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← 3-1 · A drifting clock accumulates daily error · Elapsed Time and Base-Sixty Regrouping

A drifting clock accumulates daily error · 12 practice problems

3.MD.A.13.OA.A.3

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 answer: 8:52:48 a.m.

There is a clock that loses 3 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. 6 days from now?

Show solution

Understand

A clock runs slow, loses 3 seconds for each hour that passes. It is set correctly at 9:00 a.m. today. I need to find what time it displays 6 day(s) later when the true time is again 9:00 a.m.

Givens
  • The clock loses 3 seconds every hour.
  • It is set to the exact time at 9:00 a.m. today.
  • We look at it 6 day(s) later at the true time 9:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 6 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
6 full day(s) pass between the two 9:00 a.m. moments. Each day has 24 hours.
6×24=144 hours6 \times 24 = 144 \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 3 seconds each hour, and 144 hours pass, so multiply the drift rate by the number of hours.
144×3=432 seconds144 \times 3 = 432 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 432 seconds: 432 = 420 + 12, and 420 seconds is 7 minute(s).
432 s=7 min 12 s432 \text{ s} = 7 \text{ min } 12 \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 7 min 12 s from 9:00:00 a.m.
9:00:00−0:07:12=8:52:489:00:00 - 0{:}07{:}12 = 8{:}52{:}48
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 8:52:48 a.m.

Review

A drift of about 7 min 12 s over 6 day(s) is tiny compared to 144 hours, so the clock should read just a few minutes before 9:00 a.m. - and 8:52:48 a.m. matches.

Look for a pattern: the clock drifts 3 s/hour = 72 s/day, so over 6 day(s) it drifts 6 x 72 = 432 s = 7 min 12 s, giving the same 8:52:48 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 6 days x 24 hours and 144 hours x 3 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 432 seconds into 7 min 12 s and combining it with 9:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 2 answer: 9:57:36 a.m.

There is a clock that loses 3 seconds every hour. If this clock is set to the exact time at 10:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 10:00 a.m. 2 days from now?

Show solution

Understand

A clock runs slow, loses 3 seconds for each hour that passes. It is set correctly at 10:00 a.m. today. I need to find what time it displays 2 day(s) later when the true time is again 10:00 a.m.

Givens
  • The clock loses 3 seconds every hour.
  • It is set to the exact time at 10:00 a.m. today.
  • We look at it 2 day(s) later at the true time 10:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 2 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
2 full day(s) pass between the two 10:00 a.m. moments. Each day has 24 hours.
2×24=48 hours2 \times 24 = 48 \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 3 seconds each hour, and 48 hours pass, so multiply the drift rate by the number of hours.
48×3=144 seconds48 \times 3 = 144 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#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 minute(s).
144 s=2 min 24 s144 \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 10:00:00 a.m.
10:00:00−0:02:24=9:57:3610:00:00 - 0{:}02{:}24 = 9{:}57{:}36
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 9:57:36 a.m.

Review

A drift of about 2 min 24 s over 2 day(s) is tiny compared to 48 hours, so the clock should read just a few minutes before 10:00 a.m. - and 9:57:36 a.m. matches.

Look for a pattern: the clock drifts 3 s/hour = 72 s/day, so over 2 day(s) it drifts 2 x 72 = 144 s = 2 min 24 s, giving the same 9:57:36 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 2 days x 24 hours and 48 hours x 3 seconds to get the total drift.
  • 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 combining it with 10:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 3 answer: 10:50:24 a.m.

There is a clock that loses 12 seconds every hour. If this clock is set to the exact time at 11:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 11:00 a.m. 2 days from now?

Show solution

Understand

A clock runs slow, loses 12 seconds for each hour that passes. It is set correctly at 11:00 a.m. today. I need to find what time it displays 2 day(s) later when the true time is again 11:00 a.m.

Givens
  • The clock loses 12 seconds every hour.
  • It is set to the exact time at 11:00 a.m. today.
  • We look at it 2 day(s) later at the true time 11:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 2 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
2 full day(s) pass between the two 11:00 a.m. moments. Each day has 24 hours.
2×24=48 hours2 \times 24 = 48 \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 12 seconds each hour, and 48 hours pass, so multiply the drift rate by the number of hours.
48×12=576 seconds48 \times 12 = 576 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 576 seconds: 576 = 540 + 36, and 540 seconds is 9 minute(s).
576 s=9 min 36 s576 \text{ s} = 9 \text{ min } 36 \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 9 min 36 s from 11:00:00 a.m.
11:00:00−0:09:36=10:50:2411:00:00 - 0{:}09{:}36 = 10{:}50{:}24
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 10:50:24 a.m.

