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What is the effective annual rate if the stated rate is 12% compounded quarterly?
A)
57.35%.
B)
12.55%.
C)
12.00%.



EAR = (1 + 0.12 / 4)4 – 1 = 12.55%

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As the number of compounding periods increases, what is the effect on the annual percentage rate (APR) and the effective annual rate (EAR)?
A)
APR increases, EAR increases.
B)
APR remains the same, EAR increases.
C)
APR increases, EAR remains the same.



The APR remains the same since the APR is computed as (interest per period) × (number of compounding periods in 1 year). As the frequency of compounding increases, the interest rate per period decreases leaving the original APR unchanged. However, the EAR increases with the frequency of compounding.

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A local bank advertises that it will pay interest at the rate of 4.5%, compounded monthly, on regular savings accounts. What is the effective rate of interest that the bank is paying on these accounts?
A)
4.59%.
B)
4.65%.
C)
4.50%.



(1 + 0.045 / 12)12 − 1 = 1.0459 − 1 = 0.0459.

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As the number of compounding periods increases, what is the effect on the EAR? EAR:
A)
increases at a decreasing rate.
B)
increases at an increasing rate.
C)
does not increase.



There is an upper limit to the EAR as the frequency of compounding increases. In the limit, with continuous compounding the EAR = eAPR –1. Hence, the EAR increases at a decreasing rate.

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In 10 years, what is the value of $100 invested today at an interest rate of 8% per year, compounded monthly?
A)
$222.
B)
$216.
C)
$180.



N = 10 × 12 = 120; I/Y = 8/12 = 0.666667; PV = –100; PMT = 0; CPT → FV = 221.96.

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If $1,000 is invested at the beginning of the year at an annual rate of 48%, compounded quarterly, what would that investment be worth at the end of the year?
A)
$1,574.
B)
$1,048.
C)
$4,798.



N = 1 × 4 = 4; I/Y = 48/4 = 12; PMT = 0; PV = –1,000; CPT → FV = 1,573.52.

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Given: an 11% annual rate compounded quarterly for 2 years; compute the future value of $8,000 today.
A)
$8,962.
B)
$9,939.
C)
$9,857.



Divide the interest rate by the number of compound periods and multiply the number of years by the number of compound periods. I = 11 / 4 = 2.75; N = (2)(4) = 8; PV = 8,000.

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If $2,500 were put into an account at the end of each of the next 10 years earning 15% annual interest, how much would be in the account at the end of ten years?
A)
$41,965.
B)
$27,461.
C)
$50,759.



N = 10; I = 15; PMT = 2,500; CPT → FV = $50,759.

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An investor will receive an annuity of $5,000 a year for seven years. The first payment is to be received 5 years from today. If the annual interest rate is 11.5%, what is the present value of the annuity?
A)
$15,000.
B)
$13,453.
C)
$23,185.



With PMT = 5,000; N = 7; I/Y = 11.5; value (at t = 4) = 23,185.175. Therefore, PV (at t = 0) = 23,185.175 / (1.115)4 = $15,000.68.

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What is the present value of a 10-year, $100 annual annuity due if interest rates are 0%?
A)
$900.
B)
$1,000.
C)
No solution.



When I/Y = 0 you just sum up the numbers since there is no interest earned.

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