Chapter 6.3, Problem 27IP

a.

To determine

To find: The money saved by A in 2011.

Answer to Problem 27IP

Money saved by A in 2011 is $530.

Explanation of Solution

Given information: A saved $500 in 2010 and she plans to save 6% more than her previous year’s savings for the next several years.

Formula used: $a\%\text{ of }b=\frac{a}{100}\times b$ .

Calculation:

Money saved by A in 2011 $=\$500+6\%\text{ of }\$500$

  $\begin{array}{l}=500+\frac{6}{100}\times 500\\ =500+30\\ =\$530\end{array}$

b.

To determine

To make: A table using the x values for 2010-2015.

Explanation of Solution

Given information: x represent the year and y represent the amount of money A has saved.

Formula used: $a\%\text{ of }b=\frac{a}{100}\times b$ .

Calculation:

x	y
2010	$\$500$
2011	$\$530$
2012	$\$530+\frac{6}{100}\times 530=\$561.8$
2013	$\$561.8+\frac{6}{100}\times 561.8=\$595.508$
2014	$\$595.508+\frac{6}{100}\times 595.508=\$631.23848$
2015	$\$631.23848+\frac{6}{100}\times 631.23848=\$669.1127888$

c.

To determine

To analyze: Whether A’s savings increase by constant amount each year.

Explanation of Solution

Difference between savings in 2010 and 2011 $=530-500=\$30$

Difference between savings in 2011 and 2012 $=561.8-530=\$31.8$

Difference between savings in 2012 and 2013 $=595.508-561.8=\$33.708$

Difference between savings in 2013 and 2014 $=631.23848-595.508=\$35.73048$

Difference between savings in 2014 and 2015 $=669.1127888-631.23848=\$37.8743088$

Thus, A’s savings doesn’t increase by constant amount each year.