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

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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]
01 Dec 2009, 10:58

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 tried to solve this in another way.

1) 3|x^2 -4| = y - 2 if (x^2 -4) is positive, we can rewrite above as 3(x^2 -4) = y - 2 => 3x^2-y = 10 -> Eqn. 1 if (x^2 -4) is negative, we can rewrite above as 3(4-x^2) = y - 2 => -3x^2-y = -14 -> Eqn. 2 Adding equation 1 and 2, we get -2y = -4 or y = 2. So (A) as the answer is tempting.

I know this is not correct and carries the assumption that y is an integer which is not the case here.

If y indeed were an integer in the question, do you think the above approach had any problems ? I am a little confused because every inequality problem appears to have a different method for solving it!

Re: Inequality and absolute value questions from my collection [#permalink]
01 Dec 2009, 11:57

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kaptain wrote:

Bunuel, I tried to solve this in another way.

1) 3|x^2 -4| = y - 2 if (x^2 -4) is positive, we can rewrite above as 3(x^2 -4) = y - 2 => 3x^2-y = 10 -> Eqn. 1 if (x^2 -4) is negative, we can rewrite above as 3(4-x^2) = y - 2 => -3x^2-y = -14 -> Eqn. 2 Adding equation 1 and 2, we get -2y = -4 or y = 2. So (A) as the answer is tempting.

I know this is not correct and carries the assumption that y is an integer which is not the case here.

If y indeed were an integer in the question, do you think the above approach had any problems ? I am a little confused because every inequality problem appears to have a different method for solving it!

Thanks

This approach is not correct not because we are not told that y is an integer, but because you can not add inequalities like you did.

3(x^2 -4) = y - 2 OR 3(4-x^2) = y - 2, in fact these equation are derived from one and from them only one is right. It's not that we have 3(x^2 -4) = y - 2 AND 3(4-x^2) = y - 2 and we are asked to solve fro unknowns. If it were then your solution would be right.

Re: Inequality and absolute value questions from my collection [#permalink]
01 Dec 2009, 23:05

Bunuel, you are correct. The key is understanding that the two equations are an 'OR' (either one is true depending on whether x^2-4 is positive or negative) and not an 'AND' (both are correct).

You mentioned that inequalities cannot be added 'the way' I did. I believe you are not saying that we cannot add inequalities. I saw an interesting discussion here - > http://www.beatthegmat.com/combining-in ... 21610.html (Sorry for the cross posting, but this may be of use to someone confused like me!)

Re: Inequality and absolute value questions from my collection [#permalink]
02 Dec 2009, 01:05

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kaptain wrote:

Bunuel, you are correct. The key is understanding that the two equations are an 'OR' (either one is true depending on whether x^2-4 is positive or negative) and not an 'AND' (both are correct).

You mentioned that inequalities cannot be added 'the way' I did. I believe you are not saying that we cannot add inequalities. I saw an interesting discussion here - > http://www.beatthegmat.com/combining-in ... 21610.html (Sorry for the cross posting, but this may be of use to someone confused like me!)

+1 from me.

cheers

Sure when I said that you can not add this way I meant: in this case. _________________

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, 13:09

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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.

1 and 2 are insuff for the above mentioned (by Bunuel) reasons, now taking 1 and 2 together, here's another approach: 2x-2y=1 2(x-y)=1 (x-y)=1/2

y(x/y -1)=1/2 now we know that from option 2 x/y>1

so, y(x/y -1)=1/2====> y[(a value >1) - 1]=1/2 so, [y*(+ve value)]=1/2 hence y= +ve

now since y=+ve and (x/y)>1, we have x=+ve therefore, option c. _________________

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.

gmatclubot

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