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Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 10:33

79

This post received KUDOS

Expert's post

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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

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

I got C. Can you please explain why this is incorrect?

Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy

Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong? _________________

Re: Inequality and absolute value questions from my collection [#permalink]
21 Feb 2013, 20:23

Bunuel wrote:

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

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8 y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

Answer: C.

Bunuel, I think I need some conceptual help. Why should we not solve statement 1 by rewriting the two statements and then adding them together? (Besides the fact that it's time consuming....) I rewrote them and found 3x^2 -10 = y for the positive absolute vlaue, and -3x^2+14=y for the negative abs value. From this, I added them together and got y=4..

Can you please explain what I'm getting wrong conceptually? Thanks so much!!!! I appreciate your kindness. _________________

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

I got C. Can you please explain why this is incorrect?

Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy

Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?

The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 00:27

1

This post received KUDOS

Expert's post

JJ2014 wrote:

Bunuel wrote:

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

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8 y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

Answer: C.

Bunuel, I think I need some conceptual help. Why should we not solve statement 1 by rewriting the two statements and then adding them together? (Besides the fact that it's time consuming....) I rewrote them and found 3x^2 -10 = y for the positive absolute vlaue, and -3x^2+14=y for the negative abs value. From this, I added them together and got y=4..

Can you please explain what I'm getting wrong conceptually? Thanks so much!!!! I appreciate your kindness.

|x^2-4|=x^2-4 when x^2-4>0; |x^2-4|=-(x^2-4) when x^2-4<=0.

So, the two equations you'll get from the original are relevant for different ranges of x. Hence, you cannot consider them as two separate equations and solve.

To put it simply: we cannot get the single value of y from 3|x^2 -4| = y - 2. Consider y=2 and x=2 OR y=11 and x=1.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

I got C. Can you please explain why this is incorrect?

Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy

Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?

The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a.

So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 06:26

Bunuel wrote:

JJ2014 wrote:

Bunuel wrote:

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

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8 y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

Answer: C.

Bunuel, I think I need some conceptual help. Why should we not solve statement 1 by rewriting the two statements and then adding them together? (Besides the fact that it's time consuming....) I rewrote them and found 3x^2 -10 = y for the positive absolute vlaue, and -3x^2+14=y for the negative abs value. From this, I added them together and got y=4..

Can you please explain what I'm getting wrong conceptually? Thanks so much!!!! I appreciate your kindness.

|x^2-4|=x^2-4 when x^2-4>0; |x^2-4|=-(x^2-4) when x^2-4<=0.

So, the two equations you'll get from the original are relevant for different ranges of x. Hence, you cannot consider them as two separate equations and solve.

To put it simply: we cannot get the single value of y from 3|x^2 -4| = y - 2. Consider y=2 and x=2 OR y=11 and x=1.

So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again.

What are you trying to get when setting "them" equal? Anyway, you won't be able to solve two equations with three unknowns. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
28 Feb 2013, 05:35

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

The solution seem confusing to me as I see four cases: a] x<-2, y<-2 b]x>-2, y>-2 c] x<-2, y>-2 d]x>-2, y<-2

case [a] and [b] support x=y while case [c] and [d] support x+y=-4

when xy<0, the case [c]or[d] always do not apply, for example: x=-3 and y=3 would come under case[c] and x=-1 and y=3 would come under case [b] , so it is insufficient.

when x>2 , y<2, we have a case [b] with x=3, y=-1 and a case [d] with x=3,y=-3. So insufficient

when we combine(1)+(2) , we have a case as shown above , it is also insufficient.

Re: Inequality and absolute value questions from my collection [#permalink]
28 Feb 2013, 05:42

1

This post received KUDOS

Expert's post

piealpha wrote:

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

The solution seem confusing to me as I see four cases: a] x<-2, y<-2 b]x>-2, y>-2 c] x<-2, y>-2 d]x>-2, y<-2

case [a] and [b] support x=y while case [c] and [d] support x+y=-4

when xy<0, the case [c]or[d] always do not apply, for example: x=-3 and y=3 would come under case[c] and x=-1 and y=3 would come under case [b] , so it is insufficient.

when x>2 , y<2, we have a case [b] with x=3, y=-1 and a case [d] with x=3,y=-3. So insufficient

when we combine(1)+(2) , we have a case as shown above , it is also insufficient.

So my answer choice would be E.

Can somebody help if I am wrong.

Please read the thread: 11 pages of good discussion.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Hi Bunnel Solved the 1st statement like this - \((x + y)^2 = 9a\) Since \(x^2 + y^2 >= 2xy\) \(x^2 + y^2 + x^2 + y^2 >= 9a\) \(2(x^2 + y^2) >= 9a\) \(x^2 + y^2 >= 4.5a\) Now this would have been sufficient if a is not = 0 had been given in the stem Is this approach to the problem alright?? Is this st sufficient if it is given that a is not equal to 0 Thanks

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Hi Bunnel Solved the 1st statement like this - \((x + y)^2 = 9a\) Since \(x^2 + y^2 >= 2xy\) \(x^2 + y^2 + x^2 + y^2 >= 9a\) \(2(x^2 + y^2) >= 9a\) \(x^2 + y^2 >= 4.5a\) Now this would have been sufficient if a is not = 0 had been given in the stem Is this approach to the problem alright?? Is this st sufficient if it is given that a is not equal to 0 Thanks

Yes, but we don't know whether x is 0. _________________

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

hi Bunuel,

Thank you very much for all the explanations. I have a query on this one

Combining both we get x^2+y^2=5a or x,y,a = 0

aren't those sufficient to answer the question is x^2+y^2>4a

Is the first case where x^2+y^2=5a, the answer is yes

Second case where x,y,a=0, the answer is no

Kindly do elaborate. Thanks _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

hi Bunuel,

Thank you very much for all the explanations. I have a query on this one

Combining both we get x^2+y^2=5a or x,y,a = 0

aren't those sufficient to answer the question is x^2+y^2>4a

Is the first case where x^2+y^2=5a, the answer is yes

Second case where x,y,a=0, the answer is no

Kindly do elaborate. Thanks

First of all when we combine we get that x^2+y^2=5a. If \(xya\neq{0}\), then the answer is YES but if \(xya={0}\), then the answer is NO.

Next, it's a YES/NO DS question. In a Yes/No Data Sufficiency question, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

Re: Inequality and absolute value questions from my collection [#permalink]
22 Jun 2013, 10:48

Expert's post

hariunplugs wrote:

Bunuel wrote:

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

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur [b]if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0. B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.[/b]

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Hi,Can u explain the logic behind bolded part?...i tried my best..but not able comprehend much..thanks in advance.

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