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← 3-1 · Distance grows in proportion to time · Multiplicative Comparison and Unit Rate

Distance grows in proportion to time · 9 practice problems

4.MD.A.14.MD.A.23.OA.A.3

Generated variants — 9

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 answer: 30 mi 1500 yd

A car travels 18 miles 900 yards18 \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.
18×1760+900=31680+900=32580 yd18 \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÷15=2172 yd per minute32580 \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.
2172×25=54300 yd2172 \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÷1760=30 mi remainder 1500 yd54300 \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. The car adds 2172 yd every minute at this steady speed.

Use the time ratio: scale the 32580 yd covered in 15 min by 25/15 to get 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 the new time 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!
Variant 2 answer: 23 mi 20 yd

A car travels 15 miles 600 yards15 \text{ miles } 600 \text{ yards} in 12 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 18 minutes?

Show solution

Understand

A car goes 15 mi 600 yd in 12 minutes at a steady speed. We must find how far it goes in 18 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 18, then convert back to miles and yards.

Givens
  • Distance in 12 min: 15 mi 600 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 18 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, 15 mi = 15 times 1760 = 26400 yd. Add the extra 600 yd to get the total in yards.
15×1760+600=26400+600=27000 yd15 \times 1760 + 600 = 26400 + 600 = 27000 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 27000 yd in 12 minutes at a steady speed, so divide by 12 to get how far it goes each minute.
27000÷12=2250 yd per minute27000 \div 12 = 2250 \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 18 minutes, multiply the per-minute distance by 18.
2250×18=40500 yd2250 \times 18 = 40500 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 40500 yd by 1760 to find whole miles: 23 miles use 23 times 1760 = 40480 yd, leaving 40500 - 40480 = 20 yd. So the distance is 23 mi 20 yd.
40500÷1760=23 mi remainder 20 yd40500 \div 1760 = 23 \text{ mi remainder } 20 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 23 mi 20 yd

Review

18 minutes is more than 12 minutes, so the distance should be larger than 15 mi 600 yd, and 23 mi 20 yd is indeed larger. The car adds 2250 yd every minute at this steady speed.

Use the time ratio: scale the 27000 yd covered in 12 min by 18/12 to get 40500 yd = 23 mi 20 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 the new time 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!
Variant 3 answer: 30 mi 0 yd

A car travels 12 miles 0 yards12 \text{ miles } 0 \text{ yards} in 10 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 12 mi 0 yd in 10 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 10 min: 12 mi 0 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, 12 mi = 12 times 1760 = 21120 yd. Add the extra 0 yd to get the total in yards.
12×1760+0=21120+0=21120 yd12 \times 1760 + 0 = 21120 + 0 = 21120 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 21120 yd in 10 minutes at a steady speed, so divide by 10 to get how far it goes each minute.
21120÷10=2112 yd per minute21120 \div 10 = 2112 \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.
2112×25=52800 yd2112 \times 25 = 52800 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 52800 yd by 1760 to find whole miles: 30 miles use 30 times 1760 = 52800 yd, leaving 52800 - 52800 = 0 yd. So the distance is 30 mi 0 yd.
52800÷1760=30 mi remainder 0 yd52800 \div 1760 = 30 \text{ mi remainder } 0 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 30 mi 0 yd

Review

25 minutes is more than 10 minutes, so the distance should be larger than 12 mi 0 yd, and 30 mi 0 yd is indeed larger. The car adds 2112 yd every minute at this steady speed.

Use the time ratio: scale the 21120 yd covered in 10 min by 25/10 to get 52800 yd = 30 mi 0 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 the new time 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!
Variant 4 answer: 16 mi 880 yd

A car travels 6 miles 1056 yards6 \text{ miles } 1056 \text{ yards} in 6 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 15 minutes?

Show solution

Understand

A car goes 6 mi 1056 yd in 6 minutes at a steady speed. We must find how far it goes in 15 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 15, then convert back to miles and yards.

Givens
  • Distance in 6 min: 6 mi 1056 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 15 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, 6 mi = 6 times 1760 = 10560 yd. Add the extra 1056 yd to get the total in yards.
6×1760+1056=10560+1056=11616 yd6 \times 1760 + 1056 = 10560 + 1056 = 11616 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 11616 yd in 6 minutes at a steady speed, so divide by 6 to get how far it goes each minute.
11616÷6=1936 yd per minute11616 \div 6 = 1936 \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 15 minutes, multiply the per-minute distance by 15.
1936×15=29040 yd1936 \times 15 = 29040 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 29040 yd by 1760 to find whole miles: 16 miles use 16 times 1760 = 28160 yd, leaving 29040 - 28160 = 880 yd. So the distance is 16 mi 880 yd.
29040÷1760=16 mi remainder 880 yd29040 \div 1760 = 16 \text{ mi remainder } 880 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 16 mi 880 yd

Review

15 minutes is more than 6 minutes, so the distance should be larger than 6 mi 1056 yd, and 16 mi 880 yd is indeed larger. The car adds 1936 yd every minute at this steady speed.

