AMC 8 · 2016 · #6

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangegraph-reading systematic-enumerationidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticgraph-reading

📏 Short solution 💡 2 insights 📊 Diagram

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Problem

The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#1 Draw a Diagram 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 $7, 3, 1, 4, 4$.

$$\text{length } 3 \to 7,\; 4 \to 3,\; 5 \to 1,\; 6 \to 4,\; 7 \to 4$$

💡 Reading a scaled bar graph and pulling out the count for each category is a Grade 3 data-display skill.

#7 Identify Subproblems 6.SP.B.5 Step 2

Confirm the total.
Add the five bar heights to make sure they sum to the stated $19$ people.

$$7 + 3 + 1 + 4 + 4 = 19$$

💡 Reporting the number of observations is the first habit of summarizing a data set.

#7 Identify Subproblems 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.

$$\dfrac{19+1}{2} = 10 \;\Rightarrow\; \text{median} = \text{10th value in order}$$

💡 The median is a measure of center that names a position in the ordered list, not a computed average.

#2 Make a Systematic List 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.

$$\begin{array}{l|l} \text{through length } 3 & \text{positions } 1\text{-}7 \\ \text{through length } 4 & \text{positions } 8\text{-}10 \\ \text{through length } 5 & \text{position } 11 \\ \text{through length } 6 & \text{positions } 12\text{-}15 \\ \text{through length } 7 & \text{positions } 16\text{-}19 \end{array}$$

💡 A cumulative-count list lets you locate any ranked position without writing all $19$ names.

#2 Make a Systematic List 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 $4$.

$$\text{median} = 4 \;\Rightarrow\; \textbf{(B)}$$

💡 The position you found lands inside one category, and that category's value is the median.

[1] #1 3.MD.B.3 Read the bar graph. The horizontal axis labels name lengths $3, 4, 5, 6, 7$ and

[2] #7 6.SP.B.5 Confirm the total. Add the five bar heights to make sure they sum to the stated

[3] #7 6.SP.A.3 Locate the median position. With $19$ values (odd), the median is the single mid

[4] #2 6.SP.B.5 Build a cumulative tally in order of length. After the length-$3$ group we have

[5] #2 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.3 Draw 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.3 Recognize 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.5 Summarize 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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