GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 10 Dec 2018, 00:47

# Dec 10th is GMAT Club's BDAY :-)

Free GMAT Club Tests & Quizzes for 24 hrs to celebrate together!

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free lesson on number properties

December 10, 2018

December 10, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
• ### Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.

# Louie takes out a three-month loan of $1000. The lender  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 51055 Re: Compound Interest - Lender Charges [#permalink] ### Show Tags 12 Sep 2013, 02:05 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

_________________
Intern
Joined: 06 Aug 2012
Posts: 16

### Show Tags

27 Nov 2013, 02:57
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

Sorry, what should I correct?
_________________
SVP
Joined: 06 Sep 2013
Posts: 1728
Concentration: Finance
Re: Compound Interest - Lender Charges  [#permalink]

### Show Tags

02 Feb 2014, 08:17
1
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 Manager Joined: 26 May 2013 Posts: 52 Re: Louie takes out a three-month loan of$1000. The lender  [#permalink]

### Show Tags

01 Aug 2014, 20:46
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).
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1826
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

### Show Tags

14 Aug 2014, 19:50
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
_________________

Please +1 KUDO if my post helps. Thank you.

Manager
Joined: 23 Sep 2013
Posts: 134
Concentration: Strategy, Marketing
WE: Engineering (Computer Software)

### Show Tags

09 Nov 2014, 05:27
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 Intern Joined: 08 Dec 2013 Posts: 32 Re: Louie takes out a three-month loan of$1000. The lender  [#permalink]

### Show Tags

14 Nov 2014, 09:54
in this question, how can we substitute values in this formula..
CI = P(1+r/100)^t
kindly help. thanx.
Manager
Joined: 20 Jul 2013
Posts: 56

### Show Tags

27 Apr 2015, 13:30
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
Intern
Joined: 25 May 2014
Posts: 37

### Show Tags

29 Apr 2015, 05:39
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

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
_________________

Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Manager
Joined: 10 Jun 2015
Posts: 118
Re: Louie takes out a three-month loan of $1000. The lender [#permalink] ### Show Tags 11 Jun 2015, 20:48 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

1331/(1.21+1.1+1)
Manager
Joined: 13 Jun 2016
Posts: 111
Location: United States
Concentration: Finance, Technology

### Show Tags

11 Sep 2016, 06:11
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. Intern Joined: 13 Feb 2015 Posts: 12 Louie takes out a three-month loan of$1000. The lender  [#permalink]

### Show Tags

20 Nov 2016, 04:12
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
_________________

Before getting into the options, have an idea of what you seek

Thanks & Regards,
Vipul Chhabra

Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5

### Show Tags

07 Sep 2017, 03:32
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 yes, it may seem very obvious naturally .... but can you show it on paper ...? For example, say, Mr X provides a loan amounting$100 to Mr Y in January at 10% interest compounded per month. At the end of January, the amount will turn out to be $110, at the end of February, the amount will be$121, and finally, at the end of March, the amount will be $133... here you can see....133 = 100 x 1.1 x 1.1 x 1.1... thus, the multiplier (1.1) is multiplied three times ...... so, how can you multiply$1000 with (1.1) only two times.....? it may seem very obvious, but I am in trouble ....

anybody out there to help me understand this picture, please ....