AMC 8 · 2005 · #12
Grade 6 arithmeticProblem
Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
Pick an answer.
AMC 8 2005 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: Big Al the ape ate $100$ bananas across the five days from May 1 to May 5. Each day he ate $6$ more than the day before. How many bananas did he eat on May 5?
Givens: Five days: May 1, May 2, May 3, May 4, May 5; Each day's count is $6$ more than the previous day's; The five daily counts add to $100$; Answer choices: (A) $20$, (B) $22$, (C) $30$, (D) $32$, (E) $34$
Unknowns: The number of bananas eaten on May 5
Understand
Restated: Big Al the ape ate $100$ bananas across the five days from May 1 to May 5. Each day he ate $6$ more than the day before. How many bananas did he eat on May 5?
Givens: Five days: May 1, May 2, May 3, May 4, May 5; Each day's count is $6$ more than the previous day's; The five daily counts add to $100$; Answer choices: (A) $20$, (B) $22$, (C) $30$, (D) $32$, (E) $34$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #11 Find an Invariant
The phrase "each day, six more than the previous day" is the textbook signal for Tool #5 (Look for a Pattern) — the five counts form an evenly-spaced list, going up by $6$ each step. With five terms spaced evenly, Tool #11 (Find an Invariant) catches a quiet but powerful fact: the middle term (May 3) is always the average of all five terms. That invariant turns the sum directly into the middle-day count, and from there May 5 is just two steps of $+6$ away. Picking Tool #5 plus Tool #11 keeps the work at Grade 5-6 arithmetic and avoids Tool #13 (Set Up an Equation) entirely.
Execute — Answer: D
4.OA.C.5 Step 1 - List the pattern.
- Let the five daily counts be the May 1 count, then $+6$, $+6$, $+6$, $+6$ to reach May 5.
- The list goes up by the same $6$ each step, so the five counts are evenly spaced.
💡 Writing out the evenly-spaced list is the Grade 4 "generate a pattern from a rule" move. The constant gap of $6$ is the whole pattern.
6.SP.B.5 Step 2 - Use the invariant: for an evenly-spaced list with an odd number of terms, the middle term equals the average of the whole list.
- With five terms, May 3 is the middle, so May 3 $=$ (total $\div$ count of days) $= 100 \div 5$.
💡 In an evenly-spaced list, every step above the middle is matched by an equal step below, so the middle term is the average — a Grade 6 mean idea.
5.NBT.B.5 Step 3 - Walk forward from the middle to May 5.
- May 5 is two days after May 3, and each day adds $6$, so May 5 $=$ May 3 $+ 6 + 6$.
💡 Adding two more $6$s keeps walking along the same arithmetic pattern — just continuing the rule from the middle to the end.
4.OA.C.5 List the pattern. Let the five daily counts be the May 1 count, then $+6$, $+6$, 6.SP.B.5 Use the invariant: for an evenly-spaced list with an odd number of terms, the mi 5.NBT.B.5 Walk forward from the middle to May 5. May 5 is two days after May 3, and each d Review
Reasonableness: Check the daily list against the total. Working out from the middle: May 1 $= 20 - 12 = 8$, May 2 $= 14$, May 3 $= 20$, May 4 $= 26$, May 5 $= 32$. Sum: $8 + 14 + 20 + 26 + 32 = 100$, exactly the total in the problem. Every count is a positive whole number, the gap stays $6$, and the answer $32$ matches choice (D). A magnitude check also passes: if Big Al had eaten the same amount every day, he would eat $100 \div 5 = 20$ daily, so May 5 — the largest day — must be more than $20$. Among the choices, $22$ is barely above $20$ (gap too small), $30$ is close but gives sum $90$, $34$ is too large (sum $110$), and $32$ is the unique fit.
Alternative: Tool #6 (Guess and Check) on the choices: try (D) $32$ for May 5. Then May 4 $= 26$, May 3 $= 20$, May 2 $= 14$, May 1 $= 8$, and $8 + 14 + 20 + 26 + 32 = 100$ matches the total. Try (C) $30$ instead: $6 + 12 + 18 + 24 + 30 = 90 \ne 100$. Try (E) $34$: $10 + 16 + 22 + 28 + 34 = 110 \ne 100$. Only (D) works.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number pattern that follows a given rule (Writing out the five daily banana counts as an evenly-spaced list with constant step $6$.)6.SP.B.5Summarize numerical data sets, including reporting measures of center (Using the fact that the middle term of an evenly-spaced list equals the mean of the list, so May 3 $= 100 \div 5 = 20$.)5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Carrying the pattern forward from May 3 to May 5 by adding $6 + 6$ to get $32$ bananas.)
⭐ Five days going up by the same $6$ each time form an evenly-spaced list, and the middle day (May 3) is automatically the average $100 \div 5 = 20$. From there, May 5 is just two $+6$ steps away, giving $32$ — that one "middle equals mean" idea turns this AMC 8 problem into a Grade 6 mean exercise.
⭐ Five days going up by the same $6$ each time form an evenly-spaced list, and the middle day (May 3) is automatically the average $100 \div 5 = 20$. From there, May 5 is just two $+6$ steps away, giving $32$ — that one "middle equals mean" idea turns this AMC 8 problem into a Grade 6 mean exercise.
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