Rise over run word problems express a constant rate of change by comparing vertical change to horizontal change. In coordinate language, “rise” is the change in \(y\), “run” is the change in \(x\), and the ratio \(\Delta y / \Delta x\) is the slope of the line modeling the situation.
Rise, run, slope. Rise corresponds to \(\Delta y = y_2 - y_1\). Run corresponds to \(\Delta x = x_2 - x_1\). The slope is \[ m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}. \]
Percent grade. In many rise over run word problems (ramps, hills, stairs), percent grade is the slope written as a percent: \[ \text{grade} = 100 \cdot \frac{\text{rise}}{\text{run}}\%. \]
Language-to-algebra translation
| Phrase in the scenario | Algebraic meaning | Typical units |
|---|---|---|
| “Goes up 3 meters for every 12 meters forward” | \(m=\dfrac{3}{12}=\dfrac{1}{4}\) | meters per meter (dimensionless ratio), often reported as a percent |
| “Drops 2 feet per 5 feet” | \(m=\dfrac{-2}{5}\) (negative slope because height decreases) | feet per foot |
| “Climbs 180 m over 2.4 km” | \(m=\dfrac{180}{2400}=0.075\) | convert run to meters for a clean ratio |
| “Constant rate of change” | linear model; graph is a straight line | depends on context (height vs. distance, cost vs. items, etc.) |
Geometric meaning on a graph
Worked context examples
Ramp example (units preserved). A wheelchair ramp rises 0.75 m over a horizontal run of 9 m. The slope is \[ m=\frac{0.75}{9}=\frac{1}{12}\approx 0.0833. \] The percent grade is \(100m\%\approx 8.33\%\). With the base at \((0,0)\) and horizontal distance \(x\) measured in meters, the height model is \[ y=\frac{1}{12}x. \]
Trail example (unit conversion inside the model). A trail gains 180 m of elevation over 2.4 km of horizontal distance. Converting \(2.4\text{ km}=2400\text{ m}\) gives \[ m=\frac{180}{2400}=0.075, \] so the grade is \(7.5\%\). A linear model for elevation gain above the starting point is \(y=0.075x\) when both \(x\) and \(y\) are measured in meters.
Linear-equation forms that fit rise over run word problems
Rise over run word problems describe a constant rate, so a linear equation is the natural algebraic model. Two common equivalent forms are:
\[ y-y_1=m(x-x_1) \qquad\text{and}\qquad y=mx+b. \]
The point-slope form \(y-y_1=m(x-x_1)\) aligns with data described by two points (two distances and two heights). The slope-intercept form \(y=mx+b\) highlights the rate \(m\) and the initial value \(b\) at \(x=0\).
Common pitfalls
Negative slope appears when the quantity modeled by \(y\) decreases as \(x\) increases (a descent rather than a climb), so rise is negative even if the “amount of drop” is stated as a positive number. Mixed units distort slope, so rise and run use the same base length unit (meters with meters, feet with feet). Percent grade is not the slope itself; it is \(100\cdot m\), so a slope of \(0.08\) corresponds to an \(8\%\) grade.