Question 641·Hard·Algebra

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A passenger train traveling at a constant speed covers 180 miles in the same time that a freight train covers 135 miles. The passenger train travels 15 miles per hour faster than the freight train. What is the speed, in miles per hour, of the freight train?

Your answer

A speed is distance traveled per unit of time. “In the same time” is the cue to compare how much farther the faster train traveled. Divide that extra distance by the difference in speeds to find the shared travel time. Then divide the freight train’s distance by that time. Don’t mistake extra miles for miles per hour.

Hints

Hint 1
Because both trains traveled for the same number of hours, subtract their distances. That tells you how many extra miles the passenger train covered, not how fast either train traveled.
Hint 2
A speed difference of $15$ miles per hour means the passenger train gains $15$ miles each hour. How can you use its extra miles to find the shared travel time?
Hint 3
Speed is distance divided by time. Once you know the shared number of hours, use the freight train’s distance to find the speed the question asks for.

Step-by-step

Approach 1: Use the extra distance to find the time
Step 1
Find the passenger train’s extra distance
The trains traveled for the same time, so subtract their distances to express how far ahead the passenger train got:
$180-135\text{ miles}.$
That’s extra distance, not a speed in miles per hour.
Step 2
Find the shared travel time
The passenger train gains $15$ miles each hour. So extra distance divided by extra speed gives the shared time. Type $\frac{180-135}{15}$ into Desmos; it shows $3$ . Both trains traveled for $3$ hours.
Step 3
Find the freight train’s speed
The freight train covered $135$ miles in those $3$ hours. Speed is distance divided by time, so type $\frac{135}{3}$ on the next Desmos line. It shows $45$ , the freight train’s speed in miles per hour. Grid in 45.

Approach 2: Compare distances and speeds as ratios
Step 1
Match the distance ratio to the speed ratio
Let $x$ be the freight train’s speed in miles per hour. The passenger train’s speed is $x+15$ . Because both travel for the same time, that time cancels when you divide one train’s distance by the other’s. So their distance ratio, a comparison by division, equals their speed ratio:
$\frac{135}{180}=\frac{x}{x+15}.$
Step 2
Solve for the freight speed
Type $\frac{135}{180}\sim\frac{x_1}{x_1+15}\{x_1>0\}$ into Desmos. Here $x_1$ means the freight speed, $\sim$ asks Desmos to match the two ratios, and the restriction keeps the speed positive. Under PARAMETERS, Desmos shows $x_1=45$ . The freight train travels $45$ miles per hour. Grid in 45.

Question 641·Hard·Algebra

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