Practice · Sum distances and times across legs · Sensim Math

Variant 1 answer: Total distance: 24 mi; Total time: 1 hr 31 min

In a duathlon, athletes run $2\ \text{mi}$ , then bike $20\ \text{mi}$ , and finally run another $3520\ \text{yd}$ . One competitor's times are $15 min 5 sec$ for the first run, $50 min 25 sec$ for the bike leg, and $25 min 30 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 2 mi, bike 20 mi, and run 3520 yd. The times are 15 min 5 sec, 50 min 25 sec, and 25 min 30 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 2 mi, 20 mi, and 3520 yd
Leg times: 15 min 5 sec, 50 min 25 sec, 25 min 30 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 3520 yd by 1760 to express the last leg in miles.

3520 \div 1760 = 2 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 2 mi, 20 mi, and 2 mi. Add them for the total distance.

2 + 20 + 2 = 24 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 5 + 25 + 30 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

5 + 25 + 30 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 15 + 50 + 25 + 1 = 91 min, which carries 1 hour(s) and leaves 31 min. Add the hours plus that carry: 0 + 0 + 0 + 1 = 1 hr.

15 + 50 + 25 + 1 = 91 \text{ min}, \quad 0 + 0 + 0 + 1 = 1 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 24 mi; Total time: 1 hr 31 min

Review

24 miles for a run-bike-run event is reasonable, and 1 hr 31 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 5460 s, which equals 1 hr 31 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 3520 yd into 2 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 2 answer: Total distance: 63 mi; Total time: 4 hr 16 min

In a duathlon, athletes run $6\ \text{mi}$ , then bike $50\ \text{mi}$ , and finally run another $12320\ \text{yd}$ . One competitor's times are $35 min 40 sec$ for the first run, $2 hr 30 min 10 sec$ for the bike leg, and $1 hr 10 min 10 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 6 mi, bike 50 mi, and run 12320 yd. The times are 35 min 40 sec, 2 hr 30 min 10 sec, and 1 hr 10 min 10 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 6 mi, 50 mi, and 12320 yd
Leg times: 35 min 40 sec, 2 hr 30 min 10 sec, 1 hr 10 min 10 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 12320 yd by 1760 to express the last leg in miles.

12320 \div 1760 = 7 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 6 mi, 50 mi, and 7 mi. Add them for the total distance.

6 + 50 + 7 = 63 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 40 + 10 + 10 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

40 + 10 + 10 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 35 + 30 + 10 + 1 = 76 min, which carries 1 hour(s) and leaves 16 min. Add the hours plus that carry: 0 + 2 + 1 + 1 = 4 hr.

35 + 30 + 10 + 1 = 76 \text{ min}, \quad 0 + 2 + 1 + 1 = 4 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 63 mi; Total time: 4 hr 16 min

Review

63 miles for a run-bike-run event is reasonable, and 4 hr 16 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 15360 s, which equals 4 hr 16 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 12320 yd into 7 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 3 answer: Total distance: 37 mi; Total time: 2 hr 38 min

In a duathlon, athletes run $4\ \text{mi}$ , then bike $30\ \text{mi}$ , and finally run another $5280\ \text{yd}$ . One competitor's times are $22 min 12 sec$ for the first run, $1 hr 30 min 18 sec$ for the bike leg, and $45 min 30 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 4 mi, bike 30 mi, and run 5280 yd. The times are 22 min 12 sec, 1 hr 30 min 18 sec, and 45 min 30 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 4 mi, 30 mi, and 5280 yd
Leg times: 22 min 12 sec, 1 hr 30 min 18 sec, 45 min 30 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 5280 yd by 1760 to express the last leg in miles.

5280 \div 1760 = 3 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 4 mi, 30 mi, and 3 mi. Add them for the total distance.

4 + 30 + 3 = 37 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 12 + 18 + 30 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

12 + 18 + 30 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 22 + 30 + 45 + 1 = 98 min, which carries 1 hour(s) and leaves 38 min. Add the hours plus that carry: 0 + 1 + 0 + 1 = 2 hr.

