AMC 10 · 2009 · #1
Easy mode Grade 4Kim's plane took off at 10:34 AM and landed at 1:18 PM on the same day. The clocks in both cities show the same time. Write the length of the flight as h hours and m minutes, where m is greater than 0 and less than 60. What is h+m?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A flight leaves at 10:34 AM and arrives at 1:18 PM on the same day, with no time-zone change. Write the length of the flight as $h$ hours and $m$ minutes, where $m$ is a whole number of minutes with $0 < m < 60$, and report the value of $h + m$.
Givens: Departure clock reading: 10:34 AM; Arrival clock reading: 1:18 PM; Newark and Miami are in the same time zone, so both clocks show the same instant the same way; The flight length is written as $h$ hours and $m$ minutes with $0 < m < 60$
Unknowns: The value of $h + m$
Understand
Restated: A flight leaves at 10:34 AM and arrives at 1:18 PM on the same day, with no time-zone change. Write the length of the flight as $h$ hours and $m$ minutes, where $m$ is a whole number of minutes with $0 < m < 60$, and report the value of $h + m$.
Givens: Departure clock reading: 10:34 AM; Arrival clock reading: 1:18 PM; Newark and Miami are in the same time zone, so both clocks show the same instant the same way; The flight length is written as $h$ hours and $m$ minutes with $0 < m < 60$
Plan
Primary tool: #8 Analyze the Units
Secondary: #15 Organize Information in More Ways, #7 Identify Subproblems, #11 Work Backwards
Two units are fighting here: hours and minutes, on a clock that restarts at 12 and relabels itself AM/PM. Mixed units are where the mistakes live, so I first drop to a single unit — minutes counted from midnight — where subtraction is ordinary subtraction and the AM/PM relabelling cannot bite. That gives one number of minutes. The second, separate subproblem is turning that number back into $h$ hours and $m$ minutes. That step is the one that actually needs care: $h + m$ has no meaning at all unless exactly one pair $(h, m)$ fits the rule $0 < m < 60$, so I show that only one pair does rather than assuming it.
Execute — Answer: A
3.MD.A.1 Step 1 Check the clock is trustworthy
- Elapsed time equals the difference of two clock readings only when both clocks run on the same scale and nothing resets in between.
- The problem says the two cities share a time zone, so the clocks agree.
- Departure is AM, arrival is PM, and PM follows AM within one day, so no midnight rollover happens either.
- That licenses treating this as one subtraction on a single day.
- Rewrite the readings on the 24-hour clock so that later times carry larger numbers: 10:34 AM stays 10:34, and 1:18 PM becomes 13:18 because PM hours sit $12$ past their AM labels.
💡 A clock only measures duration honestly when it is the same clock the whole way through.
4.MD.A.1 Step 2 Put both times on one number line
- A time written as hours-and-minutes is really two numbers glued together, and gluing makes subtraction awkward.
- Convert each reading into a single count of minutes after midnight.
- An hour is $60$ minutes, so $10$ hours and $34$ minutes is $10 \times 60 + 34 = 634$ minutes, and $13$ hours and $18$ minutes is $13 \times 60 + 18 = 798$ minutes.
- Both times are now plain whole numbers on one line.
💡 One unit beats two: once everything is in minutes, clock rules stop mattering and only arithmetic is left.
4.NBT.B.4 Step 3 Subtract to get the flight length
- On a number line, the gap between two points is the larger minus the smaller.
- The flight lasted $798 - 634 = 164$ minutes.
- Notice what did not happen: no borrowing across a 12, no worrying about whether $18$ minutes is 'less than' $34$ minutes.
- That trouble was traded away in the previous step.
💡 Duration is a gap between two positions, and gaps are just differences.
4.OA.A.3 Step 4 Split 164 into hours and minutes — only one way
- Now write $164$ minutes as $h$ hours and $m$ minutes, meaning $164 = 60h + m$.
- Peel off whole hours: $164 - 60 = 104$, and $104 - 60 = 44$.
- A third hour is impossible because $44 < 60$.
- So $164 = 60 \times 2 + 44$, giving $h = 2$ and $m = 44$.
- This pair is the only one allowed.
- If $h$ were smaller, the leftover $m$ would be $104$ or $164$, both at least $60$; if $h$ were larger, $m$ would be negative.
