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There are three courses to be taken and 32 students on average in each

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There are three courses to be taken and 32 students on average in each  [#permalink]

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New post Updated on: 19 Jan 2019, 03:57
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  45% (medium)

Question Stats:

68% (01:32) correct 32% (01:56) wrong based on 64 sessions

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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

Originally posted by mrr0821 on 18 Jan 2019, 03:01.
Last edited by Gladiator59 on 19 Jan 2019, 03:57, edited 1 time in total.
Edited OA.
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There are three courses to be taken and 32 students on average in each  [#permalink]

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New post Updated on: 20 Jan 2019, 03:24
by drawing, I found the answer to be 85 (D)
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Originally posted by Mahmoudfawzy83 on 19 Jan 2019, 00:37.
Last edited by Mahmoudfawzy83 on 20 Jan 2019, 03:24, edited 1 time in total.
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Re: There are three courses to be taken and 32 students on average in each  [#permalink]

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New post 19 Jan 2019, 00:55
Sum of individual - (present in exactly two) - 2*(present in all three) = total unique.

This can be seen from the fact that: those who are common to exactly two, are counted twice when we add up the individual ( and hence we need to subtract once)
And those who are common to all three get added thrice. ( Hence subtract twice)

32*3 - 5 - 3*2
96 - 5 - 6 =85

85 seems to be the unique students. Hence (D) should be the correct answer.

mrr0821, what is the source of the question?
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Re: There are three courses to be taken and 32 students on average in each  [#permalink]

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New post 19 Jan 2019, 03:38
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.

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SetsThree_1_23Sept-2.jpg
SetsThree_1_23Sept-2.jpg [ 20.19 KiB | Viewed 435 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]

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New post 19 Jan 2019, 05:11
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]

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New post 29 Jan 2019, 11:04
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]

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New post 29 Jan 2019, 11:22
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] 29 Jan 2019, 11:22
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