AMC 8 · 2019 · #10

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition

mean-median-mode-rangegraph-reading identify-subproblems ↑ Prerequisites: mean-median-mode-rangegraph-reading

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually $21$ participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?

$\textbf{(A) }$ The mean increases by $1$ and the median does not change.

$\textbf{(B) }$ The mean increases by $1$ and the median increases by $1$ .

$\textbf{(C) }$ The mean increases by $1$ and the median increases by $5$ .

$\textbf{(D) }$ The mean increases by $5$ and the median increases by $1$ .

$\textbf{(E) }$ The mean increases by $5$ and the median increases by $5$ .

Pick an answer.

(A)

The mean increases by 1 and the median does not change.

(B)

The mean increases by 1 and the median increases by 1.

(C)

The mean increases by 1 and the median increases by 5.

(D)

The mean increases by 5 and the median increases by 1.

(E)

The mean increases by 5 and the median increases by 5.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A bar chart shows the number of students at soccer practice for each of the five weekdays: Monday $20$, Tuesday $26$, Wednesday $16$, Thursday $22$, Friday $16$. The mean and median were computed from these five values, but then Coach realizes Wednesday's count was actually $21$, not $16$. Which option correctly describes how the mean and the median each change after fixing that single Wednesday value?

Givens: Original five values (Mon-Fri): $20, 26, 16, 22, 16$; Corrected Wednesday value: $21$ (replaces the original $16$); The other four weekday values are unchanged; Answer choices describe paired changes to the mean and the median (each either $0$, $1$, or $5$)

Unknowns: How much the mean changes (and in which direction) after the correction; How much the median changes (and in which direction) after the correction

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities

The question bundles two independent sub-questions: "what happens to the mean?" and "what happens to the median?". Tool #7 (Identify Subproblems) lets us answer each in isolation. For the mean, we use a shortcut: the sum increases by $21 - 16 = 5$, spread over $5$ data points, so the mean rises by $5 \div 5 = 1$. For the median, Tool #15 (Organize Information in More Ways) tells us to re-sort the list (a different ordering of the same data) so the $3$rd value pops out — once before the fix, once after. Tool #3 (Eliminate Possibilities) is the multiple-choice cleanup: once we know both changes are $+1$, only choice (B) survives.

Execute — Answer: B

#15 Organize Information in More Ways 3.MD.B.3 Step 1

Read the five values directly off the bar chart: Monday $20$, Tuesday $26$, Wednesday $16$, Thursday $22$, Friday $16$.
This gives the original dataset.

$$\{20,\ 26,\ 16,\ 22,\ 16\}$$

💡 Reading a value off a scaled bar graph is the Grade 3 "draw and interpret bar graphs" skill.

#15 Organize Information in More Ways 6.SP.B.5 Step 2

Subproblem 1a — find the original median.
Re-sort the five values from smallest to largest; the median of $5$ numbers is the $3$rd one in that sorted list.

sorted: $\{16,\ 16,\ \boxed{20},\ 22,\ 26\}\;\Rightarrow\; \text{median}_{\text{old}} = 20$

💡 Sorting the data and picking the middle value is exactly how Grade 6 "measures of center" defines the median.

#7 Identify Subproblems 6.SP.B.5 Step 3

Subproblem 1b — find the original mean.
Add the five values, then divide by $5$.

$$\dfrac{20+26+16+22+16}{5} = \dfrac{100}{5} = 20\;\Rightarrow\; \text{mean}_{\text{old}} = 20$$

💡 Summing the values and dividing by the count is the Grade 6 definition of the mean.

#15 Organize Information in More Ways 6.SP.B.5 Step 4

Subproblem 2 — replace Wednesday's $16$ with $21$, then re-sort.
The new sorted list shows that the $3$rd entry is now $21$, so the new median is $21$.
The median therefore increased by $21 - 20 = 1$.

new sorted: $\{16,\ 20,\ \boxed{21},\ 22,\ 26\}\;\Rightarrow\; \text{median}_{\text{new}} = 21,\ \Delta\text{median} = +1$

💡 Reorganizing the corrected data by size lets us read off the middle value just like before.

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

Use the mean-change shortcut instead of recomputing the whole sum.
Wednesday's value went up by $21 - 16 = 5$, so the total sum went up by $5$.
Spread over $n = 5$ data points, the mean goes up by $5 \div 5 = 1$.
(Sanity check: new sum $= 105$, new mean $= 105 / 5 = 21$, matching.)

$$\Delta\text{mean} = \dfrac{\Delta\text{sum}}{n} = \dfrac{+5}{5} = +1$$

💡 Because the mean is sum/$n$, a change in one value changes the mean by (that change)$\,/\,n$ — a clean Grade 6 measures-of-center insight.

#3 Eliminate Possibilities 6.SP.B.5 Step 6

Combine the two subproblem results and match the choices.
Mean increases by $1$, median increases by $1$ — that is exactly option (B).
Choices (A), (C), (D), (E) are eliminated because they require the wrong change in at least one of the two statistics.

$$\Delta\text{mean} = +1,\ \Delta\text{median} = +1\;\Rightarrow\; \textbf{(B)}$$

💡 Once both changes are known, the four wrong options fail at least one condition, leaving (B).

[1] #15 3.MD.B.3 Read the five values directly off the bar chart: Monday $20$, Tuesday $26$, Wedn

[2] #15 6.SP.B.5 Subproblem 1a — find the original median. Re-sort the five values from smallest

[3] #7 6.SP.B.5 Subproblem 1b — find the original mean. Add the five values, then divide by $5$.

[4] #15 6.SP.B.5 Subproblem 2 — replace Wednesday's $16$ with $21$, then re-sort. The new sorted

[5] #7 6.SP.B.5 Use the mean-change shortcut instead of recomputing the whole sum. Wednesday's v

[6] #3 6.SP.B.5 Combine the two subproblem results and match the choices. Mean increases by $1$,

Review

Reasonableness: Bumping just one of five values up by $5$ should pull the mean up by exactly $5 / 5 = 1$, which we got. The change moved that value from $16$ (a low end) past the old middle $20$ to $21$, so the sorted order shifts and the new middle reads $21$ — also $+1$. Both shifts are in the same direction (the data set got slightly bigger overall) and modest in size, consistent with changing only one of five points.

Alternative: Tool #3 (Eliminate Possibilities) alone, used as a shortcut: the mean change must be $(\text{change in one value}) / n = 5 / 5 = 1$, immediately killing choices (D) and (E), which claim $\Delta\text{mean} = 5$. Among the remaining (A), (B), (C), only (B) gives a median change ($+1$) compatible with moving Wednesday from below the old median ($16 < 20$) to above it ($21 > 20$): the median has to step up, so (A)'s "no change" and (C)'s "$+5$" are both wrong. Answer (B).

CCSS standards used (min grade 6)

3.MD.B.3 Draw and interpret scaled picture graphs and bar graphs (Reading the five weekday values $20, 26, 16, 22, 16$ off the bar chart in the problem.)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Computing and comparing the mean and the median of the original and corrected five-value dataset, including the "sum changes by $\Delta$, mean changes by $\Delta / n$" shortcut.)

⭐ This AMC 8 problem only needs Grade 6 mean and median you already know — plus the handy shortcut that when one value goes up by $5$, the mean of $5$ numbers goes up by just $1$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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