# 1. Suppose that a $100,000 investment grows 3% during the first year and 4% during the second...

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2. Rework Exercise 1 for the case in which the investment earns 4% during the first year and 3% during the second year. Is the answer to Exercise 2 greater than, less than, or equal to the answer to Exercise 1?

3. Consider an annuity in which $100,000 is invested and $10,000 is withdrawn at the end of each year. Suppose that the interest rate is 3% during the first year and 4% during the second year. What is the balance at the end of the second year?

4. Rework Exercise 3, where the interest rate is 4% during the first year and 3% during the second year. Is the balance at the end of the second year greater than, less than, or equal to the answer to Exercise 3? Explain why your answer to this question makes sense.

5. Show that, if an investment of P dollars declines by 4% during a year, the balance at the end of the year is P ( (1 - .04)-that is, P ( (.96).

6. Show that, if an investment of P dollars declines by r% during a year, the balance at the end of the year is P ( (1 - r/100). We say that the investment earned -r%.

7. Show that, if an investment of P dollars earns r% one year and s% the following year, then the balance after the 2-year period is P(1 + r/100)(1 + s/100). Conclude that the order of the two numbers r and s does not affect the balance after 2 years. (Note: The numbers r and s can be positive or negative.)

8. Which of the following two statements is true?

(a) Suppose that an investment of P dollars earns 4% one year and loses 4% the next year. Then the value of the investment at the end of the 2-year period will be P dollars.

(b) Suppose that investment A earns 4% one year and then loses 3% the next year, and investment B loses 3% one year and then gains 4% the next year. Then the balances of the two investments will be the same after the 2-year period.

9. Show that, if $1000 is invested at 8% interest compounded annually, then it will double in about 9 years. (9 = 72/8)

10. Show that, if $1000 is invested for 6 years, then it will approximately double in that time if it appreciates at 12% per year. (12 = 72/6.)

Rule of 72: If money is invested at r% interest compounded annually, then it will double in about 72/r years. Alternatively, if money is invested for n years, then it will double during that time if it appreciates by about (72/n) % per year.

11. Conclude from the Rule of 72 that, for an interest rate of r%, (1 + r/100)72/r ( 2. Also, conclude that for a number of years, n, (1 + .72/n)n ( 2.

We needn't restrict ourselves to doubling. For instance, the Rule of 114 says that, if money is invested at r % interest, then it will triple in about 114/r years. Table 1 gives the numbers associated with several different multiples. Let us denote the number associated with m by Nm. That is, N2 = 72 and N3 = 114.

12. Notice that N6 = N2 + N3. Explain why this makes sense.

13. Explain why Nn(p = Nn + Np for any positive numbers n and p.

14. Find N10, and then use that number to estimate the amount of time required for money to increase tenfold if invested at 8% interest compounded annually. Check your answer by raising (1.08) to that power.

15. Use the values of N2 and N3 to find N1.5, the number associated with money increasing by one-half. Estimate the amount of time required for $1000 to grow to $1500 when invested at 7% interest compounded annually.

Annuity

An annuity is a series of equal payment made at equal intervals during a period of time. In other words annuity is a contract between insurer and insurance company in which insurer make a lump-sum payment or a series of payment and, in return,...

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**Related Book For**

## Finite Mathematics and Its Applications

**ISBN:** 978-0134768632

12th edition

**Authors:** Larry J. Goldstein, David I. Schneider, Martha J. Siegel, Steven Hair