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?

Name: 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?
Uploaded: 2022-02-23T17:22:57+00:00
Description: 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?

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poleglit3 · Accepted Answer

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 dxdt. Every day, the fraction 2500 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 2500(500000000−x). These are replaced. We conclude that dxdt=2500(500000000−x) To avoid typing such large numbers, let x=500000000y. Then dydt=500000000dxdt, and quickly we arrive at dydt=1250(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&#039;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 dzdt=−1250z This is the familiar differential equation of exponential decay. It has the solution z=z(0)e−t250. We have z(0)=1, and want to find the time t such that z(t)=0.10. So we solve the equation e−t250=0.1. Take logarithms to the base e. We get −t250=ln⁡(0.1). I prefer to note that ln⁡(110)=−ln⁡(10), which tells us that t=250ln⁡(10). Finally, calculate. We get something like t≈575.

otoplilp1 · Accepted Answer

Let&#039;s say that on say n the proportion of new bills in the population is 0≤P(n)≤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. ThereforeP(n+1)=P(n)+f(1−P(n))which can be rearranged toP(n+1)=(1−f)P(n)+fIf 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 getc=(1−f)c+ffrom which you can deduce c=1. We know that initially there are no new bills in circulation (i.e. P(0)=0) so we have0=1+A(1−f)0=1+Aand hence A=−1, so the general solution isP(n)=1−(1−f)nThe question wants to know when this is equal to 90%. We have f=4⋅1071010=0.004 and hence 1−f=0.996, so0.9=1−0.996nand hence n=log⁡(0.1)log⁡(0.996)=574.5 days.

Suppose that Canada has a total of $10 billion in $20 bills in circulation, and each day $40 million

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