# math

Barry heard in his Personal Finance class that he should start investing as soon as possible. He had always thought that it would be smart to start investing after he finishes college and his salary is high enough to pay the bills and to have money left over. He projects that will 50-10 years from now. Barry wants to compare the difference between investing now and investing later. A financial advisor who spoke to Barry suggested that a Roth IRA (Individual Retirement Account) would be a more profitable investment over long term than a regular IRA, so Barry wants to seriously consider the Roth IRA. When table values do not include the information you need use the FV= \$1(1 +R)^N where R is the period rate and N is the number of periods.

1. If Barry purchases a \$2,000 Roth IRA when he is 25. Years old and expects to earn an average of 6% per year compounded annually over 35 years (until he is 60), how much will accumulate in the investment?

2. If Barry doesn’t put the money in the IRA until he is 35 years old, how much money will accumulate in the account by the time he is 60 years old? How much less will he earn because he invested 10 years later?
3. Interest rate is critical to the speed at which your investment grows. If \$1 is invested at 2%, it takes approximately 34.9 years to double. If \$ 1 is invested at 5%, it takes approximately 14.2 years to double. Use table 13-1 to determine how many years it takes \$1 to double if invested at 10; at 12%.

4. at what interest rate would you need to invest to have your money double in 10 years?

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1. 1. P1 = Po(1+r)^n

Po = \$2,000

r = 0.06

n = 35yrs * 1comp./yr. = 35 Compounding
periods.

P1 = 2000(1.06)^35 = \$15,372.17

2. Use same Eq as above with a 25-yr.
investment(60-35).

P2 = Po(1+r)^n

Difference = P1-P2.

3. Use your table and compare with calculated values below.

P = 1(1.10)^n = 2
n*Log1.1 = Log 2
n = Log 2/Log1.1 = 7.2725409 Compounding
periods.

T=7.273comp. periods * 1yr/comp. period = 7.273 years.

P = 1(1.12)^n = 2
n*Log 1.12 = Log 2
n = Log 2/Log 1.12 = 6.11 compounding
periods. T = 6.11 years.

4. P = \$1(1+r)^10 == 2
10*Log(1+r) = Log 2
Log(1+r) = Log 2/10 = 0.03010
1+r = 10^0.0301
1+r = 1.07177
r = 0.07177 or 7.18%.

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posted by Henry

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