AMC 8 · 2005 · #12

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

sequences-arithmeticmean-median-mode-rangelinear-equations-one-var identify-subproblemsconvert-to-algebra ↑ Prerequisites: sequences-arithmetic

📏 Short solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

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.

1STEP 1

Start at May 1 and add 6 each day to get five counts that climb by the same step — an evenly spaced list.

May 1, May 1 + 6, May 1 + 12, May 1 + 18, May 1 + 24

2STEP 2

For an evenly-spaced list with an odd count of terms, the middle equals the average, so May 3 = 100 ÷ 5 = 20.

May 3 =

\frac{100}{5}

= 20 bananas

3STEP 3

May 5 is two days past May 3, so add 6 twice: 20 + 6 + 6 = 32, which is choice (D).

May 5 = 20 + 6 + 6 = 32 → (D)

Answer

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 ÷ 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.

💡Key takeaway

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 ÷ 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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