Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

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!

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
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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
_________________

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

Show Tags

24 Apr 2016, 14:22

1

This post was BOOKMARKED

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.

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

Show Tags

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
_________________

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

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
_________________

It's the journey that brings us happiness not the destination.

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

Show Tags

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?

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, \(0^{-1} = \frac{1}{0}=undefined\).

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

Show Tags

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, \(0^{-1} = \frac{1}{0}=undefined\).

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.
_________________