Review

A drift of about 9 min 36 s over 2 day(s) is tiny compared to 48 hours, so the clock should read just a few minutes before 11:00 a.m. - and 10:50:24 a.m. matches.

Look for a pattern: the clock drifts 12 s/hour = 288 s/day, so over 2 day(s) it drifts 2 x 288 = 576 s = 9 min 36 s, giving the same 10:50:24 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 2 days x 24 hours and 48 hours x 12 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 576 seconds into 9 min 36 s and combining it with 11:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 4 answer: 8:04:00 a.m.

There is a clock that gains 5 seconds every hour. If this clock is set to the exact time at 8:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 8:00 a.m. 2 days from now?

Show solution

Understand

A clock runs fast, gains 5 seconds for each hour that passes. It is set correctly at 8:00 a.m. today. I need to find what time it displays 2 day(s) later when the true time is again 8:00 a.m.

Givens
  • The clock gains 5 seconds every hour.
  • It is set to the exact time at 8:00 a.m. today.
  • We look at it 2 day(s) later at the true time 8:00 a.m.
Unknowns
  • The time the fast clock shows (in hours, minutes, seconds) after 2 day(s).
Constraints
  • 1 day = 24 hours; 1 minute = 60 seconds.
  • A fast clock shows an LATER 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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
2 full day(s) pass between the two 8:00 a.m. moments. Each day has 24 hours.
2×24=48 hours2 \times 24 = 48 \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 gains 5 seconds each hour, and 48 hours pass, so multiply the drift rate by the number of hours.
48×5=240 seconds48 \times 5 = 240 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 240 seconds: 240 = 240 + 0, and 240 seconds is 4 minute(s).
240 s=4 min 0 s240 \text{ s} = 4 \text{ min } 0 \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 fast clock shows a later time, so add 4 min 0 s to 8:00:00 a.m.
8:00:00+0:04:00=8:04:008:00:00 + 0{:}04{:}00 = 8{:}04{:}00
Counting add from the true time gives the fast clock's display - everyday elapsed-time arithmetic.
Answer: 8:04:00 a.m.

Review

A drift of about 4 min 0 s over 2 day(s) is tiny compared to 48 hours, so the clock should read just a few minutes after 8:00 a.m. - and 8:04:00 a.m. matches.

Look for a pattern: the clock drifts 5 s/hour = 120 s/day, so over 2 day(s) it drifts 2 x 120 = 240 s = 4 min 0 s, giving the same 8:04:00 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 2 days x 24 hours and 48 hours x 5 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 240 seconds into 4 min 0 s and combining it with 8:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 5 answer: 6:58:00 a.m.

There is a clock that loses 1 second every hour. If this clock is set to the exact time at 7:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 7:00 a.m. 5 days from now?

Show solution

Understand

A clock runs slow, loses 1 seconds for each hour that passes. It is set correctly at 7:00 a.m. today. I need to find what time it displays 5 day(s) later when the true time is again 7:00 a.m.

Givens
  • The clock loses 1 seconds every hour.
  • It is set to the exact time at 7:00 a.m. today.
  • We look at it 5 day(s) later at the true time 7:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 5 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
5 full day(s) pass between the two 7:00 a.m. moments. Each day has 24 hours.
5×24=120 hours5 \times 24 = 120 \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 1 seconds each hour, and 120 hours pass, so multiply the drift rate by the number of hours.
120×1=120 seconds120 \times 1 = 120 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 120 seconds: 120 = 120 + 0, and 120 seconds is 2 minute(s).
120 s=2 min 0 s120 \text{ s} = 2 \text{ min } 0 \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 0 s from 7:00:00 a.m.
7:00:00−0:02:00=6:58:007:00:00 - 0{:}02{:}00 = 6{:}58{:}00
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 6:58:00 a.m.

Review

A drift of about 2 min 0 s over 5 day(s) is tiny compared to 120 hours, so the clock should read just a few minutes before 7:00 a.m. - and 6:58:00 a.m. matches.

Look for a pattern: the clock drifts 1 s/hour = 24 s/day, so over 5 day(s) it drifts 5 x 24 = 120 s = 2 min 0 s, giving the same 6:58:00 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 5 days x 24 hours and 120 hours x 1 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 120 seconds into 2 min 0 s and combining it with 7:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 6 answer: 8:05:36 a.m.

