Problem
Three full days pass between the two 9:00 a.m. moments. Each day has 24 hours.
Multiplying days by 24 to get hours is the same '24 hours in a day' fact a third grader uses for elapsed time.
3.OA.A.3Identify SubproblemsThe clock loses 2 seconds each hour, and 72 hours pass, so multiply the loss rate by the number of hours.
Loss-per-hour times hours gives total loss, a direct 'rate times amount' multiplication within third-grade reach.
3.OA.A.3Analyze The UnitsOver the three days the clock falls a total of 144 seconds behind true time.
Across the 72 hours that pass, each hour quietly takes the same 2 seconds off the clock, so the total lag is 72 groups of 2 seconds, which is 72 times 2 equals 144.
The 72 hours come from three full days that each hold 24 hours, so the number of hours is three equal groups of 24.
Counting three equal groups of 24 hours is exactly what 3 times 24 means, and that gives 72.
The whole amount the clock is behind is just every hour's small loss added up, with no hour skipped and none counted twice.
Since each of those 72 hours loses the identical 2 seconds, stacking the losses is 72 equal groups of 2, which is what 72 times 2 counts.
Since 60 seconds make 1 minute, regroup 144 seconds: 144 = 120 + 24, and 120 seconds is 2 minutes.
Trading 60 seconds for 1 minute is the base-sixty regrouping kids learn for telling time.
3.MD.A.1Analyze The UnitsA slow clock shows an earlier time, so subtract 2 min 24 s from 9:00:00 a.m. Borrow 1 minute (60 s): 9:00:00 becomes 8:59:60, then subtract.
Counting backward from a true time to find a slow clock's display is everyday elapsed-time subtraction.
3.MD.A.1Identify SubproblemsThis only needs Grade 3 multiplication and time-telling: find total seconds lost, then count backward!