AMC 8 · 2016 · #6
Easy mode Grade 6Problem
Picture 19 people standing in a room. Each person's name has a certain number of letters.
The bar graph below shows how many people have each name length. (The bottom axis is "name length," and the height of each bar tells how many people have that length.)
When you line up all 19 name lengths from shortest to longest, what is the middle one?
Pick an answer.
AMC 8 2016 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 how many of $19$ people have a name of each length from $3$ to $7$ letters. The bars give the counts $7, 3, 1, 4, 4$ for lengths $3, 4, 5, 6, 7$. We need the median name length — the middle value once all $19$ name lengths are lined up in order.
Givens: $19$ people in total; From the bar graph: length $3 \to 7$ people, length $4 \to 3$, length $5 \to 1$, length $6 \to 4$, length $7 \to 4$; Total checks out: $7 + 3 + 1 + 4 + 4 = 19$; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The median of the $19$ name lengths
Understand
Restated: A bar graph shows how many of $19$ people have a name of each length from $3$ to $7$ letters. The bars give the counts $7, 3, 1, 4, 4$ for lengths $3, 4, 5, 6, 7$. We need the median name length — the middle value once all $19$ name lengths are lined up in order.
Givens: $19$ people in total; From the bar graph: length $3 \to 7$ people, length $4 \to 3$, length $5 \to 1$, length $6 \to 4$, length $7 \to 4$; Total checks out: $7 + 3 + 1 + 4 + 4 = 19$; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The data come from a picture, so Tool #1 (Draw a Diagram) first — read each bar and write down the count per name length. Then Tool #7 (Identify Subproblems) splits the task into three small pieces: (a) confirm the total is $19$, (b) find which position is the median, (c) walk the counts until we reach that position. Tool #2 (Make a Systematic List) is the workhorse: build a cumulative tally — "through length $3$ we have $7$, through length $4$ we have $10$, ..." — and the median position drops out by inspection. No algebra needed.
Execute — Answer: B
3.MD.B.3 Step 1 - Read the bar graph.
- The horizontal axis labels name lengths $3, 4, 5, 6, 7$ and each bar's height is the number of people with that length.
- The bar heights are $7, 3, 1, 4, 4$.
💡 Reading a scaled bar graph and pulling out the count for each category is a Grade 3 data-display skill.
6.SP.B.5 Step 2 - Confirm the total.
- Add the five bar heights to make sure they sum to the stated $19$ people.
💡 Reporting the number of observations is the first habit of summarizing a data set.
6.SP.A.3 Step 3 - Locate the median position.
- With $19$ values (odd), the median is the single middle one — the $\tfrac{19+1}{2} = 10$th value when the lengths are written in order from smallest to largest.
💡 The median is a measure of center that names a position in the ordered list, not a computed average.
6.SP.B.5 Step 4 - Build a cumulative tally in order of length.
- After the length-$3$ group we have used positions $1$ through $7$; the length-$4$ group then fills positions $8, 9, 10$.
- So the $10$th value sits inside the length-$4$ group.
💡 A cumulative-count list lets you locate any ranked position without writing all $19$ names.
6.SP.A.3 Step 5 - Read off the answer.
- The $10$th value is in the length-$4$ block, so the median name length is $4$.
💡 The position you found lands inside one category, and that category's value is the median.
3.MD.B.3 Read the bar graph. The horizontal axis labels name lengths $3, 4, 5, 6, 7$ and 6.SP.B.5 Confirm the total. Add the five bar heights to make sure they sum to the stated 6.SP.A.3 Locate the median position. With $19$ values (odd), the median is the single mid 6.SP.B.5 Build a cumulative tally in order of length. After the length-$3$ group we have 6.SP.A.3 Read off the answer. The $10$th value is in the length-$4$ block, so the median Review
Reasonableness: The smallest length ($3$) already accounts for $7$ of $19$ people — more than a third — so the median should sit near the low end. Length $4$ is the very next category, which pushes the cumulative count from $7$ to $10$. Since $10$ is exactly the median position for $n = 19$, the median has to be $4$. A median of $5$ or $6$ would require fewer short names at the front, which the graph contradicts.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: the $7$ length-$3$ names take positions $1$-$7$, so positions $8, 9, 10$ are inside the length-$4$ block. That immediately kills (A) $3$ (too early), and (C), (D), (E) (too late, since the cumulative count only reaches $11$ at length $5$). Only (B) $4$ contains the $10$th position.
CCSS standards used (min grade 6)
3.MD.B.3Draw and interpret scaled picture graphs and bar graphs (Reading the heights of the five bars to extract the counts $7, 3, 1, 4, 4$ for name lengths $3, 4, 5, 6, 7$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Using the definition of the median as the middle value of an ordered data set, and identifying it as the $10$th value when $n = 19$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Reporting the total count ($n = 19$) and building a cumulative tally to locate the median position inside the grouped data.)
⭐ Even with $19$ names, you never have to write them all out — a short cumulative-count list shows the $10$th name lands in the length-$4$ group.
⭐ Even with $19$ names, you never have to write them all out — a short cumulative-count list shows the $10$th name lands in the length-$4$ group.
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