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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
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Cosmas
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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

How do you calculate (1.01)^12 under exam conditions. Is there an easier way?

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.
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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

How do you calculate (1.01)^12 under exam conditions. Is there an easier way?

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.

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.
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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.

To practice similar questions please follow the links in my post above.
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Thanks mate. I appreciate.
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Peter invests $100,000 in an account that pays 12% annual interest: the interest is paid once, at the end of the year. Martha invests $100,000 in an account that pays 12% annual interest, compounding monthly at the end of each month. At the end of one full year, compared to Peter's account, approximately how much more does Martha’s account have?

A. Zero
B. $68.25
C. $682.50
D. $6825.00
E. $68250.00

Peters interest = $100,000*0.12 = $12,000 or $1,000 each month.

Martha’s interest, 12%/12 = 1% each month:
For the 1st month = $100,000*0.01 = $1,000;
For the 2nd month = $1,000 + 1% of 1,000 = $1,010, so we would have interest earned on interest (very small amount);
For the 3rd month = $1,010 + 1% of 1,010 = ~$1,020;
For the 4th month = $1,020 + 1% of 1,020 = ~$1,030;
...
For the 12th month = $1,100 + 1% of 1,100 = ~$1,110.

The difference between Peters interest and Martha’s interest = ~(10 + 20 + ... + 110) = $660.

Answer: C.

Similar questions to practice:
john-deposited-10-000-to-open-a-new-savings-account-that-135825.html
on-the-first-of-the-year-james-invested-x-dollars-at-128825.html
marcus-deposited-8-000-to-open-a-new-savings-account-that-128395.html
jolene-entered-an-18-month-investment-contract-that-127308.html
alex-deposited-x-dollars-into-a-new-account-126459.html
michelle-deposited-a-certain-sum-of-money-in-a-savings-138273.html

Hope it helps.

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?
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Bunuel
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Peter invests $100,000 in an account that pays 12% annual interest: the interest is paid once, at the end of the year. Martha invests $100,000 in an account that pays 12% annual interest, compounding monthly at the end of each month. At the end of one full year, compared to Peter's account, approximately how much more does Martha’s account have?

A. Zero
B. $68.25
C. $682.50
D. $6825.00
E. $68250.00

Peters interest = $100,000*0.12 = $12,000 or $1,000 each month.

Martha’s interest, 12%/12 = 1% each month:
For the 1st month = $100,000*0.01 = $1,000;
For the 2nd month = $1,000 + 1% of 1,000 = $1,010, so we would have interest earned on interest (very small amount);
For the 3rd month = $1,010 + 1% of 1,010 = ~$1,020;
For the 4th month = $1,020 + 1% of 1,020 = ~$1,030;
...
For the 12th month = $1,100 + 1% of 1,100 = ~$1,110.

The difference between Peters interest and Martha’s interest = ~(10 + 20 + ... + 110) = $660.

Answer: C.

Similar questions to practice:
john-deposited-10-000-to-open-a-new-savings-account-that-135825.html
on-the-first-of-the-year-james-invested-x-dollars-at-128825.html
marcus-deposited-8-000-to-open-a-new-savings-account-that-128395.html
jolene-entered-an-18-month-investment-contract-that-127308.html
alex-deposited-x-dollars-into-a-new-account-126459.html
michelle-deposited-a-certain-sum-of-money-in-a-savings-138273.html

Hope it helps.

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?

Your way: 0.01*$101,000 = $1,010.
My way: $1,000 + 1% of 1,000 = $1,010.
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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.
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Is this question a 700 lvl question?
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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.
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Cosmas
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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

How do you calculate (1.01)^12 under exam conditions. Is there an easier way?


use binomial expansion to get approx ans
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Bunuel
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Peter invests $100,000 in an account that pays 12% annual interest: the interest is paid once, at the end of the year. Martha invests $100,000 in an account that pays 12% annual interest, compounding monthly at the end of each month. At the end of one full year, compared to Peter's account, approximately how much more does Martha’s account have?