22 + 30 + 45 + 1 = 98 \text{ min}, \quad 0 + 1 + 0 + 1 = 2 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 37 mi; Total time: 2 hr 38 min

Review

37 miles for a run-bike-run event is reasonable, and 2 hr 38 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 9480 s, which equals 2 hr 38 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 5280 yd into 3 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 4 answer: Total distance: 28 mi; Total time: 2 hr 1 min

In a duathlon, athletes run $3\ \text{mi}$ , then bike $22\ \text{mi}$ , and finally run another $5280\ \text{yd}$ . One competitor's times are $17 min 22 sec$ for the first run, $1 hr 5 min 18 sec$ for the bike leg, and $38 min 20 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 3 mi, bike 22 mi, and run 5280 yd. The times are 17 min 22 sec, 1 hr 5 min 18 sec, and 38 min 20 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 3 mi, 22 mi, and 5280 yd
Leg times: 17 min 22 sec, 1 hr 5 min 18 sec, 38 min 20 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 5280 yd by 1760 to express the last leg in miles.

5280 \div 1760 = 3 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 3 mi, 22 mi, and 3 mi. Add them for the total distance.

3 + 22 + 3 = 28 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 22 + 18 + 20 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

22 + 18 + 20 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 17 + 5 + 38 + 1 = 61 min, which carries 1 hour(s) and leaves 1 min. Add the hours plus that carry: 0 + 1 + 0 + 1 = 2 hr.

17 + 5 + 38 + 1 = 61 \text{ min}, \quad 0 + 1 + 0 + 1 = 2 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 28 mi; Total time: 2 hr 1 min

Review

28 miles for a run-bike-run event is reasonable, and 2 hr 1 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 7260 s, which equals 2 hr 1 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 5280 yd into 3 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 5 answer: Total distance: 51 mi; Total time: 3 hr 38 min

In a duathlon, athletes run $5\ \text{mi}$ , then bike $40\ \text{mi}$ , and finally run another $10560\ \text{yd}$ . One competitor's times are $26 min 16 sec$ for the first run, $2 hr 8 min 20 sec$ for the bike leg, and $1 hr 3 min 24 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 5 mi, bike 40 mi, and run 10560 yd. The times are 26 min 16 sec, 2 hr 8 min 20 sec, and 1 hr 3 min 24 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 5 mi, 40 mi, and 10560 yd
Leg times: 26 min 16 sec, 2 hr 8 min 20 sec, 1 hr 3 min 24 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 10560 yd by 1760 to express the last leg in miles.

10560 \div 1760 = 6 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 5 mi, 40 mi, and 6 mi. Add them for the total distance.

5 + 40 + 6 = 51 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 16 + 20 + 24 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

16 + 20 + 24 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 26 + 8 + 3 + 1 = 38 min, which carries 0 hour(s) and leaves 38 min. Add the hours plus that carry: 0 + 2 + 1 + 0 = 3 hr.

26 + 8 + 3 + 1 = 38 \text{ min}, \quad 0 + 2 + 1 + 0 = 3 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 51 mi; Total time: 3 hr 38 min

Review

51 miles for a run-bike-run event is reasonable, and 3 hr 38 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 13080 s, which equals 3 hr 38 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 10560 yd into 6 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 6 answer: Total distance: 36 mi; Total time: 2 hr 26 min

In a duathlon, athletes run $4\ \text{mi}$ , then bike $28\ \text{mi}$ , and finally run another $7040\ \text{yd}$ . One competitor's times are $20 min 0 sec$ for the first run, $1 hr 24 min 0 sec$ for the bike leg, and $42 min 0 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 4 mi, bike 28 mi, and run 7040 yd. The times are 20 min 0 sec, 1 hr 24 min 0 sec, and 42 min 0 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 4 mi, 28 mi, and 7040 yd
Leg times: 20 min 0 sec, 1 hr 24 min 0 sec, 42 min 0 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 7040 yd by 1760 to express the last leg in miles.

7040 \div 1760 = 4 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 4 mi, 28 mi, and 4 mi. Add them for the total distance.

4 + 28 + 4 = 36 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 0 + 0 + 0 = 0 sec. That carries 0 minute(s) and leaves 0 second(s).

