AMC 10 · 2008 · #1
Grade 4 arithmeticA basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A player scores exactly 5 baskets, and each basket is worth either 2 or 3 points. Counting every way the baskets could split between 2-pointers and 3-pointers, how many different total point values are possible?
Givens: There are exactly 5 baskets in total; Every basket is worth either 2 points or 3 points; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: How many different total point values the 5 baskets can produce
Understand
Restated: A player scores exactly 5 baskets, and each basket is worth either 2 or 3 points. Counting every way the baskets could split between 2-pointers and 3-pointers, how many different total point values are possible?
Givens: There are exactly 5 baskets in total; Every basket is worth either 2 points or 3 points; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #14 Extreme Principle, #5 Look for a Pattern
The phrase "how many different numbers" is the classic signal for Tool #2 (Make a Systematic List): with only 5 baskets there are just a few splits, so list them all instead of guessing. Tool #14 (Extreme Principle) pins the two ends of the list — all 2-pointers gives the smallest total, all 3-pointers the largest — and Tool #5 (Look for a Pattern) shows the totals climb by exactly $1$ each step, so nothing in between is skipped or repeated.
Execute — Answer: E
4.OA.A.3 Step 1 List the six possible splits
- The only thing that can change is how many of the 5 baskets are worth 3 points.
- That count can be $0, 1, 2, 3, 4,$ or $5$, and the rest are worth 2 points.
- Writing them out as (number of 3-pointers, number of 2-pointers) gives six cases: $(0,5), (1,4), (2,3), (3,2), (4,1), (5,0)$.
💡 With the number of baskets fixed, the only choice is how many are 3-pointers, so just list every count from 0 to 5.
3.OA.D.8 Step 2 Add up each split's total
- For each case, multiply and add.
- $(0,5)$: $0\cdot3+5\cdot2=10$.
- $(1,4)$: $3+8=11$.
- $(2,3)$: $6+6=12$.
- $(3,2)$: $9+4=13$.
- $(4,1)$: $12+2=14$.
- $(5,0)$: $15+0=15$.
- The six totals are $10, 11, 12, 13, 14, 15$.
💡 Each total is just points-from-threes plus points-from-twos.
4.OA.A.3 Step 3 Check the extremes
- The smallest total comes from the extreme of all 2-pointers: $5\cdot2=10$.
- The largest comes from all 3-pointers: $5\cdot3=15$.
- So every possible total must sit between $10$ and $15$, which matches the ends of the list.
💡 Pushing every basket to its lowest value gives the floor, and to its highest value gives the ceiling.
4.OA.C.5 Step 4 Count the distinct totals
- Turning one 2-point basket into a 3-point basket raises the total by exactly $1$, so the totals march up one at a time: $10, 11, 12, 13, 14, 15$.
- No value is skipped and none repeats, so there are $6$ different totals.
- That is choice (E).
💡 Because each swap adds exactly $1$, the totals form an unbroken run, so counting them is just counting from 10 up to 15.
4.OA.A.3 The only thing that can change is how many of the 5 baskets are worth 3 points. 3.OA.D.8 For each case, multiply and add. $(0,5)$: $0\cdot3+5\cdot2=10$. $(1,4)$: $3+8=11 4.OA.A.3 The smallest total comes from the extreme of all 2-pointers: $5\cdot2=10$. The l 4.OA.C.5 Turning one 2-point basket into a 3-point basket raises the total by exactly $1$ Review
Reasonableness: The totals run from $10$ to $15$ with no gaps, and the count of whole numbers from $10$ to $15$ inclusive is $15-10+1=6$, which agrees with the six listed totals. The answer $6$ is exactly the number of baskets plus one, which makes sense: there are 6 choices for how many baskets are 3-pointers (0 through 5), and each choice gives a different total. Smaller choices like $5$ come from forgetting one of the endpoint cases.
Alternative: Introduce a variable: let $k$ be the number of 3-point baskets, so the total is $2(5-k)+3k=10+k$. As $k$ runs through $0,1,2,3,4,5$, the total $10+k$ takes six different values, giving the same answer (E) in one line without listing every case.
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Organizing the five baskets into the six possible 3-pointer/2-pointer splits and bounding the totals with the all-2 and all-3 extremes.)3.OA.D.8Solve two-step word problems using four operations within 100 (Computing each split's total as points-from-threes plus points-from-twos, e.g. $3+8=11$.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that each 2-to-3 swap adds exactly 1, so the totals form the unbroken run $10,11,12,13,14,15$.)
⭐ When only one thing can change, list every case in order — here the totals climb from 10 to 15 by ones, so there are 6 of them.
⭐ When only one thing can change, list every case in order — here the totals climb from 10 to 15 by ones, so there are 6 of them.
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