AMC 10 · 2010 · #2
Easy mode Grade 4A ferry carries tourists to an island. It leaves once every hour, starting at 10 AM, and its last trip leaves at 3 PM. The 10 AM trip carries 100 tourists. Each trip after that carries 1 tourist fewer than the trip before it. How many tourists did the ferry carry to the island that day?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A ferry leaves for an island once an hour. Its first trip leaves at 10 AM and its last trip leaves at 3 PM. The 10 AM trip carries $100$ tourists, and every later trip carries exactly one fewer tourist than the trip before it. Find the total number of tourists carried to the island that day.
Givens: Trips leave once every hour, on the hour; The first trip leaves at 10 AM and the last trip leaves at 3 PM; The 10 AM trip carries $100$ tourists; Each trip after the first carries exactly $1$ fewer tourist than the trip just before it; Answer choices: (A) $585$, (B) $594$, (C) $672$, (D) $679$, (E) $694$
Unknowns: The total number of tourists the ferry carried to the island across all of that day's trips
Understand
Restated: A ferry leaves for an island once an hour. Its first trip leaves at 10 AM and its last trip leaves at 3 PM. The 10 AM trip carries $100$ tourists, and every later trip carries exactly one fewer tourist than the trip before it. Find the total number of tourists carried to the island that day.
Givens: Trips leave once every hour, on the hour; The first trip leaves at 10 AM and the last trip leaves at 3 PM; The 10 AM trip carries $100$ tourists; Each trip after the first carries exactly $1$ fewer tourist than the trip just before it; Answer choices: (A) $585$, (B) $594$, (C) $672$, (D) $679$, (E) $694$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #5 Look for a Pattern, #14 Extreme Principle, #3 Eliminate Possibilities
The total splits cleanly into two questions that can be answered one at a time — how many trips were there, and how many tourists rode each trip — so Tool #7 (Identify Subproblems) sets the shape of the work. The split matters because the two halves are not equally hard. The adding is routine; the trip count is the entire difficulty, and a wrong count silently corrupts every load that follows it. Tool #2 (Make a Systematic List) settles that count by writing the departures out and checking them against a rule for counting a run with both ends included. Tool #5 (Look for a Pattern) turns the step-by-step 'one fewer' rule into a direct formula for the load on any given trip, which is what pins the last trip at $95$. Tool #14 (Extreme Principle) brackets the total between the smallest and largest possible six-load days before any exact adding happens. Tool #3 (Eliminate Possibilities) finishes with a test that does not repeat the addition at all: a divisibility check that every six-trip total must pass.
Execute — Answer: A
3.MD.A.1 Step 1 Count the trips, not the hours
- Put the departures on a $24$-hour clock, where 3 PM is $15$ o'clock.
- The boat leaves at $10, 11, 12, 13, 14, 15$ — a run of consecutive whole hours from $10$ to $15$.
- The number of items in such a run is $15-10+1=6$, not $15-10=5$.
- The $+1$ is not a rule to memorize; it can be seen.
- The hour-long waits between consecutive departures are the five gaps $10\to 11$, $11\to 12$, $12\to 13$, $13\to 14$, $14\to 15$, and a chain of five gaps is held up by six posts, one at each end of the chain plus the four joints inside.
- The posts are the departures.
- Writing them in ordinary time confirms it: 10 AM, 11 AM, 12 PM, 1 PM, 2 PM, 3 PM.
- Six trips.
💡 Five one-hour gaps are fenced off by six posts, and it is the posts, not the gaps, that carry tourists.
4.OA.C.5 Step 2 Get each trip's load
- Number the trips $k=0,1,2,3,4,5$, where $k$ counts how many trips already left before this one.
- The rule takes away $1$ tourist each time one trip passes, so after $k$ trips the load has dropped by exactly $k$, giving $a_k = 100-k$.
- The six loads are $100, 99, 98, 97, 96, 95$.
- The last trip is $k=5$, so it carries $100-5=95$, not $94$.
- This is the same fencepost as before, wearing different clothes: the drop happens once per gap between trips, and six trips have only five gaps, so the last boat is five lighter than the first rather than six.
💡 The load falls once per gap between trips, not once per trip, so five drops separate the first boat from the last.
4.OA.A.3 Step 3 Bracket the total before adding
- Every trip carries at least $95$ and at most $100$, so the day's total $T$ sits between $6\times 95 = 570$ and $6\times 100 = 600$, and strictly between them because the loads are not all equal.
- That single sweep of the two extremes kills three choices at once: $672$, $679$, and $694$ all exceed $600$.
- Two choices remain, $585$ and $594$, so the bracket alone is not enough.
- It is also worth noticing what the bracket rests on.
- With seven trips the window would run from $7\times 94 = 658$ to $700$, and $679$ would sit comfortably inside it.
- The estimate is only as trustworthy as the trip count from Step 1.
💡 Six boatloads of roughly a hundred must land a little under six hundred.
