AMC 8 · 2015 · #5
Easy mode Grade 6Problem
Billy's basketball team has played games so far. Their points in those games were:
In their game, the team scores points.
After the new score is added, which of these statistics gets bigger than before?
Pick an answer.
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
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
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
6.SP.B.5 Step 1 - Check the **range** (max $-$ min).
- The max stays $73$.
- The min drops from $42$ to $40$, so the range grows.
💡 A new minimum that is smaller than the old minimum can only widen the spread.
6.SP.B.5 Step 2 - 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.
💡 The cluster of three $58$s straddles the middle, so the median stays anchored at $58$.
5.NBT.B.7 Step 3 - Check the **mean**.
- Add up the $11$ scores, then add $40$ for the $12^\text{th}$.
💡 A new value below the current mean must drag the mean down.
6.SP.B.5 Step 4 - Check the **mode** (most frequent value).
- The score $58$ already appears $3$ times; $40$ appears only once.
💡 A single new score cannot displace a value that already led with three copies.
6.SP.B.5 Step 5 - Check the **mid-range** ($\tfrac{\max + \min}{2}$).
- Max stays $73$, min drops to $40$.
💡 Lowering only the minimum lowers the average of the two extremes.
6.SP.B.5 Step 6 - Eliminate the four choices that did not increase.
- Median and mode were unchanged; mean and mid-range decreased.
- Only **range** increased.
💡 The Tool #3 verdict: of the five candidates, four are eliminated, so the remaining one is forced.
6.SP.B.5 Check the **range** (max $-$ min). The max stays $73$. The min drops from $42$ t 6.SP.B.5 Check the **median**. With $11$ values the median is the $6^\text{th}$ value; wi 5.NBT.B.7 Check the **mean**. Add up the $11$ scores, then add $40$ for the $12^\text{th}$ 6.SP.B.5 Check the **mode** (most frequent value). The score $58$ already appears $3$ tim 6.SP.B.5 Check the **mid-range** ($\tfrac{\max + \min}{2}$). Max stays $73$, min drops to 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.5Summarize 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.7Add, 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.3Recognize 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!
⭐ This AMC 8 problem only needs Grade 6 data sense — mean, median, mode, range, mid-range — that you already know!