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# m08-q31

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m08-q31 [#permalink]

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01 Feb 2009, 07:18
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The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b
II. c/a/b=10/6/2
III. c^2>a^2+b^2

A. I and III only
B. II and III only
C. I only
D. II only
E. III only

Source: GMAT Club Tests - hardest GMAT questions

Last edited by Bunuel on 30 Oct 2013, 07:17, edited 1 time in total.
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Re: m08-q31 [#permalink]

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01 Nov 2012, 07:28
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ConkergMat wrote:
If $$A$$ , $$B$$ , and $$C$$ are points on the plane, is $$AB \gt 15$$ ?

1. $$BC + AC \gt 14$$
2. Area of triangle $$ABC \lt 1$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

OA does not explain how you could get an area < 1, with side > 15?

This problem was replaced by the following question:

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b
II. c/a/b=10/6/2
III. c^2>a^2+b^2

A. I and III only
B. II and III only
C. I only
D. II only
E. III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them).

Answer: E.

Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.
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Senior Manager
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Concentration: Finance, General Management
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Re: m08-q31 [#permalink]

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30 Oct 2013, 08:11
ConkergMat wrote:
The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b
II. c/a/b=10/6/2
III. c^2>a^2+b^2

A. I and III only
B. II and III only
C. I only
D. II only
E. III only

Source: GMAT Club Tests - hardest GMAT questions

In this question you just have to start with I and II. If there are both wrong you know its answer E!

For the detailed explanation, Brunel has (as usual) done the job here. His explainations are really good!
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Re: m08-q31 [#permalink]

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21 Nov 2013, 07:41
Bunuel wrote:
ConkergMat wrote:
If $$A$$ , $$B$$ , and $$C$$ are points on the plane, is $$AB \gt 15$$ ?

1. $$BC + AC \gt 14$$
2. Area of triangle $$ABC \lt 1$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

OA does not explain how you could get an area < 1, with side > 15?

This problem was replaced by the following question:

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b
II. c/a/b=10/6/2
III. c^2>a^2+b^2

A. I and III only
B. II and III only
C. I only
D. II only
E. III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them).

Answer: E.

Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.

Is it true for all cases the side opposite to the angle >90 in a triangle will follow the same rule c^2>a^2 + b^2 ?

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Re: m08-q31 [#permalink]

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21 Nov 2013, 07:42
rango wrote:
Bunuel wrote:
ConkergMat wrote:
If $$A$$ , $$B$$ , and $$C$$ are points on the plane, is $$AB \gt 15$$ ?

1. $$BC + AC \gt 14$$
2. Area of triangle $$ABC \lt 1$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

OA does not explain how you could get an area < 1, with side > 15?

This problem was replaced by the following question:

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b
II. c/a/b=10/6/2
III. c^2>a^2+b^2

A. I and III only
B. II and III only
C. I only
D. II only
E. III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them).

Answer: E.

Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.

Is it true for all cases the side opposite to the angle >90 in a triangle will follow the same rule c^2>a^2 + b^2 ?

_______________
Yes, it's true.
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Re: m08-q31   [#permalink] 21 Nov 2013, 07:42
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# m08-q31

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