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# M04 Q34

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Manager
Joined: 22 Feb 2009
Posts: 136

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Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)

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07 May 2009, 11:11
1
This post was
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There is a certain triangle with sides 7, 10, and $$X$$ . If it is known that $$X$$ is an integer, how many different values are there of $$X$$ ?

(A) 8
(B) 10
(C) 13
(D) 14
(E) 16

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Didn't get the explanation.

Kudos [?]: 149 [0], given: 10

Intern
Joined: 25 Mar 2009
Posts: 28

Kudos [?]: 10 [6], given: 1

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07 May 2009, 11:32
6
KUDOS
Let me try answering this, I am not too sure:

As per the Logic any side of a triangle cannot be more than sum of two side and cannot be less than the difference of two sides.
Hence as per the logic: 10-7<X<10+7 i.e. 3<X<17
Hence X can have at the most 13 possible values. The answer is c.

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Manager
Joined: 29 Nov 2009
Posts: 107

Kudos [?]: 37 [2], given: 5

Location: United States

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04 Dec 2009, 07:26
2
KUDOS
Mankhu8, that is correct. That rule is known as the Triangle Inequality Rule. I can't post URLs yet, but search for triangle third side rule on Google and you'll get a link to a SparkNotes SAT page with all useful triangle rules.

Kudos [?]: 37 [2], given: 5

SVP
Affiliations: HEC
Joined: 28 Sep 2009
Posts: 1635

Kudos [?]: 689 [3], given: 432

Concentration: Economics, Finance
GMAT 1: 730 Q48 V44

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04 Dec 2009, 08:21
3
KUDOS
It's also important to not become careless and simply subtract 3 from 17. That would give us the wrong answer.

INCORRECT:
17 - 3 = 14

CORRECT:
Since 3 < x < 17, we list all the possible integers between these two extremes: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and 16. There are 13 possibilities.
_________________

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SVP
Joined: 29 Aug 2007
Posts: 2471

Kudos [?]: 857 [0], given: 19

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04 Dec 2009, 17:36
bandit wrote:
There is a certain triangle with sides 7, 10, and $$X$$ . If it is known that $$X$$ is an integer, how many different values are there of $$X$$ ?

(A) 8
(B) 10
(C) 13
(D) 14
(E) 16

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Didn't get the explanation.

Remember the third side rule:

perimeter-of-triangle-abc-87112.html#p654725
traingle-sides-85784.html#p643924
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 857 [0], given: 19

Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 389

Kudos [?]: 47 [0], given: 46

Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)

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13 Dec 2010, 06:59
3<x<17

+1 for C
_________________

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Manager
Joined: 23 Oct 2010
Posts: 84

Kudos [?]: 33 [1], given: 6

Location: India

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13 Dec 2010, 07:51
1
KUDOS
C.

Any side of a triangle cannot be greater than the sum of other two sides and cannot be less than the difference of two sides. As soon as this condition is broken, the only way to draw the triangle is by making it a straight line.
Hence boundaries for X < 10 + 7 = LT17 (max) and 10 - 7 = GT3 (min)
17 - 3 - 1 = 13

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Senior Manager
Joined: 18 Sep 2009
Posts: 352

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13 Dec 2010, 10:05
Rule: Any side in a triangle must be smaller than the sum of the other sides, and greater than the difference of the other sides.

By considering above rule : 3<x<17. Different values are: 4,5,6,7,8,9,10,11,12,13,14,15,16. There are 13 different values.Answer is C

Kudos [?]: 601 [0], given: 2

Senior Manager
Joined: 01 Nov 2010
Posts: 282

Kudos [?]: 87 [0], given: 44

Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)

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13 Dec 2010, 12:17
the ans will be 13.
sol as explained by Gmatters in above post.
_________________

kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.

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Manager
Joined: 31 May 2010
Posts: 94

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13 Dec 2010, 18:46

As :: 3 < X < 17 ,
_________________

Kudos if any of my post helps you !!!

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Manager
Status: I rest, I rust.
Joined: 04 Oct 2010
Posts: 121

Kudos [?]: 128 [0], given: 9

Schools: ISB - Co 2013
WE 1: IT Professional since 2006

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13 Dec 2010, 19:41
silasaaa2 wrote:
A

_________________

Respect,
Vaibhav

PS: Correct me if I am wrong.

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Intern
Joined: 25 Apr 2010
Posts: 6

Kudos [?]: 1 [0], given: 0

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15 Dec 2010, 05:20
x is btwn (10-7) and (10+7) so 3<x<17

Ans C

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Manager
Joined: 20 Jan 2011
Posts: 63

Kudos [?]: 1 [0], given: 8

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15 Dec 2011, 08:48
Good question testing the famous triangle concept

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Manager
Joined: 21 Nov 2010
Posts: 127

Kudos [?]: 5 [0], given: 12

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16 Dec 2011, 13:01
The sum of any two sides of a triangle must be greater than the third side.

Answer is 3< x < 17
Since answer must be integers, they are 4-16 which equals 13.

Kudos [?]: 5 [0], given: 12

Manager
Joined: 05 Sep 2012
Posts: 78

Kudos [?]: 17 [0], given: 17

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17 Dec 2012, 06:34
This cannot be the hardest one for sure...

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Intern
Joined: 10 May 2010
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06 Jan 2014, 10:58
The third side of a triangle is smaller than the sum of the other two sides and greater than the difference of the other two sides.

Lengths of the sides of the triangle are 7, 10, x

(10-7)<x<(10+7)
3<x<17

There are 13 possible integer values for x in this range - the answer is C

Kudos [?]: 11 [0], given: 0

Re: M04 Q34   [#permalink] 06 Jan 2014, 10:58
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# M04 Q34

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