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Louie takes out a three-month loan of $1000. The lender [#permalink] ### Show Tags 22 Sep 2010, 08:39 47 This post was BOOKMARKED 00:00 Difficulty: 95% (hard) Question Stats: 33% (02:38) correct 67% (01:51) wrong based on 1327 sessions ### HideShow timer Statistics 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
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Re: Compound Interest - Lender Charges [#permalink]

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22 Sep 2010, 08:47
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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. 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 _________________ If you like my answers please +1 kudos! Math Expert Joined: 02 Sep 2009 Posts: 36582 Followers: 7086 Kudos [?]: 93268 [14] , given: 10555 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 22 Sep 2010, 08:53 14 This post received KUDOS Expert's post 20 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.

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: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$

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Re: Compound Interest - Lender Charges [#permalink]

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22 Sep 2010, 09:29
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$$CI = P(1+\frac{r}{100})^t$$

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 [#permalink]

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22 Sep 2010, 10:13
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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. 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 !! Intern Joined: 27 Jun 2010 Posts: 40 Followers: 0 Kudos [?]: 129 [0], given: 7 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 22 Sep 2010, 10:20 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 !!
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Re: Compound Interest - Lender Charges [#permalink]

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22 Sep 2010, 10:27
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. _________________ If you like my answers please +1 kudos! Intern Joined: 18 Jun 2012 Posts: 41 Followers: 1 Kudos [?]: 7 [1] , given: 15 Louie takes out a three-month loan of$1000. The lender [#permalink]

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07 Jul 2012, 04:15
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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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06 Feb 2013, 02:05
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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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20 Apr 2013, 11:26
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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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21 Apr 2013, 00:02
Bunuel, Can you give links to similar problem? It would be great help. Thanks
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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 three-month loan of $1000. The lender [#permalink] ### Show Tags 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 Intern Joined: 01 Feb 2013 Posts: 12 Location: India Followers: 0 Kudos [?]: 6 [0], given: 4 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 11 Sep 2013, 02:59 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: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$

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 Math Expert Joined: 02 Sep 2009 Posts: 36582 Followers: 7086 Kudos [?]: 93268 [0], given: 10555 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 11 Sep 2013, 04:29 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: $$(1,210-2.1x)*1.1-x=0$$ --> $$1331=3.31x$$ --> $$x\approx{402}$$

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? _________________ Intern Joined: 01 Feb 2013 Posts: 12 Location: India Followers: 0 Kudos [?]: 6 [0], given: 4 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 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
Re: Compound Interest - Lender Charges   [#permalink] 11 Sep 2013, 06:20

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