Depth · one concept from first principles · step 2 of 4
In a duathlon, athletes run , then bike , and finally run another . One competitor's times are min sec for the first run, hr min sec for the bike leg, and hr min sec for the final run. Find the total distance this competitor travels and the total time it takes. (Note: .)
Since 1 mi = 1760 yd, divide 10560 yd by 1760 to express the last leg in miles.
Putting every distance in the same unit (miles) lets us add them directly.
4.MD.A.1Analyze The UnitsThe final 10560-yard run covers the same distance as 6 miles.
One mile is defined as 1760 yards, so the real question is how many whole 1760-yard miles fit inside 10560 yards.
Finding how many equal 1760-yard groups make up 10560 yards is the reverse of piling 1760 yards together some number of times, so we divide 10560 by 1760 and get 6.
That count of 6 holds up because 6 equal groups of 1760 yards is 6 times 1760, which is exactly 10560 yards.
Now all three legs are in miles: 5 mi, 40 mi, and 6 mi. Add them for the total distance.
With one unit, the total distance is just a plain sum.
4.MD.A.2Identify SubproblemsAdd the seconds: 16 + 20 + 24 = 60 sec. That equals exactly 1 minute, so carry 1 minute and leave 0 seconds.
Time is base-60, so 60 seconds rolls over into one minute.
4.MD.A.2Analyze The UnitsAdd 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.
Combining each time column separately, then carrying, gives a clean elapsed total.
3.MD.A.1Identify SubproblemsMake every distance the same unit and treat seconds and minutes like base-60 carrying, and adding up the legs becomes easy!
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The same shape shows up later on the AMC, as Arithmetic Expression Decomposition, Path Length Comparison & Shortest Path, Small-Domain Constrained Search. Why the shapes repeat →