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# Inequality and absolute value questions from my collection

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Inequality and absolute value questions from my collection [#permalink]

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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) $$x^3< 0$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

2. If y is an integer and $$y = |x| + x$$, is $$y = 0$$?
(1) $$x < 0$$
(2) $$y < 1$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

3. Is $$x^2 + y^2 > 4a$$?
(1) $$(x + y)^2 = 9a$$
(2) $$(x – y)^2 = a$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

4. Are x and y both positive?
(1) $$2x-2y=1$$
(2) $$\frac{x}{y}>1$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

6. If x and y are integer, is y > 0?
(1) $$x +1 > 0$$
(2) $$xy > 0$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747

8. $$a*b \neq 0$$. Is $$\frac{|a|}{|b|}=\frac{a}{b}$$?
(1) $$|a*b|=a*b$$
(2) $$\frac{|a|}{|b|}=|\frac{a}{b}|$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

9. Is n<0?
(1) $$-n=|-n|$$
(2) $$n^2=16$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

10. If n is not equal to 0, is |n| < 4 ?
(1) $$n^2 > 16$$
(2) $$\frac{1}{|n|} > n$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

11. Is $$|x+y|>|x-y|$$?
(1) $$|x| > |y|$$
(2) $$|x-y| < |x|$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

12. Is r=s?
(1) $$-s \leq r \leq s$$
(2) $$|r| \geq s$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

13. Is $$|x-1| < 1$$?
(1) $$(x-1)^2 \leq 1$$
(2) $$x^2 - 1 > 0$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
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08 Aug 2011, 20:54
manishgeorge 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
First, devide the whole equation by y, and
6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3
(1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient
(2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient
B

There is only one answer from the equation that you formed above from the question stem.
x^2 - 6xy +9 = 0

X =3

There for statement 1 is sufficient

Answer is A

read Bunuel's and ImJun's post for this problem.

He writes that we can't divide both sides by 'y' since that would mean that we are assuming that y is not equal to 0.

The question prompt asks us the value of xy

Sol. would be:

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

6*x*y = x^2*y + 9*y
x^2*y - 6*x*y + 9*y = 0
y (x^2 - 6*x + 9) = 0
y (x - 3)^2 = 0

Thus y could be = 0 or some other no.
X = -3 and +3

(1) y – x = 3

If x = 3, then y = 6
x = -3, then y = 0

thus two solutions, xy = 18 and xy = 0

Insuff

(2) x^3< 0

if x^3 < 0, we know that x = -3. Thus y = 0

Ans: xy = 0

Stmt 2 suff

Ans B

Hope this helps.
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08 Aug 2011, 21:12
I understand now. Thanks for explaining
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15 Aug 2011, 05:09
great effort..thanks
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25 Aug 2011, 20:53
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

Ans:

From the initial equation we get: y*(x^2-6x+9)=0 i.e y=0 or (x-3)^2=0
means either y=0 or (x-3)^2=0 or both
A does not give any idea about
B states that x^3<0 which means that x is not equal to 3.hence in that case y needs to be zero. so statement 2 itself is sufficient to answer the question.
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07 Sep 2011, 20:47
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

Hi Bunuel,

Different approach and I got the answer C. Please correct if I am wrong!

6*x*y= x^2*y + 9*y

(1) y - x = 3 => y = x+3
Substitue y into stem, we have:
6*x*(x+3) = x^2*(x+3) + 9*(x+3)
6*x^2 + 18*x = x^3 + 3*x^2 + 9*x +27
x^3 - 3*x^2 - 9*x + 27 = 0
(x-3)^2*(x+3) = 0
x = 3 or x = -3 (Insufficient)

(2) x^3 <0

(1) + (2), we have x = -3, y =0, then x*y = 0

Then the answer must be C
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16 Dec 2011, 20:24
Awesome questions Bunuel...thanks a ton for sharing...
Wonderful explanations as well..
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Re: Inequality and absolute value questions from my collection [#permalink]

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24 Jan 2012, 08:01
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.
(...)

Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!
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24 Jan 2012, 12:47
metallicafan 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 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.
(...)

Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!

We have $$|x+2|=|y+2|$$.
If both absolute values expand with + or - sign we'll get: $$x+2=y+2$$ (notice that $$-(x+2)=-(y+2)$$ is exactly the same);
If they will expand with different signs we'll get: $$-(x+2)=y+2$$ (notice that $$x+2=-(y+2)$$ is exactly the same).

So only two forms.

Hope it's clear.
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10 Feb 2012, 02:40
For question 3:

I think I found a different answer. My answer is C (DATA 1 AND 2 are SUFFICIENT together).

x^2+y^2 >4a ?

From (1), (x+y)^2=9a means that x^2+y^2 = 9a-2xy

To satisfy our equation, must 5a > 2xy

From (2), x^2+y^2 = a +2xy

To satisfy our equation, must a + 2xy > 4a, therefore we must have 2xy > 3a

Hence, 3a < 2xy < 5a (aggregating both inequalities 1 and 2)

Then we know that 9a-2xy = a +2xy (from 1 and 2)

and this gives us 2a = xy, or 2xy = 4a

Conclusion: 2xy=4a satisfies our constraint, hence X^2+Y^2 > 4a

Let me know if you see a mistake here, thanks.
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10 Feb 2012, 06:14
grotily90 wrote:
For question 3:

I think I found a different answer. My answer is C (DATA 1 AND 2 are SUFFICIENT together).

x^2+y^2 >4a ?

From (1), (x+y)^2=9a means that x^2+y^2 = 9a-2xy

To satisfy our equation, must 5a > 2xy

From (2), x^2+y^2 = a +2xy

To satisfy our equation, must a + 2xy > 4a, therefore we must have 2xy > 3a

Hence, 3a < 2xy < 5a (aggregating both inequalities 1 and 2)

Then we know that 9a-2xy = a +2xy (from 1 and 2)

and this gives us 2a = xy, or 2xy = 4a

Conclusion: 2xy=4a satisfies our constraint, hence X^2+Y^2 > 4a

Let me know if you see a mistake here, thanks.

Welcome to GMAT Club. Let me assist you with this set of problems: first post on each page contains links to the detailed solutions of the questions with OA's.

OA for the 3rd question is E, not C. Please refer to this post for explanation: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

Ask if anything remains unclear.
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23 Feb 2012, 10:41
Quote:
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.

In the first statement from 2x-2y=1 --> we can sat x-y=1/2

So it cud be 8.5-8 or 0.25 - (-0.25)
HOw can we say both x and y are positive?

similarily statement 2

x/y>1
=>x>y

how can we be sure x and y have the same sign
we can have 8>7
or
8>-8

Bunel can you pls xplain..or am i missing sumtin fundamental?
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24 Feb 2012, 19:01
Thanks a lot buneuel..You are just too good +1.

Linking inequalities and co-ordinate geometry...wonderful

This is a great take away!
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25 Feb 2012, 06:51
Quote:
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=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.

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.

Answer: D.

when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

Thanks in advance.
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25 Feb 2012, 08:06
shankar245 wrote:
Quote:
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=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.

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.

Answer: D.

when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

Thanks in advance.

When we consider ranges to expand an absolute value we should put equal sign (=) in either of the range. For our question we put equal sign for the first range (case A) when we are analyzing the case when x and y are both $$\geq{-2}$$ than -2 OR both $$\leq{-2}$$. So the scenario when $$x=y=-2$$ ($$x+y=-2-2=-4$$) is included in case A.

