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Is x>0?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6826
GMAT 1: 760 Q51 V42
GPA: 3.82

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17 Apr 2018, 01:09
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Difficulty:

75% (hard)

Question Stats:

51% (02:12) correct 49% (02:18) wrong based on 88 sessions

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[GMAT math practice question]

Is $$x>0$$?

$$1) x^3+x^2+x = 1$$
$$2) x^2 – 4x - 5 > 0$$

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"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" Retired Moderator Joined: 25 Feb 2013 Posts: 1220 Location: India GPA: 3.82 Is x>0? [#permalink] Show Tags 17 Apr 2018, 02:37 1 MathRevolution wrote: [GMAT math practice question] Is $$x>0$$? $$1) x^3+x^2+x = 1$$ $$2) x^2 – 4x - 5 > 0$$ Statement 1: $$x^3+x^2+x=1$$, implies that $$x≠0$$, because in that case LHS=0 and RHS=1, which is not possible. Also if $$-1≤x<0$$, then $$x^2>x$$ but $$|x|>|x^2|$$ which will mean that $$x^3+x^2+x<0$$ (for eg. if $$x=-0.9$$, then $$x^2=0.81$$ and $$x^3=-0.729$$, so net sum will be negative), which is not possible. Now we need to check whether $$x<-1$$ or $$x>0$$ $$x^3+x^2+x=1$$, add $$1$$ to both sides of the equation to get $$x^3+x^2+x+1=1+1$$ $$x^2(x+1)+x+1=2=>(x+1)*(x^2+1)=2$$, now $$x^2+1>0$$, hence $$x+1>0$$, because product of two numbers is $$2$$ which is positive Hence $$x>-1$$. But as we have seen above that $$x≠0$$ and $$x$$ cannot be between $$-1$$ and $$0$$, therefore $$x>0$$. Sufficient Statement 2: $$x^2 – 4x - 5 > 0=>x^2-5x+x-5>0=>(x-5)*(x+1)>0$$ therefore $$x>5$$ or $$x<-1$$. Hence Insufficient Option A Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6826 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is x>0? [#permalink] Show Tags 19 Apr 2018, 01:46 1 1 => 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. Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1): $$x^3+x^2+x = 1$$ $$=> x(x^2+x+1) = 1$$ $$=> x =\frac{1}{(x^2+x+1)}$$ $$=> x = \frac{1}{(x^2+x+1)} > 0$$since $$x^2+x+1 > 0.$$ Thus, condition 1) is sufficient. Condition 2): $$x^2 – 4x -5 > 0$$ $$=> (x+1)(x-5) > 0$$ $$=> x < 1 or x > 5$$ Since the solution set of the inequality, $$x > 0$$, from the question does not include the solution set of the inequality from condition 2), $$x < 1$$ or $$x > 5$$, condition 2) is not sufficient. Therefore, A is the answer. Answer: A Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E. _________________ 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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Joined: 19 Sep 2013
Posts: 21
Location: India
GMAT 1: 720 Q47 V41
GPA: 2.98
WE: General Management (Energy and Utilities)

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20 Apr 2018, 22:45
MathRevolution wrote:
=>
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.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1):
$$x^3+x^2+x = 1$$
$$=> x(x^2+x+1) = 1$$
$$=> x =\frac{1}{(x^2+x+1)}$$
$$=> x = \frac{1}{(x^2+x+1)} > 0$$since $$x^2+x+1 > 0.$$

Thus, condition 1) is sufficient.

Condition 2):
$$x^2 – 4x -5 > 0$$
$$=> (x+1)(x-5) > 0$$
$$=> x < 1 or x > 5$$

Since the solution set of the inequality, $$x > 0$$, from the question does not include the solution set of the inequality from condition 2), $$x < 1$$ or $$x > 5$$, condition 2) is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

i have a doubt here..
As x=-1 satisfies the first equation, then the the condition X>0 is not sufficient...
Retired Moderator
Joined: 25 Feb 2013
Posts: 1220
Location: India
GPA: 3.82

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20 Apr 2018, 22:58
dewdrops909 wrote:
MathRevolution wrote:
=>
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.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1):
$$x^3+x^2+x = 1$$
$$=> x(x^2+x+1) = 1$$
$$=> x =\frac{1}{(x^2+x+1)}$$
$$=> x = \frac{1}{(x^2+x+1)} > 0$$since $$x^2+x+1 > 0.$$

Thus, condition 1) is sufficient.

