Louie takes out a three-month loan of $1000...

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

Louie takes out a three-month loan of $1000...

by AAPL » Wed Mar 14, 2018 7:48 am

Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compounded 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

The OA is C.

I solved this PS question as follow,

For 3 months, the amount will be 1000(1+10/100)^3 = 1000(1.1)^3 = 1000*1.331 = 1331.

Louie will have to pay 1331/3 = 443, so I think it's D.

I'm not sure! I appreciate if any expert explains it to me. Thanks!

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Source: Beat The GMAT — Problem Solving | Wed Mar 14, 2018 7:48 am

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ErikaPrepScholar: Legendary Member; Posts: 503; Joined: Thu Jul 20, 2017 9:03 am; Thanked: 86 times; Followed by:15 members; GMAT Score:770

by ErikaPrepScholar » Wed Mar 14, 2018 8:16 am

This is a tricky question - we have to consider the fact that the loan is accumulating interest as Louie is paying it off. So there is less money in the account each month for Louie to pay interest on.

We'll approach it by breaking it down by month.

Month 1
The balance of the loan is 1000 dollars at the start of the month. He is charged 10% interest. Louie makes a payment of x dollars.
1000 * 1.1 - x

Month 2
The balance of the loan is (1000 - x)*1.1 dollars at the start of the month. He is charged 10% interest. Louie makes a payment of x dollars.
(1000 * 1.1 - x) * 1.1 - x

Month 3
The balance of the loan is ((1000 - x)*1.1 - x)*1.1 dollars at the start of the month. He is charged 10% interest. Louie makes a payment of x dollars and pays off the account.
((1000 * 1.1 - x) * 1.1 - x) * 1.1 - x

So ((1000 * 1.1 - x) * 1.1 - x) * 1.1 - x is the balance of the loan at the end of month 3, which should be $0, since the loan is paid off. This means that ((1000 * 1.1 - x) * 1.1 - x) * 1.1 - x = 0. We can then solve for x:
((1000 * 1.1 - x) * 1.1 - x) * 1.1 - x = 0
((1100 - x) * 1.1 - x) * 1.1 - x = 0
(1210 - 1.1x - x) * 1.1 - x = 0
(1210 - 2.1x) * 1.1 - x = 0
(1210 - 2.1x) * 1.1 - x = 0
1331 - 2.31x - x = 0
1331 - 3.31x = 0
1331 = 3.31x
x = about $402

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sun Jun 02, 2019 4:24 pm

AAPL wrote:Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compounded 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

We can let p be his monthly payment.

At the end of the first month, the loan will accrue 0.1(1000) = 100 dollars of interest, and the total amount is $1100. Since he will pay p dollars to that amount, he has 1100 - p dollars left to pay.

At the end of the second month, the loan will accrue 0.1(1100 - p) dollars of interest, and the total amount is 1.1(1100 - p) dollars. Since he will pay p dollars to that amount, he has 1.1(1100 - p) - p = 1210 - 2.1p dollars left to pay.

At the end of the third month, the loan will accrue 0.1(1210 - 2.1p) dollars of interest, and the total amount is 1.1(1210 - 2.1p) dollars. Since he will pay p dollars to that amount, he has 1.1(1210 - 2.1p) - p = 1331 - 3.31p dollars left to pay. However, since we assume that he will pay off this loan in 3 months, it must be true that what he has left to pay is $0. That is,

1331 - 3.31p = 0

1331 = 3.31p

p = 1331/3.31 â‰ˆ 402

Answer: C

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