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

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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

Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 12:55

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 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=-2. 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.

When we have scenario A, xy will be positive only. Hence if xy is not positive 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.

Answer: D.

hey Bunuel!! first i would like to thank you for posting such wonderful questions..

regarding a question that you posted above, i got a small doubt..

|x+2|=|y+2| so lets say |x+2|=|y+2|=k (some 'k')

now |x+2|=k =====> x+2=+/- k and x+2= +k, iff x>-2 x+2= -k, iff x<-2

also we have |y+2|=k =====> y+2=+/- k and y+2= +k, iff y>-2 y+2= -k, iff y<-2

so x+2=y+2 ===> x=y , iff (x>-2 and y>-2) or (x<-2 and y<-2)--------eq1 and x+y=-4, iff (x<-2 and y>-2) or (x>-2 and y<-2)-------------------eq2

now coming to the options, 1) xy<0 i.e., (x=-ve and y=+ve) or (x=+ve and y=-ve) (x=-ve and y=+ve): this also means that x and y can have values, x=-1 and y= some +ve value. so eq2 cannot be applied, x+y#-4. if x=-3 and y=some +ve value, x+y=-4. two cases. data insuff. (x=+ve and y=-ve): this also means that x and y can have values, x=+ve value and y=-1.so eq2 cannot be applied, x+y#-4. if x=some +ve value and y=-3, x+y=-4. two cases. data insuff.

2)x>2,y>2 for this option too we cannot judge the value of x+y, with the limits of x and y being different in the question and the answer stem. so data insuff.

so i have a doubt that, why the answer cannot be E??

plz point out if i made any mistake.. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 17:50

Hi Bunuel, Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0.

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.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 18:09

Expert's post

lionslion wrote:

Hi Bunuel, Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0.

I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 18:43

Hi, Thank ou very much for your clarification. that helps..

Bunuel wrote:

lionslion wrote:

Hi Bunuel, Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0.

I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 19:31

Expert's post

logan wrote:

hey Bunuel!! first i would like to thank you for posting such wonderful questions..

regarding a question that you posted above, i got a small doubt..

|x+2|=|y+2| so lets say |x+2|=|y+2|=k (some 'k')

now |x+2|=k =====> x+2=+/- k and x+2= +k, iff x>-2 x+2= -k, iff x<-2

also we have |y+2|=k =====> y+2=+/- k and y+2= +k, iff y>-2 y+2= -k, iff y<-2

so x+2=y+2 ===> x=y , iff (x>-2 and y>-2) or (x<-2 and y<-2)--------eq1 and x+y=-4, iff (x<-2 and y>-2) or (x>-2 and y<-2)-------------------eq2

now coming to the options, 1) xy<0 i.e., (x=-ve and y=+ve) or (x=+ve and y=-ve) (x=-ve and y=+ve): this also means that x and y can have values, x=-1 and y= some +ve value. so eq2 cannot be applied, x+y#-4. if x=-3 and y=some +ve value, x+y=-4. two cases. data insuff. (x=+ve and y=-ve): this also means that x and y can have values, x=+ve value and y=-1.so eq2 cannot be applied, x+y#-4. if x=some +ve value and y=-3, x+y=-4. two cases. data insuff.

2)x>2,y>2 for this option too we cannot judge the value of x+y, with the limits of x and y being different in the question and the answer stem. so data insuff.

so i have a doubt that, why the answer cannot be E??

plz point out if i made any mistake..

Hi, Logan. Good way of thinking. Though I think that your solution is not correct.

Consider this:

We have |x+2|=|y+2|

(1) xy<0, hence x and y have opposite signs. Let's take x negative and y positive (obviously it doesn't matter which one we pick as equation is symmetric).

If y is positive RHS |y+2| will be positive as well and we can expend it as |y+2|=y+2.

Now for |x+2| we can have to cases: A. -2<x<0 --> |x+2|=x+2=y+2 --> x=y. BUT: it's not valid solution as x and y have opposite signs and they can not be equal to each other.