Use the time ratio: scale the 11616 yd covered in 6 min by 15/6 to get 29040 yd = 16 mi 880 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 the new time 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!
Variant 5 answer: 72 mi 0 yd

A car travels 30 miles 0 yards30 \text{ miles } 0 \text{ yards} in 25 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 60 minutes?

Show solution

Understand

A car goes 30 mi 0 yd in 25 minutes at a steady speed. We must find how far it goes in 60 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 60, then convert back to miles and yards.

Givens
  • Distance in 25 min: 30 mi 0 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 60 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, 30 mi = 30 times 1760 = 52800 yd. Add the extra 0 yd to get the total in yards.
30×1760+0=52800+0=52800 yd30 \times 1760 + 0 = 52800 + 0 = 52800 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 52800 yd in 25 minutes at a steady speed, so divide by 25 to get how far it goes each minute.
52800÷25=2112 yd per minute52800 \div 25 = 2112 \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 60 minutes, multiply the per-minute distance by 60.
2112×60=126720 yd2112 \times 60 = 126720 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 126720 yd by 1760 to find whole miles: 72 miles use 72 times 1760 = 126720 yd, leaving 126720 - 126720 = 0 yd. So the distance is 72 mi 0 yd.
126720÷1760=72 mi remainder 0 yd126720 \div 1760 = 72 \text{ mi remainder } 0 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 72 mi 0 yd

Review

60 minutes is more than 25 minutes, so the distance should be larger than 30 mi 0 yd, and 72 mi 0 yd is indeed larger. The car adds 2112 yd every minute at this steady speed.

Use the time ratio: scale the 52800 yd covered in 25 min by 60/25 to get 126720 yd = 72 mi 0 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 the new time 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!
Variant 6 answer: 30 mi 1100 yd

A car travels 8 miles 1320 yards8 \text{ miles } 1320 \text{ yards} in 10 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 35 minutes?

Show solution

Understand

A car goes 8 mi 1320 yd in 10 minutes at a steady speed. We must find how far it goes in 35 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 35, then convert back to miles and yards.

Givens
  • Distance in 10 min: 8 mi 1320 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 35 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, 8 mi = 8 times 1760 = 14080 yd. Add the extra 1320 yd to get the total in yards.
8×1760+1320=14080+1320=15400 yd8 \times 1760 + 1320 = 14080 + 1320 = 15400 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 15400 yd in 10 minutes at a steady speed, so divide by 10 to get how far it goes each minute.
15400÷10=1540 yd per minute15400 \div 10 = 1540 \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 35 minutes, multiply the per-minute distance by 35.
1540×35=53900 yd1540 \times 35 = 53900 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 53900 yd by 1760 to find whole miles: 30 miles use 30 times 1760 = 52800 yd, leaving 53900 - 52800 = 1100 yd. So the distance is 30 mi 1100 yd.
53900÷1760=30 mi remainder 1100 yd53900 \div 1760 = 30 \text{ mi remainder } 1100 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 30 mi 1100 yd

Review

35 minutes is more than 10 minutes, so the distance should be larger than 8 mi 1320 yd, and 30 mi 1100 yd is indeed larger. The car adds 1540 yd every minute at this steady speed.

Use the time ratio: scale the 15400 yd covered in 10 min by 35/10 to get 53900 yd = 30 mi 1100 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 the new time 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!
Variant 7 answer: 30 mi 0 yd

A car travels 20 miles 0 yards20 \text{ miles } 0 \text{ yards} in 20 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 30 minutes?

Show solution

Understand

A car goes 20 mi 0 yd in 20 minutes at a steady speed. We must find how far it goes in 30 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 30, then convert back to miles and yards.

