Peter invests $100,000 in an account that pays 12%

Yes, that approach is not useful on the exam. Check here: peter-invests-100-000-in-an-account-that-pays-167793.html#p1335128

Hope it helps.

Your way: 0.01*$101,000 = $1,010.
My way: $1,000 + 1% of 1,000 = $1,010.

No easy way to solve this.
12% percent compounded monthly gives (12/12) interest rate per month.

1st month- 1% of 100000 = 0.01*100000 = 1000
2nd Month- 1% of 100000 + 1% of 1000 = 1000 + 10 = 1010
3rd month- 1% of 100000 + 1% of 2010 (add previous 2 interests) = 1000 + 20.10 = 1020.10
4th Month- 1% of 100000 + 1% of (1000 + 1010 + 1020.10= 3030.10) = 1000 + 30.301 = 1030.301
5th Month- 1% of 100000 + 1% of (1000+1010+1020.10+1030.301= 4060.401) = 1040.604

Here, we can see a pattern= Interest is increasing per month only by 10 and some number post decimal. Hence to save time, we can assume the value before decimal only.

Hence total interest is = (1000 +1010 + 1020 +1030 +1040.........1110)= 12000 + approx. 660
Hence total is approx. 12660.
By simple interest formula, the interest is coming as 12000. Hence the difference is 12660-12000= approx. 660
Closest option is C ($682.5), Hence C.

During test day conditions, I would approach a question of this level by leveraging the answer choices.

First we calculate that Peter earns 12% of $100,000 = $12,000 interest for the year and $1000 per month.

Martha's Interest, however, compounds 12% / 12 months and therefore, 1 % per month.
We can quickly gauge that Martha makes $10 more than Peter's after the second month (ie. Martha's interest month 1 = $1000, month 2 = 1.01*(1000) = $1010)

With this information, we can stop calculations and leverage the answer choices considering each choice differs by large factors.

A. none (disproved after month 2)
B. $68.25 (impossible, using the delta interest between month 1&2 (10$) Martha would surpass this amount after only 5-6 months)
C. $682.50 (looks reasonable and within our scope)
D. $6825.00 (way too big considering we're only adding an extra 1% of $1000 per month)

I would stop here, and select C .

Perhaps this method is inelegant, but I think it works!

At the end of 1 year Peter gets an interest of - Rs 12000

At the end of 1 year Martha gets an interest of = 100000(1+0.12/12)^12=100000(1.01^12)= 112682.50 -100000 = 12682.50 $

Therefore Martha has 12682.50-12000 = 682.50 $ more than Peter at the end of 1 year.
Option (C)


The compound interest formula for those who dont know - https://qrc.depaul.edu/studyguide2009/no ... terest.htm

Dear Bunuel, In the second month of Martha why didn't you pick the primary amount which is 100,000 . I mean at first month her interest is on 100,000 and second month it should be on 101,000 ? But you wrote on 1000. I am not clear actually. Can you please give a brief?

Thanks Bunuel. I just wanted to tell Manofsteel that in as much as its a correct formula, that approach will waste a lot of time. Thanks for alternative, fast approach Bunuel.

Peter --> Simple Interest (P(1+r)^t)

100,000(1+0.12)^1 = 112,000

Martha --> Compounded Interest (P[1+(r/n)]^nt)

100,000(1+(0.12/12))^[12(1)] = 112682.50

112682.50-112,000 = 682.50

C.

Hi Bunuel, I tried your method of calculating Compound Interest on Martha's Investment. However i just got lost and took me long time to calcuate 1% of interest for 12 months. is there any easy way to solvethis question or if something i am doing wrong. Pls correct me. Thanks

To practice similar questions please follow the links in my post above.

Thanks mate. I appreciate.

Is this question a 700 lvl question?

You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty: 700 Level.

use binomial expansion to get approx ans

Why is the the interest rate distributed by 1 % for each of the 12 compounding periods? Thanks in advance

What a great method
Thank you very much, expert.

I have been sitting on this for a while, I understand what happens in the first two months, but how are we getting the third month?

Why is it 1010 + 1010 x 1%?

Am I getting this right?

At the end of the 1st month ...100000 + (1% x 100000) = 100000 + 1000

At the end of the 2nd month you have accumulated 100000 + (1% x 100000) + (1% x 1000) = 100000 + 1000 + 10

At the end of the 3rd month it would now be 100000 + (1% x 100000) + (1% x 100000) + (1% x 1000) + (1% x 10) = 100000 + 1000 + 1000 + 10 + trivial amount