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

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Senior Manager
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Peter invests \$100,000 in an account that pays 12% [#permalink]

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21 Feb 2014, 07:11
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95% (hard)

Question Stats:

50% (01:41) correct 50% (01:58) wrong based on 440 sessions

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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
[Reveal] Spoiler: OA

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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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21 Feb 2014, 09:33
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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 - http://qrc.depaul.edu/studyguide2009/no ... terest.htm

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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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21 Feb 2014, 09:58
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guerrero25 wrote:
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.
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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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22 Feb 2014, 01:04
Manofsteel wrote:
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 - http://qrc.depaul.edu/studyguide2009/no ... terest.htm

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

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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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22 Feb 2014, 03:18
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Expert's post
Cosmas wrote:
Manofsteel wrote:
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 - http://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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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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22 Feb 2014, 03:51
Bunuel wrote:
Cosmas wrote:
Manofsteel wrote:
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 - http://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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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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22 Feb 2014, 03:55
Cosmas wrote:
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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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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22 Feb 2014, 03:57
Thanks mate. I appreciate.

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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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13 Jan 2015, 15:31
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Bunuel wrote:
guerrero25 wrote:
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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Re: Peter invests \$100,000 in an account that pays 12% [#permalink]

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14 Jan 2015, 02:03
Expert's post
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Salvetor wrote:
Bunuel wrote:
guerrero25 wrote:
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 invests \$100,000 in an account that pays 12% [#permalink]

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01 Dec 2016, 19:18
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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Peter invests \$100,000 in an account that pays 12%   [#permalink] 01 Dec 2016, 19:18
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# Peter invests \$100,000 in an account that pays 12%

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