Measurement & Time

Problem

Sum distances and times across legs

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.
Measurement & data
Your answer
How to solve
Strategy Analyze the Units — 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.
1STEP 1

Convert the final run to miles

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

10560 ÷ 1760 = 6 \text{ mi}
2STEP 2

Add the distances

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}
3STEP 3

Add the seconds and carry

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}
4STEP 4

Add the minutes (with the carry) and the hours

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}, 2 + 1 = 3 \text{ hr}
Answer
Total distance: 51 mi; Total time: 3 hr 38 min
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.
Takeaway

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

  • Convert the final run to miles
  • Add the distances
  • Add the seconds and carry
  • Add the minutes (with the carry) and the hours
Where next?
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