There is a clock that gains 2 seconds every hour. If this clock is set to the exact time at 8:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 8:00 a.m. 7 days from now?

Show solution

Understand

A clock runs fast, gains 2 seconds for each hour that passes. It is set correctly at 8:00 a.m. today. I need to find what time it displays 7 day(s) later when the true time is again 8:00 a.m.

Givens
  • The clock gains 2 seconds every hour.
  • It is set to the exact time at 8:00 a.m. today.
  • We look at it 7 day(s) later at the true time 8:00 a.m.
Unknowns
  • The time the fast clock shows (in hours, minutes, seconds) after 7 day(s).
Constraints
  • 1 day = 24 hours; 1 minute = 60 seconds.
  • A fast clock shows an LATER 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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
7 full day(s) pass between the two 8:00 a.m. moments. Each day has 24 hours.
7×24=168 hours7 \times 24 = 168 \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 gains 2 seconds each hour, and 168 hours pass, so multiply the drift rate by the number of hours.
168×2=336 seconds168 \times 2 = 336 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 336 seconds: 336 = 300 + 36, and 300 seconds is 5 minute(s).
336 s=5 min 36 s336 \text{ s} = 5 \text{ min } 36 \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 fast clock shows a later time, so add 5 min 36 s to 8:00:00 a.m.
8:00:00+0:05:36=8:05:368:00:00 + 0{:}05{:}36 = 8{:}05{:}36
Counting add from the true time gives the fast clock's display - everyday elapsed-time arithmetic.
Answer: 8:05:36 a.m.

Review

A drift of about 5 min 36 s over 7 day(s) is tiny compared to 168 hours, so the clock should read just a few minutes after 8:00 a.m. - and 8:05:36 a.m. matches.

Look for a pattern: the clock drifts 2 s/hour = 48 s/day, so over 7 day(s) it drifts 7 x 48 = 336 s = 5 min 36 s, giving the same 8:05:36 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 7 days x 24 hours and 168 hours x 2 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 336 seconds into 5 min 36 s and combining it with 8:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 7 answer: 9:08:00 a.m.

There is a clock that gains 10 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. 2 days from now?

Show solution

Understand

A clock runs fast, gains 10 seconds for each hour that passes. It is set correctly at 9:00 a.m. today. I need to find what time it displays 2 day(s) later when the true time is again 9:00 a.m.

Givens
  • The clock gains 10 seconds every hour.
  • It is set to the exact time at 9:00 a.m. today.
  • We look at it 2 day(s) later at the true time 9:00 a.m.
Unknowns
  • The time the fast clock shows (in hours, minutes, seconds) after 2 day(s).
Constraints
  • 1 day = 24 hours; 1 minute = 60 seconds.
  • A fast clock shows an LATER 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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
2 full day(s) pass between the two 9:00 a.m. moments. Each day has 24 hours.
2×24=48 hours2 \times 24 = 48 \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 gains 10 seconds each hour, and 48 hours pass, so multiply the drift rate by the number of hours.
48×10=480 seconds48 \times 10 = 480 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 480 seconds: 480 = 480 + 0, and 480 seconds is 8 minute(s).
480 s=8 min 0 s480 \text{ s} = 8 \text{ min } 0 \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 fast clock shows a later time, so add 8 min 0 s to 9:00:00 a.m.
9:00:00+0:08:00=9:08:009:00:00 + 0{:}08{:}00 = 9{:}08{:}00
Counting add from the true time gives the fast clock's display - everyday elapsed-time arithmetic.
Answer: 9:08:00 a.m.

Review

A drift of about 8 min 0 s over 2 day(s) is tiny compared to 48 hours, so the clock should read just a few minutes after 9:00 a.m. - and 9:08:00 a.m. matches.

Look for a pattern: the clock drifts 10 s/hour = 240 s/day, so over 2 day(s) it drifts 2 x 240 = 480 s = 8 min 0 s, giving the same 9:08:00 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 2 days x 24 hours and 48 hours x 10 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 480 seconds into 8 min 0 s and combining it with 9:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 8 answer: 7:06:00 a.m.

There is a clock that gains 15 seconds every hour. If this clock is set to the exact time at 7:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 7:00 a.m. 1 day from now?

Show solution

Understand

A clock runs fast, gains 15 seconds for each hour that passes. It is set correctly at 7:00 a.m. today. I need to find what time it displays 1 day(s) later when the true time is again 7:00 a.m.

