AMC 8 · 2015 · #5

Easy mode Grade 6

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Problem

Billy's basketball team has played $11$ games so far. Their points in those games were:

$42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73$

In their $12^{th}$ game, the team scores $40$ points.

After the new score is added, which of these statistics gets bigger than before?

Pick an answer.

(A)

range

(B)

median

(C)

mean

(D)

mode

(E)

mid-range

AMC 8 2015 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.

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Toolkit + CCSS Solution

Understand

Restated: Billy's basketball team has $11$ game scores: $42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73$. In game $12$ they score $40$. Which one of these five statistics — range, median, mean, mode, mid-range — is larger for the $12$-game data than for the $11$-game data?

Givens: Original $11$ scores (sorted): $42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73$; New $12^\text{th}$ score: $40$; Answer choices: (A) range, (B) median, (C) mean, (D) mode, (E) mid-range

Unknowns: Which one of the five statistics strictly increases when the $40$ is added

Understand

Givens: Original $11$ scores (sorted): $42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73$; New $12^\text{th}$ score: $40$; Answer choices: (A) range, (B) median, (C) mean, (D) mode, (E) mid-range

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #7 Identify Subproblems

The question lists five named statistics and asks which one increases — a textbook setup for tool #3 (Eliminate Possibilities): compute each statistic before and after, then eliminate any that stayed the same or went down. Tool #7 (Identify Subproblems) makes the work manageable by treating each of the five statistics as its own mini-problem with a clear formula. Two structural facts cut the work in half: $40$ is the new minimum (so it lowers anything that depends on the minimum) and $40$ is below the old mean (so it pulls the mean down), so before computing anything we already expect mean and mid-range to decrease — leaving range as the prime suspect.

Execute — Answer: A

#7 Identify Subproblems 6.SP.B.5 Step 1

Check the range

Check the **range** (max $-$ min).
The max stays $73$.
The min drops from $42$ to $40$, so the range grows.

$\text{old range} = 73 - 42 = 31$, $\quad \text{new range} = 73 - 40 = 33$. $\;33 > 31$, so range **increases**.

💡 A new minimum that is smaller than the old minimum can only widen the spread.

#3 Eliminate Possibilities 6.SP.B.5 Step 2

Check the median

Check the **median**.
With $11$ values the median is the $6^\text{th}$ value; with $12$ values it is the average of the $6^\text{th}$ and $7^\text{th}$.
Adding $40$ at the bottom shifts every original value up by one position.

Old sorted list, position $6$: $58$. New sorted list: $40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73$. Positions $6$ and $7$ are both $58$, so new median $= \tfrac{58 + 58}{2} = 58$. **Unchanged.**

💡 The cluster of three $58$s straddles the middle, so the median stays anchored at $58$.

#3 Eliminate Possibilities 5.NBT.B.7 Step 3

Check the mean

Check the **mean**.
Add up the $11$ scores, then add $40$ for the $12^\text{th}$.

Old sum $= 42+47+53+53+58+58+58+61+64+65+73 = 632$. Old mean $= \tfrac{632}{11} \approx 57.45$. New sum $= 632 + 40 = 672$. New mean $= \tfrac{672}{12} = 56$. $\;56 < 57.45$, so mean **decreases**.

💡 A new value below the current mean must drag the mean down.

#3 Eliminate Possibilities 6.SP.B.5 Step 4

Check the mode

Check the **mode** (most frequent value).
The score $58$ already appears $3$ times; $40$ appears only once.

Old mode $= 58$ (count $3$). New mode $= 58$ (count $3$, the new $40$ has count $1$). **Unchanged.**

💡 A single new score cannot displace a value that already led with three copies.

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

Check the mid-range

Check the **mid-range** ($\tfrac{\max + \min}{2}$).
Max stays $73$, min drops to $40$.

Old mid-range $= \tfrac{73 + 42}{2} = 57.5$. New mid-range $= \tfrac{73 + 40}{2} = 56.5$. $\;56.5 < 57.5$, so mid-range **decreases**.

💡 Lowering only the minimum lowers the average of the two extremes.

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

Eliminate what did not rise

Eliminate the four choices that did not increase.
Median and mode were unchanged; mean and mid-range decreased.
Only **range** increased.

$$\Rightarrow \textbf{(A) range}$$

💡 The Tool #3 verdict: of the five candidates, four are eliminated, so the remaining one is forced.

[1] #7 6.SP.B.5 Check the **range** (max $-$ min). The max stays $73$. The min drops from $42$ t

[2] #3 6.SP.B.5 Check the **median**. With $11$ values the median is the $6^\text{th}$ value; wi

[3] #3 5.NBT.B.7 Check the **mean**. Add up the $11$ scores, then add $40$ for the $12^\text{th}$

[4] #3 6.SP.B.5 Check the **mode** (most frequent value). The score $58$ already appears $3$ tim

[5] #3 6.SP.B.5 Check the **mid-range** ($\tfrac{\max + \min}{2}$). Max stays $73$, min drops to

[6] #3 6.SP.B.5 Eliminate the four choices that did not increase. Median and mode were unchanged

Review

Reasonableness: Two structural checks confirm the answer. (1) The new score $40$ is the new minimum, and only two statistics depend on the minimum: the range and the mid-range. Of those two, range $= \max - \min$ goes **up** when $\min$ goes down, while mid-range $= \tfrac{\max + \min}{2}$ goes **down** when $\min$ goes down — so range is the only one that can increase. (2) The mean drops (new value below old mean), and the median/mode are pinned by the three $58$s, so range is the only candidate left.

Alternative: Tool #16 (Change Focus / Complement): instead of checking all five, ask which statistics *could* go up when adding a value smaller than every existing score. Median can only stay or decrease. Mean must decrease. Mode keeps the dominant value (unless the new value ties or surpasses count $3$, which $40$ doesn't). Mid-range must decrease since min drops. That leaves range as the only statistic that can possibly increase — and the calculation $33 > 31$ confirms it.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize and describe numerical data sets (measures of center and variability) (Computing and comparing range, median, mode, and mid-range of the $11$-score and $12$-score data sets to decide which one increased.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Calculating the old mean $\tfrac{632}{11} \approx 57.45$ and new mean $\tfrac{672}{12} = 56$ and comparing the two decimal values.)
6.SP.A.3 Recognize that a measure of center summarizes data values with a single number, while a measure of variation describes spread (Distinguishing the role of each statistic (mean/median/mode summarize the center, range/mid-range describe the spread or extremes) so we can predict the direction each one moves when a new low score is added.)

⭐ This AMC 8 problem only needs Grade 6 data sense — mean, median, mode, range, mid-range — that you already know!

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Check the range

Check the median

Check the mean

Check the mode

Check the mid-range

Eliminate what did not rise

Review

CCSS standards used (min grade 6)

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