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

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Re: Compound Interest - Lender Charges [#permalink]

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New post 12 Sep 2013, 03:05
ishanbhat455 wrote:
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


You won't get some very tough numbers to manipulate with or there will be some shortcut approach available.

Theory on Percent and Interest Problems: math-number-theory-percents-91708.html

All DS Percent and Interest Problems to practice: search.php?search_id=tag&tag_id=33
All PS Percent and Interest Problems to practice: search.php?search_id=tag&tag_id=54

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 27 Nov 2013, 03:43
Dear Brunel ,

Please correct as we have formula

A=P(1+R/n)^nt

= 1000(1+ 10/100*3)^3/4

n=3
t=1/4

Kindly correct

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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Re: Compound Interest - Lender Charges [#permalink]

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New post 02 Feb 2014, 09:17
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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.


It's so frustrating to get to the 3.31x = 1331 and then get the answer wrong. I did heavy division shortcut but still the answer choices are a bit close. Any suggestion other than long division to better approximate this division?

Cheers
J

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 01 Aug 2014, 21:46
I wonder why Pascal's Triangle approach is not so famous..... Have a look at below problem solved using Pascal's Triangle

Pascal's Triangle in following way:
Number of Years (N)
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
. 1 .... .... ... ... 1
Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1
We did this three times b'cos N=3.
Step 2:
Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-
The coefficents - 1,3,3,1 are lifted from the pascal's triangle above.
Step 3:
C.Interest after 3 years = 3*100 + 3*10 + 3*1 = Rs.331/- (leaving out first term in step
2)
If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-
CI = 2 * 100 + 1* 10 = Rs. 210/-
This method is extendable for any 'N' and it avoids calculations involving higher
powers on 'N' altogether!
A variant to this short cut can be applied to find depreciating value of some
property. (Example, A property worth 100,000 depreciates by 10% every year, find
its value after 'N' years).

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 13 Aug 2014, 23:28
archit wrote:
Dear Brunel ,

Please correct as we have formula

A=P(1+R/n)^nt

= 1000(1+ 10/100*3)^3/4

n=3
t=1/4

Kindly correct


\(Amount = Principal (1 + \frac{Rate Of Int}{100})^{No Of Years}\)
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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 14 Aug 2014, 20:50
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'm still not convinced ! Even though he paid off the loan at the end of the 3 month, but still !he kept the money for 2 month and 29 days !
Do we have to assume that those 29 days he won't be charged ?
what about if he borrowed the 1000$ at a 12% yearly interest rate and he paid it off after 3 month, would I put zero interest in my calculation?
do we have to have the same assumption on the other official gmat questions?
:? :?
thank u for the clarifications :P
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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 04 Sep 2014, 08:02
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A slightly easier approach. :)

As we know compound interest is slightly greater than the simple interest calculated for the same period.

So instead calculate SI for this period.
Rate per month is 10%. Therefore rate per annum is 12*10 = 120%
Simple interest to be paid at the end = 1000* (120/100)= 1200

Simple Interest to be paid each month of 3 month pay back period =1200/3= 400

Now since Compound interest is slightly greater than Simple interest i.e. answer is 402.

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 09 Nov 2014, 06:27
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



Hi, your method for some reason isn't working on this question a-sum-of-3310-is-to-be-paid-back-in-3-equal-installments-how-much-is-185380.html?fl=similar

It is conceptually the same....Please avise

thanks :)

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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in this question, how can we substitute values in this formula..
CI = P(1+r/100)^t
kindly help. thanx.

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 22 Jan 2015, 18:26
Since the loan = 1000, and this amount were to be charged 10% interest per month, compounded monthly ...

1000/3 = 333

1 --- 333 * 1.1 = 366
2 --- 366 * 1.1 = 402
3 --- 402 * 1.1 = 442

Sum the 3 months: 366 + 402 + 442 = 1210 total owed

3 equal installments: 1210/3 = roughly 403

The closest answer (sans the slight rounding): Choice C.

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 27 Apr 2015, 14:30
I have a very silly question here

the question is solved by assuming that first the interest rate willl be applied and then he will pay the first installment.

the case can be like this also "he paid the installment on 15th of first month and interest was applied on 30th of the month"

Please correct me if i am wrong

thanks

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 27 Apr 2015, 14:59
solitaryreaper wrote:
A slightly easier approach. :)

As we know compound interest is slightly greater than the simple interest calculated for the same period.

So instead calculate SI for this period.
Rate per month is 10%. Therefore rate per annum is 12*10 = 120%
Simple interest to be paid at the end = 1000* (120/100)= 1200

Simple Interest to be paid each month of 3 month pay back period =1200/3= 400

Now since Compound interest is slightly greater than Simple interest i.e. answer is 402.




Hi

This i think is calculating obly intrest bt we need the amount paid.

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 29 Apr 2015, 06:39
Radhika11 wrote:
I have a very silly question here

the question is solved by assuming that first the interest rate willl be applied and then he will pay the first installment.

the case can be like this also "he paid the installment on 15th of first month and interest was applied on 30th of the month"

Please correct me if i am wrong

thanks


Hi Radhika11,

The question talks about payment of loan in monthly installments i.e. the first installment would be paid at the end of 1 month from the date on which loan was taken (unless it is specified otherwise). Hence the installment can't be paid on the 15th day of the month or 15 days from the date on which loan was taken.

Hence in this question the charging of interest and the payment of installment would happen every 1 month from the day the loan was taken.

As the best practice, any assumption which does not have any logical backing in the question statement should be avoided.

Hope its clear :) Let me know in case you have any other doubt.

Regards
Harsh
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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 11 Jun 2015, 21:48
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

answer is (C)
1331/(1.21+1.1+1)

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 04 Aug 2016, 18:07
damn it! the trick here is to know there is only interest accumulated twice not thrice

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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 11 Sep 2016, 07:11
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

Hi,
Why have you ignored the third month? Kindly, provide logic behind your method.

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Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 20 Nov 2016, 05:12
Whenever you get problems pertaining to 'loan' or 'borrowed amount' to be paid in equal installments, just apply the following formula, it is easy to remember:

Value of each equal annual installments = a
Rate of interest = r% p.a.
No. of installments per year = n
No. of years = t
Therefore, total no. of installments = n*t = N
Borrowed amount(or loan taken) = B

Then:

a[{100/(100+r)} + {100/(100+r)}^2 + ......... + {100/(100+r)}^N] = B

Now, in the above question(three equal installments to be paid in three months):

a[{100/110} + {100/110}^2 + {100/110}^3] = 1000

Solve the above equation and you will get 'a' as equal to 402.11
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Re: Louie takes out a three-month loan of $1000. The lender [#permalink]

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New post 29 Jun 2017, 03:08
I think it is not a high quality questions as it does not clearly specify when the payments start.

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Re: Louie takes out a three-month loan of $1000. The lender   [#permalink] 29 Jun 2017, 03:08

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