Chapter ISG, Problem 7.2.5P

a.

To determine

To write an expression.

Answer to Problem 7.2.5P

  $A=n(1+0.04t)$ is the expression

Explanation of Solution

Given:

Rate of interest=4%

Formula used:

  $A=P(1+rt)$ Where, P is the principle amount, tis time in years, r is rate of interest.

  $r=\frac{R}{100}$ Where, R is rate of year as percent.

Calculation:

Substituting values,

  $A=n(1+0.04t)$ n is dollars invested.

b.

To determine

To find the value of investment changed

Answer to Problem 7.2.5P

  $24\%$ is the change in investment.

Explanation of Solution

Given:

Time₁= 2 years

Time₂= 8 years

Calculation:

By the end of two years,

  $A=n(1+0.04t)$

Substituting values,

  $A=n(1+0.04(2))$

Solving,

  $A=1.08n$   (1)

By the end of 8 years,

  $A=n(1+0.4t)$

Substituting values,

  $A=n(1+0.04(8))$

Solving,

  $A=1.32n$   (2)

Change In investment, subtract (1) from (2)

Change= $1.32n-1.08n$

Solving,

  = $(1.32-1.08)n$

So,

  = $(0.24)n$

Percentage change in investment= $\frac{change}{initial}\times 100$

Substituting values,

  = $\frac{0.24n}{n}\times 100$

Solving,

  = $24\%$

Hence, change in investment is $24\%$