AMC 8 · 2015 · #5

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

mean-median-mode-range caseworksystematic-enumeration ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

$text{range}$

(B)

$text{median}$

(C)

$text{mean}$

(D)

$text{mode}$

(E)

$text{mid-range}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

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.

1STEP 1

The max stays 73 but the min drops from 42 to 40, so the range climbs from 31 to 33.

old range = 73 - 42 = 31, new range = 73 - 40 = 33. 33 ＞ 31, so range increases.

2STEP 2

Adding 40 at the bottom shifts positions, but the 6th and 7th values are both 58, so the median stays 58.

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 = (58 + 58)/2 = 58. Unchanged.

3STEP 3

The sum grows to 672 over 12 games, so the mean drops from about 57.45 to 56.

Old sum = 42+47+53+53+58+58+58+61+64+65+73 = 632. Old mean =

\frac{632}{11}

≈ 57.45. New sum = 632 + 40 = 672. New mean =

\frac{672}{12}

= 56. 56 ＜ 57.45, so mean decreases.

4STEP 4

58 already appears three times while 40 appears once, so the mode stays 58.

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

5STEP 5

With the max at 73 and the min now 40, the mid-range falls from 57.5 to 56.5.

Old mid-range =

\frac{73 + 42}{2}

= 57.5. New mid-range =

\frac{73 + 40}{2}

= 56.5. 56.5 ＜ 57.5, so mid-range decreases.

6STEP 6

Median and mode held, mean and mid-range fell — the only statistic that rose is the range.

→ (A) range

Answer

text{range}

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 =

\frac{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 58s, so range is the only candidate left.

💡Key takeaway

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

More like this