pipes to pump in 60 gallons \(= \frac{work}{rate} = \frac{60}{15} = 4\) hours.

Remaining volume = 180-60 = 120 gallons.
Since the leak reduces the rate by 1/3, the resulting rate \(= \frac{2}{3}*15 = 10\) gallons per hour.
Since the new rate = 10 gallons per hour, the time for the remaining 120 gallons \(= \frac{work}{rate} = \frac{120}{10} = 12\) hours.

Total time to fill the tank = (4 hours for the first 1/3 of the tank) + (12 hours for the remaining volume) = 4+12 = 16 hours.


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Let's assume the size of the tank to be LCM(20,30) = 60 units.
The individual rates of the two pipes filling are 3 units & 2 units respectively.

Together the tanks fill 5 units in an hour. When the tank is \(\frac{1}{3}\)rd full(20 units have been filled). The leak develops
causing the tank to fill at \(\frac{2}{3}\)rds of its usual rate. The time taken to fill the first 20 units is \(\frac{20}{5} = 4\) hours.

For the remaining 40 units, the tanks will fill \(\frac{10}{3}(5*\frac{2}{3})\) units in an hour.
Because of the leak. the pipe takes \(\frac{40}{\frac{10}{3}} = \frac{120}{10} = 12\) hours to fill the remaining tank.

Therefore, the total time taken for the two pipes to fill the tank is 4 + 12 = 16 hours(Option C)

Pipe A - can fill the tank in 20 hrs. Hence, in 1 hr it can fill 1/20th of the tank.
Pipe B - can fill the tank in 30 hrs. Hence, in 1 hr it can fill 1/30th of the tank.

Therefore, when both pipes are opened, (1/20 + 1/30 =) 1/12th of the tank is filled in an hr. It'll take both the pipes 12 hrs to fill the tank completely.

When 1/3rd of the tank is filled, both the pipes would have been opened for (1/3 = 4/12 = 4*(1/12)) 4 hrs.
Now, 2/3rd (= 8/12) of the tank remains to be filled.

However, at this stage due to the leak in the tank, 1/3rd of the water supplied by the pipes leaks.
So, now only (2/3 * 1/12 = ) 1/18th of the tank is filled in an hour instead of 1/12th earlier.
At this rate it would have taken 18 hrs to fill the tank if it were completely empty.
However, since only 8/12th of the tank needs to be filled, it'd take another (8/12 * 18 =) 12 hrs to completely fill the tank.

Therefore, the tank would be completely filled in (4 + 12 =) 16 hrs.

Dear GMATGuruNY

I do not understand the reasoning behind the leak in the tank and the be 2/3 of the combines rate of both pumps?

does not the leak produce 1/3 of the water supplied to lost, lowering the water to 40 (1/3 * 60 =20 so remaining water to be 40)?

Can you please elaborate more? I'm confused.

Thanks

Is there any specic reason to take tak capacity as 180 gallons?(LCM formula?)

Prompt:
When the tank is 1/3 full, a leak develops through which 1/3 of the water supplied by both the pipes leaks out.
This wording is intended to convey the following:
Once the tank is 1/3 full, a leak develops.
This leak decreases the input rate, as follows:
For every 3 gallons supplied by the two pipes, the leak removes 1/3 of the 3 gallons.
In other words, the leak removes 1 gallon for every 3 gallons supplied by the two pipes, reducing the input rate by 1/3.
The resulting input rate is thus 2/3 of the original input rate.
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The capacity must be divided by the two given times (20 hours and 30 hours) and then by 3 (since the rate is reduced by 1/3).
To get a good value for the capacity, I multiplied the LCM of 20 and 30 by 3:
LCM (20,30) * 3 = 60*3 = 180
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Yes, this is the only reason this value has been chosen. You could also take 60 (LCM of 20 & 30), but that would become a fraction after initial division. So, choose LCM of 20 and 30 and multiply it by 3.

Means 60 * 3 = 180

This way you can get rid of fractions and can increase you calculation speed.
_________________

Work rate formula: AB/(A+B)=T(for both to work together)
20x30/(20+30)=12
While working simultaneously, it takes the two pipes 12 hours to fill the tank.
Given 1/3 of the water filled by the pipes are lost, meaning 12x(1/3) of time is wasted, =4 hours.
So, it takes an additional 4 hours to fill the tank: 12+4=16hours.

