Question

PLEASE HELP AND SHOW WORK!!

Answer 1

Answer:

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

Step-by-step explanation: