interest

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interest

by ketkoag » Fri Jun 19, 2009 4:45 pm
A total of $1000 was invested for one year. Annual interest rate is r, compound interest is counted semiannually.If the total interest earned by $1000 for that year was $80.56, what is the value of r?
A. 4<r<5;
B. 5<r<6;
C. 6<r<7;
D. 7<r<8;
E. 8<r<9

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by abhinav85 » Fri Jun 19, 2009 7:28 pm
IMO A

What is the OA?

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by rahulg83 » Sat Jun 20, 2009 3:30 am
Total amount after n years if interest is compounded q times a year

A = P(1 + r/q)^nq

1080.56 = 1000(1+r/(100*2))^1*2

solve for r, between 7 and 8

D

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Re: interest

by Stuart@KaplanGMAT » Sat Jun 20, 2009 10:38 am
ketkoag wrote:A total of $1000 was invested for one year. Annual interest rate is r, compound interest is counted semiannually.If the total interest earned by $1000 for that year was $80.56, what is the value of r?
A. 4<r<5;
B. 5<r<6;
C. 6<r<7;
D. 7<r<8;
E. 8<r<9
The quickest way to solve is to use some common sense and work with the choices. The total interest for the year is just over $80; at simple interest 8% would get you $80, so let's start with an 8% annual rate.

If we were to earn 8% annually, compounded semi-annually, it would be 4% twice. So:

1000(.04) = $40
1040(.04) = $41.60

Total interest earned: $81.60

Since 8% gives us a tiny bit too much interest, we know the rate is going to be a bit less than 8%... choose (D).
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by BlindVision » Sat Jun 20, 2009 3:01 pm
rahulg83 wrote:Total amount after n years if interest is compounded q times a year

A = P(1 + r/q)^nq

1080.56 = 1000(1+r/(100*2))^1*2

solve for r, between 7 and 8

D
Can you please help me understand why there is "(100*2)" in your denominator? I thought that it should simply have a "2" for number of compounds. Thanks!
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by Stuart@KaplanGMAT » Sat Jun 20, 2009 6:23 pm
BlindVision wrote:
rahulg83 wrote:Total amount after n years if interest is compounded q times a year

A = P(1 + r/q)^nq

1080.56 = 1000(1+r/(100*2))^1*2

solve for r, between 7 and 8

D
Can you please help me understand why there is "(100*2)" in your denominator? I thought that it should simply have a "2" for number of compounds. Thanks!
If you leave r as a percent instead of a decimal, you divide by 100 to change its form. If you already wrote r as a decimal, there's no need to divide by 100.

Since the interest is compounded semi-annually, you divide the annual interest by 2 to get the semi-annual rate.
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by rahulg83 » Sat Jun 20, 2009 11:27 pm
BlindVision wrote:
rahulg83 wrote:Total amount after n years if interest is compounded q times a year

A = P(1 + r/q)^nq

1080.56 = 1000(1+r/(100*2))^1*2

solve for r, between 7 and 8

D
Can you please help me understand why there is "(100*2)" in your denominator? I thought that it should simply have a "2" for number of compounds. Thanks!
as Stuart said, i assumed r as a percentage and not decimal. If u leave r as as a decimal, the value will come out between 0.07 and 0.08

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by ketkoag » Mon Jun 22, 2009 3:11 am
thanks a lot Stuart for you explanation..