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21 Nov 2019, 03:28
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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:
55%
(hard)
Question Stats:
59% (01:56) correct
41%
(01:46)
wrong
based on 282
sessions
History
Date
Time
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Not Attempted Yet
23 students appeared for the Geography paper. What is the sum of marks obtained by them?
(1) The average mark obtained by the students equals the median mark obtained by the students
(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks
Are You Up For the Challenge: 700 Level Questions
(1) The average mark obtained by the students equals the median mark obtained by the students
(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks
Are You Up For the Challenge: 700 Level Questions
21 Nov 2019, 04:38
Kudos
Bookmarks
Need to find:
\(S_1\) + ... + \(S_2_3\) = ?
Assuming \(S_1\)... \(S_2_3\) to be in ascending order.
Statement 1:
\(S_1\) + ... + \(S_2_3\) = \(23S_1_2\) ->(1)
Not sufficient
Statement 2:
\(S_1\) + ... + \(S_1_2\) = 744 -> (2)
\(S_2_3\) + ... + \(S_1_2\) = 1008 ->(3)
Adding (2) and (3)
\(S_1\) + .. + 2\(S_1_2\) .. + \(S_2_3\) = 1752
Not sufficient
From (1)
\((24)S_1_2\) = 1752
\(S_1_2\) = 73
Therefore, \(S_1\) + ... + \(S_2_3\) = 1679
Ans: C
\(S_1\) + ... + \(S_2_3\) = ?
Assuming \(S_1\)... \(S_2_3\) to be in ascending order.
Statement 1:
\(S_1\) + ... + \(S_2_3\) = \(23S_1_2\) ->(1)
Not sufficient
Statement 2:
\(S_1\) + ... + \(S_1_2\) = 744 -> (2)
\(S_2_3\) + ... + \(S_1_2\) = 1008 ->(3)
Adding (2) and (3)
\(S_1\) + .. + 2\(S_1_2\) .. + \(S_2_3\) = 1752
Not sufficient
From (1)
\((24)S_1_2\) = 1752
\(S_1_2\) = 73
Therefore, \(S_1\) + ... + \(S_2_3\) = 1679
Ans: C
22 Nov 2019, 07:26
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Bunuel
average=sum/n
(1) The average mark obtained by the students equals the median mark obtained by the students insufic
average=median; median = 12th highest mark
(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks insufic
average of marks from 1 to 12 = 62; sum without median = 12(62)-m
average of marks from 12 to 23 = 84; sum without median = 12(84)-m
(1&2) sufic
average=sum/n
m=[12(62)-m+12(84)-m+m]/23
23m=12(62+84)-m
24m=12(146)
m=12(146)/24=146/2=73
sum=average•n=73•23=1679
Ans (C)
