AMC 10 · 2017 · #5

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
mean-median-mode-rangedecimal-arithmeticinterval-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: mean-median-mode-range
📏 Short solution 💡 2 insights
Problem
The sorted data set 6, 19, 33, 33, 39, 41, 41, 43, 51, 57 has first quartile 33 and third quartile 43. The interquartile range is the third quartile minus the first. A value counts as an outlier if it sits more than 1.5 interquartile ranges below the first quartile, or more than 1.5 above the third. Count how many values in the set are outliers.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4
How to solve
Strategy Convert to Algebra

Tool #13 (Convert to Algebra): the outlier rule is a long sentence, and the whole difficulty is turning it into two plain inequalities, x ＜ Q₁ - 1.5 IQR and x ＞ Q₃ + 1.5 IQR, so that checking a value becomes a comparison with a single number. Tool #7 (Identify Subproblems): build those two numbers in order — first the IQR, then 1.5 times it, then subtract from Q₁ and add to Q₃. Tool #14 (Extreme Principle): the list is already sorted, so if the smallest value fails the low test, no value can pass it, and the same holds at the top — that turns a ten-value scan into a look at the ends.

1STEP 1

Compute the interquartile range

The gap between the quartiles is the interquartile range.

IQR = Q₃ - Q₁ = 43 - 33 = 10
2STEP 2

Scale it by 1.5

Scale it by 1.5.

1.5 × IQR = 3/2 × 10 = 15
3STEP 3

Turn the rule into two cutoffs

The rule becomes two cutoffs, low and high.

x ＜ 33 - 15 = 18 or x ＞ 43 + 15 = 58
4STEP 4

Check the ends of the sorted list

Since it is sorted, only the two ends need checking.

6 ＜ 18 ✓ 19 ≮ 18 57 ≯ 58
5STEP 5

Count the outliers

Exactly 1 value is caught.

outliers = {6} → 1 → (B)
Answer
1
The answer 1 matches what the list looks like: nine of the ten values sit inside the band from 19 to 57, while 6 is stranded 13 below its nearest neighbor — a visible gap. The near misses are worth checking, because they are what the wrong choices are built from. The value 19 is 33 - 19 = 14 below Q₁, and 57 is 57 - 43 = 14 above Q₃; both are 14, just short of the required 15, so neither is an outlier. Had the rule used 1.5 times a slightly smaller IQR, all three would have counted and the answer would have been 3. Since the fences are strict at 18 and 58, only 6 survives the test, so 1 is right and 0, 2, 3, 4 are not.
💡Key takeaway

Turn the outlier sentence into two numbers — 33 - 15 = 18 and 43 + 15 = 58 — then only 6 falls outside them, so there is 1 outlier, choice (B).

  • Compute the interquartile range
  • Scale it by 1.5
  • Turn the rule into two cutoffs
  • Check the ends of the sorted list
  • Count the outliers

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