0 + 0 + 0 = 0 \text{ sec} = 0 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 20 + 24 + 42 + 0 = 86 min, which carries 1 hour(s) and leaves 26 min. Add the hours plus that carry: 0 + 1 + 0 + 1 = 2 hr.

20 + 24 + 42 + 0 = 86 \text{ min}, \quad 0 + 1 + 0 + 1 = 2 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 36 mi; Total time: 2 hr 26 min

Review

36 miles for a run-bike-run event is reasonable, and 2 hr 26 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 8760 s, which equals 2 hr 26 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 7040 yd into 4 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 7 answer: Total distance: 58 mi; Total time: 3 hr 54 min

In a duathlon, athletes run $7\ \text{mi}$ , then bike $45\ \text{mi}$ , and finally run another $10560\ \text{yd}$ . One competitor's times are $33 min 40 sec$ for the first run, $2 hr 15 min 10 sec$ for the bike leg, and $1 hr 5 min 10 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 7 mi, bike 45 mi, and run 10560 yd. The times are 33 min 40 sec, 2 hr 15 min 10 sec, and 1 hr 5 min 10 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 7 mi, 45 mi, and 10560 yd
Leg times: 33 min 40 sec, 2 hr 15 min 10 sec, 1 hr 5 min 10 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 10560 yd by 1760 to express the last leg in miles.

10560 \div 1760 = 6 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 7 mi, 45 mi, and 6 mi. Add them for the total distance.

7 + 45 + 6 = 58 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 40 + 10 + 10 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

40 + 10 + 10 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 33 + 15 + 5 + 1 = 54 min, which carries 0 hour(s) and leaves 54 min. Add the hours plus that carry: 0 + 2 + 1 + 0 = 3 hr.

33 + 15 + 5 + 1 = 54 \text{ min}, \quad 0 + 2 + 1 + 0 = 3 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 58 mi; Total time: 3 hr 54 min

Review

58 miles for a run-bike-run event is reasonable, and 3 hr 54 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 14040 s, which equals 3 hr 54 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 10560 yd into 6 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 8 answer: Total distance: 32 mi; Total time: 2 hr 11 min

In a duathlon, athletes run $3\ \text{mi}$ , then bike $25\ \text{mi}$ , and finally run another $7040\ \text{yd}$ . One competitor's times are $18 min 30 sec$ for the first run, $1 hr 12 min 15 sec$ for the bike leg, and $40 min 15 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 3 mi, bike 25 mi, and run 7040 yd. The times are 18 min 30 sec, 1 hr 12 min 15 sec, and 40 min 15 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 3 mi, 25 mi, and 7040 yd
Leg times: 18 min 30 sec, 1 hr 12 min 15 sec, 40 min 15 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 7040 yd by 1760 to express the last leg in miles.

7040 \div 1760 = 4 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 3 mi, 25 mi, and 4 mi. Add them for the total distance.

3 + 25 + 4 = 32 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 30 + 15 + 15 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

30 + 15 + 15 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 18 + 12 + 40 + 1 = 71 min, which carries 1 hour(s) and leaves 11 min. Add the hours plus that carry: 0 + 1 + 0 + 1 = 2 hr.

18 + 12 + 40 + 1 = 71 \text{ min}, \quad 0 + 1 + 0 + 1 = 2 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 32 mi; Total time: 2 hr 11 min

Review

32 miles for a run-bike-run event is reasonable, and 2 hr 11 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 7860 s, which equals 2 hr 11 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 7040 yd into 4 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 9 answer: Total distance: 45 mi; Total time: 3 hr 4 min

In a duathlon, athletes run $5\ \text{mi}$ , then bike $35\ \text{mi}$ , and finally run another $8800\ \text{yd}$ . One competitor's times are $28 min 50 sec$ for the first run, $1 hr 45 min 5 sec$ for the bike leg, and $50 min 5 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 5 mi, bike 35 mi, and run 8800 yd. The times are 28 min 50 sec, 1 hr 45 min 5 sec, and 50 min 5 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 5 mi, 35 mi, and 8800 yd
Leg times: 28 min 50 sec, 1 hr 45 min 5 sec, 50 min 5 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 8800 yd by 1760 to express the last leg in miles.

