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LMP
Joined: 29 Jun 2018
Last visit: 15 Mar 2022
Posts: 49
Given Kudos: 62
Location: India
Schools: Johnson '20 HEC Montreal '21
GPA: 4
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
81% (02:12) correct
19%
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wrong
based on 130
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X persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 minutes, what is X?
(a) 5
(b) 7
(c) 9
(d) 4
(e) 11
(a) 5
(b) 7
(c) 9
(d) 4
(e) 11
17 Aug 2018, 10:35
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Avinasht123
Avinasht123
I am not expert, I can try to explain
Let there be \(n\) person standing in a circle. Then we have to form teams with 2 people each, such that these 2 people selected in a team should not be adjacent to each other in the circle.
For first person , we can \(n\) choices, we can select anybody out of \(n\) persons.
For second person, we have \(n-3\) [\(n- 1\)(person already selected)\(-1\)(Person to the right of already selected person)\(-1\)(Person to the left of already selected person)]
Total number of such 2 person team:\(\frac{n(n-3)}{2}\)
Division by \(2\) is required as expression \(n(n-3)\) counts AB,BA as two pairs , whereas they are only one pair
Now Question says each team formed should sing a 2 minute song.
Total time spent by \(\frac{n(n-3)}{2}\) teams singing=\(2* \frac{n(n-3)}{2}\)minutes
Now in question , it is given that \(n(n-3)=28\)
This leads to \(n(n-3)=28\)
\(n^2-3n-28=0\)
\(n^2-7n+4n-28=0\)
\(n(n-7)+4(n-7)=0\)
\((n+4)(n-7)=0\)
\(n=7\) neglecting \(n=-4\)
17 Aug 2018, 11:24
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Bulusuchaitanya
A way to look at it..
When you join these X person, a polygon is formed. And as we do not count the neighbors, it is basically asking us to find DIAGONAL..
Number of diagonals = X(x-3)/2=28/2....X(X-3)=28=7*4..
Thus X is 7
another way...
A person cannot make pair with himself and his neighbours so 3 out of X gone..
So he can make pair with X-3 person..
Thus total X*X-3 but this includes where each combination is counted twice A with C as a different way from C with A..
So total =X(X-3)/2
Total minutes=28, so pairs = 28/2=14..
Thus X (X-3)/2=14....X(X-3)=28=7*4
X=7
B
