A drifting clock accumulates daily error

3.MD.A.13.OA.A.3 · take · grade 3

Archetype: Elapsed Time and Base-Sixty Regrouping · step in a 10-type progression

There is a clock that loses 2 seconds every hour. If this clock is set to the exact time at 9:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 9:00 a.m. three days from now?

Show solution

Understand

A clock runs slow, losing 2 seconds for each hour that passes. It is set correctly at 9:00 a.m. today. I need to find what time it displays three days later when the true time is again 9:00 a.m.

Givens

The clock loses 2 seconds every hour.
It is set to the exact time at 9:00 a.m. today.
We look at it 3 days later at the true time 9:00 a.m.

Unknowns

The time the slow clock shows (in hours, minutes, seconds) after 3 days.

Constraints

1 day = 24 hours; 1 minute = 60 seconds.
A slow clock shows an EARLIER time than the true time.

Plan

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

Break the work into small pieces: first find how many hours pass, then how many seconds are lost, then convert those seconds into minutes and seconds, then subtract from the true time. Tracking units (hours -> seconds -> minutes) keeps the loss rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3

Three full days pass between the two 9:00 a.m. moments. Each day has 24 hours.

3 \times 24 = 72 \text{ hours}

Multiplying days by 24 to get hours is the same '24 hours in a day' fact a third grader uses for elapsed time.

#8 Analyze the Units 3.OA.A.3

The clock loses 2 seconds each hour, and 72 hours pass, so multiply the loss rate by the number of hours.

72 \times 2 = 144 \text{ seconds}

Loss-per-hour times hours gives total loss, a direct 'rate times amount' multiplication within third-grade reach.

#8 Analyze the Units 3.MD.A.1

Since 60 seconds make 1 minute, regroup 144 seconds: 144 = 120 + 24, and 120 seconds is 2 minutes.

144 \text{ s} = 2 \text{ min } 24 \text{ s}

Trading 60 seconds for 1 minute is the base-sixty regrouping kids learn for telling time.

#7 Identify Subproblems 3.MD.A.1

A 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.

9{:}00{:}00 - 0{:}02{:}24 = 8{:}57{:}36

Counting backward from a true time to find a slow clock's display is everyday elapsed-time subtraction.

Answer: 8:57:36 a.m.

Review

A loss of about 2 minutes 24 seconds over three days is tiny compared to 72 hours, so the clock should read just a few minutes before 9:00 a.m. - and 8:57:36 a.m. is indeed only 2 min 24 s early, which matches.

Look for a pattern (tool 5): the clock loses 2 s/hour = 48 s/day, so over 3 days it loses 3 x 48 = 144 s = 2 min 24 s, giving the same 8:57:36 a.m.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 3 days x 24 hours and 72 hours x 2 seconds to get the total loss.
3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 144 seconds into 2 min 24 s and subtracting it from 9:00:00 a.m.

💡 This only needs Grade 3 multiplication and time-telling: find total seconds lost, then count backward!