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Re: Is X Negative? [#permalink]
29 Dec 2013, 02:26
is x>0?
1. \(x^2=positive.\)
NS. x could be both a positive number and a negative number.
2 x|y| is not a positive number
different scenarios unfold from this expression. x|y|= -ve number, 0, or a positive rational number.
case a) |y| is not zero and x is negative. The answer would be yes b) if |y| is zero the value of x is concealed (x is either positive or negative). c) x can be a positive rational number and |y| a positive integer or a positive rational number. d) x can be a negative rational number and |y| zero.
NS
1+2) combining the informations provided doesn't help to define a clear answer. NS and the answer is E. _________________
learn the rules of the game, then play better than anyone else.
Re: Is x a negative number? [#permalink]
08 Jan 2015, 07:44
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Is x a negative number? [#permalink]
26 Aug 2015, 05:46
Hi there,
I did reach the correct answer but i am not sure If my reasoning is fully correct (specifically statement 2). I assumed the following:
(1) x^2 is a positive number.
Statement 1 is obviously not sufficient since x could be either positive or negative and the result when squaring both types would still be positive. Since there might be two possible cases for x, this statement is Insufficient.
(2) x*|y| is not a positive number.
Statement 2 can be broken down into two cases since we are mainly looking at the absolute value of "y" (|y|).
Case 1: The absolute value of |y| is positive. If "y" is positive, and "x" being multiplied with "+y" 1. Assuming that "x" is positive will give us a positive product. (Contradicting to the original statement which states that " x*|y| is not a positive number) 2. Assuming that "x" is negative will give us a negative product. (Lets assume that at the moment it tells you that "x" is negative)
Case 2: The absolute value of |y| is negative. If "y" is negative, "x" being multiplied with "-y" 1. Assuming that "x" is positive will give us a negative product. 2. Assuming that "x" is negative will give us a positive product. (Again totally contradicting with the original statement which states that " x*|y| is not a positive number) & also If x were negative, it should have matched with our 1st case second assumption)
Again "x" could have either of these two signs, we cannot be sure if "x" is negative or positive, Hence this case also does not help us in determining the value for x.
Looking at the above reasoning, statement 2 itself is not sufficient.
Combining these two statements also does not help us determining the sign of "x" since statement (1) does not add any additional information and statement (2) as already described does not help us either. Therefore both statements are not sufficient to answer the question. Answer E.
Is x a negative number? [#permalink]
26 Aug 2015, 05:59
HarveyKlaus wrote:
Hi there,
I did reach the correct answer but i am not sure If my reasoning is fully correct (specifically statement 2). I assumed the following:
(1) x^2 is a positive number.
Statement 1 is obviously not sufficient since x could be either positive or negative and the result when squaring both types would still be positive. Since there might be two possible cases for x, this statement is Insufficient.
(2) x*|y| is not a positive number.
Statement 2 can be broken down into two cases since we are mainly looking at the absolute value of "y" (|y|).
Case 1: The absolute value of |y| is positive. If "y" is positive, and "x" being multiplied with "+y" 1. Assuming that "x" is positive will give us a positive product. (Contradicting to the original statement which states that " x*|y| is not a positive number) 2. Assuming that "x" is negative will give us a negative product. (Lets assume that at the moment it tells you that "x" is negative)
Case 2: The absolute value of |y| is negative. If "y" is negative, "x" being multiplied with "-y" 1. Assuming that "x" is positive will give us a negative product. 2. Assuming that "x" is negative will give us a positive product. (Again totally contradicting with the original statement which states that " x*|y| is not a positive number) & also If x were negative, it should have matched with our 1st case second assumption)
Again "x" could have either of these two signs, we cannot be sure if "x" is negative or positive, Hence this case also does not help us in determining the value for x.
Looking at the above reasoning, statement 2 itself is not sufficient.
