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Difficulty:
95%
(hard)
Question Stats:
18% (02:08) correct
82%
(02:31)
wrong
based on 39
sessions
History
Date
Time
Result
Not Attempted Yet
The 12 districts of Bradenton reported on average 7.5 traffic accidents yesterday, and no district reported more than 10 accidents. More districts reported six accidents than any other number of accidents. The median number of accidents for the 12 districts yesterday is one of how many possible numbers?
A) 1
B) 2
C) 3
D) 4
E) 5
A) 1
B) 2
C) 3
D) 4
E) 5
ankushsambare
Joined: 13 Jun 2022
Last visit: 05 Sep 2026
Posts: 313
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Given Kudos: 230
Location: India
Schools: ISB '27 Wharton '26 INSEAD Aug '26
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04 Apr 2026, 20:24
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Total = 90, 12 districts, max 10, and 6 is the most frequent.
Median = average of 6th and 7th terms.
Key restriction: since 6 appears more than any other number, count of 6 must be strictly highest.
Now test possibilities:
1. Median = 6
This works when both 6th and 7th terms are 6 → requires enough 6s. Valid.
2. Median = 5.5
This happens if 6th = 5 and 7th = 6 → so number of 6s is limited, but still must be the most frequent. This is also possible.
3. Median = 6.5
Would need 6th = 6 and 7th = 7 → but then 7 would start competing with 6, and with sum constraint + max 10, you cannot keep 6 as strictly most frequent. Not possible.
So only two medians work: 5.5 and 6.
Answer: 2
Option: B
Median = average of 6th and 7th terms.
Key restriction: since 6 appears more than any other number, count of 6 must be strictly highest.
Now test possibilities:
1. Median = 6
This works when both 6th and 7th terms are 6 → requires enough 6s. Valid.
2. Median = 5.5
This happens if 6th = 5 and 7th = 6 → so number of 6s is limited, but still must be the most frequent. This is also possible.
3. Median = 6.5
Would need 6th = 6 and 7th = 7 → but then 7 would start competing with 6, and with sum constraint + max 10, you cannot keep 6 as strictly most frequent. Not possible.
So only two medians work: 5.5 and 6.
Answer: 2
Option: B
04 Apr 2026, 22:21
Kudos
Bookmarks
This is pretty tough.
I am getting 5 medians.
6,6,6,6,6,8,8,8,9,9,9,9
Median = 8
Subtract 1 before the midline and add 1 after midline:
6,6,6,6,6,7,8,8,9,9,9,10
Median = 7.5
Repeat it again:
6,6,6,6,6,6,8,8,9,9,10,10
Median = 7
6,6,6,6,6,6,7,8,9,10,10,10
Median = 6.5
Edit:
6,6,6,6,6,6,6,8,10,10,10,10
Median = 6 is also possible.
Answer should be 5.
Bunuel what do you think?
I am getting 5 medians.
6,6,6,6,6,8,8,8,9,9,9,9
Median = 8
Subtract 1 before the midline and add 1 after midline:
6,6,6,6,6,7,8,8,9,9,9,10
Median = 7.5
Repeat it again:
6,6,6,6,6,6,8,8,9,9,10,10
Median = 7
6,6,6,6,6,6,7,8,9,10,10,10
Median = 6.5
Edit:
6,6,6,6,6,6,6,8,10,10,10,10
Median = 6 is also possible.
Answer should be 5.
Bunuel what do you think?
kevincan
