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# x > 0

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x > 0 [#permalink]  17 Dec 2017, 12:52
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Question Stats:

60% (01:00) correct 39% (01:01) wrong based on 28 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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Re: x > 0 [#permalink]  18 Dec 2017, 01:59
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.
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Re: x > 0 [#permalink]  18 Dec 2017, 11:48
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Re: x > 0 [#permalink]  18 Dec 2017, 18:41
Carcass wrote:

Can I ask you what is the right reasoning?
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Re: x > 0 [#permalink]  19 Dec 2017, 13:35
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Expert's post
Popo wrote:
Carcass wrote:

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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Re: x > 0 [#permalink]  19 Dec 2017, 21:30
1
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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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Re: x > 0 [#permalink]  04 Jun 2018, 19:58
sandy wrote:
Popo wrote:
Carcass wrote:

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
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Re: x > 0 [#permalink]  05 Jun 2018, 12:51
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@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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Re: x > 0   [#permalink] 05 Jun 2018, 12:51
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