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# A tank is normally filled in 6 hrs but takes 4 hrs more

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Joined: 02 Aug 2009
Posts: 7946
A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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16 Apr 2016, 09:40
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A tank is normally filled in 6 hrs but takes 4 hrs more to fill due to a leakage in bottom. In how much time will the tank get empty if it is $$\frac{2}{3}$$rd full?
(A) 5
(B) 10
(C) 15
(D) 16
(E) 20

OA in 2 days
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Re: A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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17 Apr 2016, 07:52
2
this is a good one. Actually there are two things going on simultaneously. One fills and one takes out, the net results of this process is a rate that is 1/10
1/10 = 1/6-1/x
x - is the time is takes the tank to get empty.
solving this equation we get x = 15. so the rate is 1/15. time = work/rate. (2/3)/(1/15) = 10 hours.
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A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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17 Apr 2016, 22:47
I think I get this one if we view them as flow rates. Using the MGMAT "Choose Smart Numbers" strategy you can perform the following:

If in 6 hours a tank is full, normally, then assume the tank has 60 gallons (multiple of 6). This gives:

60gal/6hrs = 10 gal/hr flow rate

Now calculate the new flow rate of the tank after it has sprung a leak. If the new time to fill the tank is 6+4 hours then t=10 and the volume remains the same:

60gal/10hrs = 6 gal/hr flow rate.

Taking the difference of flow rates you get: 10gph - 6gph = 4gph which allows you to say that the tank is leaking at a rate of 4 gal/hr.

If the tank is 2/3 full then:

(2/3)*60 = 40 gallons which defines the new volume.

To eliminate gallons and solve for the time is a simple calculation using the leaking flow rate of 4gph and the new volume:

40 gallons / 4 gal/hr =
10 hours
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Re: A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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18 Apr 2016, 01:03
Plugging in is the best way to solve the problem:

Tank consists of 6 liters
If it takes 6 hours to fill this tank, it takes 1 liter per hour
However, if it takes four more hours than original, it means that the tank is now filled at 6/(6+4)=0.6 liters per hour
1 liter per hour - 0.6 liters per hour means you are losing 0.4 liters per hour

If the tank is now 2/3 full, you just multiply the 6 liter tank by (2/3), giving you a total of 4 liters.

If a four liter tank loses water at a rate of 0.4 liters per hour, it will be empty in ten hours.
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Re: A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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18 Apr 2016, 09:23
time to fill without leak*(time to fill with leak/time to empty)*(2/3)rd of leak

=6*(10/4)*(2/3)
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A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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18 Apr 2016, 11:13
2
chetan2u wrote:
A tank is normally filled in 6 hrs but takes 4 hrs more to fill due to a leakage in bottom. In how much time will the tank get empty if it is $$\frac{2}{3}$$rd full?
(A) 5
(B) 10
(C) 15
(D) 16
(E) 20

OA in 2 days

Let the capacity of the tank be 60

Efficiency of the pipe filling the tank {without leakage in the tank} is 10 units/hour {60/6}

Efficiency of the pipe filling the tank {with leakage in the tank} is 6 units/hour {60/4}

The difference in efficiency of 4 units/hour is due to the leakage in the tank = Efficiency of the outlet/leakage

Given the tank is 2/3rd full or 2/3*60 => 40 units

So, the time required for the outlet pipe to empty will be 40/4 = 10 Hours.

Actually this problem will seem easier if one consider 2 pipes , one filling the tank and another pipe empting the tank.

chetan2u : good question sir !!! Regards
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A tank is normally filled in 6 hrs but takes 4 hrs more  [#permalink]

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10 Feb 2019, 15:18
chetan2u wrote:
A tank is normally filled in 6 hrs but takes 4 hrs more to fill due to a leakage in bottom. In how much time will the tank get empty if it is $$\frac{2}{3}$$rd full?
(A) 5
(B) 10
(C) 15
(D) 16
(E) 20

OA in 2 days

if one tank is filled in 6 hours,
in 10 hours 1 2/3 tanks should be filled
thus the excess 2/3 tank leaked out in 10 hours.
B
A tank is normally filled in 6 hrs but takes 4 hrs more   [#permalink] 10 Feb 2019, 15:18
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