Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that Canada has a total of \$10 billion in \$20 bills in circulation, and each day \$40 million of these \$20 bills passes through one bank or another. A new harder-to-forge version of the \$20 bill is developed, and the banks replace the old bills with new ones whenever they can. How long does it take for the new bills to reach 90% of the total number of $20 bills in circulation?

share|improve this question
add comment

2 Answers

Let's say that on say $n$ the proportion of new bills in the population is $0\leq P(n)\leq 1$. Then each day a fraction $f$ of all the bills go through the banks. $1-P(n)$ of those will be old bills, which will immediately be replaced with new. Therefore

$$P(n+1) = P(n) + f(1-P(n))$$

which can be rearranged to

$$P(n+1) = (1-f) P(n) + f$$

If we solve the homogeneous part, we find that $P(n)=A(1-f)^n$ for some constant $A$. Now if we guess that $P(n)=c$ solves the whole equation, we get

$$c = (1-f)c + f$$

from which you can deduce $c=1$. We know that initially there are no new bills in circulation (i.e. $P(0)=0$) so we have

$0 = 1 + A(1-f)^0 = 1 + A$

and hence $A = -1$, so the general solution is

$$P(n) = 1 - (1-f)^n$$

The question wants to know when this is equal to 90%. We have $f=4\cdot 10^7 / 10^{10} = 0.004$ and hence $1-f=0.996$, so

$$0.9 = 1 - 0.996^n$$

and hence $n=\log(0.1)/\log(0.996) = 574.5$ days.

share|improve this answer
add comment

Since the tag is differential equations, we use a continuous model. Ten billion in $20$ dollar bills is $500$ million bills. Of these, $2$ million pass daily through a bank.

Let $x=x(t)$ be the number of new bills in circulation after time $t$. We will find an expression for $\dfrac{dx}{dt}$.

Every day, the fraction $\dfrac{2}{500}$ of bills in circulation passes though a bank. If there are $x$ new bills in circulation, then the number of old bills that pass through the bank is $$\frac{2}{500}(500000000 -x).$$ These are replaced. We conclude that $$\frac{dx}{dt}=\frac{2}{500}(500000000-x).$$ To avoid typing such large numbers, let $x=500000000y$. Then $\dfrac{dy}{dt}=500000000\dfrac{dx}{dt}$, and quickly we arrive at $$\frac{dy}{dt}=\frac{1}{250}(1-y).$$ Initially, $y=0$. We want to find out when $y$ reaches $0.9$. Not that the above differential equation looks very much like the one we get in Newton's Law of Cooling.

Make the change of variable $z=1-y$. Then $z$ is the proportion of old bills after time $t$. We get $$\frac{dz}{dt}=-\frac{1}{250}z.$$ This is the familiar differential equation of exponential decay. It has the solution $$z=z(0)e^{-t/250}.$$ We have $z(0)=1$, and want to find the time $t$ such that $z(t)=0.10$. So we solve the equation $$e^{-t/250}=0.1.$$ Take logarithms to the base $e$. We get $-\dfrac{t}{250}=\ln(0.1)$. I prefer to note that $\ln(1/10)=-\ln(10)$, which tells us that $$t=250\ln(10).$$ Finally, calculate. We get something like $t\approx 575$.

Remark: It is important to note that we were working with a model of reality. That model cannot be fully accurate. The round numbers supplied are obviously only approximate, and the total money supply may grow. And old $20$ dollar bills may be hidden in socks and circulate only every few years. And there are light days and heavy days at banks. And a new bill put into circulation today unlikely to return to the bank tomorrow. And we are using continuous modelling for a discrete situation. So it would be unreasonable to treat the answer of $575$ as anything more than a ballpark figure.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.