- So the rule $0 \le m < 60$ leaves exactly one pair, and here $m = 44$ is not $0$, so the problem's stricter rule $0 < m < 60$ is satisfied too.
- That matters: if the flight had lasted a whole number of hours, no pair would obey $0 < m$ and $h + m$ would not exist.
- The condition $0 < m < 60$ is precisely what makes the answer a single number.
💡 Taking away 60 as many times as you can leaves a remainder smaller than 60, and there is no second way to stop.
4.NBT.B.4 Step 5 Add the two counts
- The question does not ask for a duration; it asks for $h + m$, which adds an hour count to a minute count.
- That sum is a code for the pair $(h, m)$, not a length of time, which is exactly why the previous step had to pin the pair down uniquely.
- With $h = 2$ and $m = 44$, the value is $2 + 44 = 46$, so the answer is $\textbf{(A)}\ 46$.
💡 Once the pair is forced, adding its two parts is the last and easiest move.
3.MD.A.1 Elapsed time equals the difference of two clock readings only when both clocks r 4.MD.A.1 A time written as hours-and-minutes is really two numbers glued together, and gl 4.NBT.B.4 On a number line, the gap between two points is the larger minus the smaller. Th 4.OA.A.3 Now write $164$ minutes as $h$ hours and $m$ minutes, meaning $164 = 60h + m$. P 4.NBT.B.4 The question does not ask for a duration; it asks for $h + m$, which adds an hou Review
Reasonableness: Run the claim forward instead of backward: leave at 10:34 AM, add $2$ hours to reach 12:34 PM, then add $44$ minutes. $34 + 44 = 78$ minutes, which is $60 + 18$, so the hour rolls over to 1 PM with $18$ minutes left: 1:18 PM. That is the stated arrival, so $(h, m) = (2, 44)$ is confirmed independently of the subtraction. The size is sensible too — Newark to Miami is a bit under three hours. The wrong choices are all reachable by mishandling exactly one of the two counts. Choice $\textbf{(B)}\ 47$ is the classic double count: the hour hand moves from 10 to 1, which looks like $3$ hours, and adding $44$ minutes on top counts the partial hour twice; the true hour count is $2$ because the third hour is only $44$ minutes deep. Choices $\textbf{(C)}\ 50$, $\textbf{(D)}\ 53$, and $\textbf{(E)}\ 54$ all require $h + m \ne 46$, which by the uniqueness argument in step 4 means either an hour count other than $2$ or a minute count other than $44$ — and both are ruled out by the forward check above.
Alternative: Split the trip at noon instead of measuring from midnight, which never forms the number $164$ at all. From 10:34 AM to 11:34 AM is $1$ hour, and from 11:34 AM to noon is $26$ minutes, since $60 - 34 = 26$. From noon to 1:18 PM is $1$ hour and $18$ minutes. Adding the pieces gives $1 + 1 = 2$ hours and $26 + 18 = 44$ minutes, and $44 < 60$ so no further carrying is needed. Same pair, same sum $46$. The two routes differ in kind: this one adds up signed pieces of the timeline and depends on splitting at a landmark, while the main route assigns every instant an absolute coordinate and subtracts. They agree, which is good evidence the noon boundary and the AM/PM relabelling were both handled correctly.
CCSS standards used (min grade 4)
3.MD.A.1Tell and write time to the nearest minute and solve elapsed time problems (Justifying that the flight time is the difference of the two clock readings, given one shared time zone and no midnight rollover.)4.MD.A.1Know relative sizes of measurement units and convert larger to smaller units (Converting each clock reading into a single count of minutes after midnight using $1$ hour $= 60$ minutes.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $798 - 634 = 164$ and the final sum $2 + 44 = 46$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Splitting $164$ minutes into whole hours plus a remainder and interpreting that remainder as the unique $m$ with $0 < m < 60$.)
⭐ Turn both clock times into minutes after midnight, subtract, and then take away 60 as many times as you can — the hours you removed and the leftover under 60 are the only pair that fits.
⭐ Turn both clock times into minutes after midnight, subtract, and then take away 60 as many times as you can — the hours you removed and the leftover under 60 are the only pair that fits.
More like this
Same archetype — closest grade level first.