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22 Sep 2010, 09:39
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32% (02:39) correct 68% (01:51) wrong based on 1360 sessions

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 [Reveal] Spoiler: OA Current Student Status: What's your raashee? Joined: 12 Jun 2009 Posts: 1846 Location: United States (NC) Concentration: Strategy, Finance Schools: UNC (Kenan-Flagler) - Class of 2013 GMAT 1: 720 Q49 V39 WE: Programming (Computer Software) Followers: 26 Kudos [?]: 252 [9] , given: 52 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 22 Sep 2010, 09:47 9 This post received KUDOS 2 This post was BOOKMARKED sachinrelan 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

Couldn't solve by a systematic approach.

ok, there is an interest formula that i forget but lets do it another way:
so basically he is getting 10% interest per month for TWO month since he pays off in 3 months.
so 1000*1.1*1.1 = 1210
now divide by 3 = ~403.333
C
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Re: Compound Interest - Lender Charges [#permalink]

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22 Sep 2010, 09:53
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sachinrelan 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 Couldn't solve by a systematic approach. Let the monthly payment be $$x$$. After the 1st month there will be $$1,000*1.1-x$$ dollars left to repay; After the 2nd month there will be $$(1,000*1.1-x)*1.1-x=1,210-2.1x$$ dollars left to repay; After the 3rd month there should be 0 dollars left to repay: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$ Answer: C. _________________ Senior Manager Joined: 20 Jul 2010 Posts: 269 Followers: 2 Kudos [?]: 85 [2] , given: 9 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 22 Sep 2010, 10:29 2 This post received KUDOS $$CI = P(1+\frac{r}{100})^t$$ Assume he pays off entire amount in 3rd month or interest is accrued for 2 months. Find the amount at end of 3 months and divide by 3 to know monthly EMI _________________ If you like my post, consider giving me some KUDOS !!!!! Like you I need them Intern Joined: 27 Jun 2010 Posts: 40 Followers: 0 Kudos [?]: 134 [1] , given: 7 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 22 Sep 2010, 11:13 1 This post received KUDOS Bunuel wrote: sachinrelan 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

Couldn't solve by a systematic approach.

Let the monthly payment be $$x$$.

After the 1st month there will be $$1,000*1.1-x$$ dollars left to repay;
After the 2nd month there will be $$(1,000*1.1-x)*1.1-x=1,210-2.1x$$ dollars left to repay;
After the 3rd month there should be 0 dollars left to repay: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$

This is the same method i have used to solve the question, but can you suggest some short cut to solve this ques as i felt this approach in the exam would take lot of time to solve !!
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Re: Compound Interest - Lender Charges [#permalink]

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22 Sep 2010, 11:20
shaselai wrote:
sachinrelan 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 Couldn't solve by a systematic approach. ok, there is an interest formula that i forget but lets do it another way: so basically he is getting 10% interest per month for TWO month since he pays off in 3 months. so 1000*1.1*1.1 = 1210 now divide by 3 = ~403.333 C I Couldnt get why interest would be paid for 2 months, as per me 1. 1st month at the end monthly interest would be Accrued and monthly installment would be deducted from that amount. 2. For 2nd month start amount would be remaining amt of 1st month and at the end of 2nd month, monthly interest would be Accrued and thereafter again monthly installment would be deducted 3. For the 3rd month start amt would again be the remaning amt of 2nd month and at the end of 3rd month monthly interest would be accrued which should be equal to monthly installment. So as per this interest was paid thrice ..request you to please clarify !! Current Student Status: What's your raashee? Joined: 12 Jun 2009 Posts: 1846 Location: United States (NC) Concentration: Strategy, Finance Schools: UNC (Kenan-Flagler) - Class of 2013 GMAT 1: 720 Q49 V39 WE: Programming (Computer Software) Followers: 26 Kudos [?]: 252 [0], given: 52 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 22 Sep 2010, 11:27 sachinrelan wrote: shaselai wrote: sachinrelan 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

Couldn't solve by a systematic approach.

ok, there is an interest formula that i forget but lets do it another way:
so basically he is getting 10% interest per month for TWO month since he pays off in 3 months.
so 1000*1.1*1.1 = 1210
now divide by 3 = ~403.333
C

I Couldnt get why interest would be paid for 2 months, as per me

1. 1st month at the end monthly interest would be Accrued and monthly installment would be deducted from that amount.
2. For 2nd month start amount would be remaining amt of 1st month and at the end of 2nd month, monthly interest would be Accrued and thereafter again monthly installment would be deducted
3. For the 3rd month start amt would again be the remaning amt of 2nd month and at the end of 3rd month monthly interest would be accrued which should be equal to monthly installment.

So as per this interest was paid thrice ..request you to please clarify !!

this is because you are paying off in the third and last months. This is assuming the interest rate is calculated at the end of the month. So it is assumed you paid off the balance at the end of third month so 0 balance. Like CC statements - if you didnt pay off your statement by end of month you get charged interest - you dont get charged interest throughout.
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24 Aug 2012, 12:54
Can anyone tell me the source of the question as this question is a simple example of EMI (Equal Monthly installment). As far as the logic is concerned it's ok but i don't think such kind of questions do appear in gmat.
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09 Feb 2013, 18:43
why its not 443 (1331/3)
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20 Apr 2013, 23:28
summer101 wrote:
Why are we assuming he pays from the 3rd month? The question does not specify that, it just says he has to pay in 3 installments.

Why not this way?
Total Loan disbursed in 3 months = 1.1 * 1.1* 1.1* 1000 = 1331
Repaid in 3 months, hence per month = 1331/3 = 443

Because he pays each month. He doesn't have to pay interest on the amount that he has already paid.

IE. If he is paying $402 a month, then at the end of the first month his balance will be (1000 * 1.1) - 402 =$698, so going into the second month that 10% interest is only accruing on $698 rather than on the full$1000.
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21 Apr 2013, 04:15
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atalpanditgmat wrote:
Bunuel, Can you give links to similar problem? It would be great help. Thanks

Check here: search.php?search_id=tag&tag_id=191

Hope it helps.
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24 May 2013, 08:18
1000 * 1.1 = 1100 month one plus compounded interest
1100 - 402 = 698 first months payment @ "correct" answer
698 * 1.1 = 767.80 month 2 balance plus interest
767.80 - 402 = 365.80 payment deducted for month two
365.8 * 1.1 = 402.38
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Re: Compound Interest - Lender Charges [#permalink]

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11 Sep 2013, 03:59
Bunuel wrote:
sachinrelan 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 Couldn't solve by a systematic approach. Let the monthly payment be $$x$$. After the 1st month there will be $$1,000*1.1-x$$ dollars left to repay; After the 2nd month there will be $$(1,000*1.1-x)*1.1-x=1,210-2.1x$$ dollars left to repay; After the 3rd month there should be 0 dollars left to repay: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$ Answer: C. I get a different answer by using the Compound Interest formula, i.e- P[1 +(r)/100n]^nt Since this formula uses annualized figures, so: r = 10% per month = 120% per year n = 12 (as interest is compounded monthly) t = 3 months = 3/12 years Using the formula for compound interest, I get: P + C.I = 1000(1.1)^3 = 1331 So, EMI = 1331/3 = 443.66 which is ~$444

What's wrong with this approach?

Thanks,
Ishan
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Re: Compound Interest - Lender Charges [#permalink]

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11 Sep 2013, 05:29
ishanbhat455 wrote:
Bunuel wrote:
sachinrelan 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 Couldn't solve by a systematic approach. Let the monthly payment be $$x$$. After the 1st month there will be $$1,000*1.1-x$$ dollars left to repay; After the 2nd month there will be $$(1,000*1.1-x)*1.1-x=1,210-2.1x$$ dollars left to repay; After the 3rd month there should be 0 dollars left to repay: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$ Answer: C. I get a different answer by using the Compound Interest formula, i.e- P[1 +(r)/100n]^nt Since this formula uses annualized figures, so: r = 10% per month = 120% per year n = 12 (as interest is compounded monthly) t = 3 months = 3/12 years Using the formula for compound interest, I get: P + C.I = 1000(1.1)^3 = 1331 So, EMI = 1331/3 = 443.66 which is ~$444

What's wrong with this approach?

Thanks,
Ishan

Since he pays after each month, then after the firs month (after the first payment) the interest is calculated on reduced balance.

Does this make sense?
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Re: Compound Interest - Lender Charges [#permalink]

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11 Sep 2013, 07:20
Bunuel wrote:
ishanbhat455 wrote:

I get a different answer by using the Compound Interest formula, i.e- P[1 +(r)/100n]^nt

Since this formula uses annualized figures, so:
r = 10% per month = 120% per year
n = 12 (as interest is compounded monthly)
t = 3 months = 3/12 years

Using the formula for compound interest, I get:
P + C.I = 1000(1.1)^3 = 1331

So, EMI = 1331/3 = 443.66 which is ~ $444 What's wrong with this approach? Thanks, Ishan Since he pays after each month, then after the firs month (after the first payment) the interest is calculated on reduced balance. Does this make sense? Bunuel, Thanks for clarifying. What if the problem was such that the loan tenure were 2 years, interest rate was 10% per annum and compounded annually? How do I compute EMI then? In such a scenario, won't the monthly approach of computation be very lengthy? I am just trying to get a clearer picture on EMI questions. Thanks, Ishan Re: Compound Interest - Lender Charges [#permalink] 11 Sep 2013, 07:20 Go to page 1 2 Next [ 39 posts ] Similar topics Replies Last post Similar Topics: 2 Nathan took out a student loan for 1200$ at 10 percent annual interest 1 22 Dec 2015, 21:18
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# Louie takes out a three-month loan of \$1000. The lender

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