B. x<=-2 --> |x+2|=-x-2=y+2 --> x+y=-4. Already clear and sufficient.

If we go one step further to see for which x and y is this solution is valid, we'll get: As x+y=-4, x<=-2 and y>0 --> x must be <-4. If you substitute values of x<-4 you'll receive the values of y>0 and their sum will always be -4.

(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,

I kind of disagree with your conclusion when you combined both the stmts. If x,y, and a all are 0 then the actual question (x^2+y^2 > 4a) itself will become whether 0 > 0 ?....so I would say that the answer should be C.

(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,

I kind of disagree with your conclusion when you combined both the stmts. If x,y, and a all are 0 then the actual question (x^2+y^2 > 4a) itself will become whether 0 > 0 ?....so I would say that the answer should be C.

hi xyztroy,

i think i can answer ur question.

in question, no limits for x and y are given, like x&y are integers or x&y are real numbers. so x and y can assume any values, including 0. but we have to conclusively show that (x^2+y^2 > 4a). as you see, 1&2 are individually insufficient. combining 1&2 we have (x^2+y^2 = 5a), which is definitely greater than 4a. when you substitute values for x,y and a, all values of x,y and a which satisfy (x^2+y^2 = 5a) also satisfies (x^2+y^2 > 4a), except the values x=y=a=0. so two cases arise. hence insufficient.

Re: Inequality and absolute value questions from my collection [#permalink]
23 Dec 2009, 01:44

Bunuel wrote:

Hi, Logan. Good way of thinking. Though I think that your solution is not correct.

Consider this:

We have |x+2|=|y+2|

(1) xy<0, hence x and y have opposite signs. Let's take x negative and y positive (obviously it doesn't matter which one we pick as equation is symmetric).

If y is positive RHS |y+2| will be positive as well and we can expend it as |y+2|=y+2.

Now for |x+2| we can have to cases: A. -2<x<0 --> |x+2|=x+2=y+2 --> x=y. BUT: it's not valid solution as x and y have opposite signs and they can not be equal to each other.

B. x<=-2 --> |x+2|=-x-2=y+2 --> x+y=-4. Already clear and sufficient.

If we go one step further to see for which x and y is this solution is valid, we'll get: As x+y=-4, x<=-2 and y>0 --> x must be <-4. If you substitute values of x<-4 you'll receive the values of y>0 and their sum will always be -4.

The same approach works for (2) as well.

Hope it's clear.

thanx 4 d quick reply Bunuel...

yeah, i think the blue colored line of ur's helped me clear my doubt..

nice question...keep up the good work.. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
18 Jan 2010, 12:43

Question 3

I think we are not concerned with the value of x,y because question asks for whether x2+y2>4a.once we get x2+y2=5a, it is confirmed that x2+y2>4a.One think more we should consider here is question is not giving any clue about x,y and a as if it is not concerned with these variables.

Re: Inequality and absolute value questions from my collection [#permalink]
18 Jan 2010, 12:49

Expert's post

GMAT10 wrote:

Question 3

I think we are not concerned with the value of x,y because question asks for whether x2+y2>4a.once we get x2+y2=5a, it is confirmed that x2+y2>4a.One think more we should consider here is question is not giving any clue about x,y and a as if it is not concerned with these variables.

Whats the OA.

Answers (OA) and solutions are given in my posts on previous pages. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
19 Jan 2010, 01:22

sriharimurthy wrote:

Quote:

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

Question Stem : x > 0 ; y > 0 ?

St. (1) : 2x -2y = 1 x = y + 0.5 Equation can be satisfied for both positive and negative values of x and y. Hence Insufficient.

St. (2) : x/y > 1 Equation can be satisfied when both x and y are either positive or negative. Hence Insufficient.

