# Section 4B

...

### Basic skills. Evaluate or simplify (cf. p.199)

powers: Exercises 16,18,20,22,24,26 p. 214

$$3^4=81$$

$$3^{-2} = \frac{1}{9}$$

$$81^{1/2} = \left( 9^2 \right)^{1/2} = 9^{2 \times 1/2} = 9^1=9$$

$$2^3 \times 2^5 = 2^{3+5}=2^8=256$$

$$6^2 \times 6^{-2} = 6^{2-2} = 6^0 = 1$$

$$3^3+2^3 = 27+8=35$$

### Basic skills. Algebra review (cf. p.210)

solving equations: Exercises 28,30,32,34,36,38,46,50 p. 214

$$y+4=7$$ ...yields... $$y=3$$

$$2x=8$$ ...yields... $$x=4$$

$$4y+2=18$$ ...yields... $$y=4$$

$$1-6y=13$$ ...yields... $$y=-2$$

$$5-4s=6s-5$$ ...yields... $$s=1$$

$$3n-16=53$$ ...yields... $$n=23$$

$$p^{1/3}=3$$ ...yields... $$p=27$$

$$v^3+4=68$$ ...yields... $$v=4$$

### Simple interest. Exercise 52, p.214

You deposit $$\1,200$$ in an account with an annual interest rate of 3%. Calculate the amount of money you will have after 5 years, assuming that you earn simple interest.

Solution. Every year you earn $$0.03 \times 1200 = \36$$. After 5 years you will earn $$5 \times 36=\ 180$$. Your accumulated balance will be $$1200 + 180 = \1380$$.

### Compound interest. Exercise 58, p.214

Use the compound interest formula to compute the balance if $$\ 10,000$$ is invested for 20 years in an account with an APR of 2.5% compounded annually

Solution. The formula
$$A = P \times (1+APR)^Y$$ gives us
$$A=10000 \times (1+0.025)^{20} = \ 16386.16$$

### Compound interest. Exercise 64, p.214

What is the accumulated balance if $$\2000$$ is invested for 5 years with an APR of 3% and daily compounding?

Solution. The compound interest formula for interest paid $$n$$ times a year is
$$A = P \times (1+\frac{APR}{n})^{(nY)}$$
In this case, n=365, and
$$A = 2000 \times (1+\frac{0.03}{365})^{(365 \times 5)} = \ 2323.65$$ <=== answer

To compare with continuous compounding
$$A = P \times e^{(APR \times Y)}$$ $$= 2000 \times e^{(0.03 \times 5)} = \ 2323.67$$

### Annual Percentage Yield (APY). Exercise 72, p.214

Find the APY if a bank offers an APR of 3.2% compounded monthly

Solution. Invest $$\100$$, and after 1 year you will have
$$A = P \times (1+\frac{APR}{n})^{(nY)} = 100 \times (1+ \frac{0.032}{12})^{(12 \times 1)} = \ 103.247$$
The relative difference is $$\frac{103.247-100}{100} =0.03247 =3.247 \%$$

### Continuous compounding. Exercise 76, p.215

Compute the balance after 1,5 and 20 years for a $$\2000$$ deposit in an account with APR of 3.1%, and continuous compounding

Solution. Continuous compounding formula
$$A = P\times e^{(APR \times Y)}$$
gives us
$$A = 2000\times e^{(0.031 \times Y)}$$.
After 1 year: $$A = 2000\times e^{(0.031 \times 1)} = \ 2062.97$$
After 5 years: $$A = 2000\times e^{(0.031 \times 5)} = \ 2335.32$$
After 20 year: $$A = 2000\times e^{(0.031 \times 20)} = \ 3717.86$$

### Planning ahead. Exercise 82, p.215

How much must you deposit today into an account with quarterly compounding and an APR of 4.5% in order to have $$\ 25,000$$ in 8 years?

Solution. Compound interest formula for interest paid $$n$$ times a year
$$A = P \times (1+\frac{APR}{n})^{(nY)}$$
solved for the principal
$$P= \frac{A}{(1+\frac{APR}{n})^{(nY)} }$$
gives us
$$P= \frac{25000}{(1+\frac{0.045}{4})^{(4\times 8)} } = \17477$$

### College fund. Exercise 86, p.215

How much must you deposit today into an account with an APR of 5.5% compounded daily in order to have a $$\120,000$$ college fund in 15 years?

Solution. Compound interest formula for interest paid $$n$$ times a year
$$A = P \times (1+\frac{APR}{n})^{(nY)}$$
solved for the principal
$$P= \frac{A}{(1+\frac{APR}{n})^{(nY)} }$$
gives us
$$P= \frac{120000}{(1+\frac{0.055}{365})^{(365\times 15)} } = \52,591.47$$

##### Small rate difference Exercise 90, p.215

Jose invests $$\1500$$ into an account that earns $$5.6\%$$ compounded annually. Marta invests $$\1500$$ into an account that earns $$5.7\%$$ compounded annually. Compute the balance after $$10$$ and $$30$$ years.

Solution. Compound interest formula for interest paid once a year is
$$A = P \times (1+APR)^Y$$.

After $$10$$ years:

Jose: $$1500 \times (1+0.056)^{10} = \ 2,586.6$$

Marta: $$1500 \times (1+0.057)^{10} = \ 2,611.2$$

##### Small rate difference Exercise 90, p.215 ... continued

Jose invests $$\1500$$ into an account that earns $$5.6\%$$ compounded annually. Marta invests $$\1500$$ into an account that earns $$5.7\%$$ compounded annually. Compute the balance after $$10$$ and $$30$$ years.

Solution. Compound interest formula for interest paid once a year is
$$A = P \times (1+APR)^Y$$.

After $$30$$ years:
Jose: $$1500 \times (1+0.056)^{30} = \ 7,691.5$$
Marta: $$1500 \times (1+0.057)^{30} = \ 7.913$$

##### Small rate difference Exercise 90, p.215 ... continued

Jose invests $$\1500$$ into an account that earns $$5.6\%$$ compounded annually. Marta invests $$\1500$$ into an account that earns $$5.7\%$$ compounded annually. Compare the balance after $$10$$ years.

Comparison.

After $$10$$ years Jose has $$\ 2,586.6$$, while Marta has $$\ 2,611.2$$.
Relative difference $$=\frac{2,611.2 - 2,586.6}{2,586.6}$$ $$= \frac{24.6}{2,586.6}$$ $$\approx 0.0095 = 0.95\%$$
After $$10$$ years, Marta has $$\24.6$$ or $$0.95\%$$ more than Jose.

##### Small rate difference Exercise 90, p.215 ... continued

Jose invests $$\1500$$ into an account that earns $$5.6\%$$ compounded annually. Marta invests $$\1500$$ into an account that earns $$5.7\%$$ compounded annually. Compare the balance after $$30$$ years.

Comparison.

After $$30$$ years Jose has $$\ 7,691.5$$, while Marta has $$\ 7.913$$.
Relative difference $$=\frac{7913 - 7691.5}{7691.5}$$ $$= \frac{221.5}{7691.5}$$ $$\approx 0.029 = 2.9\%$$
After $$30$$ years, Marta has $$\ 221.5$$ or $$2.9\%$$ more than Jose.