Givens
  • Distance in 20 min: 20 mi 0 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 30 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, 20 mi = 20 times 1760 = 35200 yd. Add the extra 0 yd to get the total in yards.
20×1760+0=35200+0=35200 yd20 \times 1760 + 0 = 35200 + 0 = 35200 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 35200 yd in 20 minutes at a steady speed, so divide by 20 to get how far it goes each minute.
35200÷20=1760 yd per minute35200 \div 20 = 1760 \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 30 minutes, multiply the per-minute distance by 30.
1760×30=52800 yd1760 \times 30 = 52800 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 52800 yd by 1760 to find whole miles: 30 miles use 30 times 1760 = 52800 yd, leaving 52800 - 52800 = 0 yd. So the distance is 30 mi 0 yd.
52800÷1760=30 mi remainder 0 yd52800 \div 1760 = 30 \text{ mi remainder } 0 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 30 mi 0 yd

Review

30 minutes is more than 20 minutes, so the distance should be larger than 20 mi 0 yd, and 30 mi 0 yd is indeed larger. The car adds 1760 yd every minute at this steady speed.

Use the time ratio: scale the 35200 yd covered in 20 min by 30/20 to get 52800 yd = 30 mi 0 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 the new time 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!
Variant 8 answer: 11 mi 1100 yd

A car travels 5 miles 500 yards5 \text{ miles } 500 \text{ yards} in 10 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 22 minutes?

Show solution

Understand

A car goes 5 mi 500 yd in 10 minutes at a steady speed. We must find how far it goes in 22 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 22, then convert back to miles and yards.

Givens
  • Distance in 10 min: 5 mi 500 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 22 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, 5 mi = 5 times 1760 = 8800 yd. Add the extra 500 yd to get the total in yards.
5×1760+500=8800+500=9300 yd5 \times 1760 + 500 = 8800 + 500 = 9300 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 9300 yd in 10 minutes at a steady speed, so divide by 10 to get how far it goes each minute.
9300÷10=930 yd per minute9300 \div 10 = 930 \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 22 minutes, multiply the per-minute distance by 22.
930×22=20460 yd930 \times 22 = 20460 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 20460 yd by 1760 to find whole miles: 11 miles use 11 times 1760 = 19360 yd, leaving 20460 - 19360 = 1100 yd. So the distance is 11 mi 1100 yd.
20460÷1760=11 mi remainder 1100 yd20460 \div 1760 = 11 \text{ mi remainder } 1100 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 11 mi 1100 yd

Review

22 minutes is more than 10 minutes, so the distance should be larger than 5 mi 500 yd, and 11 mi 1100 yd is indeed larger. The car adds 930 yd every minute at this steady speed.

Use the time ratio: scale the 9300 yd covered in 10 min by 22/10 to get 20460 yd = 11 mi 1100 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 the new time 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!
Variant 9 answer: 60 mi 0 yd

A car travels 24 miles 0 yards24 \text{ miles } 0 \text{ yards} in 16 minutes. If the car keeps driving at the same speed, how many miles and yards can it travel in 40 minutes?

Show solution

Understand

A car goes 24 mi 0 yd in 16 minutes at a steady speed. We must find how far it goes in 40 minutes. We convert the distance to a single unit (yards), find the distance per minute, multiply by 40, then convert back to miles and yards.

Givens
  • Distance in 16 min: 24 mi 0 yd
  • The car keeps the same (constant) speed
  • 1 mi = 1760 yd
Unknowns
  • The distance traveled in 40 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, 24 mi = 24 times 1760 = 42240 yd. Add the extra 0 yd to get the total in yards.
24×1760+0=42240+0=42240 yd24 \times 1760 + 0 = 42240 + 0 = 42240 \text{ yd}
Working in a single unit (yards) makes the speed easy to compute.
#7 Identify Subproblems 3.OA.A.3
The car covers 42240 yd in 16 minutes at a steady speed, so divide by 16 to get how far it goes each minute.
42240÷16=2640 yd per minute42240 \div 16 = 2640 \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 40 minutes, multiply the per-minute distance by 40.
2640×40=105600 yd2640 \times 40 = 105600 \text{ yd}
More time at the same speed means proportionally more distance.
#8 Analyze the Units 4.MD.A.1
Divide 105600 yd by 1760 to find whole miles: 60 miles use 60 times 1760 = 105600 yd, leaving 105600 - 105600 = 0 yd. So the distance is 60 mi 0 yd.
105600÷1760=60 mi remainder 0 yd105600 \div 1760 = 60 \text{ mi remainder } 0 \text{ yd}
Regrouping yards back into miles plus leftover yards gives the answer in the requested units.
Answer: 60 mi 0 yd

Review

40 minutes is more than 16 minutes, so the distance should be larger than 24 mi 0 yd, and 60 mi 0 yd is indeed larger. The car adds 2640 yd every minute at this steady speed.

Use the time ratio: scale the 42240 yd covered in 16 min by 40/16 to get 105600 yd = 60 mi 0 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 the new time 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!