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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 ?  [#permalink]

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13 Sep 2019, 22:14
1
If x is an integer, is x|x| < $$2^x$$ ?

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

Edit: Formating

Given: x is an integer

Asked: Is x|x| < $$2^x$$ ?

(1) x < 0
|x| = -x
-x^2 < 2^x
-x^2 < 0
2^x >0
-x^2 < 2^x
$$x|x| < 2^x$$
SUFFICIENT

(2) x = -10
|x| = 10
x|x| = -10 * 10 = -100
$$2^x = 2^{-10} = \frac{1}{1024}$$
$$-100 < \frac{1}{1024}$$
$$x|x| < 2^x$$
SUFFICIENT

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

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22 Feb 2020, 03:30
Bunuel 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.

Hi Bunuel,

Please correct me if I am wrong

1) x|x|<2^x
x=-1
so,

-1|1|<2^-1
-1<1/2

which will always be true for any negative value of x and also with positive values it will remain true and sufficient

2) condition 2 is already giving fix value so it will be sufficient
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Re: If x is an integer, is x|x|<2^x ?  [#permalink]

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22 Feb 2020, 13:25
2
Farina wrote:
Bunuel 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.

Hi Bunuel,

Please correct me if I am wrong

1) x|x|<2^x
x=-1
so,

-1|1|<2^-1
-1<1/2

which will always be true for any negative value of x and also with positive values it will remain true and sufficient

2) condition 2 is already giving fix value so it will be sufficient

Hi Farina,

Most of what you wrote is correct. This DS question is based on a couple of Number Properties:

when X is NEGATIVE....
X|X| will ALWAYS be NEGATIVE
2^(X) will ALWAYS be POSTIIVE

This means that the answer to the question is ALWAYS YES and Fact 1 is SUFFICIENT.

However, what you noted about POSITIVE values is not true. For example....

when X = 1...
1|1| = 1
2^1 = 2
so the answer to the question "is X|X| < 2^X?".... is YES.

when X = 2...
2|2| = 4
2^2 = 4
so the answer to the question "is X|X| < 2^X?".... is NO.

This means that when X is POSITIVE, the answer to the question changes and this situation would be INSUFFICIENT.

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

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25 May 2020, 02:29
BrentGMATPrepNow 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

Hi Brent, you have mentioned So, x|x| = (NEGATIVE)(|NEGATIVE|) = (NEGATIVE)(POSITIVE) = NEGATIVE

but isn't |x| = -x when x<0 ? based on this logic it should be Negative * Negative = + ive isn't it ??

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

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25 May 2020, 02:54
Bunuel 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.

Hi @Bunuel - Opening up an old thread but need some clarification.

In this question you have taken |x| as |-x| = x when x<0

However, in another question - https://gmatclub.com/forum/is-x-1-1-x-1 ... 34652.html

you have taken |x| as -x when x<0 - can you please clarify the conditions wherein we have to take two different things ?
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Posts: 65187
Re: If x is an integer, is x|x|<2^x ?  [#permalink]

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25 May 2020, 03:01
Anuragjn wrote:
Bunuel 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.

Hi @Bunuel - Opening up an old thread but need some clarification.

In this question you have taken |x| as |-x| = x when x<0

However, in another question - https://gmatclub.com/forum/is-x-1-1-x-1 ... 34652.html

you have taken |x| as -x when x<0 - can you please clarify the conditions wherein we have to take two different things ?

Not sure I understand from where you've taken the red part.

When x < 0, then |x| = -x, so in this case $$x*|x|=x*(-x)=-x^2$$.
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Re: If x is an integer, is x|x|<2^x ?  [#permalink]

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25 May 2020, 03:09
Bunuel wrote:
Anuragjn wrote:
Bunuel 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.

Hi [b]Bunuel[/b] - Opening up an old thread but need some clarification.

In this question you have taken |x| as |-x| = x when x<0

However, in another question - https://gmatclub.com/forum/is-x-1-1-x-1 ... 34652.html

you have taken |x| as -x when x<0 - can you please clarify the conditions wherein we have to take two different things ?

Not sure I understand from where you've taken the red part.

When x < 0, then |x| = -x, so in this case $$x*|x|=x*(-x)=-x^2$$.

Hi Bunuel - Thanks for quick response.

I have taken it from this explanation - https://gmatclub.com/forum/if-x-is-an-i ... ml#p703553

You have mentioned when x<0; x|x| as Negative * positive = -x^2

is x<0, should it be Negative * negative = X^2 ?

Can you please clarify further ?
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Posts: 65187
Re: If x is an integer, is x|x|<2^x ?  [#permalink]

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25 May 2020, 03:17
Anuragjn wrote:
Bunuel wrote:
Anuragjn wrote:
Bunuel 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.

Hi [b]Bunuel[/b] - Opening up an old thread but need some clarification.

In this question you have taken |x| as |-x| = x when x<0

However, in another question - https://gmatclub.com/forum/is-x-1-1-x-1 ... 34652.html

you have taken |x| as -x when x<0 - can you please clarify the conditions wherein we have to take two different things ?

Not sure I understand from where you've taken the red part.

When x < 0, then |x| = -x, so in this case $$x*|x|=x*(-x)=-x^2$$.

Hi Bunuel - Thanks for quick response.

I have taken it from this explanation - https://gmatclub.com/forum/if-x-is-an-i ... ml#p703553

You have mentioned when x<0; x|x| as Negative * positive = -x^2

is x<0, should it be Negative * negative = X^2 ?

Can you please clarify further ?

An absolute value is never negative (NEVER).

When x < 0, then |x| = -x, but -x, since x is negative, is -(negative), so still POSITIVE (-x = -negative = positive). So, again, when x < 0, x|x| = x*(-x) = negative*positive = -x^2 = negative.
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Re: If x is an integer, is x|x|<2^x ?  [#permalink]

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10 Jun 2020, 10:39
1
cucrose wrote:
If x is an integer, is x|x| < 2^x?

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

x|x|<0 if x<0

x|x|>=0 if x>=0

In x|x| < 2^x

RHS will be always positive i.e 2^x>0 for any value of x

So We need to know the sign of x in order to solve this question,

1.x<0 Sufficient x|x| < 2^x is true
2.x=-10 Sufficient x|x| < 2^x is true

Re: If x is an integer, is x|x|<2^x ?   [#permalink] 10 Jun 2020, 10:39

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