Identify the problem type - compound interest problem.

Setup (Write what you know, and the info the problem gives.)

Known - compound interest is A = P(1+r)^t where r is the interest rate, t is the time, P is principal, and A is the accumulated amount at the end of the compounding period.

Given - this problem adds another element, some equal monthly payments. Since we don't know what those payments are, call them x. Each month, subtract x from the accumulated amount each month.

Month 1: Louie pays x: A = 1000(1+.1)^1 - x = 1100-x
Month 2: The principal amount this month is the 1100-x: A = (1100-x)(1+.1)-x = 1210-1.1x -x = 1210-2.1x
Month 3: The principal amount this month is the 1210-2.1x from last month. When this month ends, the accumulated amount has to = 0, because Louie has to pay off the loan at the end of this 3rd month: (1210-2.1x)*(1+.1) - x = 0

Make Sure : Double check that you are answering the right question, and have all the facts correct.

Execute: Solve the problem.

(1210-2.1x)*(1+.1) - x = 0
1331-2.31x -x = 0
1331-3.31x = 0
1331 = 3.31x
402.11 = x (approximately)

There is a formula for installments paid while interest accrues. Most people don't know the formula. Memorize it.

x =(P*r)/[1-(1/(1+r))^n]

x: each installment
r: rate
n: number of installments
P: Principal amount borrowed by borrower.

For this problem:

r = .10
n = 3 installments
P = 1000
x = (1000*.10)/[1-(1/1.1)^3] = 402

guerrero wrote:Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compunded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month?

A. 333
B. 383
C. 402
D. 433
E. 483

OA C

At the end of each month, 10% interest accrues, followed by a monthly payment.
We can plug in the answers, which represent the monthly payment.
We should start with C, the middle value.
If C doesn't work, it will be easy to determine whether the monthly payment must increase or decrease.

Answer choice C: 402
To make the math easier, let's round the monthly payment to 400.

Amount at the end of the first month = 1000 + .1(1000) = 1100.
Remaining amount after the monthly payment = 1100-400 = 700.
Amount at the end of the second month = 700 + .1(700) = 770.
Remaining amount after the monthly payment = 770-400 = 370.
Amount at the end of the third month = 370 + .1(370) = 407.
Remaining amount after the monthly payment = 407-400 = 7.
Since only a few dollars remain, the monthly payment must be just a bit more than 400, implying that 402 is the required monthly payment.

The correct answer is C.