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Math Expert V
Joined: 02 Sep 2009
Posts: 56366
Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 77% (01:29) correct 23% (01:45) wrong based on 74 sessions

### HideShow timer Statistics Is $$x > 1$$?

(1) $$x^2 + x + 2 > 8$$

(2) $$8(x – 4) > 4(x – 2)$$

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examPAL Representative P
Joined: 07 Dec 2017
Posts: 1073
Re: Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)  [#permalink]

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Bunuel wrote:
Is $$x > 1$$?

(1) $$x^2 + x + 2 > 8$$

(2) $$8(x – 4) > 4(x – 2)$$

Instead of solving explicitly, we'll look for simple numbers that contradict the given statements.
This is an Alternative approach.

(1)
Let's try a number that gives us a YES. If we make x very large, say x = 100, then 100^2+100+2 is definitely larger than 8.
Not let's try to contradict this by getting a NO. If we make x very small, say x = -100 then (-100)^2 - 100 + 2 is 10,000 -100 + 2 which is also larger than 8.
We have found both x > 1 and x < 1 which gives a correct statement (1) so this is not enough.
Insufficient!

(2)
Since it's hard to guess what numbers to pick, we'll first simplify a bit.
8x - 32 > 4x - 8
4x > 24
x > 6.
In this case, there is no need to pick numbers at all!
Sufficient.

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Re: Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)  [#permalink]

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Bunuel wrote:
Is $$x > 1$$?

(1) $$x^2 + x + 2 > 8$$

(2) $$8(x – 4) > 4(x – 2)$$

The first statement tells us that
$$x^2 + x + 2 > 8$$ or
$$x^2 + x - 6 > 0$$,
$$(x+3)(x-2)>0$$.
So, from this, we know that $$x>2$$ and $$x<-3$$. Insufficient.

If we solve the second inequality, then we end up with $$x>6$$ which is sufficient.

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Tulkin.
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Joined: 11 Mar 2015
Posts: 32
Re: Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)  [#permalink]

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1) From 1 we get X>2 and X<-3.
2) From 2 we get X>6.

Intern  B
Joined: 08 Jul 2018
Posts: 23
Location: India
Concentration: General Management, Marketing
GPA: 4
Re: Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)  [#permalink]

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Tulkin987 wrote:
Bunuel wrote:
Is $$x > 1$$?

(1) $$x^2 + x + 2 > 8$$

(2) $$8(x – 4) > 4(x – 2)$$

The first statement tells us that
$$x^2 + x + 2 > 8$$ or
$$x^2 + x - 6 > 0$$,
$$(x+3)(x-2)>0$$.
So, from this, we know that $$x>2$$ and $$x<-3$$. Insufficient.

If we solve the second inequality, then we end up with $$x>6$$ which is sufficient.

Hi,

How did you get from
x+3>0 to x<-3

I couldn't understand it.
VP  D
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1028
WE: Supply Chain Management (Energy and Utilities)
Re: Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)  [#permalink]

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Bunuel wrote:
Is $$x > 1$$?

(1) $$x^2 + x + 2 > 8$$

(2) $$8(x – 4) > 4(x – 2)$$

Question stem:- Is x>1 ?

St1:- $$x^2 + x + 2 > 8$$
Or, $$x^2+x-6>0$$
Or, $$x^2-2x+3x-6>0$$
Or, $$x(x-2)+3(x-2)>0$$
Or, $$\left(x-2\right)\left(x+3\right)>0$$
Cut off points:- x=2, -3
Applying wavy-curve method(figure enclosed), the region of the curve above horizontal axis(since the product is positive),
x<-3 , x>2
So, x may or mayn't be greater than 1.
Insufficient.

St2:- $$8(x – 4) > 4(x – 2)$$
Or, 8x-32>4x-8
Or, 8x-4x>-8+32
Or, 4x > 24
Or, $$x> \frac{24}{4}$$
Or, x > 6
Sufficient.

Ans. (B)
Attachments wavy.JPG [ 44.31 KiB | Viewed 443 times ]

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Rise above the storm, you will find the sunshine Re: Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)   [#permalink] 14 Aug 2018, 09:58
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