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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]
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, 02:15

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gmat620 wrote:

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.

These questions are from various sources. Couple of questions might be from MGMAT CAT or Gmat Prep, but not more than that.

I'll provide OA in a day or two, after discussions. Tell me if you want the answers for the specific questions earlier than that and I'll mail you. _________________

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, 06:38

1

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3) I) (x+y)^2=9a x^2+y^2=9a-2xy NS II) (x-y)^2=a x^2+y^2=a+2xy NS Together 2(x^2+y^2)=10a x^2+y^2=5a If either x or y are larger than 0, the stem would be true, but if they’re both zero the stem is false, hence E

4) I don’t get the two clues; they seem to be mutually exclusive

5) I) 3|x^2-4|=y-2 either y=3x^2-10 or y=14-3x^2 NS II) |3-y|=11 either y=-8 or y=14 NS Together -8=3x^2-10 so 3x^2=2 ok 14=3x^2-10 so 3x^2=28 ok, hence E

6) I) x+1>0 so x={0, 1, 2, …} NS II) xy>0 so x and y have the same sign and none of them is zero NS Together, x={1, 2, 3, ..} and y has the same sign, hence C

7) |x+2|=|y+2| either x+2=y+2 or x+2=-y-2 (the other two combinations can be transformed into these by multiplying by -1) Reordering: x-y=0 or x+y=-4 I)xy<0, hence x and y have different signs and none of them is zero. The only possibility is x+y=-4 S II) x>2, y<2 hence x#y. The only possibility is x+y=-4 S, therefore D

8)a*b#0, hence a and b are both non-zero I) |a*b|=a*b a and b have the same sign and the stem is always true S II) |a|/|b|=|a/b| this is true regardless of the values of a and b, and nothing can be said about the stem NS, therefore A

9) I) –n=|-n| n<=0 NS II) n^2=16 n=+/-4 NS Together n=-4 therefore C

10)n#0 I) n^2>16, so |n|>4 S II) 1/|n|>n true for n<-1 NS, therefore A

11) Plugging in numbers I get B, but there’s no rime or reason to my solution

12) I) –s<=r<=s obviously NS. Since s>=-s, s is either positive or zero II)|r|>=s obviously NS Together: I) tells us that s>=0; II) tells us that r>=s or r<=-s. The only case in which I and II are simultaneously satisfied is r=s, therefore C

13) x=(0:2) with 0 and 2 excluded I) (x-1)^2<=1, hence x=[0:2] with 0 and 2 included, hence NS II) x^2-1>0 x<-1 or x>1. For x=1.5 the stem is true, for x=3 it is false, hence NS Together, for x=1.5 the stem is true, for x=2 it is false, hence E

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:07

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Expert's post

Marco83 wrote:

4) I don’t get the two clues; they seem to be mutually exclusive

Yes there was a typo in 4. Edited. Great job Marco83. Even though not every answer is correct, you definitely know how to deal with this kind of problems. _________________

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, 09:34

1

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Bunuel wrote:

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

Statement 1:

Two equations, two unknowns... INSUFFICIENT

Statement 2:

|3 - y| = 11 (3-y)=11 or (3-y)=-11 y=-8, 14

INSUFFICIENT

Statements 1 and 2:

y must be 14 because 3|x^2 -4| can never be a negative value (no matter what you plug in for x, you will get a positve value because of the absolute value signs).

SUFFICIENT

ANSWER: C.

Last edited by h2polo on 17 Nov 2009, 09:54, edited 1 time in total.

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

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