Math story Problem answer ASAP!?

Question

A toxic subatance deteriorates at a rate such that one-third of it is converted to a non-toxic product every 10 years. How long does it take for less than 10kg of the material to remain from and original amount of 100kg?

Answer 1

Socialist had the right idea, but I think he missed something.

The equation for exponential decay is indeed:

A(t) = A(0)e^(kt)

Where A(0) is the initial amount.

But the problem says that after 10 years, one third of the substance is converted. So actually after 10 years, you'd still have two thirds of it left:

A(10) = 2/3(100kg) = A(100kg)e^(10k)

Divide both sides by 100kg:

2/3 = e^(10k)

Take ln of both sides:

ln 2/3 = 10k
k = 1/10 ln 2/3

Now, you need to solve for t when A(t) = 10kg

A(t) = 10kg = 100kg e^(1/10 ln 2/3 t)
10kg/100kg = 0.1 = e^(1/10 ln 2/3 t)

Take ln of both sides:

ln 0.1 = 1/10 ln 2/3 t

Solve for t:

t = 10 ln 0.1/ ln(2/3) = about 56.789 years.

Answer 2

Ok, let's see...this problem looks like a series one:

t(0) = 100 kg Original Toxic
t(1) = 1/3 (100) = 33.33 Non-toxic 66.67 Toxic
t(2) = 1/3 (66.67) = 22.22 Non Toxic 66.67-22.22 = 44.44
t(3) = 1/3 (44.45) = 14.81 Non Toxic 44.45-14.81 = 29.64
t(4) = 1/3 (29.64) = 9.88 Non Toxic 29.64 - 9.88 = 19.76
t(5) = 1/3 (19.76) = 6.58 Non Toxic 19.76-6.58 = 13.18
t(6) = 1/3 (13.18) = 4.39 Non Toxic 13.18 - 4.39 = 8.79

So, it took 6 periods, or 60 years for less than 10kg of toxic materiel to remain as part of the 100kg toxic sample.

You can try the series representation as a solution of a differential equation...but I am sure the above may suffice. Good luck.

Answer 3

If their sum is ninety 4 and one extensive style is n, then the different is ninety 4 - n. So call n the better one. Translate the words into an equation. "is" potential =, 5 decrease than potential subtract 5 from. So n = 8(ninety 4 - n) - 5 n = 752 - 8n - 5 n = 747 - 8n 9n = 747 n = 747 ÷ 9 n = 80 3 so use that to respond to the question.

Answer 4

This link has a web caclulator for exponential growth/decay.
Use - (minus) 33.33333333333333 for percentage as you want decay.
The answer will need to be multiplied by ten as you need a step decay period of 10 units (years) and not 1 as default in the calculator.

I make the answer 56.78873587267572 years for resulting 10KG.

http://library.thinkquest.org/11771/english/hi/math/calcs/growdeca.html

David

Answer 5

this is almost the same as a half-life problem, except they gave you a 1/3-life problem.

y=Ae^(kt)
and you know 1/3 = e^(10k)
solve for k

now A = 100 and you already found k. Determine what t should be in order for there to be 10 kgs left.

Answer 6

10 years 1/3 deterioration leaves 66.7kg
10 more years leaves 44.5
10 more leaves 29.7
10 more leaves 19.8
10 more leaves 13.2
10 more leaves 8.8 - so you need to figure where it falls below 10

somewhere between 50-60