AMC 8 · 2011 · #11
Easy mode Grade 6Problem
The bar graph shows how many minutes Asha and Sasha each studied on Monday, Tuesday, Wednesday, Thursday, and Friday. Asha's bars are black and Sasha's bars are grey.
Use the graph to find the total minutes each girl studied that week. Then find the daily average for each girl.
On average, how many more minutes per day did Sasha study than Asha?
Pick an answer.
AMC 8 2011 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 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.
Givens: Mon: Asha $60$, Sasha $70$; Tue: Asha $90$, Sasha $80$; Wed: Asha $100$, Sasha $120$; Thu: Asha $80$, Sasha $110$; Fri: Asha $70$, Sasha $50$; Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $10$, (E) $12$
Unknowns: Average daily difference (Sasha minus Asha), in minutes per day
Understand
Restated: 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.
Givens: Mon: Asha $60$, Sasha $70$; Tue: Asha $90$, Sasha $80$; Wed: Asha $100$, Sasha $120$; Thu: Asha $80$, Sasha $110$; Fri: Asha $70$, Sasha $50$; Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $10$, (E) $12$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Complement
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.
Execute — Answer: A
3.MD.B.3 Step 1 - Read each day off the graph and compute Sasha $-$ Asha.
- Some days are negative, which is fine — "how many more" is a signed quantity here.
💡 Reading a scaled bar graph one category at a time to answer a comparison question is exactly the Grade 3 bar-graph standard.
4.NBT.B.4 Step 2 - Add the five daily differences.
- Group the positives and negatives so the arithmetic is easy: positives $10+20+30=60$; negatives $(-10)+(-20)=-30$.
💡 Adding multi-digit whole numbers (and their opposites) fluently is the Grade 4 NBT skill.
6.SP.B.5 Step 3 - Divide the total difference by the number of days to get the mean.
- There are $5$ days, so the average daily extra is $30 \div 5$.
💡 The mean of a small data set $=$ sum $\div$ count is the Grade 6 "measure of center" definition.
3.MD.B.3 Read each day off the graph and compute Sasha $-$ Asha. Some days are negative, 4.NBT.B.4 Add the five daily differences. Group the positives and negatives so the arithme 6.SP.B.5 Divide the total difference by the number of days to get the mean. There are $5$ Review
Reasonableness: 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.
Alternative: Tool #16 (Change Focus): instead of averaging the differences, compute each girl's weekly total first. Asha $=60+90+100+80+70=400$ min; Sasha $=70+80+120+110+50=430$ min. Per-day means are $400/5=80$ and $430/5=86$, so Sasha's daily mean exceeds Asha's by $86-80=6$ — same answer (A). This works because the mean of a difference equals the difference of the means.
CCSS standards used (min grade 6)
3.MD.B.3Draw a scaled bar graph and solve one- and two-step "how many more" problems using information from the graph (Reading the five paired bar heights off the graph to extract Asha's and Sasha's minutes for each weekday.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing each daily difference and summing $10+(-10)+20+30+(-20)=30$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures of center (Dividing the total difference of $30$ minutes by the $5$ days to get the mean daily difference of $6$ minutes.)
⭐ This AMC 8 problem only needs Grade 6 "mean = sum $\div$ count" — averaging is just adding and dividing!
⭐ This AMC 8 problem only needs Grade 6 "mean = sum $\div$ count" — averaging is just adding and dividing!
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