Combining these two statements also does not help us determining the sign of "x" since statement (1) does not add any additional information and statement (2) as already described does no help us either. Therefore both statements are not sufficient to answer the question. Answer E.
Your reasoning is correct but a bit more time consuming. A quicker way is to realise |y| \(\geq\)0 for all y. Thus for statement 2, you get 2 cases;
Case 1: When y = 0 , x*|y|\(\leq\)0 , then x be either negative or positive and hence not sufficient. you could have stopped here as you get 2 possible answers for x.
Case 2: when y \(\neq\)0, then x MUST be <0 for x*|y| \(\leq\)0. Although this case does give a unique "yes" to "is x<0?" but with case 2, statement 2 is not sufficient.
Had it been the case that the question stem mentioned that either y=0 or y\(\neq\)0 , then we could have eliminated 1 case and showed that statement 2 is indeed sufficient.
Is x a negative number? [#permalink]
28 Aug 2015, 09:35
Engr2012 wrote:
HarveyKlaus wrote:
Hi there,
I did reach the correct answer but i am not sure If my reasoning is fully correct (specifically statement 2). I assumed the following:
(1) x^2 is a positive number.
Statement 1 is obviously not sufficient since x could be either positive or negative and the result when squaring both types would still be positive. Since there might be two possible cases for x, this statement is Insufficient.
(2) x*|y| is not a positive number.
Statement 2 can be broken down into two cases since we are mainly looking at the absolute value of "y" (|y|).
Case 1: The absolute value of |y| is positive. If "y" is positive, and "x" being multiplied with "+y" 1. Assuming that "x" is positive will give us a positive product. (Contradicting to the original statement which states that " x*|y| is not a positive number) 2. Assuming that "x" is negative will give us a negative product. (Lets assume that at the moment it tells you that "x" is negative)
Case 2: The absolute value of |y| is negative. If "y" is negative, "x" being multiplied with "-y" 1. Assuming that "x" is positive will give us a negative product. 2. Assuming that "x" is negative will give us a positive product. (Again totally contradicting with the original statement which states that " x*|y| is not a positive number) & also If x were negative, it should have matched with our 1st case second assumption)
Again "x" could have either of these two signs, we cannot be sure if "x" is negative or positive, Hence this case also does not help us in determining the value for x.
Looking at the above reasoning, statement 2 itself is not sufficient.
Combining these two statements also does not help us determining the sign of "x" since statement (1) does not add any additional information and statement (2) as already described does no help us either. Therefore both statements are not sufficient to answer the question. Answer E.
Your reasoning is correct but a bit more time consuming. A quicker way is to realise |y| \(\geq\)0 for all y. Thus for statement 2, you get 2 cases;
Case 1: When y = 0 , x*|y|\(\leq\)0 , then x be either negative or positive and hence not sufficient. you could have stopped here as you get 2 possible answers for x.
Case 2: when y \(\neq\)0, then x MUST be <0 for x*|y| \(\leq\)0. Although this case does give a unique "yes" to "is x<0?" but with case 2, statement 2 is not sufficient.
Had it been the case that the question stem mentioned that either y=0 or y\(\neq\)0 , then we could have eliminated 1 case and showed that statement 2 is indeed sufficient.
Hope this helps.
X is not a negative number. Does this mean X can be 0, which is non negative, or X can be a positive number?
Pls. Help. I went with B, which has two cases.
Case 1: Where x=0, I'm getting y as a non negative. Case 2: is clear that x is 0. _________________
Is x a negative number? [#permalink]
26 Sep 2015, 09:35
1
This post received KUDOS
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longfellow wrote:
hogann wrote:
Is x a negative number?
(1) x^2 is a positive number.
(2) x*|y| is not a positive number.
Help needed!
I have tried solving modulus questions but I tend to get stuck in every other question.
Please let me know where am I going wrong.
1. x^2 is positive: this implies that x is not equal to 0.
