# Loan Repayment Question From Pre Calculus Test (4 questions witihn) [closed]

Edit: A farmer borrows $80,000 to purchase new machinery. The interest is calculated monthly at the rate of 2% per month, and is compounded each month. The farmer intends to pay the loan with interest in two equal installments of$M at the end of the first and second years.

i) How much does the farmer owe at the end of the first month?

ii) Write an expression involving M for the total amount owed by the farmer after 12 months, just after the first installment of $M has been paid. iii)Find an expression for the amount owed at the end of the second year and deduce that:$M = \frac{80000(1.02)^{24}}{(1.02)^{12} + 1}$iv) What is the total interest over the two year period? Any help will be appreciated, but I was particularly stuck with (ii) and (iii) ## closed as off-topic by John Doe, Davide Giraudo, John B, Namaste, JMPJun 4 '17 at 19:11 This question appears to be off-topic. The users who voted to close gave this specific reason: • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Doe, Davide Giraudo, John B, Namaste, JMP If this question can be reworded to fit the rules in the help center, please edit the question. ## 3 Answers The original loan was$\$80000$, and you probably figured out that after a month the farmers owes $\$80000\times 1.02$. After 12 month, the total owed is$ \$80000\times 1.02^{12}$. At this point the farmers pays $M$, so the answer to (ii) is $\$80000\times 1.02^{12}-M.$At the start of the second year, the amount owed is the answer to part (ii). With interest, at the end of the second year, the amount is$ (\$80000\times 1.02^{12}-M)\times 1.02^{12}$. Remember that you were told that this is the same as $M$, since you have two equal payments. Therefore $$(80000\times 1.02^{12}-M)\times 1.02^{12}=M$$

Expanding the parenthesis, and moving the terms with $M$ on one side will yield the answer to part (iii)

Recall that the formula for compound interest is $A = P(1+i)^n$,

where:

• $A$ is the final amount
• $P$ is the initial amount
• $i$ is the interest rate (in decimals)
• $n$ is the time that has elapsed.

For question (I):

Since the initial amount is $80 000$,

$A = 80000(1+0.02)^1$

$A = 81,600$

For question (II):

$A = 80000(1+0.02)^{12} - m$

$A = 101,459 - m$

Using this information, can you see what (III) is asking?

Let $a_n$ be the money owed in the $n^{th}$ month and let $r = 1.02$ be the rate at which the money owed increases.

Notice that $$a_n = ra_{n-1} \space\text{ and } \space a_0 = \80,000 \space \text{ for } \space 0\le n \le 12$$ We can find an explicit formula from the recursion by noting the pattern (and proving it with induction if you feel the need) to get

$$a_n = a_0 r^n \space \text{ for } \space 0 \le n \le 12$$ After a year, then, $$a_{12} = a_0 r^{12} \implies a_{12} = a_0 r^{12} - M \space \text{ after paying the annual installment }$$

Now define $b_n = b_{n-1}r$ with $b_0 = a_0 r^{12} - M$

After another year and payment, we get $$b_{12} = b_0r^{12}- M = (a_0 r^{12} - M)r^{12} - M = 0$$

Solve for $M$ to get $$M = \frac {a_0r^{24}}{r^{12} + 1}$$

• Very complicated way of looking at it, but it works. Maybe the concepts aren't so complicated, but the way you've written it looks foreign to me. – user432114 Jun 4 '17 at 2:05
• @Dodsy: That's a fair perspective. Personally, I will do any problem using sequences that can be done that way. It typically is the most intuitive and comfortable approach for me. I also prefer to derive formulas from scratch rather than use preexisting ones in many cases (at least when first learning it). I despise how mathematics is handled in grade school, where they often just throw formulas at you unjustifiably. – infinitylord Jun 4 '17 at 2:15
• I think your answer is more than adequate, and agree with your comment fully and completely! – user432114 Jun 4 '17 at 2:24