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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.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
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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!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
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.BlindVision wrote: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!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
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 0.08BlindVision wrote: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!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
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