2. x*|y| is not a positive number: this means that x<=0. Not sufficient as x can be zero or negative.
both combined state that x is not zero but negative.
Hence C.
Also, can someone please advice how to approach DS when it comes to absolute values.
Your analysis of statement 2 is not correct. When its given x|y| \(\neq\) positive ---> x|y| \(\leq\) 0 ---> This MAY or MAY NOT mean that x \(\leq\) 0.
Consider 2 cases:
Case 1: when x|y| <0 ---> as |y| can not be <0 ---> x<0
Case 2: when x|y| =0 ---> both x and |y| separately or together be = 0, you can not say anything about sign of 'x'. x|y| =0 when |y|=0 for all x or when x=0 for all |y|.
Thus when you combine the 2 statements, you still might get a case with |y|=0 for x=-3, this satisfies both statements and will give you a "yes" for the question asked.
but, with |y|=0 for x=3, this satisfies both statements and will give you a "no" for the question asked. Thus E is the correct answer. There are other cases as well but you dont have to spend time on ALL possible cases. As soon as you get 2 contradictory answers in a DS question, mark E and move on.
As for your question on how to approach modulus questions: it does not depend on whether the question is asked in PS or DS. The logic remains the same.
3 points you need to remember:
1. For any modulus or absolute value of 'x' = |x| , |x|\(\geq\) 0 (ALWAYS)
2. |x-a| means the distance of 'x' from 'a'
3. for solving modulus questions, |x-a|=0, you need to look at 2 cases:
i) when x-a < 0 --->|x-a| = -(x-a) ii) when x-a \(\geq\) 0 --->|x-a| = (x-a)
You can remember this by assuming a=0 and x=-5, then |x| = 5 = -(-5)=-x but if x=5, then |x|=5=x
Re: Is x a negative number? [#permalink]
26 Sep 2015, 17:08
Engr2012 wrote:
longfellow wrote:
hogann wrote:
Is x a negative number?
(1) x^2 is a positive number.
(2) x*|y| is not a positive number.
Help needed!
I have tried solving modulus questions but I tend to get stuck in every other question.
Please let me know where am I going wrong.
1. x^2 is positive: this implies that x is not equal to 0.
2. x*|y| is not a positive number: this means that x<=0. Not sufficient as x can be zero or negative.
both combined state that x is not zero but negative.
Hence C.
Also, can someone please advice how to approach DS when it comes to absolute values.
Your analysis of statement 2 is not correct. When its given x|y| \(\neq\) positive ---> x|y| \(\leq\) 0 ---> This MAY or MAY NOT mean that x \(\leq\) 0.
Consider 2 cases:
Case 1: when x|y| <0 ---> as |y| can not be <0 ---> x<0
Case 2: when x|y| =0 ---> both x and |y| separately or together be = 0, you can not say anything about sign of 'x'. x|y| =0 when |y|=0 for all x or when x=0 for all |y|.
Thus when you combine the 2 statements, you still might get a case with |y|=0 for x=-3, this satisfies both statements and will give you a "yes" for the question asked.
but, with |y|=0 for x=3, this satisfies both statements and will give you a "no" for the question asked. Thus E is the correct answer. There are other cases as well but you dont have to spend time on ALL possible cases. As soon as you get 2 contradictory answers in a DS question, mark E and move on.
As for your question on how to approach modulus questions: it does not depend on whether the question is asked in PS or DS. The logic remains the same.
3 points you need to remember:
1. For any modulus or absolute value of 'x' = |x| , |x|\(\geq\) 0 (ALWAYS)
2. |x-a| means the distance of 'x' from 'a'
3. for solving modulus questions, |x-a|=0, you need to look at 2 cases:
i) when x-a < 0 --->|x-a| = -(x-a) ii) when x-a \(\geq\) 0 --->|x-a| = (x-a)
You can remember this by assuming a=0 and x=-5, then |x| = 5 = -(-5)=-x but if x=5, then |x|=5=x
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