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# If x is not equal to 0, is –1 ≤ x ≤ 1 ?

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Joined: 03 Mar 2018
Posts: 205
If x is not equal to 0, is –1 ≤ x ≤ 1 ?  [#permalink]

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Updated on: 16 Apr 2018, 23:10
2
00:00

Difficulty:

65% (hard)

Question Stats:

49% (01:40) correct 51% (01:35) wrong based on 53 sessions

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If x is not equal to 0, is –1 ≤ x ≤ 1 ?

(1) $$x^3 < x^2$$

(2) $$x^2 < x$$

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Originally posted by itisSheldon on 16 Apr 2018, 12:18.
Last edited by Bunuel on 16 Apr 2018, 23:10, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 26 Mar 2013
Posts: 1921
Re: If x is not equal to 0, is –1 ≤ x ≤ 1 ?  [#permalink]

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16 Apr 2018, 13:19
itisSheldon wrote:
If x is not equal to 0 , is –1 ≤x≤ 1 ?

1) $$x^3$$ < $$x^2$$
2) $$x^2$$ < x

1) $$x^3$$ < $$x^2$$

$$x^2$$ is positive so that we can divide both of inequalities safely without changing the sign.

$$x < 1$$

For example,

x could be $$\frac{1}{2}$$.....answer is Yes, or

x could be -10. .....answer is No

Insufficient

2) $$x^2$$ < x

The only viable examples are in range 0 < x <1.......Answer is Yes

Sufficient

Algebraic way:

x - $$x^2$$ > 0

x (1-x) > 0

x > 0 & 1-x > 0
x > 0 & x < 1
The range is 0<x<1

or

x <0 & 1-x<0
x <0 & 1<x ........ how come x is positive and negative in same time...Not viable option...So dicard

We have one range: 0<x<1............Answer is always yes

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Re: If x is not equal to 0, is –1 ≤ x ≤ 1 ?  [#permalink]

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17 Apr 2018, 07:58
itisSheldon wrote:
If x is not equal to 0, is –1 ≤ x ≤ 1 ?

(1) $$x^3 < x^2$$

(2) $$x^2 < x$$

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 and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
$$x^3 < x^2$$
$$⇔ x^3 - x^2 < 0$$
$$⇔ x^2 ( x - 1 ) < 0$$
$$⇔ x - 1 < 0$$ since $$x^2 > 0$$
$$⇔ x < 1$$
In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
Since the solution set of the question does not include that of condition 1), it is not sufficient.

Condition 2)
$$x^2 < x$$
$$⇔ x^2 - x < 0$$
$$⇔ x ( x - 1 ) < 0$$
$$⇔ 0 < x < 1$$
In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
Thus condition 2) is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: If x is not equal to 0, is –1 ≤ x ≤ 1 ?  [#permalink]

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01 May 2018, 08:36
I) possible cases when x is -ve
or x is fraction
So x could be between -1 and 1 and beyond that also

Not sufficient

II)only possible when x is +ve fraction
So x is between -1 and 1

Sufficient

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Re: If x is not equal to 0, is –1 ≤ x ≤ 1 ? &nbs [#permalink] 01 May 2018, 08:36
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