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Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)

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

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New post 17 Apr 2018, 01:04
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

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

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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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New post 17 Apr 2018, 02:29
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.

(B) is our answer.
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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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New post 17 Apr 2018, 04:25
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.

Answer: B
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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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New post 17 Apr 2018, 05:00
1) From 1 we get X>2 and X<-3.
2) From 2 we get X>6.

Hence B is my answer.
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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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New post 14 Aug 2018, 07:20
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.

Answer: B




Hi,

Please explain

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


I couldn't understand it.
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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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New post 14 Aug 2018, 09:58
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)
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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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