Condition 2):
$$x^2 – 4x -5 > 0$$
$$=> (x+1)(x-5) > 0$$
$$=> x < 1 or x > 5$$

Since the solution set of the inequality, $$x > 0$$, from the question does not include the solution set of the inequality from condition 2), $$x < 1$$ or $$x > 5$$, condition 2) is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

i have a doubt here..
As x=-1 satisfies the first equation, then the the condition X>0 is not sufficient...

Hi dewdrops909

x=-1 will not satisfy the equation $$x^3+x^2+x=1$$ because at x=-1 it will become

-1+1-1=-1 which is not possible.

You may also refer to my solution above to understand the boundary condition.
Intern
Joined: 19 Sep 2013
Posts: 21
Location: India
GMAT 1: 720 Q47 V41
GPA: 2.98
WE: General Management (Energy and Utilities)

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21 Apr 2018, 03:47
niks18 wrote:
dewdrops909 wrote:
MathRevolution wrote:
=>
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.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1):
$$x^3+x^2+x = 1$$
$$=> x(x^2+x+1) = 1$$
$$=> x =\frac{1}{(x^2+x+1)}$$
$$=> x = \frac{1}{(x^2+x+1)} > 0$$since $$x^2+x+1 > 0.$$

Thus, condition 1) is sufficient.

Condition 2):
$$x^2 – 4x -5 > 0$$
$$=> (x+1)(x-5) > 0$$
$$=> x < 1 or x > 5$$

Since the solution set of the inequality, $$x > 0$$, from the question does not include the solution set of the inequality from condition 2), $$x < 1$$ or $$x > 5$$, condition 2) is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

i have a doubt here..
As x=-1 satisfies the first equation, then the the condition X>0 is not sufficient...

Hi dewdrops909

x=-1 will not satisfy the equation $$x^3+x^2+x=1$$ because at x=-1 it will become

-1+1-1=-1 which is not possible.

You may also refer to my solution above to understand the boundary condition.

yeah my mistake...
thanks.
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Joined: 03 Feb 2016
Posts: 68
Location: India
Concentration: Technology, Marketing
GMAT 1: 650 Q48 V32
GPA: 4
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23 Apr 2018, 01:11
x2+x+1 --> Contains imaginary cube roots of unity so always positive... D < 0 and a> 0
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GMAT1 650 Q48 V32.

DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1434
Location: India

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23 Apr 2018, 04:03
1
MathRevolution wrote:
[GMAT math practice question]

Is $$x>0$$?

$$1) x^3+x^2+x = 1$$
$$2) x^2 – 4x - 5 > 0$$

We can also apply the quadratics here, i.e., knowledge of the shapes of the graph that a quadratic expression will take.

(1) x^3 + x^2 + x = 1 or x(x^2 + x + 1) = 1.
Now lets look at x^2 + x + 1 here. Its a quadratic expression with coefficient of x^2 as positive, so upward facing parabola, and its discriminant is negative, so the parabola will not cross the x axis at all. This means the parabola, or the graph of x^2 + x + 1 will lie entirely above the x axis, thus x^2 + x + 1 will always have positive values only. Knowing this, now we can apply that since x^2 + x + 1 is positive, and product of 'x' and 'x^2 + x + 1' is '1', which is a positive number, it can only happen when x is also positive. So we conclude that x > 0. Sufficient.

(2) x^2 - 4x - 5 > 0 or (x+1)(x-5) > 0. Here x can take positive value say 6, but it can also take negative value say -2. Not sufficient.

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Joined: 26 Mar 2013
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23 Apr 2018, 14:05
MathRevolution wrote:
[GMAT math practice question]

Is $$x>0$$?

$$1) x^3+x^2+x = 1$$
$$2) x^2 – 4x - 5 > 0$$

$$2) x^2 – 4x - 5 > 0$$

$$2) (x-5)(x+1) > 0$$

Draw number to test ranges

.......Ok.....-1.......Not ok.......5.........Ok.........

$$x$$ could be positive or negative

Insufficient

$$1) x^3+x^2+x = 1$$

Let's solve conceptually on number line

Let $$x=0$$......This is invalid as 0\neq{1}.......So x can't be 0

Let $$x$$=-$$\frac{1}{2}$$..... it will yield negative value

Let $$x=-1$$ or less.......... it will yield negative value

Clearly x can't be negative value (no need to calculate aggressively as it negative values will make x^2 more lower)

There is only choice left which is x >0

Sufficient

Re: Is x>0? &nbs [#permalink] 23 Apr 2018, 14:05
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