Question 69382

Let x= amount of toxic waste released 
Let t= the time it takes Plant II to release x amount  
Then (t/2)= the time it takes Plant I to release x amount 

So Plant II releases toxic waste at the rate of x/t cu units per hour

And Plant I releases toxic waste at the rate of x/(t/2) or 2x/t cu units per hour

Now we are told that in 3 hours, Plant I and Plant II together can release the x amount of toxic waste

Sooooo
(x/t+2x/t)*3=x first, lets get rid of parens
3x/t+6x/t=x
9x/t=x   divide both sides by x

9/t=1  multiply both sides by t

t=9 hours  -------------------Time it takes Plant II (the slower Plant) to release x amount of toxic waste.

(t/2)=9/2=4.5 hours------------Time it takes Plant I to release x amount of waste

CK
Plant I releases at the rate of x/9

Plant II releases at the rate of x/4.5

Together they release at the rate of x/9+x/4.5.  This equals:

x/9+x/4.5=x/9+2x/9=3x/9=x/3 

Or in 3 hours they together have released the certain amount.


Hope this helps---ptaylor