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Senior Manager  Joined: 10 Apr 2012
Posts: 266
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)
Peter invests \$100,000 in an account that pays 12%  [#permalink]

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1
42 00:00

Difficulty:   95% (hard)

Question Stats: 45% (02:20) correct 55% (02:33) wrong based on 536 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
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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12
30
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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Posts: 46
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Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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2
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
Intern  Joined: 10 Apr 2012
Posts: 38
Concentration: Finance
Schools: Goizueta '19 (I)
WE: Analyst (Commercial Banking)
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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2
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?
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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3
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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Intern  Joined: 10 Apr 2012
Posts: 38
Concentration: Finance
Schools: Goizueta '19 (I)
WE: Analyst (Commercial Banking)
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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1
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.
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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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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Intern  Joined: 10 Apr 2012
Posts: 38
Concentration: Finance
Schools: Goizueta '19 (I)
WE: Analyst (Commercial Banking)
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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Thanks mate. I appreciate.
Intern  B
Joined: 23 Dec 2014
Posts: 47
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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1
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?
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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1
1
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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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.
Intern  B
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Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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Is this question a 700 lvl question?
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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JouviMat wrote:
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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Joined: 23 Oct 2013
Posts: 44
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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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?

use binomial expansion to get approx ans
Intern  B
Joined: 03 Oct 2016
Posts: 12
Re: Peter invests \$100,000 in an account that pays 12%  [#permalink]

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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:
http://gmatclub.com/forum/john-deposite ... 35825.html
http://gmatclub.com/forum/on-the-first- ... 28825.html
http://gmatclub.com/forum/marcus-deposi ... 28395.html
http://gmatclub.com/forum/jolene-entere ... 27308.html
http://gmatclub.com/forum/alex-deposite ... 26459.html
http://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 Re: Peter invests \$100,000 in an account that pays 12%   [#permalink] 03 Aug 2019, 22:55
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