8800 \div 1760 = 5 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 5 mi, 35 mi, and 5 mi. Add them for the total distance.

5 + 35 + 5 = 45 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 50 + 5 + 5 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

50 + 5 + 5 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 28 + 45 + 50 + 1 = 124 min, which carries 2 hour(s) and leaves 4 min. Add the hours plus that carry: 0 + 1 + 0 + 2 = 3 hr.

28 + 45 + 50 + 1 = 124 \text{ min}, \quad 0 + 1 + 0 + 2 = 3 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 45 mi; Total time: 3 hr 4 min

Review

45 miles for a run-bike-run event is reasonable, and 3 hr 4 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 11040 s, which equals 3 hr 4 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 8800 yd into 5 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Variant 10 answer: Total distance: 76 mi; Total time: 5 hr 1 min

In a duathlon, athletes run $8\ \text{mi}$ , then bike $60\ \text{mi}$ , and finally run another $14080\ \text{yd}$ . One competitor's times are $40 min 25 sec$ for the first run, $3 hr 0 min 20 sec$ for the bike leg, and $1 hr 20 min 15 sec$ for the final run. Find the total distance this competitor travels and the total time it takes. (Note: $1\ \text{mi} = 1760\ \text{yd}$ .)

Show solution

Understand

A duathlon has three legs: run 8 mi, bike 60 mi, and run 14080 yd. The times are 40 min 25 sec, 3 hr 0 min 20 sec, and 1 hr 20 min 15 sec. We add the distances (converting yards to miles) and add the times (carrying over 60s and 60min) to get the total distance and total time.

Givens

Leg distances: 8 mi, 60 mi, and 14080 yd
Leg times: 40 min 25 sec, 3 hr 0 min 20 sec, 1 hr 20 min 15 sec
1 mi = 1760 yd

Unknowns

The total distance traveled (in miles)
The total time taken (in hours, minutes, seconds)

Constraints

Distances must share one unit (miles) before adding
Times must be carried: 60 sec make a minute, 60 min make an hour

Plan

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

The hard part is matching units: the final run is in yards while the others are in miles, and the times mix hours, minutes, and seconds. Once units agree, the problem splits into two simple subproblems -- sum the distances and sum the times with carrying.

Execute

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

Since 1 mi = 1760 yd, divide 14080 yd by 1760 to express the last leg in miles.

14080 \div 1760 = 8 \text{ mi}

Putting every distance in the same unit (miles) lets us add them directly.

#7 Identify Subproblems 4.MD.A.2

Now all three legs are in miles: 8 mi, 60 mi, and 8 mi. Add them for the total distance.

8 + 60 + 8 = 76 \text{ mi}

With one unit, the total distance is just a plain sum.

#8 Analyze the Units 4.MD.A.2

Add the seconds: 25 + 20 + 15 = 60 sec. That carries 1 minute(s) and leaves 0 second(s).

25 + 20 + 15 = 60 \text{ sec} = 1 \text{ min } 0 \text{ sec}

Time is base-60, so every 60 seconds rolls over into one minute.

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute(s): 40 + 0 + 20 + 1 = 61 min, which carries 1 hour(s) and leaves 1 min. Add the hours plus that carry: 0 + 3 + 1 + 1 = 5 hr.

40 + 0 + 20 + 1 = 61 \text{ min}, \quad 0 + 3 + 1 + 1 = 5 \text{ hr}

Combining each time column separately, then carrying, gives a clean elapsed total.

Answer: Total distance: 76 mi; Total time: 5 hr 1 min

Review

76 miles for a run-bike-run event is reasonable, and 5 hr 1 min for that distance is a sensible pace. The base-60 carries were handled column by column, so no hidden carry was missed.

Convert every time fully to seconds (tool 8): the three legs total 18060 s, which equals 5 hr 1 min, confirming the column method.

Standards · min grade 4

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units — Converting 14080 yd into 8 mi using 1 mi = 1760 yd
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Summing the distances and the time amounts across the three legs
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Carrying 60 seconds into a minute and adding the hour/minute columns

💡 Make every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!

Sum distances and times across legs · 10 practice problems

Generated variants — 10

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Standards · min grade 4