2007 CEMC Pascal Problems/Problem 5

Problem

The table shows the pay Leona earned for two different shifts at the same fixed hourly rate. How much will she earn for a five hour shift at this rate?

$\text{ (A) }\ $43.75 \qquad\text{ (B) }\ $46.25 \qquad\text{ (C) }\ $38.75 \qquad\text{ (D) }\ $36.25 \qquad\text{ (E) }\ $41.25$

Solution 1

Since the hourly rate is fixed, we can represent this as a linear function. Let $p(h)$ be the total pay at $h$ hours. We then have:

$p(h) = mh + b$ , where $m$ is the rate and $b$ is the y-intercept

Using $m = \frac{y_2 - y_1}{x_2 - x_1}$ , we have:

$m = \frac{49.50 - 24.75}{6 - 3} = \frac{24.75}{3} = $8.25$ per hour

This means:

$p(h) = 8.25h + b$

We can now plug in $h = 3$ to get:

$p(h) = 8.25 \times 3 + b = 24.75$

$24.75 + b = 24.75$

$b = $0$

We then have:

$p(h) = 8.25h$

Plugging in $h = 5$ , we have:

$p(5) = 8.25 \times 5 = \boxed {\textbf {(E) } $41.25}$

~anabel.disher

Solution 1.1

We can just simply subtract the slope from the amount at $6$ hours to find the amount of $5$ hours because exactly $1$ hour passed.

Doing this, we get $49.50 - 8.25 = \boxed {\textbf {(E) } $41.25}$

~anabel.disher

Solution 1.2

Like solution 1.1, we can find the slope. However, we can add the slope twice to the amount at $3$ hours because $2$ hours passed.

Doing this, we get $24.75 + 8.25 \times 2 = 24.75 + 16.50 = \boxed {\textbf {(E) } $41.25}$

~anabel.disher

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2007 CEMC Pascal Problems/Problem 5

Contents

Problem

Solution 1

Solution 1.1

Solution 1.2