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20 Jun 2019, 02:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
28% (03:18) correct
72%
(02:45)
wrong
based on 178
sessions
History
Date
Time
Result
Not Attempted Yet
In a survey on the popularity of magazines A,B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
A) 10
B) 12
C) 20
D) 32
E) 48
A) 10
B) 12
C) 20
D) 32
E) 48
20 Jun 2019, 11:44
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It's a very good problem.
Let 'x' pupil don't read any magazine and 'y' pupil read all 3 magazines
100-x= 39+36+45-14-11-5-2y
x=10+2y
We need to maximize the value of 'y' in order to maximize the value of x
y can be maximize when 11-y=0 or y=11
{If y were greater than 11, the pupil who read only magazine would be negative} You can clearly see that in the diagram below
The maximum number of pupils who do not like any of the magazines= 10+2*11=32
Let 'x' pupil don't read any magazine and 'y' pupil read all 3 magazines
100-x= 39+36+45-14-11-5-2y
x=10+2y
We need to maximize the value of 'y' in order to maximize the value of x
y can be maximize when 11-y=0 or y=11
{If y were greater than 11, the pupil who read only magazine would be negative} You can clearly see that in the diagram below
The maximum number of pupils who do not like any of the magazines= 10+2*11=32
Attachments
dgdg.png [ 11.42 KiB | Viewed 7571 times ]
General Discussion
20 Jun 2019, 05:47
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Arvind42
\(None_{max} =\) 100 - (39+36+45-14-11-5-2*0(all three)) = 10
Answer: Option A
