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# Given that the length of each side of a quadrilateral is a distinct

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Senior Manager
Joined: 22 Feb 2018
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Given that the length of each side of a quadrilateral is a distinct  [#permalink]

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02 May 2018, 08:33
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19
00:00

Difficulty:

95% (hard)

Question Stats:

19% (01:45) correct 81% (02:04) wrong based on 64 sessions

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Given that the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7,How many different possible combinations of side lengths are there?

[A] 21
[B] 24
[C] 32
[D] 34
[E] 35
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Re: Given that the length of each side of a quadrilateral is a distinct  [#permalink]

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02 May 2018, 09:02
4
That's really a nice question:

The concept used :

like the triangle inequality theorem, the sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.

Since the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, all four sides must be taken from the set {1, 2, 3, 4, 5, 6, 7}. How many different sets of four could be selected? $$7C4$$= 7*6*5/(3!) = 7*5 = 35 Of those 35, the only ones that don’t work for sides of a quadrilateral are the ones in which the sum of the three smallest sides are equal to or less than the longest side. These would be

1 + 2 + 3 < 7

1 + 2 + 3 = 6

1 + 2 + 4 = 7

Those are the only invalid combinations. The other 32 work.

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Given that the length of each side of a quadrilateral is a distinct  [#permalink]

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02 May 2018, 09:13
2
The simple proof of the concept used in this question :

"The sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side."

Attachment:

gmatbusters2.jpg [ 12.4 KiB | Viewed 3839 times ]

Join any two vertices(Say AC)

Now from triangle inequality j+k>n
Also n+m>l

So, from the above two equations we have j+k+m > l

Hence,

The sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.

It is analogous to triangle inequality theorem, which states that sum of two sides of a triangle is greater than third side.
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Re: Given that the length of each side of a quadrilateral is a distinct  [#permalink]

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11 Jun 2018, 23:03
Hi,

Sorry but why have you factored 1 + 2 + 3 two times? IMO the only combinations that dont satisfy the rule are 1,2,3 and 1,2,4. Thus 35 - 2 = 33 possible combinations.

gmatbusters wrote:
That's really a nice question:

The concept used :

like the triangle inequality theorem, the sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.

Since the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, all four sides must be taken from the set {1, 2, 3, 4, 5, 6, 7}. How many different sets of four could be selected? $$7C4$$= 7*6*5/(3!) = 7*5 = 35 Of those 35, the only ones that don’t work for sides of a quadrilateral are the ones in which the sum of the three smallest sides are equal to or less than the longest side. These would be

1 + 2 + 3 < 7

1 + 2 + 3 = 6

1 + 2 + 4 = 7

Those are the only invalid combinations. The other 32 work.

Retired Moderator
Joined: 27 Oct 2017
Posts: 1319
Location: India
GPA: 3.64
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Re: Given that the length of each side of a quadrilateral is a distinct  [#permalink]

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12 Jun 2018, 00:29
1
The three sets of side lengths that do not work are:

{1, 2, 3, 6}

{1, 2, 3, 7}

{1, 2, 4, 7}

For each of these sets, the sum of the 3 smallest sides is NOT greater than the length of the longest side. Therefore, these 3 side lengths don’t work.

ColonelRed wrote:
Hi,

Sorry but why have you factored 1 + 2 + 3 two times? IMO the only combinations that dont satisfy the rule are 1,2,3 and 1,2,4. Thus 35 - 2 = 33 possible combinations.

gmatbusters wrote:
That's really a nice question:

The concept used :

like the triangle inequality theorem, the sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.

Since the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, all four sides must be taken from the set {1, 2, 3, 4, 5, 6, 7}. How many different sets of four could be selected? $$7C4$$= 7*6*5/(3!) = 7*5 = 35 Of those 35, the only ones that don’t work for sides of a quadrilateral are the ones in which the sum of the three smallest sides are equal to or less than the longest side. These would be

1 + 2 + 3 < 7

1 + 2 + 3 = 6

1 + 2 + 4 = 7

Those are the only invalid combinations. The other 32 work.

Posted from my mobile device
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Re: Given that the length of each side of a quadrilateral is a distinct  [#permalink]

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27 Sep 2019, 13:39
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Re: Given that the length of each side of a quadrilateral is a distinct   [#permalink] 27 Sep 2019, 13:39
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