Givens
  • The clock gains 15 seconds every hour.
  • It is set to the exact time at 7:00 a.m. today.
  • We look at it 1 day(s) later at the true time 7:00 a.m.
Unknowns
  • The time the fast clock shows (in hours, minutes, seconds) after 1 day(s).
Constraints
  • 1 day = 24 hours; 1 minute = 60 seconds.
  • A fast clock shows an LATER 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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
1 full day(s) pass between the two 7:00 a.m. moments. Each day has 24 hours.
1×24=24 hours1 \times 24 = 24 \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 gains 15 seconds each hour, and 24 hours pass, so multiply the drift rate by the number of hours.
24×15=360 seconds24 \times 15 = 360 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 360 seconds: 360 = 360 + 0, and 360 seconds is 6 minute(s).
360 s=6 min 0 s360 \text{ s} = 6 \text{ min } 0 \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 fast clock shows a later time, so add 6 min 0 s to 7:00:00 a.m.
7:00:00+0:06:00=7:06:007:00:00 + 0{:}06{:}00 = 7{:}06{:}00
Counting add from the true time gives the fast clock's display - everyday elapsed-time arithmetic.
Answer: 7:06:00 a.m.

Review

A drift of about 6 min 0 s over 1 day(s) is tiny compared to 24 hours, so the clock should read just a few minutes after 7:00 a.m. - and 7:06:00 a.m. matches.

Look for a pattern: the clock drifts 15 s/hour = 360 s/day, so over 1 day(s) it drifts 1 x 360 = 360 s = 6 min 0 s, giving the same 7:06:00 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 1 days x 24 hours and 24 hours x 15 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 360 seconds into 6 min 0 s and combining it with 7:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 9 answer: 5:50:24 a.m.

There is a clock that loses 6 seconds every hour. If this clock is set to the exact time at 6:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 6:00 a.m. 4 days from now?

Show solution

Understand

A clock runs slow, loses 6 seconds for each hour that passes. It is set correctly at 6:00 a.m. today. I need to find what time it displays 4 day(s) later when the true time is again 6:00 a.m.

Givens
  • The clock loses 6 seconds every hour.
  • It is set to the exact time at 6:00 a.m. today.
  • We look at it 4 day(s) later at the true time 6:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 4 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
4 full day(s) pass between the two 6:00 a.m. moments. Each day has 24 hours.
4×24=96 hours4 \times 24 = 96 \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 6 seconds each hour, and 96 hours pass, so multiply the drift rate by the number of hours.
96×6=576 seconds96 \times 6 = 576 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 576 seconds: 576 = 540 + 36, and 540 seconds is 9 minute(s).
576 s=9 min 36 s576 \text{ s} = 9 \text{ min } 36 \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 9 min 36 s from 6:00:00 a.m.
6:00:00−0:09:36=5:50:246:00:00 - 0{:}09{:}36 = 5{:}50{:}24
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 5:50:24 a.m.

Review

A drift of about 9 min 36 s over 4 day(s) is tiny compared to 96 hours, so the clock should read just a few minutes before 6:00 a.m. - and 5:50:24 a.m. matches.

Look for a pattern: the clock drifts 6 s/hour = 144 s/day, so over 4 day(s) it drifts 4 x 144 = 576 s = 9 min 36 s, giving the same 5:50:24 a.m..

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Multiplying 4 days x 24 hours and 96 hours x 6 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 576 seconds into 9 min 36 s and combining it with 6:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 10 answer: 11:04:48 a.m.

There is a clock that gains 4 seconds every hour. If this clock is set to the exact time at 11:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 11:00 a.m. 3 days from now?

Show solution

Understand

A clock runs fast, gains 4 seconds for each hour that passes. It is set correctly at 11:00 a.m. today. I need to find what time it displays 3 day(s) later when the true time is again 11:00 a.m.

Givens
  • The clock gains 4 seconds every hour.
  • It is set to the exact time at 11:00 a.m. today.
  • We look at it 3 day(s) later at the true time 11:00 a.m.
Unknowns
  • The time the fast clock shows (in hours, minutes, seconds) after 3 day(s).
Constraints
  • 1 day = 24 hours; 1 minute = 60 seconds.
  • A fast clock shows an LATER 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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
3 full day(s) pass between the two 11:00 a.m. moments. Each day has 24 hours.
3×24=72 hours3 \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 gains 4 seconds each hour, and 72 hours pass, so multiply the drift rate by the number of hours.
72×4=288 seconds72 \times 4 = 288 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 288 seconds: 288 = 240 + 48, and 240 seconds is 4 minute(s).
288 s=4 min 48 s288 \text{ s} = 4 \text{ min } 48 \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 fast clock shows a later time, so add 4 min 48 s to 11:00:00 a.m.
11:00:00+0:04:48=11:04:4811:00:00 + 0{:}04{:}48 = 11{:}04{:}48
Counting add from the true time gives the fast clock's display - everyday elapsed-time arithmetic.
Answer: 11:04:48 a.m.

