1. Sorry to disturb the vast majority of this math community. I ran into a problem I do not under stand how to solve.

One power plant releases a certain amount of toxic waste into the atmosphere twice as fast as another plant. Together they release that amount in 3 hours. How long does it take the slower plant to release that amount?
I know this is probably the easiest question but for some reason I cannot see through the words to find my equation.

I attempted x+0.5x=(3)(2x)
but that is simply wrong.

Can someone please give me the correct equation?
I really appreciate it, thanks.

2.

3. These kind of problems are not easy. They never were for me anyway. The trick is to not just start writing an equation. Be very clear about what your equation states. In this case, your basic equation will be Rate * Time = Amount. In this case you will be better off to let the amount equal 1, as in 1 blob of pollution. So the first thing you will write down is

X * T = 1 where x is the slower rate and T is the time to release 1 blob of pollution at that rate

Then write down the equation for both plants combined. If this hint is not enough, come back and I will help you some more.

4. Well to answer the question originally asked, the slower plant contributes 1 hour.

The total amount being released is 9. The slower power plant coughs out 1 hours worth, so double that is 2.

1 + 2 = 3

Obviously.

So to reach 3 hours without the fast power plant it would take the slower power plant 9 hours.

Can any good mathmatician put what I've done into equation?

I don't know if what I've done here is right. I'm guessing it is because the fast plant gives 2/3 more than the slower power plant, so I just multiplied 3 by 3 to get 9.

Ridicule away...

5. If the output of A is twice that of B then A = 2B

Next A + B = C
Now a bit of substitution and away you go, I'll give you that because you got close.

6. It seems to me that either you are missing a number or two, or you are supposed to end up with two variables. Could you also explain the parts of your equation and why you constructed your equation the way you did?

7. Me? Did I do something right in my method? Thank you.

8. Holy four posts at once! I wanted to be helpful, but i got flooded. (sorry for this second meaningless post.)

9. OK, I'm directly thinking here for a change.

P = (st)+(ft)

p = product
st = slow power plant time
ft = fast power plant time

So rearranging:

(ft) = (st) - P

No, I'm sorry I don't have the maths skills to do this .

Wait, (ft)^2 = (st)^2 - (p)^2

So:

(ft)^2 = (sp/t)^2

Where ft = the fast plant time
Where sp = slow plant time, the t in the same bracket appyling to the same number.

So:

Oh I don't know . I don't want to give up I really don't :x I love maths but haven't got the skillz to pay the outstanding bills. Isn't there a way to get rid of the t's in the equation and make them one?

I know my above equations are incorrect, I was just seeing if I could trial and error and stumble upon the answer without knowing the true method. I haven't the resources to learn it .

10. Ah, Harold14370 thank you.

I created a quick table so I can understand the situation a bit better:

___| R | T | A |
P1 | r | a | 1 |
P2 | 2r | b | 1 |

My equations were:
(r)(a)=1
2(r)(b)=1
a+b=3

and I solved for a and b
a=2
b=1

So is 1 hour the correct answer?

11. No because it takes 3 hours with the fast plant. 1 is how much the slow plant gives to that 3 hours, it isn't enough to fill that amount. If the fast plant is 2 out of 3 hours, then 1 hour doesn't cut it. I'm sure the answer is 9.

12. No. Actually, svwillmer got the correct answer the first time, but was not able to explain how he arrived at it.

Your third equation is wrong because it says the time for the slow plant to emit 1 unit of pollution plus the time for the fast plant to emit 1 unit is equal to 3. But 3 is the time for both working together which has to be less than either a or b.

Replace the second equation with the rate of the 2 plants combined. The rate will be r+2r = 3r and the time will be 3 hours.

13. Originally Posted by svwillmer
Well to answer the question originally asked, the slower plant contributes 1 hour.
So am I correct?

14. Originally Posted by DivideByZero
Originally Posted by svwillmer
Well to answer the question originally asked, the slower plant contributes 1 hour.
So am I correct?
Yes in that 1 hour out of three hours is from the slow power plant. But don't forget the slow power plant on its own is the question. So without the fast power plant (2 hours) it will take more than 3 hours.

Consider you have three slow power plants. They will give 3 hours with 1 hour each. But you have three slow power plants now, so you need three slow power plants to make the total toxic amount. So if 3 slow power plants make the total toxic in 3 hours, 1 slow power plant will do it in 9.
Its this:

3 plants x 3 hours = 9
1 plant x 9 hours = 9

1 fast plant + 1 slow plant = 3 hours

2 fast plants = 4 hours.

15. Originally Posted by DivideByZero
Originally Posted by svwillmer
Well to answer the question originally asked, the slower plant contributes 1 hour.
So am I correct?
No. Solve this equation

combined rate * combined time = 1 unit of pollution
(r+2r) * 3 = 1

16. k, that wasn't so hard.

here are the new equations I used:

(r)(a)=x
(2r)(b)=x
(r+2r)(3)=x

then solved for a,
a=9

w00t!

I guess its all about the way you approach the problem.
Rate * Time = Amount
now that helped alot.

Thanks

17. Let the rate of emission of the slower plant be r units per hour.

Then the rate of emission of the other plant is 2r units per hour.

In 3 hours, the slower plant releases 3r units of waste and the other releases 6r units of waste. Let the total amount of waste released be V units.

Then 3r + 6r = V

∴ r = V/9

So the time taken for the slow plant alone to release V units of waste is V/r = V/(V/9) = 9 hours.

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