St. (1) and (2) together : (y + 0.5)/y > 1 1 + 0.5/y > 1 0.5/y > 1 For this to be true, y must be positive. If y is positive then x will also be positive. Hence Sufficient.

Answer : C

I think small mistake in solution although solution is right. 1+0.5/y>1 = 0.5/y>0 => y will be positive always and from x=y+0,5 => x will be positive.

Re: Inequality and absolute value questions from my collection [#permalink]
20 Jan 2010, 02:45

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.

Quote:

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

A. x+2=y+2 --> x=y. This will occur 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=-2. 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.

When we have scenario A, xy will be positive only. Hence if xy is not positive 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.

Answer: D.

Wonderful question, I solved it up to an extent but in reasoning I messed up. Question for the quoted part which talks about the scenarios to reducing to two scenarios. I can think and verfy also though the validity of it, but want to drill further and understand how could we generalise this scenario if we could ?

Like what if |x+3|=|X-2| or |x+3|=|X-2| + |X+4| etc , just crude examples I have put , How could we generalise such scenarios ? instead of imagining six scenarios or more .

Re: Inequality and absolute value questions from my collection [#permalink]
20 Jan 2010, 05:15

Bunuel wrote:

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

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement: A. s is either positive or zero, as -s<=s; B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s. But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well. Consider: s=5, r=5 --> -5<=5<=5 |5|>=5 s=5, r=-5 --> -5<=-5<=5 |-5|>=5 Both statements are true with these values. Hence insufficient.

Answer: E.

Quote:

(1) -s<=r<=s, we can conclude two things from this statement: A. s is either positive or zero, as -s<=s;

How do you conclude A here can't make out or I am tired.; anyways, it is mental marathon .. wonderful questions Bunuel. Any more link for inequality as I need some more practise .

Re: Inequality and absolute value questions from my collection [#permalink]
20 Jan 2010, 23:41

Expert's post

GMATMadeeasy wrote:

How do you conclude A here can't make out or I am tired.; anyways, it is mental marathon .. wonderful questions Bunuel. Any more link for inequality as I need some more practise .

We have \(-s<=r<=s\) --> \(-s<=s\). Now if \(s\) in negative, let's say \(-2\), then we would have \(-(-2)<=-2\) --> \(2<=-2\), which is not right. Hence \(s\) can not be negative. But \(s\) can be zero --> \(0<=0\), true. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
20 Jan 2010, 23:55

Expert's post

GMATMadeeasy 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.

Quote:

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

A. x+2=y+2 --> x=y. This will occur 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=-2. 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.

When we have scenario A, xy will be positive only. Hence if xy is not positive 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.

Answer: D.

Wonderful question, I solved it up to an extent but in reasoning I messed up. Question for the quoted part which talks about the scenarios to reducing to two scenarios. I can think and verfy also though the validity of it, but want to drill further and understand how could we generalise this scenario if we could ?

Like what if |x+3|=|X-2| or |x+3|=|X-2| + |X+4| etc , just crude examples I have put , How could we generalise such scenarios ? instead of imagining six scenarios or more .

We can count the # of scenarios for the above examples as well, but as in these examples there is only one variable we can directly solve them using the ranges, so no need to use the approach we used in solving the original question.

|x+3|=|x-2|, two check points -3 and 2, thus we'll have three ranges in which we should expand the absolute values:

A. x<-3 --> -(x+3)=-(x-2) --> -3=2, which is not true, hence in this range we have no solution; B. -3<=x<=2 --> x+3=-(x-2) --> x=-1/2, which IS in the range we are expanding so it's a valid solution; C. x>2 --> x+3=x+2 --> 3=2, which is not true, hence in this range we have no solution.

The equation |x+3|=|x-2| has only one solution x=-1/2. If you look at the scenarios A,B and C you'll see that |x+3|=|x-2| can take also only TWO forms x+3=x+2 or x+3=-(x-2), but the ranges are important here and we should consider these forms in their respective ranges.

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