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# M04 Q34

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Manager
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M04 Q34 [#permalink]  07 May 2009, 10:11
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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.
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Re: M04 Q34 [#permalink]  07 May 2009, 10:32
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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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Re: M04 Q34 [#permalink]  04 Dec 2009, 06:26
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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.
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Re: M04 Q34 [#permalink]  04 Dec 2009, 07:21
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Expert's post
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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Re: M04 Q34 [#permalink]  04 Dec 2009, 16:36
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
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Re: M04 Q34 [#permalink]  13 Dec 2010, 05:59
3<x<17

+1 for C
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Re: M04 Q34 [#permalink]  13 Dec 2010, 06:51
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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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Re: M04 Q34 [#permalink]  13 Dec 2010, 09:05
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
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Re: M04 Q34 [#permalink]  13 Dec 2010, 11:17
the ans will be 13.
sol as explained by Gmatters in above post.
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Re: M04 Q34 [#permalink]  13 Dec 2010, 17:46

As :: 3 < X < 17 ,
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Kudos if any of my post helps you !!!

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Re: M04 Q34 [#permalink]  13 Dec 2010, 18:41
silasaaa2 wrote:
A

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Respect,
Vaibhav

PS: Correct me if I am wrong.

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Re: M04 Q34 [#permalink]  15 Dec 2010, 04:20
x is btwn (10-7) and (10+7) so 3<x<17

Ans C
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Re: M04 Q34 [#permalink]  15 Dec 2011, 07:48
Good question testing the famous triangle concept
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Re: M04 Q34 [#permalink]  16 Dec 2011, 12:01
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.
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Re: M04 Q34 [#permalink]  17 Dec 2012, 05:34
This cannot be the hardest one for sure...
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Re: M04 Q34 [#permalink]  06 Jan 2014, 09:58
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
Re: M04 Q34   [#permalink] 06 Jan 2014, 09:58
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# M04 Q34

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