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M20-24

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M20-24 [#permalink]

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What is the area of rectangle \(ABCD\) ?


(1) Diagonal \(AC\) is twice as long as side \(CD\)

(2) Diagonal \(AC\) is 0.1 meters longer than side \(AD\)
[Reveal] Spoiler: OA

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Statement (1) by itself is insufficient. S1 defines the shape of the rectangle but does not specify its size.

Statement (2) by itself is insufficient. The area of the rectangle can be made arbitrarily small (as the length of side \(AD\) goes to 0).

Statements (1) and (2) combined are sufficient. Because \(AC^2 = CD^2 + AD^2\) and \(AC = 2CD\), \(4CD^2 = CD^2 + AD^2\) which simplifies to \(3CD^2 = AD^2\). It follows from here that \(AD = \frac{\sqrt{3}}{2} AC\).

This is one of the equations, the other one is \(AC = AD + 0.1\). This system of linear equations is enough to determine both sides of the rectangle.


Answer: C
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Re: M20-24 [#permalink]

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New post 26 Jul 2016, 12:21
Hi Bunuel,

Wouldn't (1) work with the triangle type x:xsqrt3:2x?

Thanks

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Re: M20-24 [#permalink]

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New post 26 Jul 2016, 12:39
dauren88 wrote:
Hi Bunuel,

Wouldn't (1) work with the triangle type x:xsqrt3:2x?

Thanks


It will
But area of the said rectangle will change I.e directly proportional to x
Hence in suff


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Re: M20-24 [#permalink]

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New post 16 Aug 2017, 07:04
Statements (1) and (2) combined are sufficient. Because \(AC^2 = CD^2 + AD^2\) and \(AC = 2CD\), \(4CD^2 = CD^2 + AD^2\) which simplifies to \(3CD^2 = AD^2\). [color=#fff200]It follows from here that \(AD = \frac{\sqrt{3}}{2} AC\).[/color]

This is one of the equations, the other one is \(AC = AD + 0.1\). This system of linear equations is enough to determine both sides of the rectangle.


Answer: C[/quote]

Hello Bunuel,
Can you please let me know how have you arrived at the highlighted text?

Thanks!

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New post 16 Aug 2017, 07:12
anirudhasopa wrote:
Statements (1) and (2) combined are sufficient. Because \(AC^2 = CD^2 + AD^2\) and \(AC = 2CD\), \(4CD^2 = CD^2 + AD^2\) which simplifies to \(3CD^2 = AD^2\). [color=#fff200]It follows from here that \(AD = \frac{\sqrt{3}}{2} AC\).[/color]

This is one of the equations, the other one is \(AC = AD + 0.1\). This system of linear equations is enough to determine both sides of the rectangle.


Answer: C

Hello Bunuel,
Can you please let me know how have you arrived at the highlighted text?

Thanks!


\(3CD^2 = AD^2\)

Take the square root from both sides: \(\sqrt{3}CD = AD\).

We know that \(AC = 2CD\), so \(\frac{AC}{2} = CD\).

Thus, \(\sqrt{3}*\frac{AC}{2} = AD\).

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: M20-24 [#permalink]

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New post 16 Oct 2017, 06:09
+1 for option C.
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Re: M20-24   [#permalink] 16 Oct 2017, 06:09
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