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24 Feb 2026, 15:48
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D
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Each page of a 120-page, two-part manuscript contains between 1 and 3 spelling mistakes, inclusive. Is the average number of mistakes per page greater than 2.0?
(1) The median number of mistakes per page is 2.5.
(2) The second part of the manuscript contains 80 pages and has an average of 2.55 mistakes per page.
(1) The median number of mistakes per page is 2.5.
(2) The second part of the manuscript contains 80 pages and has an average of 2.55 mistakes per page.
24 Feb 2026, 22:11
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kevincan
Let the number of pages which has the mistakes 1,2,3 are x, y, and z respectively.
So, x+y+z =120
We need to find: IS (x+2y+3z)/120 >2 ?
For a value greater than 2, the numerator should be at least 241 to exceed 2.
Statement 1:
The median number of mistakes per page is 2.5.
Median of 120 values is between 60 and 61.
if we consider the worst case of 60th term as 2, and from 61st term to 120th term is 3. Let the remaining values from 1 to 59 be 1.
The numerator gives a value of (3*60)+59 = 239, by adding 2 meets the 241 mark.
(59+2+3*60)/120 is greater than 2.
Hence, Sufficient
Statement 2:
The second part of the manuscript contains 80 pages and has an average of 2.55 mistakes per page.
80*2.55 yields 204. After adding a 40, we get 244.
[(80*2.55)+(40*1) ]/120 gives a value greater than 2.
Hence, Sufficient
Option D
25 Feb 2026, 04:45
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• 120 pages total (two parts)
• Each page has 1, 2, or 3 mistakes (this constraint is crucial!)
• Question: Is the average > 2.0? (Yes/No question)
Statement 1: Median = 2.5
Since mistakes per page can only be 1, 2, or 3, a median of 2.5 means:
• The 60th page (when sorted) has 2 mistakes
• The 61st page has 3 mistakes
• So at least 60 pages have 3 mistakes each
Worst case for average:
• 59 pages with 1 mistake = 59
• 1 page with 2 mistakes = 2
• 60 pages with 3 mistakes = 180
• Total = 241 mistakes
Average = 241/120 = 2.008 > 2.0 ✓
Even in the worst case, average exceeds 2.0. Answer is always YES.
Statement 1 is SUFFICIENT
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Statement 2: Part 2 has 80 pages with average 2.55 mistakes
Part 2 contributes: 80 × 2.55 = 204 mistakes
Part 1 has: 40 pages (each with 1, 2, or 3 mistakes)
Worst case for average:
• Part 1: 40 pages × 1 mistake = 40 mistakes
• Total = 204 + 40 = 244 mistakes
Average = 244/120 = 2.03 > 2.0 ✓
Even if Part 1 has minimum mistakes, the average still exceeds 2.0. Answer is always YES.
Statement 2 is SUFFICIENT
---
Answer: D
• Each page has 1, 2, or 3 mistakes (this constraint is crucial!)
• Question: Is the average > 2.0? (Yes/No question)
Statement 1: Median = 2.5
Since mistakes per page can only be 1, 2, or 3, a median of 2.5 means:
• The 60th page (when sorted) has 2 mistakes
• The 61st page has 3 mistakes
• So at least 60 pages have 3 mistakes each
Worst case for average:
• 59 pages with 1 mistake = 59
• 1 page with 2 mistakes = 2
• 60 pages with 3 mistakes = 180
• Total = 241 mistakes
Average = 241/120 = 2.008 > 2.0 ✓
Even in the worst case, average exceeds 2.0. Answer is always YES.
Statement 1 is SUFFICIENT
---
Statement 2: Part 2 has 80 pages with average 2.55 mistakes
Part 2 contributes: 80 × 2.55 = 204 mistakes
Part 1 has: 40 pages (each with 1, 2, or 3 mistakes)
Worst case for average:
• Part 1: 40 pages × 1 mistake = 40 mistakes
• Total = 204 + 40 = 244 mistakes
Average = 244/120 = 2.03 > 2.0 ✓
Even if Part 1 has minimum mistakes, the average still exceeds 2.0. Answer is always YES.
Statement 2 is SUFFICIENT
---
Answer: D
