What Your Bank Never Wants You To Find Out About Credit Cards

Algebra Level 3

Gullible Gregory got himself a credit card from the TwentyBucks Corporation. The advertisement for the card touted the low monthly minimal payment, which is $\$ 20$ if your outstanding balance is under $1000. As a further promotion, the interest rate was a low $2\%$ per month (compounded monthly), if you use an automatic minimal payment plan.

Gregory used the card to buy himself a 40 inch HD TV which costs $\$ 800$ , and signed up for the monthly automatic minimal payment plan. How many months will it take for him to pay for the TV?

80--90 months 40--50 months 60--70 months 100--110 months

1 solution

Aditya Joshi
Feb 21, 2014

Assuming that the amount first gets compounded and he pays thereafter,

Let Gullible Gregory owe the bank $f(n)$ at month $n$ .

We know that $f(1) = 800$

We can formulate the recurrence for the amount owed on the $n^{\text{th}}$ month as follows

$f(n) = f(n-1) (1 + 0.02) - 20$

This is explained as follows. Say in the first month he owes the bank $\$800$ . In the second month, he will first have to pay $2\%$ interest, therefore the amount that he owes the bank will become $800(1.02) = 816$ . Then he pays $\$ 20$ , thus, at the end of the first month, he owes the bank $816 - 20 = \$796$ .

At the beginning of the third month, the bank charges interest on the amount due, which is $796$ . The amount owed becomes $796(1.02)$ and he pays $20$ , thus he owes the bank $796(1.02) - 20$ at the end of the third month. This fits neatly in our equation.

Thus, again, $f(n) = f(n-1) (1 + 0.02) - 20$ and $f(1) = 800$

This is a recurrence and we can solve this. But I used Wolfram alpha because it looks quite complex. You can use the following code in WolframAlpha

RSolve[{a[n] = a[n-1]*((1.02)) - 20, a[1] == 800}, a, n]

We get a closed form as follows

a(n) = 1.96078^(-n) 5.^(-2. n) (1000. 1.96078^n 5.^(2. n)+803.922 2.^n 5.^(2. n)-1000. 50.^n)

which is quite terrifying to be honest. We want to find out when $a(n) = 0$ , in other words, when he owes the bank nothing.

We equate $a(n) = 0$ , which gives us $n = 82.2649$ . This means he takes $\approx 83$ months to repay the bank, and thus our answer is $\boxed{\text{80-90 months.}}$

Yes, the code works. One can also use the formula for the sum of a geometric progression..

Alexander Borisov - 7 years, 3 months ago

800 is the present value of all the equated monthly instalments

Therefore 800 = 20 + 20 / 1.02 + 20/(1.02)^2 +......n months

which is a GP a*(r^n -1) / (r-1) with r as 1/1.02, a as 20 which we have to equate to 800 and then solve for n

Sundar R - 7 years, 3 months ago

Once you have the recursive formula for the sequence you could also implement a while loop and track the number of iterations it takes for the owed amount to reach zero. In MATLAB:

month = 1;
x = 800;

while x > 0

        x = x*1.02 - 20;
        month = month + 1;
end

month

And this gets you 83.

Dale Scutti - 7 years, 3 months ago

Which is more than twice the retail price!

MJ Ewing - 7 years, 2 months ago

What Your Bank Never Wants You To Find Out About Credit Cards

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