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If x is an integer, is x|x|<2^x ?

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If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x = -10
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Walkabout wrote:
If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x= -10


If x is an integer, is x|x|<2^x ?

Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) --> \(x*|x|=x^2\)), negative when \(x<0\) (\(x<0\) --> \(x*|x|=-x^2\)) and equals to zero when \(x={0}\).

(1) x < 0. According to the above \(x*|x|<0<2^x\). Sufficient.

(2) x = -10. The same here \(x*|x|=-100<0<\frac{1}{2^{10}}\). Sufficient.

Answer: D.
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seabhi wrote:
If x is an integer, is x |x| < 2x ?

(1) x < 0
(2) x = –10

DS from OG.


OA is D for the following reasons:
When you first see a DS question, see if there is anyway to simplify the question stem
In this case, since |x| is positive, we can divide both sides by |x| giving us a new question stem --> is x < 2?

S1: x<0, therefore x must be <2 = sufficient
S2: x = -10 and -10 < 2 = sufficient

Let me know if this helps!

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Re: DS from OG [#permalink]

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New post 30 Dec 2013, 20:36
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bparrish89 wrote:
seabhi wrote:
If x is an integer, is x |x| < 2x ?

(1) x < 0
(2) x = –10

DS from OG.


OA is D for the following reasons:
When you first see a DS question, see if there is anyway to simplify the question stem
In this case, since |x| is positive, we can divide both sides by |x| giving us a new question stem --> is x < 2?

S1: x<0, therefore x must be <2 = sufficient
S2: x = -10 and -10 < 2 = sufficient

Let me know if this helps!



The OA is wrong here because of the following reasons:


(1) if x=-10 then -100<-20, on the other hand if x= -1, x<0 then the inequality changes from < to >, namely, -1 > -2 ; This statement is absolutely insufficient!


(2) This statement is obviously sufficient!


So, the correct answer is notD, but B

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If x is an integer, is x|x|<2^x ? [#permalink]

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reza52520 wrote:
If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x = -10


Question : Is x|x|<2^x ?

Statement 1: x < 0

For x to be Negative LHS i.e. x|x| will always be NEGATIVE
and 2^x will be positive for any value of x
i.e. x|x|<2^x will always be true
SUFFICIENT

Statement 1: x = -10
For x =-10 LHS i.e. x|x| will always be NEGATIVE (-100)
and 2^x will be positive for given x (1/2^10)
i.e. x|x|<2^x will always be true
SUFFICIENT

Answer: option D
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Last edited by GMATinsight on 27 Jul 2015, 05:44, edited 1 time in total.

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 27 Jul 2015, 05:44
reza52520 wrote:
If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x = -10


Hi,
we have an equation and the RHS 2^x will be positive irrespective of value of x and LHS xlxl will depend on the value of x..
1) x is -ive .. so LHS is -ive and RHS is +ive.. suff
2) x=-10... again LHS is -ive and RHS is +ive.. suff
ans D
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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 13 Dec 2015, 10:47
For,
and 2^x will be positive for any value of x

2 power x, X can be negative no?
since its x, we dont know positive or negative..
What am I missing?

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 13 Dec 2015, 11:04
paidlukkha wrote:
For,
and 2^x will be positive for any value of x

2 power x, X can be negative no?
since its x, we dont know positive or negative..
What am I missing?


You are missing a crucial thing here. Even if x is <0, \(2^x\) with x<0 = \(1/2^x\) , it is still >0...(1)

Thus with x<0, |x| = -x and hence x|x| = -\(x^2\)

As, x^2 is always \(\geq\) 0 for all x, -\(x^2\)<0 ...(2)

Thus, from (1) and (2), you get a definite "yes" for the question "is \(x|x| < 2^x\)" for x<0.

Hope this helps.

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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This question can be solved as follows.

stmt1) it says that x /x/ < 2^x. and also we are told that x< 0. So if x is zero and the abs of x is always positive then we know that x/x/ will be negative. In addition to that, we know that 2^negative number will be positive because it will be in the form of 1/2^x, it will be less than 1 but it will be greater than a negative number. So stmt1 is SUFF.

stmt2) this is a repetition of stmt1 because the left side is negative and the right side is positive. SUFF.

Answer is D!

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 24 Nov 2016, 05:03
I would say statement 1 is insufficient... Am I doing something incorrect?

\(x |x| < 2^x\)

\(|x| < \frac{2^x}{x}\)

For x = -1

\(|-1| < \frac{1}{2}/-1\)
\(1 < - \frac{1}{2}\)

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 23 Jan 2017, 08:30
1, x < 0
x = -1
-1 < \(\frac{1}{2}\)
x = -2
-4 < \(\frac{1}{4}\)
AD
2, X = -10
-100 < \(\frac{1}{2}\)^10
D
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Re: If x is an integer, is x*lxl<2^x [#permalink]

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New post 28 Jan 2017, 06:37
praveen27sinha wrote:
1. If x is an integer, is x*lxl<2^x

(1) x<0
(2) x=-10



|x| is positive so is 2^x
so if we can find whether x is positive or negative we can find our solution.

both of the option says x is negative.
so answer is D

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 17 Apr 2017, 01:20
Bunuel wrote:
Walkabout wrote:
If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x= -10


If x is an integer, is x|x|<2^x ?

Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) --> \(x*|x|=x^2\)), negative when \(x<0\) (\(x<0\) --> \(x*|x|=-x^2\)) and equals to zero when \(x={0}\).

(1) x < 0. According to the above \(x*|x|<0<2^x\). Sufficient.

(2) x = -10. The same here \(x*|x|=-100<0<\frac{1}{2^{10}}\). Sufficient.

Answer: D.


just for the sake of time saving:

(2) x = -10 --> we could test it! done.... next question
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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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Walkabout wrote:
If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x = -10


Target question: Is x|x|< 2^x ?

Given: x is an integer

Statement 1: x < 0
In other words, x is NEGATIVE
So, x|x| = (NEGATIVE)(|NEGATIVE|) = (NEGATIVE)(POSITIVE) = NEGATIVE

IMPORTANT: 2^x will be POSITIVE for all values of x.

Since x|x| must be NEGATIVE, and since 2^x must be POSITIVE, we can be certain that x|x|< 2^x
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x = -10
So, x|x| = (-10)(|-10|) = (-10)(10) = -100 = a NEGATIVE
On the other hand, 2^x = 2^(-10) = 1/(2^10) = some POSITIVE number
Since x|x| is NEGATIVE, and since 2^x must be POSITIVE, we can be certain that x|x|< 2^x
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer:
[Reveal] Spoiler:
D


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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 27 Aug 2017, 07:01
Bunuel mikemcgarry IanStewart shashankism Engr2012

I know this is an OG Q, but is it not strange that st 2 is an integral part of st 1.
During actual exam I do not think to need to evaluate S2 separately since x=-10
is always going to be x<0. Let me know if my understanding is correct
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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 22 Oct 2017, 15:14
Hi,

For statement 1 I tested values i.e. x = -1 or x = -2.This is more so a question regarding reciprocals and inequalities. If x = -2, then -2 (|-2|) = 2^-2. Then, this is equal to -2 (2) = 1/2^2. In the second step where I converted 2^-2 to 1/2^2 -- would I have to also flip the other side to become 1 / -2 (2) or is that wrong?

Thanks,

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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infinitemac wrote:
Hi,

For statement 1 I tested values i.e. x = -1 or x = -2.This is more so a question regarding reciprocals and inequalities. If x = -2, then -2 (|-2|) = 2^-2. Then, this is equal to -2 (2) = 1/2^2. In the second step where I converted 2^-2 to 1/2^2 -- would I have to also flip the other side to become 1 / -2 (2) or is that wrong?

Thanks,

infinitemac


No. The right hand side is \(2^{(-2)}\), which is the same as \(\frac{1}{2^2}\) but the left hand side (-2*|-2|) stays the same.

Negative powers:
\(a^{-n}=\frac{1}{a^n}\)
Important: you cannot rise 0 to a negative power because you get division by 0, which is NOT allowed. For example, \(0^{-1} = \frac{1}{0}=undefined\).

8. Exponents and Roots of Numbers



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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 25 Oct 2017, 15:29
Bunuel wrote:
infinitemac wrote:
Hi,

For statement 1 I tested values i.e. x = -1 or x = -2.This is more so a question regarding reciprocals and inequalities. If x = -2, then -2 (|-2|) = 2^-2. Then, this is equal to -2 (2) = 1/2^2. In the second step where I converted 2^-2 to 1/2^2 -- would I have to also flip the other side to become 1 / -2 (2) or is that wrong?

Thanks,

infinitemac


No. The right hand side is \(2^{(-2)}\), which is the same as \(\frac{1}{2^2}\) but the left hand side (-2*|-2|) stays the same.

Negative powers:
\(a^{-n}=\frac{1}{a^n}\)
Important: you cannot rise 0 to a negative power because you get division by 0, which is NOT allowed. For example, \(0^{-1} = \frac{1}{0}=undefined\).

8. Exponents and Roots of Numbers



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.



Thank you very much Bunuel!

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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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New post 28 Oct 2017, 12:31
Walkabout wrote:
If x is an integer, is x|x|<2^x ?

(1) x < 0
(2) x = -10


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Since we have 1 variables and 0 equation, D could be the answer most likely.

Condition 1)
Since x < 0 and |x|≥0, x|x|≤0.
2^x > 0
Thus x|x| < 2^x.
This is sufficient.

Condition 2)
Since x = -10, x|x| = (-10)*10 = -100 < 0
And 2^(-10) = 1/(2^10) > 0
Thus x|x| < 2^x
This is also sufficient.

Therefore, D is the answer.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.
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