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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]
22 Feb 2013, 00:27

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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.

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

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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.

Re: Inequality and absolute value questions from my collection [#permalink]
11 Jul 2013, 20:42

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Hi johncoffey , My two cents- for (1) - it is always useful to start out by factoring an expression if possible, especially when there is a variable in common ("y" in this example). Even though it does make sense to isolate the expression "xy" that we are being asked for- note that in this case that would give us more unknowns on the RHS. Hope tht helps.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Sep 2013, 05:00

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StormedBrain wrote:

Bunuel wrote:

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

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Hi Bunuel ,

I know saying (1/|n|) < n will be true for all n<0 is quite clear logically. Still I want to reach this conclusion mathematically.

I got swayed solving for n|n| < 1 .

n*|n| < 1.

If n<0, then we'll have -n^2<1 --> n^2>-1. Which is true. So, n*|n| < 1 holds true for any negative value of n. If n>0, then we'll have n^2<1 --> -1<n<1. So, n*|n| < 1 also holds true for 0<n<1.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Sep 2013, 07:07

1

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Bunuel wrote:

StormedBrain wrote:

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.

Bunuel , Can you please show how we can reach to C using graphical approach ?

4. Are x and y both positive?

The question asks whether point (x, y) is in the first quadrant.

(1) 2x-2y=1 --> draw line y=x-1/2:

Attachment:

graph.png

Not sufficient.

(2) x/y>1 --> Draf line x/y=1. The solutions is the green region:

Attachment:

graph (1).png

Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Answer: C.

Hope it helps.

Hey Bunuel,

I am a bit confused. Shouldn't the green area in 3rd quadrant be above the line and below x-axis ?

Lets take a point (-0.5,-1) in the green shaded region , then -0.5/-1 = 1/2 <1.. _________________

The First and Last time !!!

BKPL - Below Kudos Poverty Line .....Need your help.

Re: Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 19:07

Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 04:18

Bunuel wrote:

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

Not sure about this one...

First I reduced the given equation (divided out the y) and solved for x: 6*x*y = x^2*y + 9*y 6*x = x^2 + 9 0 = x^2 - 6*x + 9 0 = (x-3)^2 x = 3

Statement 1:

y-x=3 y-3=3 y=6 xy=3*6=18

SUFFICIENT

Statement 2:

x^3<0

We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 04:34

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.

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:27

12. Is r=s?

(1) -s<=r<=s

(2) |r|>=s

E – for this - both can be true or false when 0< r < 1 For example , take r as 0.8 S = 0.86 i.e. -0.86 < = 0.8 < = 0.86 |0.8|>= 0.86 i.e. 1 >= 0.86 Combining , any values can be taken , on values > =1 , both r and s will be same

3. Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a

(2) (x – y)^2 = a C is the answer

Combined both and the equation will give x^2 + y^2 = 5a _________________

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:48

4) I) 2x-2y=1 so y=x-1/2 NS II)x/y>0 so x and y have the same sign and the modulus of x has to be larger than the modulus of y NS Together, to satisfy both clues needs to be larger than 1/2 and x becomes larger than 0; the stem is true, therefore C

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:54

ichha148 wrote:

12. Is r=s?

(1) -s<=r<=s

(2) |r|>=s

E – for this - both can be true or false when 0< r < 1 For example , take r as 0.8 S = 0.86 i.e. -0.86 < = 0.8 < = 0.86 |0.8|>= 0.86 i.e. 1 >= 0.86 Combining , any values can be taken , on values > =1 , both r and s will be same

Taking the modulus does not mean rounding up to the nearest integer; it means removing the negative sign if present. |0.8|<0.86

ichha148 wrote:

3. Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a

(2) (x – y)^2 = a C is the answer

Combined both and the equation will give x^2 + y^2 = 5a

Nowhere it is said that x and y are non-zero. If x and y are zero, 5a=0, therefore a=0, and the stem is false (x^2+y^2=0)

Last edited by Marco83 on 17 Nov 2009, 09:57, edited 1 time in total.

gmatclubot

Re: Inequality and absolute value questions from my collection
[#permalink]
17 Nov 2009, 09:54

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...