Chapter ISG, Problem 7.2.6P

a.

To determine

To find meaning of decrease in investment by 5%

Answer to Problem 7.2.6P

The decrease in investment by 5% means investment value is 95 dollars now.

Explanation of Solution

Given information:

Decrease in investment

Decrease in investment=5%

5% of investment= $\frac{5}{100}\times 100$

Solving,

5% decrease of investment= investment-decrease

= 100-5

Hence, decrease in 5% of investment means loss of 5 dollars.

b.

To determine

To explain the statement

Answer to Problem 7.2.6P

The statement is correct.

Explanation of Solution

Given information:

An investment

5% decrease in investment=95 dollars.

Multiplying investment by 0.95=0.95 $\times 100$

And,

Multiplying investment by 0.95=95 dollars.

The statement is correct

c.

To determine

To write an expression

Answer to Problem 7.2.6P

The equation is

  $A=n(1+\frac{Rt}{100})$

Explanation of Solution

Given information:

Time given= $t$ years

Investment= $n$ dollars

Calculation:

The equation is

  $A=n(1+\frac{Rt}{100})$

Where $A$ is value after t year, n is initial investment, R is percent rate of decrease of increase in value.

d.

To determine

To find percentage change of investment

Answer to Problem 7.2.6P

Percentage change in investment is $\frac{7R}{50+R}\times 100$

Explanation of Solution

Given information:

  $\begin{array}{l}{t}_1^{}=2\\ t_2^{}=8\end{array}$

Calculation:

For $t=2$

Substituting value of t,

  $A=n(1+\frac{2R}{100})$

Solving,

  $A=n(1+\frac{R}{50})$

(1)

For $t=8$

Substituting values,

  $A=n(1+\frac{8R}{100})$

Solving,

  $A=n(1+\frac{4R}{25})$

Change in investment= $n(1+\frac{4R}{25})$ - $n(1+\frac{R}{50})$

Solving,

= $n(\frac{4R}{25}-\frac{R}{50})$

So,

= $n(\frac{7R}{50})$

Percentage change= $\frac{n(\frac{7R}{50})}{n(1+\frac{R}{50})}\times 100$

Solving,

= $\frac{7R}{50+R}\times 100$

Hence, percentage change in investment is $\frac{7R}{50+R}\times 100$