4.NBT.B.4 Step 4 Add by leaning on 100
- Rather than adding the six numbers one after another, write each load as $100$ minus its shortfall: $100-0$, $100-1$, $100-2$, $100-3$, $100-4$, $100-5$.
- Adding them, each of the six terms contributes a full $100$, and then the six shortfalls are handed back.
- The shortfalls are $0,1,2,3,4,5$, which total $15$.
- So $T = 6\times 100 - 15 = 600 - 15 = 585$.
- This lands inside the bracket from Step 3 and is one of the two survivors there.
💡 Pretend every boat left full at a hundred, then give back the fifteen seats that were actually missing.
4.OA.B.4 Step 5 Check the shape of the total
- There is a test that does not repeat the addition at all.
- If six trips carry six consecutive numbers of tourists $n, n+1, \dots, n+5$, their sum is $6n+15$ no matter what $n$ is.
- Two facts follow immediately: $6n$ is even and $15$ is odd, so the total is always odd; and $6n+15 = 3(2n+5)$, so the total is always a multiple of $3$.
- Test the choices against those two facts.
- Now $594$, $672$, and $694$ are even, so no such six-trip day could ever produce them.
- And $679$ is odd but $6+7+9=22$ is not a multiple of $3$, so it fails too.
- Only $585 = 3\times 195$ passes both, agreeing with Step 4.
- The near miss is worth naming, because it is the trap the problem is built around: $679 = 7\times 97$ is exactly $94+95+96+97+98+99+100$, the total of a day miscounted as seven trips, and a run of seven consecutive numbers always sums to a multiple of $7$, which $585$ is not.
- The answer is $\textbf{(A)}$.
💡 A six-in-a-row total always comes out odd and on a multiple of three, and exactly one choice does both.
3.MD.A.1 Put the departures on a $24$-hour clock, where 3 PM is $15$ o'clock. The boat le 4.OA.C.5 Number the trips $k=0,1,2,3,4,5$, where $k$ counts how many trips already left b 4.OA.A.3 Every trip carries at least $95$ and at most $100$, so the day's total $T$ sits 4.NBT.B.4 Rather than adding the six numbers one after another, write each load as $100$ m 4.OA.B.4 There is a test that does not repeat the addition at all. If six trips carry six Review
Reasonableness: Rebuild the total from the other end as an independent check. Every one of the six trips carries at least $95$, which accounts for $6\times 95 = 570$, and the amounts carried above that floor are $5,4,3,2,1,0$, which total $15$. That gives $570+15=585$ again, from the bottom rather than the top. The two ways to go wrong here are both about counting, not arithmetic. Counting seven trips gives $679$, which is offered as choice (D); counting five trips gives $100+99+98+97+96=490$, which is not offered, so only the overcount is baited. It is also worth being honest about how much each check can carry on its own. The bracket $570<T<600$ leaves both $585$ and $594$ alive, so estimation alone does not finish the problem, and the divisibility test is what removes $594$: an even number can never be a sum of six consecutive integers. Finally the reading holds up. Only outbound trips are counted, which is what 'took to the island' asks for, and the answer does not depend on whether the same tourist rode twice, since the problem reports how many were on each trip and asks for the day's total.
Alternative: Pair the ends instead of anchoring at $100$. In a list that steps down by $1$, two terms equally far from the two ends always add to the same amount, because moving one step inward from the left raises one term by $1$ exactly as moving one step inward from the right lowers the other by $1$. Here $100+95=99+96=98+97=195$. Six terms make three such pairs, so $T = 3\times 195 = 585$. The same idea stated as an average says the mean load is the midpoint of the two ends, $\frac{100+95}{2}=97.5$, so $T = 6\times 97.5 = 585$. This route never adds the six numbers individually and works for a run of any length, and it exposes the trip count from a third angle: the pairing is clean only because the number of trips is even, so a day with an odd number of trips would leave a middle term standing alone.
CCSS standards used (min grade 4)
3.MD.A.1Tell and write time to the nearest minute and solve elapsed time problems (Turning the schedule 'hourly from 10 AM through 3 PM' into the six departure times and seeing that five one-hour gaps mean six departures.)4.OA.C.5Generate a number or shape pattern following a given rule (Converting the step-by-step rule 'one fewer than the previous trip' into the direct load formula $a_k = 100-k$, which pins the last trip at $95$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Bracketing the day's total between $6\times 95$ and $6\times 100$ to throw out the three choices above $600$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the six loads as $6\times 100$ minus the shortfalls $0+1+2+3+4+5$ to get $600-15=585$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Checking each choice against the facts that a sum of six consecutive integers is odd and a multiple of $3$, and recognizing $679=7\times 97$ as the seven-trip total.)
⭐ Count the boats, not the hours between them — 10 AM through 3 PM is six departures — then pretend all six left full at a hundred and give back the fifteen missing seats.
⭐ Count the boats, not the hours between them — 10 AM through 3 PM is six departures — then pretend all six left full at a hundred and give back the fifteen missing seats.
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