Sum distances and times across legs

4.MD.A.24.MD.A.13.MD.A.1 · adapt · grade 4

Archetype: Length as Sum of Parts with Unit Matching · step in a 7-type progression

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 equals exactly 1 minute, so carry 1 minute and leave 0 seconds.

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

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

#7 Identify Subproblems 3.MD.A.1

Add the minutes plus the carried minute: 26 + 8 + 3 + 1 = 38 min, which is under 60 so no further carry. Add the hours: 2 + 1 = 3 hr.

26 + 8 + 3 + 1 = 38 \text{ min}, \quad 2 + 1 = 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 seconds added to exactly 60, cleanly making one extra minute, and 38 min stays under an hour, so no hidden carry was missed.

Convert every time fully to seconds (tool 13/8): the three legs are 1576 s, 7700 s, and 3804 s, totaling 13080 s = 3 hr 38 min 0 sec, 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!