Hi DisciplinedPrep

I hope you are doing well in GMAT prep as your nickname says.

What is the source of the question? it is really good.

let total capacity of tank = 60 ltrs
so rate of A = 60/20 ; 3 and rate of B = 60/3 ; 2
together ; 5
given 1/3 of tank is filled ; i.e 20 ltrs ; so time taken would have been 20/5 ; 4 hrs
and later leak happens and the rate decreases by 1/3 or original value or say the rate becomes 2/3 so new combined flow rate 5*2/3 ; 10/3
so for 40 ltrs balance vol the time it will take
40/10 /3 ; 12 hrs
total time taken 12 + 4 ; 16 hrs
IMO C

First post gmatclub. I guess this is a positive sign. ^^

Well, I did it in this way:

Rate of pipe A: 1/20
Rate of pipe B: 1/30

Both pipes are opened to fill the tank; so combined rate is 5/60 or 1/12

Combined rate is 1/12 so tank will be full after 12h.

When the tank is 1/3 full a leak develops in the tank through which 1/3 of the water leak out
Translated: When tank is at 4/12 all the water leaks out. (This is after 4 Hours)

So there are 4 extra hours to add because of the leak.

12h+4h = 16h

Please correct me if I made a mistake.

DM Me to know the simple method.

Let us assume capacity of the tank is 600 Liters.
Efficiency of pipe 1 - 30 L/H
Efficiency of Pipe 2 - 20 L/H

Pipe 1 + Pipe 2 = 50 L/H which means Tank will take 12 H to fill it if there is no leak.

now Opening C will leak 200 Liters from tank which means another 200/50 i.e 4 Hrs to fill the void

hence total 12 + 4 = 16 HRS.

Hope it helps!!

Vikas Singla

I have also done this in the same way as you did but don't know everyone started calculating the rate of water loss due to leak in the tank.

I assumed the leak got sealed after the tank empty due to leak.

combined rate=1/20+1/30=1/12
12/3=4 hours to fill 1/3 of tank before leak
let t=time to fill tank after leak
t*(1/12)(2/3)=2/3→
t=12 hours+4 original hours=16 total hours to fill tank
C

A: 1/20. B: 1/30. A+B: 1/12 (they are filling the tank in 12 hours).

Suppose the tank is 180 liters, then they fill 15 liters each hour.

Now, they worked (12*1/3) 4 hours and filled (180*1/3) 60 liters - we are left with (180-60) 120 liters to be filled.

From now on, their rate changes to (10*2/3) 10 liters per hour to fill 120 liters. It will take A&B (120/10) 12 hours to fill the rest of the tank.

C: 12+4 = 16.

Break into into parts
A and B work to fill the tank 1/3
A and B work together against the leak to fill 2/3

A and B together
1/3 = (1/20+1/30)t
t=4

AB against leak
2/3 = (1/20+1/30)t - 1/3(1/20+1/30)t
t=12

12+4=16

Hard to understand the wording at first, but basically what happens is

-2 pipes fill up 1/3rd of the tank

-then that water is lost

-then the 2 pipes start from scratch and fill up the tank


So, in total, they are filing up (1 full tank) + (1/3rd of a tank redo) = 4/3 of a tank

Combined rate = (1/20) + (1/30) = 1 / 12 —- (tank) / (hour)


Time taken = (Work to be done) / (Rate of Work)

(4/3) / (1/12) =

(12 * 4) / (3) =

16 hours

Posted from my mobile device

Hard to understand the wording at first, but basically what happens is

-2 pipes fill up 1/3rd of the tank

-then that water is lost

-then the 2 pipes start from scratch and fill up the tank


So, in total, they are filing up (1 full tank) + (1/3rd of a tank redo) = 4/3 of a tank

Combined rate = (1/20) + (1/30) = 1 / 12 —- (tank) / (hour)


Time taken = (Work to be done) / (Rate of Work)

(4/3) / (1/12) =

(12 * 4) / (3) =

16 hours

Posted from my mobile device