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Guys I didn't forget your request, just was collecting good questions to post.

So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers.

1. If \(6*x*y = x^2*y + 9*y\), what is the value of xy? (1) \(y – x = 3\) (2) \(x^3< 0\)

Re: Inequality and absolute value questions from my collection [#permalink]

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21 May 2010, 02:37

Bunuel wrote:

4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

I am a little confused on this one . Can the answer be E??

From A: 2x-2y=1 => x-y= 0.5 INSF

From B x/y > 1 => x > y INSF

From A & B x-y =0.5 and x > y

If x = -0.5 and y = -1 then x > y and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5 Hence both x and y can be negative

If x= 1 and y = 0.5 then x > y and x- y = 1 -0.5 = 0.5 Hence both x and y can be positive

Re: Inequality and absolute value questions from my collection [#permalink]

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21 May 2010, 03:21

Bunuel wrote:

ManishS wrote:

I am a little confused on this one . Can the answer be E??

From A: 2x-2y=1 => x-y= 0.5 INSF

From B x/y > 1 => x > y INSF

From A & B x-y =0.5 and x > y

If x = -0.5 and y = -1 then x > y and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5 Hence both x and y can be negative

If x= 1 and y = 0.5 then x > y and x- y = 1 -0.5 = 0.5 Hence both x and y can be positive

Ans = E ??

Problem with your solution is that the red part is not correct.

\(\frac{x}{y}>1\) does not mean that \(x>y\). If both x and y are positive, then \(x>y\), BUT if both are negative, then \(x<y\).

From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).

When we consider two statement together:

From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.

OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.

Re: Inequality and absolute value questions from my collection [#permalink]

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14 Jun 2010, 22:47

h2polo wrote:

Bunuel wrote:

2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1

Another way of looking at the problem is to ask, is x<0? Because if it is, then we know that y is zero. The only case in which y will not be zero is if x is positive.

Statement 1:

x<0... answers my question above.

SUFFICIENT

Statement 2:

y<1

Because y is an integer, it must be one of the following values: 0, -1, -2, -3...

BUT |x| + x can never be a negative value. The lowest value that it can be is 0.

Hence, y can never be negative and the only possible value it can be then is 0.

SUFFICIENT

ANSWER: D.

---------------------------- hear it is not give that X is also a integer,

---------------------------- hear it is not give that X is also a integer,

S1: if X=1/2 or -1/2 then also Y is integer ,

in this case Ans: B

Is i am missing somthing?

OA' s and solutions for all the problems are given in my posts on pages 2 and 3.

OA for this question is D. Below is solution for it.

2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1

Note: \(y=|x|+x\), this expression is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.

Re: Inequality and absolute value questions from my collection [#permalink]

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07 Jul 2010, 00:47

Ive got C for this question..

when both together yields x^2 + y^2 = 5a why it is E?

Also I dont understand the explanation of below, St. (1) and (2) together : x^2 + y^2 = 5a When either x or y is not 0, question stem holds true. When x and y are both 0, question stem is false.

Can somebody clarify how to solve this please...

sriharimurthy wrote:

Quote:

3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a

when both together yields x^2 + y^2 = 5a why it is E?

Also I dont understand the explanation of below, St. (1) and (2) together : x^2 + y^2 = 5a When either x or y is not 0, question stem holds true. When x and y are both 0, question stem is false.

Can somebody clarify how to solve this please...

sriharimurthy wrote:

Quote:

3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a

St. (2) : (x - y)^2 = a x^2 + y^2 - 2xy = a Insufficient.

St. (1) and (2) together : x^2 + y^2 = 5a When either x or y is not 0, question stem holds true. When x and y are both 0, question stem is false.

Hence insufficient.

Answer : E

OA' s and solutions for all the problems are given in my posts on pages 2 and 3.

OA for this question is E.

When we consider statement together we'll have: \(x^2+y^2=5a\). Now, if \(x\), \(y\) and \(a\) are different from zero (for example: \(x=3\), \(y=4\) and \(a=5\)) then \(x^2+y^2=5a>4a\) and the answer to the question is YES, but if \(x=y=a=0\), then \(x^2+y^2=5a=4a=0\) and the answer to the question is NO. Two different answers, thus not sufficient.

Re: Inequality and absolute value questions from my collection [#permalink]

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16 Jul 2010, 08:07

Hi Bunuel, I forgot to thank you for these great questions and solutions. You are the BEST!!!!

If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1

(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence . Sufficient.

