Number 35 is where I need help.
2 answers:
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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
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