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In one community college, the cost per credit hour for in-state students has been increasing by $\$3$ a year. In another community college, it has been increasing by $9\%$ per year. This year, the cost is $\$65$ at both colleges. To the nearest dollar, what was the positive difference in costs $2$ years ago?

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Percent of change: word problems

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Q. In one community college, the cost per credit hour for in-state students has been increasing by

\$3

a year. In another community college, it has been increasing by

9\%

per year. This year, the cost is

\$65

at both colleges. To the nearest dollar, what was the positive difference in costs

2

years ago?

Calculate Cost $2$ Years Ago: Calculate the cost per credit hour $2$ years ago for the first community college. $\newline$ The cost has been increasing by $\$3$ a year, so $2$ years ago, the cost would have been $\$3$ less per year for $2$ years. $\newline$ Cost $2$ years ago = Current cost - ( $2$ years * $\$3/year$ increase) $\newline$ = $\$65$ - ( $2$ * $\$3$ ) $\newline$ = $\$65$ - $\$6$ $\newline$ = $\$59$
Calculate Cost Increase: Calculate the cost per credit hour $2$ years ago for the second community college. $\newline$ The cost has been increasing by $9$ % per year. Let's denote the cost $2$ years ago as $C$ . $\newline$ We know that the cost this year is $\$65$ , which is after two $9$ % increases. $\newline$ So, we can set up the equation: $C \times (1 + 9\%)^2 = \$65$ $\newline$ $C \times (1.09)^2 = \$65$ $\newline$ $C = \$65 / (1.09)^2$ $\newline$ $C \approx \$65 / 1.1881$ $\newline$ $C \approx \$54.73$ , which we round to the nearest dollar to get \$\(55\).
Calculate Cost Difference: Calculate the positive difference in costs per credit hour between the two colleges \(2\) years ago.\(\newline\)Difference = Cost at first college \(2\) years ago - Cost at second college \(2\) years ago\(\newline\)= \(\$59\) - \(\$55\)\(\newline\)= \(\$4\)

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