A. Zero
B. $68.25
C. $682.50
D. $6825.00
E. $68250.00

Peters interest = $100,000*0.12 = $12,000 or $1,000 each month.

Martha’s interest, 12%/12 = 1% each month:
For the 1st month = $100,000*0.01 = $1,000;
For the 2nd month = $1,000 + 1% of 1,000 = $1,010, so we would have interest earned on interest (very small amount);
For the 3rd month = $1,010 + 1% of 1,010 = ~$1,020;
For the 4th month = $1,020 + 1% of 1,020 = ~$1,030;
...
For the 12th month = $1,100 + 1% of 1,100 = ~$1,110.

The difference between Peters interest and Martha’s interest = ~(10 + 20 + ... + 110) = $660.

Answer: C.

Similar questions to practice:
https://gmatclub.com/forum/john-deposite ... 35825.html
https://gmatclub.com/forum/on-the-first- ... 28825.html
https://gmatclub.com/forum/marcus-deposi ... 28395.html
https://gmatclub.com/forum/jolene-entere ... 27308.html
https://gmatclub.com/forum/alex-deposite ... 26459.html
https://gmatclub.com/forum/michelle-depo ... 38273.html

Hope it helps.

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
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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.
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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!
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Why is the the interest rate distributed by 1 % for each of the 12 compounding periods? Thanks in advance
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Bunuel
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Peter invests $100,000 in an account that pays 12% annual interest: the interest is paid once, at the end of the year. Martha invests $100,000 in an account that pays 12% annual interest, compounding monthly at the end of each month. At the end of one full year, compared to Peter's account, approximately how much more does Martha’s account have?

A. Zero
B. $68.25
C. $682.50
D. $6825.00
E. $68250.00

Peters interest = $100,000*0.12 = $12,000 or $1,000 each month.

Martha’s interest, 12%/12 = 1% each month:
For the 1st month = $100,000*0.01 = $1,000;
For the 2nd month = $1,000 + 1% of 1,000 = $1,010, so we would have interest earned on interest (very small amount);
For the 3rd month = $1,010 + 1% of 1,010 = ~$1,020;
For the 4th month = $1,020 + 1% of 1,020 = ~$1,030;
...
For the 12th month = $1,100 + 1% of 1,100 = ~$1,110.

The difference between Peters interest and Martha’s interest = ~(10 + 20 + ... + 110) = $660.

Answer: C.

Hope it helps.

What a great method
Thank you very much, expert.
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Bunuel
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Peter invests $100,000 in an account that pays 12% annual interest: the interest is paid once, at the end of the year. Martha invests $100,000 in an account that pays 12% annual interest, compounding monthly at the end of each month. At the end of one full year, compared to Peter's account, approximately how much more does Martha’s account have?

A. Zero
B. $68.25
C. $682.50
D. $6825.00
E. $68250.00

Peters interest = $100,000*0.12 = $12,000 or $1,000 each month.

Martha’s interest, 12%/12 = 1% each month:
For the 1st month = $100,000*0.01 = $1,000;
For the 2nd month = $1,000 + 1% of 1,000 = $1,010, so we would have interest earned on interest (very small amount);
For the 3rd month = $1,010 + 1% of 1,010 = ~$1,020;
For the 4th month = $1,020 + 1% of 1,020 = ~$1,030;
...
For the 12th month = $1,100 + 1% of 1,100 = ~$1,110.

The difference between Peters interest and Martha’s interest = ~(10 + 20 + ... + 110) = $660.

Answer: C.

Similar questions to practice:
https://gmatclub.com/forum/john-deposite ... 35825.html
https://gmatclub.com/forum/on-the-first- ... 28825.html
https://gmatclub.com/forum/marcus-deposi ... 28395.html
https://gmatclub.com/forum/jolene-entere ... 27308.html
https://gmatclub.com/forum/alex-deposite ... 26459.html
https://gmatclub.com/forum/michelle-depo ... 38273.html

Hope it helps.

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
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