# If both are used together, how long will it take to fill the empty tanker of an oil tanker that is filled in 4 hours by one pipe; another pipe to empty the tanker takes 6 hours.

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Let the volume of the tank is `V m^3` .

Then;

The rate of filling (input)= V/4 `m^3/h`

The rate of emptying (out put) = `V/6 m^3/h`

So;

input rate = V/4 `m^3/h`

Output rate = V/6 `m^3/h`

When both are working simultaniously;

Net input rate = input rate - output rate

= V/4-V/6 `m^3/h`

= V/12 `m^3/h`

So the rate of filling the tank is `V/12 m^3/h` .

So the time to fill = Volume /rate of filling

= `V/(V/12)`

= 12 h

**So it ill take 12 hours to fill the tank.**

An oil tanker can be filled by pipe A in 4 hours. The pipe fills the tanker at the rate of 1/4 per hour. The tanker can be emptied by pipe B in 6 hours. It is emptied at the rate of 1/6 per hour.

If the two pipes are used together, the rate at which the tanker is filled is 1/4 - 1/6 = 1/12

The time taken to fill an empty tanker at this rate is 12 hours.

**It would take 12 hours to fill an empty tanker if both the pipes are used together.**

Since In 4 hrs. a pipe (A) fills a tan

Therefore In 1 hr. it will fill 1/4 of a tank.

In 6 hrs. another pipe(let B) empties the tank

Therefore in 1hr. the pipe(B) will empty 1/6 of a tank

If both the pipes(A & B) are on then ,

in 1 hr {(1/4) - (1/6) } of the tank will be filled

1/4 -1/6 = (3 - 2)/12 = 1/12

Therefore in 1 hr. both pipes will fill 1/12 of the tank.

Since 1/12 of the tank is filled in 1 hr. when both pipes are on.

Therefore 1 tank will bw filled in 12 hrs.

**It will take 12 hrs. to fill empty tanker if both pipes are used <- Answer **