A student borrows money to pay for university tuition. He borrows 1000 at the end of each month for four years. No payments are made to repay the loan until the end of five years. The loan accumulates interest at a 6% nominal interest rate convertible monthly for the first two years and at an 8% nominal interest rate convertible monthly for the following two years.

Calculate the loan balance at the end of four years immediately following the receipt of the final 1000.

• 54,098
• 55,224
• 55,762
• 56,134
• 56,350

A loan is amortized with level monthly payments at an annual effective interest rate of 10%. The amount of principal repaid in the 6th month is 500.

Calculate the principal repaid in the 30th month.

• 500
• 555
• 605
• 705
• 805

Seth repays a 30-year loan with a payment at the end of each year. Each of the first 20 payments is 1200, and each of the last 10 payments is 900. Interest on the loan is at an annual effective rate of i, i > 0. The interest portion of the 11th payment is twice the interest portion of the 21st payment.

Calculate the interest portion of the 21st payment.

• 250
• 275
• 300
• 325
• There is not enough information to calculate the interest portion of the 21st payment.

John took out a 20-year loan of 85,000 on July 1, 2005 at an annual nominal interest rate of 6% compounded monthly. The loan was to be paid by level monthly payments at the end of each month with the first payment on July 31, 2005.

Right after the regular monthly payment on June 30, 2009, John refinanced the loan at a new annual nominal rate of 5.40% compounded monthly, and the remaining balance will be paid with monthly payments beginning July 31, 2009. The amount of each payment is 500 except for a final drop payment.

Calculate the date of John’s last payment.

• July 31, 2022
• April 30, 2030
• May 31, 2030
• April 30, 2031
• May 31, 2031

A 25-year loan is being repaid with annual payments of 1300 at an annual effective rate of interest of 7%. The borrower pays an additional 2600 at the time of the 5th payment and wants to repay the remaining balance over 15 years.

Calculate the revised annual payment.

• 1054.58
• 1226.65
• 1300.00
• 1369.38
• 1512.12

A small business takes out a 10-year loan with level end-of-quarter payments. The payments are based on an annual nominal interest rate of 12% convertible quarterly. The amount of principal repaid in the 15th payment is 10,030.27.

Calculate the amount of interest paid in the 25th payment.

• 7,521
• 8,151
• 9,467
• 10,030
• 11,601

Trish had a loan with a balance of 4000 at the beginning of month 1. Starting with month 1, and every month thereafter, she made a payment of X in the middle of the month. At the beginning of month 4, and every 6 months thereafter, she borrowed an additional 800. Trish’s loan balance became 4000 again at the end of month 36. The annual nominal interest rate for the loan is 26.4%, convertible quarterly.

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

• [[math]]\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{4}}\right)^{2n-1}}}=\sum_{n=1}^{36}{\frac{X}{\left(1+{\frac{0.2640}{4}}\right)^{\frac{n-0.5}{3}}}} [[/math]]
• [[math]]\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{12}\right)^{6n-3}}=\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{12}\right)^{n-0.5}} [[/math]]
• [[math]]4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{4}}\right)^{\frac{6n-2}{3}}}}+\sum_{n=1}^{\frac{\displaystyle4000}{4}}{\frac{X}{\left(1+{\frac{0.2640}{4}}\right)^{\frac{n}{3}}}} [[/math]]
• [[math]]4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{12}}\right)^{6n-3}}}={\frac{4000}{\left(1+{\frac{0.2640}{12}}\right)^{36}}}+\sum_{n=1}^{36}{\frac{X}{\left(1+{\frac{0.2640}{12}}\right)^{n-0.5}}} [[/math]]
• [[math]]4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{4}}\right)^{2n-1}}}={\frac{4000}{\left(1+{\frac{0.2640}{4}}\right)^{12}}}+\sum_{n=1}^{\infty}{\frac{X}{\left(1+{\frac{0.2640}{4}}\right)^{n-0.5}}} [[/math]]

A borrower takes out a 15-year loan at an annual effective interest rate of i with payments of 50 at the end of each year. The borrower decides to pay off the loan early by making extra payments of 30 with each of the sixth through tenth regularly scheduled payments. As a result, the loan will be paid off at the end of 10 years (instead of 15).

Determine which of the following equations of value is correct.

• [math]50a_{\overline{15}|}= 50a_{\overline{10}|} + 30a_{\overline{5}|}[/math]
• [math]50s_{\overline{15}|}= 50s_{\overline{10}|} + 30s_{\overline{5}|}[/math]
• [math]50v^5s_{\overline{15}|}= 50s_{\overline{10}|} + 30s_{\overline{5}|}[/math]
• [math]50s_{\overline{15}|}= 50s_{\overline{10}|} + 30v^5s_{\overline{5}|}[/math]
• [math]50s_{\overline{15}|}= 50s_{\overline{10}|} + 30(1+i)^5s_{\overline{5}|}[/math]

A borrower takes out a loan to be repaid over 20 years. The first payment is 1102 payable at the end of the first month. Each subsequent monthly payment is five more than the previous month’s payment.

Calculate the accumulated value of the payments at the end of 15 years using an annual effective interest rate of 6.5%.

• 442,031
• 443,525
• 445,578
• 447,287
• 448,547

Fund J begins with a balance of 20,000 and earns an annual effective rate of 6.5%. At the end of each year, the interest earned and an additional 1000 is withdrawn from the fund so that by the end of the 20th year, the fund is depleted.

The annual withdrawals of interest and principal are deposited into Fund K, which earns an annual effective rate of 8.25%. At the end of the 20 th year, the accumulated value of Fund K is x.

Calculate x.

• 39,332
• 54,818
• 84,593
• 86,902
• 97,631