Question

PLEASE HELP AND SHOW WORK!!

A water basin starts with 2 liters of water and fills up at rate of 0.4 liters per second.

Write and solve an equation for the number of seconds it will take for both water basins to have an equal amount of water.

Answer 1

Pipe 1 alone fills a tank in 10 min

Pipe 2 alone fills the tank in 20 min

If both are turned on together, how long will it take to fill the tank???

Pipe 1 fills at the RATE of 1/10th tank per min

Pipe 2 fills at the RATE of 1/20th tank per min

Pipe 1 + Pipe 2 fills at the rate of 1/10+1/20 or 3/20th tank per min

Let x= time to fill the tank

(3/20)(x)=1 Where 1 denotes a full tank (Multiply both sides by 20)

3x=20

x=6 2/3 min----------both working together

NOW YOUR PROBLEM

Let x= amount of time for smaller pipe to fill the tank

Then (x-5)=amount of time for larger pipe to fill the tank

NOW HERE'S WHERE THE RECIPROCAL COMES IN:

The smaller pipe fills the tank at the RATE of 1/x cu units per min

The larger tank fills the tank at the RATE of 1/(x-5) cu units per min

Together, they fill the tank at the rate of 1/x+(1/(x-5)) cu units per min

But now we are told that together the fill the tank in 11 1/9 minutes. so our equation to solve is:

(1/x+(1/(x-5))(11 1/9)=1 where 1 denotes a full tank. In fact, in each of the above reciprocals, the 1 denotes a full tank.

So, dividing both sides by 11 1/9, we get:

1/x+(1/(x-5)=1/(11 1/9)-------------- which is your equation

I assume that you have no problems solviing this eq.

Hope this helps----ptaylor

Explanation: