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Louie takes out a threemonth loan of $1000. The lender
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22 Sep 2010, 08:39
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Louie takes out a threemonth 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
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Re: Compound Interest  Lender Charges
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22 Sep 2010, 08:53
sachinrelan wrote: Louie takes out a threemonth 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 \(1,000*1.1x\) dollars left to repay; After the 2nd month there will be \((1,000*1.1x)*1.1x=1,2102.1x\) dollars left to repay; After the 3rd month there should be 0 dollars left to repay: \((1,2102.1x)*1.1x=0\) > \(1331=3.31x\) > \(x\approx{402}\) Answer: C.
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Re: Compound Interest  Lender Charges
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22 Sep 2010, 08:47
sachinrelan wrote: Louie takes out a threemonth 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
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22 Sep 2010, 09:29
\(CI = P(1+\frac{r}{100})^t\) 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
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Re: Compound Interest  Lender Charges
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22 Sep 2010, 10:13
Bunuel wrote: sachinrelan wrote: Louie takes out a threemonth 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 \(1,000*1.1x\) dollars left to repay; After the 2nd month there will be \((1,000*1.1x)*1.1x=1,2102.1x\) dollars left to repay; After the 3rd month there should be 0 dollars left to repay: \((1,2102.1x)*1.1x=0\) > \(1331=3.31x\) > \(x\approx{402}\) Answer: C. 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
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22 Sep 2010, 10:20
shaselai wrote: sachinrelan wrote: Louie takes out a threemonth 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 !!



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Re: Compound Interest  Lender Charges
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22 Sep 2010, 10:27
sachinrelan wrote: shaselai wrote: sachinrelan wrote: Louie takes out a threemonth 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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Louie takes out a threemonth loan of $1000. The lender
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07 Jul 2012, 04:15
The interest has to be calculated on a reducing balance. If monthly repayment = x At the end of the 3 month period, 1.1*[1.1*{1.1*(1000)x}x]x = 0 => 3.31x = 1331 => x ~ 402



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Re: Louie takes out a threemonth loan of $1000. The lender
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24 Aug 2012, 11:54
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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Re: Louie takes out a threemonth loan of $1000. The lender
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06 Feb 2013, 02:05
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
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Re: Louie takes out a threemonth loan of $1000. The lender
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09 Feb 2013, 17:43
why its not 443 (1331/3)



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Re: Louie takes out a threemonth loan of $1000. The lender
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20 Apr 2013, 11:26
In case of CI ,Repayment in equal installments (X) can be given as: X =P*r/ [1(100/100+r)^n]where X :each installment r: rate n: number of installments P: Principal amount borrowed by borrower. So in this case it would be 1000*10/[1(10/11)^3] = 133100/331 = 402



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Re: Louie takes out a threemonth loan of $1000. The lender
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20 Apr 2013, 22:28
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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Re: Louie takes out a threemonth loan of $1000. The lender
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21 Apr 2013, 00:02
Bunuel, Can you give links to similar problem? It would be great help. Thanks
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Re: Louie takes out a threemonth loan of $1000. The lender
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Re: Louie takes out a threemonth loan of $1000. The lender
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23 Apr 2013, 00:00
rajatr wrote: In case of CI ,Repayment in equal installments (X) can be given as: X =P*r/ [1(100/100+r)^n]where X :each installment r: rate n: number of installments P: Principal amount borrowed by borrower. So in this case it would be 1000*10/[1(10/11)^3] = 133100/331 = 402 P(1+R/100)^n Can we do this using this formula?



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Re: Louie takes out a threemonth loan of $1000. The lender
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24 May 2013, 07:18
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
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11 Sep 2013, 02:59
Bunuel wrote: sachinrelan wrote: Louie takes out a threemonth 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 \(1,000*1.1x\) dollars left to repay; After the 2nd month there will be \((1,000*1.1x)*1.1x=1,2102.1x\) dollars left to repay; After the 3rd month there should be 0 dollars left to repay: \((1,2102.1x)*1.1x=0\) > \(1331=3.31x\) > \(x\approx{402}\) 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
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11 Sep 2013, 04:29
ishanbhat455 wrote: Bunuel wrote: sachinrelan wrote: Louie takes out a threemonth 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 \(1,000*1.1x\) dollars left to repay; After the 2nd month there will be \((1,000*1.1x)*1.1x=1,2102.1x\) dollars left to repay; After the 3rd month there should be 0 dollars left to repay: \((1,2102.1x)*1.1x=0\) > \(1331=3.31x\) > \(x\approx{402}\) 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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Compound Interest  Lender Charges
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11 Sep 2013, 06:20
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




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