The number of diagonals of a polygon of n sides is given by the formul

Question

The number of diagonals of a polygon of n sides is given by the formula d=n(n-3)/2. If a polygon has twice as many diagonals as sides, how many sides does it have?

(A) 3
(B) 5 
(C) 6
(D) 7
(E) 8

Skywalker18 · Answer

d=n(n-3)

d=2*n
2n=n(n-3)
=>2=n-3
=>n=5
Answer B

GMATinsight · Answer

No. of Diagonals in an n sided polygon, d=n(n-3)

Given, n(n-3) = 2*n

Checking Options

(A) 3 i.e. n(n-3) = 3*0 = 0 which is NOT twice of no. of sides INCORRECT 
 (B) 5 i.e. n(n-3) = 5*2 = 10 which is twice of no. of sides (2*5=10) CORRECT  
(C) 6
(D) 7
(E) 8

Answer: Option B

nipunjain14 · Answer

Two equations are given :
d= n(n-3)
d= 2n

Thus we can write as :
2n= n^2-3n
5n=n^2
n=5

OME: B

v12345 · Answer

given d=n(n-3) = 2n

=> n - 3 =2 (as n is not equal to 0)
=> n = 5

Answer choice B

longfellow · Answer

Answer is 5.
2n=n(n-3)
n=5 or n=0
n=5 right answer

mohit0181 · Answer

I think the given equation should be n * (n-3) / 2. Then the answer is 7. (D)..

torreadortorment · Answer

Answer is B. Something wrong with either the question or the OA given.

Bunuel · Answer

_________________________
Edited the question. Thank you.

dcummins · Answer

Its fairly easy to plugin the numbers and test this.
d=n(n-3)/2

test 5 sides

d= 5(2)/2 = 5
Insufficient, but clearly we should try increase

d=7(7-3)/2
d= 7(4)/2
d= 14

sufficient since d = 2n

Algebraically 
2n = n(n-3)/2
4n=n^2-3n
7n=n^2
n=7

Algebraic way may not be as clear to see straight away.

Kinshook · Answer

Given: The number of diagonals of a polygon of n sides is given by the formula d=n(n-3)/2.

Asked: If a polygon has twice as many diagonals as sides, how many sides does it have?

n(n-3)/2 = 2n
n^2 - 3n = 4n
n(n-7) = 0
n=7

IMO D

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The number of diagonals of a polygon of n sides is given by the formul

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