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16 Aug 2021, 08:15
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
91% (01:45) correct
9%
(02:43)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
In a competitive examination, the average marks obtained was 45. It was later discovered that there was some error in computerization and the marks of 90 candidates had to be changed from 80 to 50, and the average came down to 40 marks. The total number of candidates appeared in the examination is:
(A) 520
(B) 525
(C) 540
(D) 550
(E) 560
(A) 520
(B) 525
(C) 540
(D) 550
(E) 560
16 Aug 2021, 10:54
Kudos
Bookmarks
Average marks =\(\frac{ sum of the marks obtained by students }{ # students}\) = 45
=> Initially, marks obtained by 90 students = 80
=> Let's say the sum of marks obtained by other students = x
=> Total students = n
=> average = \(\frac{[x + (90 * 80)]}{n}\) = 45
Now when the marks have been reduced
Average = \(\frac{[x + (90 * 50)]}{n}\) = 40
=> Solving the two equations for n = 540
OA should be C
=> Initially, marks obtained by 90 students = 80
=> Let's say the sum of marks obtained by other students = x
=> Total students = n
=> average = \(\frac{[x + (90 * 80)]}{n}\) = 45
Now when the marks have been reduced
Average = \(\frac{[x + (90 * 50)]}{n}\) = 40
=> Solving the two equations for n = 540
OA should be C
16 Aug 2021, 12:47
Kudos
Bookmarks
Bunuel
Let the total number of students be x
So, total marks before miscalculation = 45x(Since, average was 45)
There was an error of (80 - 50) = 30 marks for 1 student
There was an error of 30*90 = 2700 marks for 90 students
So, \(\frac{45x - 2700}{x} = 40\)
45x - 2700 = 40x
5x = 2700
x = 520
IMO, (C)!
