Distance grows in proportion to time

4.MD.A.23.OA.A.3 · adapt · grade 4

Archetype: Multiplicative Comparison and Unit Rate · step in a 7-type progression

A car travels $18 \text{ miles } 900 \text{ yards}$ in 15 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 25 minutes?

Show solution

Understand

A car goes 18 mi 900 yd in 15 minutes at a steady speed. We must find how far it goes in 25 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 25, then convert back to miles and yards.

Givens

Distance in 15 min: 18 mi 900 yd
The car keeps the same (constant) speed
1 mi = 1760 yd

Unknowns

The distance traveled in 25 minutes, in miles and yards

Constraints

Speed is constant, so distance grows in proportion to time
Distance must be in one unit (yards) before dividing and multiplying

Plan

#8 Analyze the Units · also uses: #7 Identify Subproblems

Constant speed means distance per minute is fixed, so the natural plan is: convert the distance to yards, find yards-per-minute, multiply by the new time, then regroup back into miles and yards. Splitting it into these unit-driven subproblems keeps the arithmetic simple for an elementary learner.

Execute

#8 Analyze the Units 4.MD.A.1

Since 1 mi = 1760 yd, 18 mi = 18 times 1760 = 31680 yd. Add the extra 900 yd to get the total in yards.

18 \times 1760 + 900 = 31680 + 900 = 32580 \text{ yd}

Working in a single unit (yards) makes the speed easy to compute.

#7 Identify Subproblems 3.OA.A.3

The car covers 32580 yd in 15 minutes at a steady speed, so divide by 15 to get how far it goes each minute.

32580 \div 15 = 2172 \text{ yd per minute}

A constant speed means the same number of yards is added every minute.

#7 Identify Subproblems 4.MD.A.2

For 25 minutes, multiply the per-minute distance by 25.

2172 \times 25 = 54300 \text{ yd}

More time at the same speed means proportionally more distance.

#8 Analyze the Units 4.MD.A.1

Divide 54300 yd by 1760 to find whole miles: 30 miles use 30 times 1760 = 52800 yd, leaving 54300 - 52800 = 1500 yd. So the distance is 30 mi 1500 yd.

54300 \div 1760 = 30 \text{ mi remainder } 1500 \text{ yd}

Regrouping yards back into miles plus leftover yards gives the answer in the requested units.

Answer: 30 mi 1500 yd

Review

25 minutes is more than 15 minutes, so the distance should be larger than 18 mi 900 yd — and 30 mi 1500 yd is indeed larger. As a rough check, 25/15 is about 1.67, and 18 mi times 1.67 is about 30 mi, matching our result.

Use the time ratio (tool 5/6): 25 min is 5/3 of 15 min, so the distance is 32580 times 5 divided by 3 = 162900 / 3 = 54300 yd = 30 mi 1500 yd, confirming the per-minute method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting between miles and yards using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Finding the distance traveled in 25 minutes at constant speed
3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing to find yards per minute and multiplying by the new time

💡 At a steady speed, find how far you go in one minute, then multiply by the new minutes — same speed, more time, more distance!