In the above problem, it is very obvious that Y is not equal to 0 from the 2nd statement. Do you think GMAT questions will have this much obvious statements?

Hi Bunuel, I forgot to thank you for these great questions and solutions. You are the BEST!!!!

If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1

(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence . Sufficient.

In the above problem, it is very obvious that Y is not equal to 0 from the 2nd statement. Do you think GMAT questions will have this much obvious statements?

The trick here is to conclude that \(y\) can not be negative, the rest is relatively easy. And yes, I think you can see such questions on GMAT.
_________________

Re: Inequality and absolute value questions from my collection [#permalink]

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20 Jul 2010, 02:34

lagomez wrote:

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

answer A because in number 2 n can be negative or a fraction

Doubt :The question is is |N|<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions for example for n =1/2 , 1/(|1/2|)=2 which is greater than n=1/2

Also for n=-1/2,1/(|-1/2|)=2 which is greater than n=-1/2

for negative integers also its true ,say we take n=-3 , 1/|n|=1/3 which is greater than n=-3

But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!!

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

answer A because in number 2 n can be negative or a fraction

Doubt :The question is is |N|<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions for example for n =1/2 , 1/(|1/2|)=2 which is greater than n=1/2

Also for n=-1/2,1/(|-1/2|)=2 which is greater than n=-1/2

for negative integers also its true ,say we take n=-3 , 1/|n|=1/3 which is greater than n=-3

But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!!

\(\frac{1}{|n|}>n\) holds true for ALL negative values of \(n\), as if \(n<0\) then \(LHS=positive>RHS=negative\). Hence we don't know whether \(-4<n<4\) is true. That's why statement (2) is not sufficient.

The complete range of \(n\) for which \(\frac{1}{|n|}>n\) holds true is \(n<1\).

P.S. OA's and solutions for all question are given in my posts on pages 2 and 3.

Re: Inequality and absolute value questions from my collection [#permalink]

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22 Jul 2010, 06:36

1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0 First, devide the whole equation by y, and 6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3 (1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient (2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient B

2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1 (1)Sufficient (2)Sufficient, y couldn't be negative, the least it could be is 0. D 3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a

(1) (x+y)^2= x^2+y^2+2x*y 9a=x^2+y^2+2x*y Still Insufficient

Sum up equations (1) and (2) , then we can get 10a=2(x^2+y^2) => 5a=x^2+y^2, and we know a is necessarily positive since (x – y)^2 = a With both factors, we can know x^2 + y^2 >4a

the answer is C

4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1)Insufficient (2)that only tells you x is greater than y Insufficient (1) and (2) together: x-y=0.5, and x is great than y, Insufficient E

5. What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

(1) we only know y must be positive, insufficient (2)y coule be -8 or 14 insufficient BOth (1) and (2): from statement (1) weve learned that y must be a positive number, along with statement (2) we know y is 14 the answer is C

6. If x and y are integer, is y > 0? (1) x +1 > 0 (2) xy > 0

(1)x>-1, we wouldnt know anything about y with that, insufficient (2)xy>0, if x is less than 0, then y too has to be less than 0 Insufficient

both (1) and (2): xy>0, x>-1, x could be 0 or any positive integer, and even if x=0, y still has to be great than 0 coz xy>0 Sufficient

C

7. |x+2|=|y+2| what is the value of x+y? (1) xy<0 (2) x>2 y<2

(1)xy<0, that means either x or y is positive and the other negative. seems like x+y= -4, coz x+2=-2-y => x+y=-4, sufficient (2) Sufficient too, same reasons as above

D

8. a*b#0. Is |a|/|b|=a/b? (1) |a*b|=a*b (2) |a|/|b|=|a/b|

(1) a*b has to be positive, a/b has to positive too. Sufficient

(2) we already know that for sure, Insufficient

A

9. Is n<0? (1) -n=|-n| (2) n^2=16

(1)since the absolute value must be positive, so we know n has to be negative, or 0 Not Sufficient (2)ok, n could be 4 or -4, so what? Insufficient BOth (1) and (2) together: ok, n could be 4 or -4, and it's less than or equal to zero, so it must be -4 Sufficient

C

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

(1) n>4 or n<-4, |n| must be greater than 4 sufficient (2)we only know n<1, |n| could be greater than 4 if n < -4 insufficient

A

11. Is |x+y|>|x-y|? (1) |x| > |y| (2) |x-y| < |x|

Just leave it for a while

12. Is r=s? (1) -s<=r<=s (2) |r|>=s

(1) Not sufficient (2) Not sufficient Together: Still Not sufficient