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Re: There are three courses to be taken and 32 students on average in each [#permalink]
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mrr0821 wrote:
There are three courses to be taken and 32 students on average in each course. 5 students take two courses only and 3 students take all three courses. How many students are there in total?

(a) 91
(b) 88
(c) 94
(d) 85
(e) 92


Since average in each course is 32, the total instances of course-student pairing is 32*3 = 96. This includes double/triple counting for those students who are taking 2/3 courses.

Each person would lie in one of the a - g regions.

Attachment:
SetsThree_1_23Sept-2.jpg
SetsThree_1_23Sept-2.jpg [ 20.19 KiB | Viewed 4937 times ]


5 students take exactly 2 courses (in d, e and f) so they have been double counted. So we should take away 5 from 96 to get the number of people.

3 students take all 3 courses (in g) so they have been triple counted. So we should take away 3*2 = 6 instances from 96 to get the number of people.

Number of students = 96 - 5 - 6 = 85

Answer (D)
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Re: There are three courses to be taken and 32 students on average in each [#permalink]
mrr0821 wrote:
There are three courses to be taken and 32 students on average in each course. 5 students take two courses only and 3 students take all three courses. How many students are there in total?

(a) 91
(b) 88
(c) 94
(d) 85
(e) 92



total = sum of individual - both- (2 all)
sum of individual = 32*3 = 96
total = 96-5-2*3
= 96-11
= 85
IMO D
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Re: There are three courses to be taken and 32 students on average in each [#permalink]
Archit3110 wrote:
mrr0821 wrote:
There are three courses to be taken and 32 students on average in each course. 5 students take two courses only and 3 students take all three courses. How many students are there in total?

(a) 91
(b) 88
(c) 94
(d) 85
(e) 92



total = sum of individual - both- (2 all)
sum of individual = 32*3 = 96
total = 96-5-2*3
= 96-11
= 85

IMO D


Why are we not using "Total=A+B+C−(AnB+AnC+BnC)+AnBnC+Neither" this method.
Assuming "neither" as zero
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Re: There are three courses to be taken and 32 students on average in each [#permalink]
likhith wrote:

Why are we not using "Total=A+B+C−(AnB+AnC+BnC)+AnBnC+Neither" this method.
Assuming "neither" as zero


we can use it. we have to go through 2 steps
1 - find the value of (AnB+AnC+BnC)
Exactly 2 = (AnB+AnC+BnC) - 3AnBnC
Exactly 2 = 5
AnBnC = 3
(AnB+AnC+BnC) = 5 + 3*3 = 14

2-finding the total (AuBuC)
Total=A+B+C−(AnB+AnC+BnC)+AnBnC
Total = 32 + 32+ 32 - 14 + 3 = 85

you can revise the equations here as well:
https://gmatclub.com/forum/formulae-for-3-overlapping-sets-69014.html
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Re: There are three courses to be taken and 32 students on average in each [#permalink]
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Re: There are three courses to be taken and 32 students on average in each [#permalink]
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