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Math Practice

Distance, Rate, and Time Practice Problems

Twelve hand-built distance-rate-time problems with full worked solutions. Each one uses a chart with one row per moving object, the formula d = rt, and a single equation relating the distances.

  1. 1.

    Two cars leave towns 420 miles apart at the same time, driving toward each other at 55 mph and 65 mph. When do they meet?

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    $$ 55t + 65t = 420 \quad\Longrightarrow\quad 120t = 420 \quad\Longrightarrow\quad t = 3.5 \text{ hours} $$

  2. 2.

    A runner and a cyclist start at the same point, going opposite directions at 6 mph and 15 mph. When are they 42 miles apart?

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    $$ 6t + 15t = 42 \quad\Longrightarrow\quad 21t = 42 \quad\Longrightarrow\quad t = 2 \text{ hours} $$

  3. 3.

    Two trains 480 miles apart travel toward each other. One is 10 mph faster than the other. They meet in 4 hours. Find each speed.

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    Let the slower be \(r\); the faster is \(r + 10\).

    $$ 4r + 4(r + 10) = 480 \quad\Longrightarrow\quad 8r + 40 = 480 \quad\Longrightarrow\quad r = 55 $$

    Speeds: \(55\) mph and \(65\) mph.

  4. 4.

    Two cyclists 60 miles apart ride toward each other. One is twice as fast as the other. They meet in 2.5 hours. Find each speed.

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    Let the slower be \(r\); the faster is \(2r\).

    $$ 2.5r + 2.5(2r) = 60 \quad\Longrightarrow\quad 7.5r = 60 \quad\Longrightarrow\quad r = 8 $$

    Speeds: \(8\) mph and \(16\) mph.

  5. 5.

    A freight train leaves a station at 30 mph. Three hours later a passenger train leaves the same station on the same track at 50 mph. How long does the passenger train travel before catching the freight train?

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    Let \(t\) be the passenger train's time; the freight train's is \(t + 3\).

    $$ 50t = 30(t + 3) \quad\Longrightarrow\quad 50t = 30t + 90 \quad\Longrightarrow\quad 20t = 90 \quad\Longrightarrow\quad t = 4.5 \text{ hours} $$

  6. 6.

    A hiker walks out at 4 mph and back over the same trail at 3 mph. The round trip takes 3.5 hours. How far out did the hiker go?

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    $$ \frac{d}{4} + \frac{d}{3} = 3.5 $$

    Multiply by 12: \(3d + 4d = 42\), so \(7d = 42\), \(d = 6\) miles.

  7. 7.

    Car A leaves at 8:00 a.m. at 50 mph. Car B leaves the same point at 10:00 a.m. at 70 mph in the same direction. When does B catch A?

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    Let \(t\) be B's travel time; A's is \(t + 2\).

    $$ 70t = 50(t + 2) \quad\Longrightarrow\quad 20t = 100 \quad\Longrightarrow\quad t = 5 \text{ hours} $$

    B catches A at \(10\):00 a.m. \(+ 5\) h \(=\) 3:00 p.m.

  8. 8.

    A family drives to a cabin at 60 mph and returns at 40 mph. The return takes 1 hour longer. How far is the cabin?

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    $$ \frac{d}{40} = \frac{d}{60} + 1 $$

    Multiply by 120: \(3d = 2d + 120\), so \(d = 120\) miles.

  9. 9.

    A boat goes 36 miles downstream in 2 hours and returns in 3 hours. Find the boat's still-water speed and the current speed.

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    $$ 2(s + c) = 36 \;\Rightarrow\; s + c = 18 \qquad 3(s - c) = 36 \;\Rightarrow\; s - c = 12 $$

    Adding: \(2s = 30\), so \(s = 15\) mph and \(c = 3\) mph.

  10. 10.

    A plane flies 900 miles with the wind in 3 hours and back against the wind in 5 hours. Find the still-air speed and the wind speed.

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    $$ 3(s + w) = 900 \;\Rightarrow\; s + w = 300 \qquad 5(s - w) = 900 \;\Rightarrow\; s - w = 180 $$

    Adding: \(2s = 480\), so \(s = 240\) mph and \(w = 60\) mph.

  11. 11.

    A jogger runs out at 5 mph. Fifteen minutes later a friend on a bike leaves the same spot at 12 mph on the same route. How far from the start does the friend catch the jogger?

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    \(15\) min \(= 0.25\) h. Let \(t\) be the friend's time; the jogger's is \(t + 0.25\).

    $$ 12t = 5(t + 0.25) \quad\Longrightarrow\quad 12t = 5t + 1.25 \quad\Longrightarrow\quad 7t = 1.25 \quad\Longrightarrow\quad t = \tfrac{1.25}{7} $$

    Distance \(= 12t = \tfrac{15}{7} \approx 2.14\) miles.

  12. 12.

    A boat's still-water speed is 12 mph. It travels 45 miles downstream and 45 miles back in a total of 8 hours. Find the current speed.

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    $$ \frac{45}{12 + c} + \frac{45}{12 - c} = 8 $$

    Multiply by \((12 + c)(12 - c)\): \(45(12 - c) + 45(12 + c) = 8(144 - c^{2})\), so \(1080 = 1152 - 8c^{2}\), giving \(8c^{2} = 72\), \(c^{2} = 9\), \(c = 3\) mph.
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Distance, Rate, and Time Practice Problems

The Distance, Rate, and Time lesson works through the four standard scenarios with a chart; this set is 12 problems to practice on, each with a full solution.

How to use this set

These problems are hand-written, not auto-generated. For each one, draw a chart with a row per moving object, fill it in, write the distance relationship as an equation, and solve. Then open the solution.

Common mistakes to watch for

Entering a known distance in the chart’s distance column. That column is always rate times time; use a known distance in the equation.

Giving both objects the same time when one has a head start.

Adding distances that should be set equal (or vice versa).

Mixing units — convert minutes to hours first.

Where to go next

Work problems use the same chart with rates of work — see the Work Problems Practice Problems. Round-trip problems that lead to a variable in a denominator connect to rational equations. For the four scenarios and worked examples, see the Distance, Rate, and Time lesson.

Frequently Asked Questions

What is the formula for distance, rate, and time problems?+

. Rearranged: , . Every motion problem applies this once per moving object, then links them with one equation about distances.

How do I set up the chart?+

One row per moving object, columns for rate, time, and distance. Fill in what you know, name one unknown with a variable, and compute distance as rate times time. The equation comes from a sentence relating the distances.

When do the two distances add, and when are they equal?+

They add when the objects move toward each other or in opposite directions (their distances make up the total gap). They are equal when they cover the same route — a round trip, or a faster object catching a slower one.

How do I handle a head start?+

Different departure times mean different time expressions. If one object left 2 hours earlier and the other travels for hours, the first travels for .

What do 'still water' and 'no wind' mean?+

The object's own speed without help or resistance. Downstream or with the wind, the effective rate is ; upstream or against the wind, it is .

Why must units match?+

only balances when rate and time use consistent units — miles per hour with hours, not minutes. Convert everything before building the chart.

More practice problems

Algebra Word Problems Practice Problems Work Rate Practice Problems Mixture Problems Practice Problems