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# Is |1 - 4k| > k? (1) k > 4x^3 (2) k < 2x x^2 - 2

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Re: Is |1 - 4k| > k? (1) k > 4x^3 (2) k < 2x x^2 - 2 [#permalink]
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GMATPrepNow wrote:
Is |1 - 4k| > k?

(1) k > 4x³
(2) k < 2x – x² - 2

*kudos for all correct solutions

Very nice and great question GMATPrepNow Brent but I suspect it is 700. I think it 750. I will try my best.

Let's try to rehearse the question stem?

Is |1 - 4k| > k? (Please note I prefer to leave 'is' in order not to lose track)

we have 2 cases:

Case 1:

Is 1 - 4k > k.........Is 1>5k........Is k<1/5

Case 2:

Is 4k - 1 > k.........Is 3k>1........Is k>1/3

The question could be: Is k<1/5 or k>1/3 ?

Graphically is k in the following accepted ranges (Green area):

------Accepted----------$$\frac{1}{5}$$-------------$$\frac{1}{3}$$------Accepted----------

(1) k > 4x³

Let x =0 then k could be $$\frac{1}{4}$$.............Answer is No

Let x =0 then k could be $$100$$.............Answer is Yes

Insufficient

(2) k < 2x – x² - 2

k < – (x² – 2x + 2).......It is not easy to find roots and it waste to use the long equation. Let's use logic to analyze (x² – 2x + 2)

If x = 0........K < -2 < 1/5 ..............Answer is Yes

If x = negative value........then (x² – 2x + 2) will be always positive......... K < – (positive value) .......K is always negative less than 1/5.....Answer is Yes

If x = positive value .....x²+2 is always positive and higher than -2x........then the magnitude of (x² – 2x + 2) will be always positive.....the k < – (positive value) .......K is always negative less than 1/5.....Answer is Yes

In all cases k lies in the left green area less than 1/5

Sufficient

I hope I got right
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Re: Is |1 - 4k| > k? (1) k > 4x^3 (2) k < 2x x^2 - 2 [#permalink]
2
Kudos
The answer to this will have to be B.

For Statement 1, k >4x^3, x can be greater than, lesser than or equal to zero. Therefore, we cannot get any further information from it.

For statement 2, we can reduce the expression on the right as k < -[(x-1)^2 +1]. As the expression on the right is always negative, k is always lesser than a negative number (or k is a negative number) and hence the modulus value in the original statement is always greater than k
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Re: Is |1 - 4k| > k? (1) k > 4x^3 (2) k < 2x x^2 - 2 [#permalink]
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Top Contributor
GMATPrepNow wrote:
Is |1 - 4k| > k?

(1) k > 4x³
(2) k < 2x – x² - 2

Target question: Is |1 - 4k| > k?

Statement 1: k > 4x³
This pretty much tells is that k can have ANY value.
For example, notice that, if x = -100, then 4x³ = -4,000,000
So, for this value of x, k can be any number greater than -4,000,000

Let's TEST some values.
There are several values of k that satisfy statement 1. Here are two:
Case a: k = 0. Here, |1 - 4k| = |1 - 4(0)| = |1| = 1. In this case, the answer to the target question is YES, it IS the case that |1 - 4k| > k
Case b: k = 0.25. Here, |1 - 4k| = |1 - 4(0.25)| = |0| = 0. In this case, the answer to the target question is NO, it is NOT the case that |1 - 4k| > k
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: k < 2x – x² - 2
Rewrite as: k < –x² + 2x - 1 - 1
Rewrite as: k < –(x² - 2x + 1) - 1
Factor to get: k < –(x - 1)² - 1
Notice that (x - 1)² is always greater than or equal to zero
So, -(x - 1)² is always LESS THAN or equal to zero
So -(x - 1)² - 1 must be NEGATIVE
So, statement 2 essentially tells us that k < some NEGATIVE number
This means k must be NEGATIVE
Since |1 - 4k| is always greater than or equal to zero, the answer to the target question is YES, it IS the case that |1 - 4k| > k
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Cheers,
Brent
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Is |1 - 4k| > k? (1) k > 4x^3 (2) k < 2x x^2 - 2 [#permalink]
BrentGMATPrepNow wrote:
Is |1 - 4k| > k?

(1) k > 4x^3
(2) k < 2x – x^2 - 2

The question is asking is |1 - 4k| > k?
Another way to write this is; is 1-4k>k or is 1-4k<-k?
After further simplification, this comes down to is k<1/5 or is k>1/3?

.........................0,2.............0,333..........................

Any value of k that lives in the pink area will give us a YES answer to the question asked.

Statement 1 tells us that k > 4x^3
We can test cases here, for example if x=1/2; k > 4*(1/2)^3, therefore k>1/2 and the answer to our question will be yes, because if k>1/2 then k is definitely greater than 1/3.
But if x=-2, k>-32, k might or might not be in the pink area.
Hence statement 1 is not sufficient.

Statement 2 tells us that k < 2x – x^2 - 2
Using the discriminant formula b^2-4ac, we can tell that the equation -x^2+2x-2=0 has no roots
And by testing values of x in the expression -x^2+2x-2 we can tell it will always be negative.
Another method of simplifying -x^2+2x-2 and recognizing that it is always negative, is writing it as following:
-x^2+2x-2= - (x^2+2x+2)= - (x^2+2x+1) -1 = - (x+1)^2 -1 which is always negative.
Therefore k< something negative.
Hence k is negative.
And if k is negative then k will always be less than 1/5.
The answer to our question will be YES and statement 2 is sufficient.