Review

A drift of about 4 min 48 s over 3 day(s) is tiny compared to 72 hours, so the clock should read just a few minutes after 11:00 a.m. - and 11:04:48 a.m. matches.

Look for a pattern: the clock drifts 4 s/hour = 96 s/day, so over 3 day(s) it drifts 3 x 96 = 288 s = 4 min 48 s, giving the same 11:04:48 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 4 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 288 seconds into 4 min 48 s and combining it with 11:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 11 answer: 9:50:24 a.m.

There is a clock that loses 8 seconds every hour. If this clock is set to the exact time at 10:00 a.m. today, what time (in hours, minutes, and seconds) will the clock show at 10:00 a.m. 3 days from now?

Show solution

Understand

A clock runs slow, loses 8 seconds for each hour that passes. It is set correctly at 10:00 a.m. today. I need to find what time it displays 3 day(s) later when the true time is again 10:00 a.m.

Givens
  • The clock loses 8 seconds every hour.
  • It is set to the exact time at 10:00 a.m. today.
  • We look at it 3 day(s) later at the true time 10:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 3 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
3 full day(s) pass between the two 10:00 a.m. moments. Each day has 24 hours.
3×24=72 hours3 \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 8 seconds each hour, and 72 hours pass, so multiply the drift rate by the number of hours.
72×8=576 seconds72 \times 8 = 576 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#8 Analyze the Units 3.MD.A.1
Since 60 seconds make 1 minute, regroup 576 seconds: 576 = 540 + 36, and 540 seconds is 9 minute(s).
576 s=9 min 36 s576 \text{ s} = 9 \text{ min } 36 \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 9 min 36 s from 10:00:00 a.m.
10:00:00−0:09:36=9:50:2410:00:00 - 0{:}09{:}36 = 9{:}50{:}24
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 9:50:24 a.m.

Review

A drift of about 9 min 36 s over 3 day(s) is tiny compared to 72 hours, so the clock should read just a few minutes before 10:00 a.m. - and 9:50:24 a.m. matches.

Look for a pattern: the clock drifts 8 s/hour = 192 s/day, so over 3 day(s) it drifts 3 x 192 = 576 s = 9 min 36 s, giving the same 9:50:24 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 8 seconds to get the total drift.
  • 3.MD.A.1 Tell and write time to the nearest minute and solve elapsed time problems — Regrouping 576 seconds into 9 min 36 s and combining it with 10:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!
Variant 12 answer: 8:57:36 a.m.

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. 3 days from now?

Show solution

Understand

A clock runs slow, loses 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 3 day(s) 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 day(s) later at the true time 9:00 a.m.
Unknowns
  • The time the slow clock shows (in hours, minutes, seconds) after 3 day(s).
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 drift, then convert those seconds into minutes and seconds, then combine with the true time. Tracking units (hours -> seconds -> minutes) keeps the drift rate honest.

Execute

#7 Identify Subproblems 3.OA.A.3
3 full day(s) pass between the two 9:00 a.m. moments. Each day has 24 hours.
3×24=72 hours3 \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 drift rate by the number of hours.
72×2=144 seconds72 \times 2 = 144 \text{ seconds}
Drift-per-hour times hours gives total drift, a direct 'rate times amount' multiplication.
#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 minute(s).
144 s=2 min 24 s144 \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.
9:00:00−0:02:24=8:57:369:00:00 - 0{:}02{:}24 = 8{:}57{:}36
Counting subtract from the true time gives the slow clock's display - everyday elapsed-time arithmetic.
Answer: 8:57:36 a.m.

Review

A drift of about 2 min 24 s over 3 day(s) 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. matches.

Look for a pattern: the clock drifts 2 s/hour = 48 s/day, so over 3 day(s) it drifts 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 drift.
  • 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 combining it with 9:00:00 a.m.
💡 This only needs Grade 3 multiplication and time-telling: find the total seconds drifted, then count forward or backward!