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Difficulty:
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Question Stats:
55% (02:20) correct
45%
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The list -1, 4, x, x^2, 11, 5 contains six numbers, where x is an integer.
Which of the following must be true?
I. The median of the list is greater than 1
II. If the range is 12, the list has exactly one mode.
III. The average of the numbers is greater than 2
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
Which of the following must be true?
I. The median of the list is greater than 1
II. If the range is 12, the list has exactly one mode.
III. The average of the numbers is greater than 2
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
19 Mar 2026, 00:55
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Let me walk through each statement for the list: -1, 4, x, x2, 11, 5, where x is an integer.
Statement I: Median > 1
With 6 numbers, the median is the average of the 3rd and 4th values when sorted. Test key cases:
- x = 0: sorted list is -1, 0, 0, 4, 5, 11. Median = (0 + 4)/2 = 2
- x = 1: sorted list is -1, 1, 1, 4, 5, 11. Median = (1 + 4)/2 = 2.5
- x = -2: sorted list is -2, -1, 4, 4, 5, 11. Median = (4 + 4)/2 = 4
- For large |x|, both x and x2 move to the extremes, so the middle stays anchored around the fixed numbers 4 and 5, giving median = 4.5.
In every case, median > 1. TRUE.
Statement II: If range = 15, then no mode.
Key Insight: The trick here is that NO integer x actually produces a range of 15!
- If x extends the minimum (x < -1), and max stays 11: 11 - x = 15 gives x = -4, but then x2 = 16 becomes the new max, making range = 16 - (-4) = 20, not 15.
- If x2 extends the max: x2 - (-1) = 15 gives x2 = 14 — not a perfect square.
- If both x and x2 extend the range: x2 - x = 15 gives no integer solutions (discriminant = 61, not a perfect square).
Since the condition 'range = 15' is NEVER true, the entire if-then statement is automatically (vacuously) true. TRUE.
Statement III: Average > 2
Average = (-1 + 4 + x + x2 + 11 + 5) / 6 = (x2 + x + 19) / 6
We need x2 + x + 19 > 12, i.e., x2 + x + 7 > 0.
The discriminant is 1 - 28 = -27 (negative), and since the x2 coefficient is positive, this expression is ALWAYS positive. TRUE.
All three statements must be true.
Answer: E
Statement I: Median > 1
With 6 numbers, the median is the average of the 3rd and 4th values when sorted. Test key cases:
- x = 0: sorted list is -1, 0, 0, 4, 5, 11. Median = (0 + 4)/2 = 2
- x = 1: sorted list is -1, 1, 1, 4, 5, 11. Median = (1 + 4)/2 = 2.5
- x = -2: sorted list is -2, -1, 4, 4, 5, 11. Median = (4 + 4)/2 = 4
- For large |x|, both x and x2 move to the extremes, so the middle stays anchored around the fixed numbers 4 and 5, giving median = 4.5.
In every case, median > 1. TRUE.
Statement II: If range = 15, then no mode.
Key Insight: The trick here is that NO integer x actually produces a range of 15!
- If x extends the minimum (x < -1), and max stays 11: 11 - x = 15 gives x = -4, but then x2 = 16 becomes the new max, making range = 16 - (-4) = 20, not 15.
- If x2 extends the max: x2 - (-1) = 15 gives x2 = 14 — not a perfect square.
- If both x and x2 extend the range: x2 - x = 15 gives no integer solutions (discriminant = 61, not a perfect square).
Since the condition 'range = 15' is NEVER true, the entire if-then statement is automatically (vacuously) true. TRUE.
Statement III: Average > 2
Average = (-1 + 4 + x + x2 + 11 + 5) / 6 = (x2 + x + 19) / 6
We need x2 + x + 19 > 12, i.e., x2 + x + 7 > 0.
The discriminant is 1 - 28 = -27 (negative), and since the x2 coefficient is positive, this expression is ALWAYS positive. TRUE.
All three statements must be true.
Answer: E
