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Moderator  Joined: 18 Apr 2015
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x > 0 [#permalink]
Expert's post 00:00

Question Stats: 62% (00:55) correct 37% (01:09) wrong based on 40 sessions
$$x > 0$$

 Quantity A Quantity B The area of a square region with diagonal of length $$\sqrt{2} x$$ The area of a circular region with diameter of length x

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Manager Joined: 25 Nov 2017
Posts: 51
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Kudos [?]: 41 , given: 5

Re: x > 0 [#permalink]
The diagonal of a square can be found by the following formula : Squroot 2 * side
Since we already know the length = Squroot 2 * side = Squroo 2 * x
We can divide the square root and we have side = x. So the area is x^2

For Quantity B we know that the diameter is x, so the radii must be x/2 and for the area it is multiplied by pi.

Lets plug some values.
For x = 2.
QA = Area is equal to 4
QB= Area is 3,14
QA is greater than B.

Lets try x =36
QA = 36
QB = 9,42

But what is bizarre it that when I plug 1 B is greater

Last edited by Popo on 18 Dec 2017, 18:41, edited 1 time in total.
Moderator  Joined: 18 Apr 2015
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Re: x > 0 [#permalink]
Expert's post
The answer is A
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Manager Joined: 25 Nov 2017
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Re: x > 0 [#permalink]
Carcass wrote:
The answer is A

Can I ask you what is the right reasoning? GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: x > 0 [#permalink]
1
KUDOS
Expert's post
Popo wrote:
Carcass wrote:
The answer is A

Can I ask you what is the right reasoning?

When you plug 1. The Quanity A is $$x^2=1$$
and Quantiy B is $$\frac{3.14}{4}x^2 = 0.785$$.

Hence quanity A is always greater.
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Sandy
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Re: x > 0 [#permalink]
1
KUDOS
Carcass wrote:
$$x > 0$$

 Quantity A Quantity B The area of a square region with diagonal of length $$\sqrt{2} x$$ The area of a circular region with diameter of length x

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.

Here let us put statement 1

Area of the square = $$\frac{{diagonal^2}}{2}$$

or $$(\sqrt{2}x)^2$$ * $$\frac{1}{2}$$
or $$\frac{2x^2}{2}$$

From statement 2

Area of circle = $$\pi * r^2$$

or $$\frac{22}{7}$$ * $$(\frac{Diameter}{2})^2$$

or $$\frac{22x}{28}$$

From the two results statement 1 > statement 2 for any positive value of x
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Intern Joined: 08 Apr 2018
Posts: 44
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Kudos [?]: 15 , given: 16

Re: x > 0 [#permalink]
sandy wrote:
Popo wrote:
Carcass wrote:
The answer is A

Can I ask you what is the right reasoning?

When you plug 1. The Quanity A is $$x^2=1$$
and Quantiy B is $$\frac{3.14}{4}x^2 = 0.785$$.

Hence quanity A is always greater.

For quantity B, why are we dividing by 4? When a diameter is given, we divide by 2 to get the radius, we should divide x by 2 to get the radius, square it and then multiply by the value of pi GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1783  , given: 397

Re: x > 0 [#permalink]
1
KUDOS
Expert's post
@Emike56

You are right if $$x$$ is the diameter then radius of circle is $$\frac{x}{2}$$.

Area of the circle is $$\pi \times radius^2= \pi \times (\frac{x}{2})^2=\pi \times \frac{x^2}{2^2}= \pi \times \frac{x^2}{4}$$.

Hope this explains the 4.
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Director Joined: 09 Nov 2018
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Re: x > 0 [#permalink]
Ans A
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Re: x > 0 [#permalink]
Expert's post
Carcass wrote:
$$x > 0$$

 Quantity A Quantity B The area of a square region with diagonal of length $$\sqrt{2} x$$ The area of a circular region with diameter of length x

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.

The area of a square region with diagonal of length $$\sqrt{2} x$$.
If the diagonal is $$x\sqrt{2}$$, the sides are x as $$side^2+side^2=(x\sqrt{2})^2=2x^2=2side^2....side=x$$
so area = $$side^2=x^2=A$$

The area of a circular region with diameter of length x
so radius = x/2.
area = B = $$\pi*(\frac{x}{2})^2=\pi*\frac{x^2}{4}=\frac{22}{28}*x^2=\frac{11}{14}*A$$.

Thus A is greater
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html Re: x > 0   [#permalink] 19 Jan 2019, 18:53
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