# T&W q4

#### Saumyojit

##### Senior Member
A tank, which usually takes 5 hours to be filled, takes 6 hours to fill because of a leak. Find the time in which the leak can empty the tank if it is filled and the inlet pipe is shut.

(Volume filled per hour - volume leaked per hour)* 6 = volume of cistern.

From here leaked per hour I will find out then ,
leaked per hour * total time taken for empty = volume of cistern

Volume of cistern= any volume assume then 30 hours .

I have assumed that inlet is much faster than leaked but can leaked be faster than inlet ?
I think if i assume so here -> ( (Volume filled per hour - volume leaked per hour)* 6 = volume of cistern.)
value will be negative thats why inlet is much faster than leaked

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#### Dr.Peterson

##### Elite Member
(Volume filled per hour - volume leaked per hour)* 6 = volume of cistern.

From here leaked per hour I will find out then ,
leaked per hour * total time taken for empty = volume of cistern

Volume of cistern= any volume assume then 30 hours .
Since there is no mention of volume, I would take the volume of the tank as 1; that is, measure volume in tankfuls. There is no data here from which to determine the actual volume.

So the rate of inflow is 1/5 tank per hour, and so on.
I have assumed that inlet is much faster than leaked but can leaked be faster than inlet ?
I think if i assume so here -> ( (Volume filled per hour - volume leaked per hour)* 6 = volume of cistern.)
value will be negative thats why inlet is much faster than leaked
If the leak were faster than inflow, then it would never fill! But you don't need to think about that, nor about whether inflow is "much" faster than the leak. Just solve the problem!

#### Jomo

##### Elite Member
I would assume that the tank holds 5 gallons, so 1 gallon per hour is entering the tank. After 6 hours the tank has 5 gallons in it which means 1 gallon was lost--every 6 hours. Continue....