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### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Wed Mar 25, 2015 8:43 am
gmat_winter wrote:
If the lengths of the sides of a triangle are a, b, and 7, which of the following could be the value of a - b?

1)4
11)7
111)12

I only
II only
I and II only
II and III only
I, II, and III

OAA
IMPORTANT RULE: If two sides of a triangle have lengths A and B, then . . .
difference between sides A and B < third side < sum of sides A and B

Plugging in we get: a - b < 7 < a + b
So, a - b < 7

RELATED RESOURCE:
We explain HOW and WHY the rule (above) works in the following video: https://www.gmatprepnow.com/module/gmat-geometry/video/860 (starting at 1:50)

Cheers,
Brent

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Last edited by Brent@GMATPrepNow on Wed Mar 23, 2016 7:13 am; edited 1 time in total

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### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
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Wed Mar 25, 2015 5:52 pm
Hi gmat_winter,

This question is based on the Triangle Inequality Theorem, as the other posters have noted. Since the three Roman Numerals are all numbers, and the question asks what COULD be the value of A-B, you can take more of a 'hands-on' approach and TEST VALUES (instead of trying to handle this with the broad rules that the Theorem is base don.

We know that one of the three sides is a 7

I. COULD the difference in the other two sides be 4?

If the sides were 11 and 7, then we'd have an isosceles triangle with sides of 7, 7 and 11. The difference COULD be 4.
Roman Numeral I is possible.

II. COULD the difference in the other two sides be 7?

If the sides were 8 and 1, then we would not have a triangle - we'd have a line "on top" of another line (since 1+7 = 8).
If the sides were 9 and 2, then we'd have the same situation.
If the sides were 10 and 3, then we'd have the same situation.
Etc.
Roman Numeral II is NOT possible

III. COULD the difference in the other two sides be 12?

If the sides were 13 and 1, then the "1" and the "7" could not meet (they're too far away from one another).
If the sides were 14 and 2, then we'd have the same situation.
If the sides were 15 and 3, then we'd have the same situation.
Etc.
Roman Numeral III is NOT possible.

Only Roman Numeral 1 is possible.

GMAT assassins aren't born, they're made,
Rich

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a.gold93 Newbie | Next Rank: 10 Posts
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Posted:
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Tue Mar 22, 2016 6:10 pm
Hi,

I'm kind of stunned by this question. I though that the rule was that the sum of any 2 sides as to be greater than the 3rd side. If you put a-b = 4, then lets say a=5, b=1, and c=7 (given). 5+1 < 7. 5 and 1 being the 2 sides, 7 being the 3rd side

I'm likely confusing 2 different topics, I would appreciate if somebody can clarify what I'm missing or doing wrong.

Thanks! Much appreciated!

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Marty Murray Legendary Member
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Tue Mar 22, 2016 6:35 pm
a.gold93 wrote:
Hi,

I'm kind of stunned by this question. I though that the rule was that the sum of any 2 sides as to be greater than the 3rd side. If you put a-b = 4, then lets say a=5, b=1, and c=7 (given). 5+1 < 7. 5 and 1 being the 2 sides, 7 being the 3rd side

I'm likely confusing 2 different topics, I would appreciate if somebody can clarify what I'm missing or doing wrong.

Thanks! Much appreciated!
What you are doing wrong is extending the question in a way that does not make sense.

The question is asking which of the DIFFERENCES between a and b are possible.

4 is possible. For instance you could have a triangle such that one side has length 4, one has length 8 and the third has length 7.

However you can't extend the idea "a difference of 4 is possible" to make it "any two numbers that differ by 4 work as lengths of the two sides."

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