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19 Feb 2026, 06:47
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
44% (02:08) correct
56%
(02:21)
wrong
based on 39
sessions
History
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7, -4, 2, 3, x
Which of the following is equal to the range of the 5 numbers above for some integer value of x ?
I. 3x – 12
II. 10
III. 2x + 25
A. I only
B. III only
C. I and III only
D. II and III only
E. I, II and III
Which of the following is equal to the range of the 5 numbers above for some integer value of x ?
I. 3x – 12
II. 10
III. 2x + 25
A. I only
B. III only
C. I and III only
D. II and III only
E. I, II and III
19 Feb 2026, 10:55
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kevincan
Since the question contains positive and negative integers, x can be placed at three locations.
1) positive extreme after 7. ( -4,2,3,7,x)
Range = x-(-4) = x+4
2) middle : ( -4,2,3,x,7) for example.
Range = 7+4 =11.
3) negative extreme beyond -4 in the number line.
(-x, -4, 2,3,7). Thus, Range = 7+x
Equate the ranges to the options:
I. 3x – 12
case 1: if 3x-12 = x+4, 2x = 16. And, x=8.
sufficient as x is greater than or equal to 7.
case 2: 3x -12 = 11, x not an integer.
case 3: 3x -12 = 7+x , we get 2x=19. Not an integer.
II. 10
All three cases do not satisfy for this option.
III. 2x + 25
case 1: 2x+25 = x+4 , x = - 21 not true. As the value x+4 is for positive extremes.
case 2: 2x+25 = 11, we get 2x = -14, and x =-7. Which is also not true. As it’s not satisfying the middle value.
case 3: 2x+25 = 7+x , we get x =-18. Which is beyond -4 on the number line. Satisfies.
Thus, I and III satisfies.
option C
19 Feb 2026, 11:00
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The key concept being tested here is range with a variable element — specifically, what happens to the range of a set when one value is unknown and can shift the maximum or minimum.
The big trap most people fall into: they assume x is just "some middle value" and the range stays fixed at 11. The whole question hinges on recognizing that x can be the new maximum, the new minimum, or somewhere in the middle — and each case gives a different range formula.
Step 1 — Identify the range for each case of x.
The known values are 7, –4, 2, 3. Without x:
- Current max = 7, current min = –4, so the "interior" range = 11.
Three cases based on where x falls:
- Case A (x ≥ 7): x is the new max → Range = x – (–4) = x + 4
- Case B (x ≤ –4): x is the new min → Range = 7 – x
- Case C (–4 < x < 7): range stays = 11
Step 2 — Test Statement I: 3x – 12
Try Case A (x ≥ 7): Set 3x – 12 = x + 4 → 2x = 16 → x = 8. Since 8 ≥ 7 and is an integer, ✓ this works.
Statement I is possible.
Step 3 — Test Statement II: 10
- Case A: x + 4 = 10 → x = 6. But 6 < 7, contradiction.
- Case B: 7 – x = 10 → x = –3. But –3 > –4, contradiction.
- Case C: Range = 11 ≠ 10.
Statement II is never possible.
Step 4 — Test Statement III: 2x + 25
Try Case B (x ≤ –4): Set 2x + 25 = 7 – x → 3x = –18 → x = –6. Since –6 ≤ –4 and is an integer, check: range = 7 – (–6) = 13, and 2(–6) + 25 = 13. ✓
Statement III is possible.
Answer: C (I and III only)
Takeaway: Whenever a set contains an unknown x, always split into three cases — x is the new max, x is the new min, or x is interior — and test each expression against each case. Skipping even one case is how you miss an answer.
The big trap most people fall into: they assume x is just "some middle value" and the range stays fixed at 11. The whole question hinges on recognizing that x can be the new maximum, the new minimum, or somewhere in the middle — and each case gives a different range formula.
Step 1 — Identify the range for each case of x.
The known values are 7, –4, 2, 3. Without x:
- Current max = 7, current min = –4, so the "interior" range = 11.
Three cases based on where x falls:
- Case A (x ≥ 7): x is the new max → Range = x – (–4) = x + 4
- Case B (x ≤ –4): x is the new min → Range = 7 – x
- Case C (–4 < x < 7): range stays = 11
Step 2 — Test Statement I: 3x – 12
Try Case A (x ≥ 7): Set 3x – 12 = x + 4 → 2x = 16 → x = 8. Since 8 ≥ 7 and is an integer, ✓ this works.
Statement I is possible.
Step 3 — Test Statement II: 10
- Case A: x + 4 = 10 → x = 6. But 6 < 7, contradiction.
- Case B: 7 – x = 10 → x = –3. But –3 > –4, contradiction.
- Case C: Range = 11 ≠ 10.
Statement II is never possible.
Step 4 — Test Statement III: 2x + 25
Try Case B (x ≤ –4): Set 2x + 25 = 7 – x → 3x = –18 → x = –6. Since –6 ≤ –4 and is an integer, check: range = 7 – (–6) = 13, and 2(–6) + 25 = 13. ✓
Statement III is possible.
Answer: C (I and III only)
Takeaway: Whenever a set contains an unknown x, always split into three cases — x is the new max, x is the new min, or x is interior — and test each expression against each case. Skipping even one case is how you miss an answer.
