The following two annuities-immediate have the same present value at an annual effective interest rate of i, i > 0.

1. A ten-year annuity with annual payments of 475.
2. A perpetuity with annual payments of 400 in years 1-5, zero in years 6-10, and 400 in years 11 and beyond.

Calculate i.

• 10.65%
• 10.75%
• 10.85%
• 10.95%
• 11.05%

A two-year loan of 100 is repaid with a payment of X at the end of the first year and 2X at the end of the second year. The annual effective interest rate charged by the lender is 8% in the first year and i in the second year. The annual effective yield rate for the lender is 10%.

Calculate i.

• 12.8%
• 12.9%
• 13.0%
• 13.1%
• 13.2%

An investor will accumulate 10,000 at the end of ten years by making level deposits of X at the beginning of each year. The deposits earn 12% simple interest at the end of every year but the interest is reinvested at an annual effective rate of 8%.

Calculate X.

• 508.79
• 541.47
• 569.84
• 597.73
• 608.42

An investment of 10,000 produces a series of 30 annual payments. The first payment of X is made one year after the investment is made. Each successive payment decreases by 5 from the previous payment.

At an annual effective interest rate of 5%, calculate X.

• 685
• 695
• 705
• 715
• 725

You are given the following information about a perpetuity with annual payments:

1. The first 15 payments are each 2500, with the first payment to be made three years from now.
2. Beginning with the 16th payment, each payment is k% larger than the previous payment.
3. Using an annual effective interest rate of 3.5%, the present value of the perpetuity is 115,000.

Calculate k

• 1.66
• 1.74
• 1.78
• 1.83
• 1.89

An insurance company sells an annuity that provides 20 annual payments, with the first payment beginning one year from today and each subsequent payment 2% greater than the previous payment. Using an annual effective interest rate of 3%, the present value of the annuity is 200,000.

Calculate the amount of the final payment from this annuity.

• 11,282
• 16,436
• 16,765
• 19,784
• 24,162

Sam deposits 5000 at the beginning of each year for ten years. The deposits earn an annual effective interest rate of i. All interest is reinvested at an annual effective interest rate of 5%. Sam has 100,000 at the end of ten years.

Calculate i.

• 11.6%
• 15.6%
• 19.1%
• 23.2%
• 27.6%

Kevin makes a deposit at the end of each year for 10 years into a fund earning interest at a 4% annual effective rate. The first deposit is equal to X, with each subsequent deposit 9.2% greater than the previous year’s deposit. The accumulated value of the fund immediately after the 10 th deposit is 5000.

Calculate X.

• 255
• 260
• 270
• 279
• 293

An investor makes deposits into an account at the end of each year for ten years. The deposit in year one is 1, year two is 2 and so forth until the final deposit of 10 in year ten. The account pays interest at an annual effective rate of i. Immediately following the final deposit, the investor uses the entire account balance to purchase a perpetuity-immediate at an annual effective interest rate of i. The perpetuity makes annual payments of 10.

Calculate the purchase price of the perpetuity.

• 68.0
• 72.4
• 76.2
• 81.3
• 91.3

An investor’s new savings account earns an annual effective interest rate of 3% for each of the first ten years and an annual effective interest rate of 2% for each year thereafter. The investor deposits an amount X at the beginning of each year, starting with year 1, so that the account balance just after the deposit in the beginning of year 26 is 100,000.

Determine which of the following is an equation of value that can be used to solve for X.

• [[math]]{\frac{1(00,000}{(1.03)^{10}(1.02)^{15}}}=X\sum_{k=1}^{11}{\frac{1}{(1.03)^{k-1}}}+X\sum_{k=12}^{26}{\frac{1}{(1.03)^{10}(1.02)^{k-11}}} [[/math]]
• [[math]]{\frac{100,000}{(1.03)^{10}(1.02)^{16}}}=X\sum_{k=1}^{10}{\frac{1}{(1.03)^{k}}}+X\sum_{k=11}^{26}{\frac{1}{(1.03)^{10}(1.02)^{k-10}}} [[/math]]
• [[math]]\frac{100,000}{\left(1.03\right)^{10}(1.02)^{16}}=X\sum_{k=1}^{11}\frac{1}{\left(1.03\right)^{k-1}}+X\sum_{k=11}^{26}\frac{1}{\left(1.03\right)^{10}(1.02)^{k-11}} [[/math]]
• [[math]]{\frac{100,000}{(1.03)^{10}(1.02)^{15}}}=X\sum_{k=1}^{11}{\frac{1}{(1.03)^{k-1}}}+X\sum_{k=12}^{26}{\frac{1}{(1.02)^{k-1}}} [[/math]]
• [[math]]{\frac{100,000}{(1.03)^{10}(1.02)^{16}}}=X\sum_{k=1}^{10}{\frac{1}{(1.03)^{k}}}+X\sum_{k=11}^{26}{\frac{1}{(1.02)^{k}}}[[/math]]