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M60-06

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6811
GMAT 1: 760 Q51 V42
GPA: 3.82

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11 Jun 2018, 01:47
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00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:22) correct 33% (01:42) wrong based on 6 sessions

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Is $$x^3 - x^2 + x < 0?$$

1) $$x < 0$$

2) $$x^5 + x < 0$$

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6811 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M60-06 [#permalink] Show Tags 11 Jun 2018, 01:47 1 Official Solution: Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question. Modifying the question: $$x^3 - x^2 + x < 0$$ ⇔ $$x(x^2 - x + 1) < 0$$ ⇔ $$x < 0$$ since $$x^2 - x + 1 > 0$$ always. So, the question becomes, 'is $$x < 0?$$'. Condition 1) is certainly sufficient. Condition 2), $$x^5 + x < 0$$, is equivalent to $$x(x^4 + 1) < 0$$ or $$x < 0$$, since $$x^4 + 1 > 0$$ is always true. So, condition 2) is also sufficient. Therefore, D is the answer. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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13 Sep 2018, 03:14
Didnt understand this part - ⇔ x<0x<0 since x2−x+1>0x2−x+1>0 always. How is x2−x+1>0 ?

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29 Sep 2018, 17:18
1
Anirudh134 wrote:
Didnt understand this part - ⇔ x<0x<0 since x2−x+1>0x2−x+1>0 always. How is x2−x+1>0 ?

x(x2-x+1)<0 implies either x or x2-x+1 has to be negative (only 1 HAS to be negative).

If x is positive, then x2-x+1 has to be negative for the equation to be true. For no value of x (x>0), x2-x+1 is <0. So x<0 & for all values of x<0, x2-x+1 is >0. Does it make sense?
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29 Sep 2018, 17:38
1
Anirudh134 wrote:
Didnt understand this part - ⇔ x<0x<0 since x2−x+1>0x2−x+1>0 always. How is x2−x+1>0 ?

Anirudh134

$$X^2 - X + 1 = X^2 - 2X + 1 + X = (X - 1)^2 + X = K ( Say)$$

Now for X >= 0 , above expression always positive.

Next, for X < 0 also it is always positive.

Plug some values and check

X = - 2 then K = 9 - 2= 7

X = -0.5 then K = 2.25 - 0.5 = 1.75

Does this help?
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Re: M60-06 &nbs [#permalink] 29 Sep 2018, 17:38
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