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Given that x ≠ 5, is x>{1/(x-5)^2}

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Given that x ≠ 5, is x>{1/(x-5)^2}

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Given that x ≠ 5, is x>{1/(x-5)^2}

Statement #1: x > 0

Statement #2: x > 10

OA B

Source: Magoosh

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BTGmoderatorDC wrote:
Given that x ≠ 5, is x > {1/(x - 5)^2}

Statement #1: x > 0

Statement #2: x > 10

OA B

Source: Magoosh
Given: x ≠ 5

Question: Is x > {1/(x - 5)^2}?

Let's take each statement one by one.

(1) x > 0

Case 1: Say x = 11

Thus, x > {1/(x - 5)^2} = 11 ? {1/(11 - 5)^2} => 11 > 1/36. The answer is Yes.

Case 2: Say x = 1/100

Thus, x > {1/(x - 5)^2} = 1/100 ? {1/(1/100 - 5)^2} => 1/100 > A very big number. The answer is No.

No unique answer. Insufficient.

(2) x > 10

We have seen that for x = 11, the answer is yes. As we increase the value of x, the value of 1/(x - 5)^2 will decrease further. Thus, the answer would be Yes.

The correct answer: B

Hope this helps!

-Jay
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Manhattan Review

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BTGmoderatorDC wrote:
Given that x ≠ 5, is x>{1/(x-5)^2}

Statement #1: x > 0

Statement #2: x > 10
Source: Magoosh
$$x\,\,\mathop > \limits^? \,\,\frac{1}{{{{\left( {x - 5} \right)}^2}}}\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,x{{\left( {x - 5} \right)}^2}\,\,\mathop > \limits^? \,\,1\,\,\,{\text{with}}\,\,{\text{x}} \ne {\text{5}}\,\,}$$
$$\left( 1 \right)\,\,\,x > 0\left\{ \matrix{
\,{\rm{Take}}\,\,{\rm{x = 1}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr
\,{\rm{Take}}\,\,{\rm{x = 5 + }}{1 \over {10}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\left[ {5.1 \cdot {1 \over {100}} < 1} \right] \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,x > 10\,\,\, \Rightarrow \,\,\,{\left( {x - 5} \right)^2} > 25\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
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