AMC 8 · 2011 · #11

Grade 6 arithmetic

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

mean-median-mode-rangegraph-reading identify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights 📊 Diagram

Problem

A bar graph shows the minutes Asha (dark bar) and Sasha (light bar) studied on each of the 5 weekdays Mon-Fri. Find the average number of extra minutes per day that Sasha studied compared to Asha.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The question "how many more minutes per day on average" is really two smaller jobs glued together: (1) for each of the 5 days, find Sasha-minus-Asha; (2) average those 5 numbers. Tool #7 (Identify Subproblems) makes that split explicit so we don't try to eyeball the whole graph at once. Tool #16 (Change Focus) gives a clean cross-check: instead of averaging differences, we can total each girl's week first and take the difference of totals divided by 5 — these two views must agree, which catches arithmetic slips.

1STEP 1

Read each day off the graph and compute Sasha minus Asha; days when Sasha studied less just count as negative.

Mon: 70-60=10, Tue: 80-90=-10, Wed: 120-100=20, Thu: 110-80=30, Fri: 50-70=-20

2STEP 2

Add the five differences; the positive days give 60 and the negative days give -30, for a total of 30.

10+(-10)+20+30+(-20)=60-30=30

3STEP 3

Divide that total by the 5 days; the quotient is the average extra minutes per day.

\frac{30}{5}

=6 → (A)

Answer

The five daily differences range from -20 to +30, so a positive average somewhere in the middle (single digits) is exactly what we should expect. A value like 12 would require Sasha to dominate every day, but she actually loses on Tue and Fri. So 6 fits the data shape, and (A) is the only single-digit positive choice — both internal and against-the-options checks pass.

💡Key takeaway

This AMC 8 problem only needs Grade 6 "mean = sum ÷ count" — averaging is just adding and dividing!

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