Hope it's clear.
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28 Mar 2012, 22:53
Nice set of questions....... superlike.......
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28 Mar 2012, 23:25
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

Tricky one......
now i know its C for sure
nice explanation bunnel

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29 Mar 2012, 00:03
vdbhamare wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

Tricky one......
now i know its C for sure
nice explanation bunnel

Actually the answer is E, not C: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

The links to the OA's with solutions are given in the initial post: inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806

Hope it helps.
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29 Mar 2012, 02:56
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

1. r is between -s & s
2. r>s or r <-s

1 & 2 together to be true r=s 0r r=-s hence E
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Re: Inequality and absolute value questions from my collection [#permalink]

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31 Mar 2012, 08:34
I got

===========================
1 - A

6*x*y = x^2*y + 9*y
can be simplified to x = 3
==================================
2 - A

y = |x| + x

for y to be = 0 ; x should be either 0 or -ve
==================================
3 - E
==================================
4 - C

(A)

(x-y) = 1/2

x can be either (0 or -1/2)
y can be either (1 or 1/2)

(B)

x/y > 1

both x & y can be either -ve
or
both x & y can be either +ve

(c)

works for both

==================================

5- E

==================================

6 - C

==================================

7 - A

|x+2|=|y+2|
we have
if -(x+2) = (y+2)
then x+y = -4

if (x+2) = -(y+2)
then x+y = -4

(A) xy<0 - sufficient data

==================================

8 - A

(A) |a*b|=a*b

if (a & b ) = +ve then |a*b|=a*b = true
|a|/|b|=a/b data sufficient

if (a & b ) = -ve then |a*b|=a*b = true
|a|/|b|=a/b data sufficient

(B)

|a|/|b|=|a/b|
if(a = -ve and b = + ve) then |a|/|b|=|a/b| = true
|a|/|b|=a/b data NOT sufficient

if(a = -ve and b = - ve) then |a|/|b|=|a/b| = true
|a|/|b|=a/b data sufficient

==================================

9 - A

(A) -n = |-n|

if ( n = -ve) then -n = |-n| is true and n < 0 true - data sufficient

if ( n = +ve) then -n = |-n| is false

(B) n^2=16
n = + or - 4 not sufficient

==================================
10 - A

(A) n^2 > 16

n > 4 therefore: |n| < 4 = no (Data sufficient)

(B) 1/|n| > n

if n = -2
then 1/|-2|> -2 true --> |-2| < 4 = true

if n = -8
then 1/|-8| > -8 true --> |-8| < 4 = false

==================================

11- E

(A) |x| > |y|

if (X = 3 && Y = 2)

then |x| > |y| = true
|x+y|>|x-y| = true

if (X = -3 && Y = 2)

then |x| > |y| = true
|x+y|>|x-y| = false

if (X = -3 && Y = -2)

then |x| > |y| = true
|x+y|>|x-y| = true

(B) |x-y| < |x|

if (X = 3 && Y = 2)

then |x-y| < |x| = true
|x+y|>|x-y| = true

if (X = -3 && Y = -2)

then |x-y| < |x| = true
|x+y|>|x-y| = true

(C)

if (X = 3 && Y = 2)

then |x| > |y| = true
then |x-y| < |x| = true
|x+y|>|x-y| = true

if (X = -3 && Y = -2)

then |x| > |y| = true
then |x-y| < |x| = true
|x+y|>|x-y| = true

=================================

12 - C

(A) -s<=r<=s

if s = 3

then -3<= r <= 3

r = [-3,-2,-1,0,1,2,3]

[not sufficient]

(B) |r|>=s
if s=3

r = {3 or-3}
[not sufficient]

(C)

if s = 3 then r should be 3

================================

13 - E

(A)

(x-1)^2 <= 1

x^2 - 2x + 1 <= 1
x^2 - 2x < = 0

x(x-2)<= 0
x<=0 or x <= 2

not sufficient

(B)

x^2 - 1 > 0

(x+1)(x-1)>0

x>-1

not sufficient

(C)

2>= x > -1

not sufficient
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Re: Inequality and absolute value questions from my collection [#permalink]

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02 Apr 2012, 19:09
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as $$y=|x|+x$$ then $$y$$ is never negative. For $$x>{0}$$ then $$y=x+x=2x$$ and for $$x\leq{0}$$ then (when x is negative or zero) then $$y=-x+x=0$$.

(1) $$x<0$$ --> $$y=|x|+x=-x+x=0$$. Sufficient.

(2) $$y<1$$, as we concluded y is never negative, and we are given that $$y$$ is an integer, hence $$y=0$$. Sufficient.

Answer: D.

Answer should be A.

for evaluating 2nd Option- y<1, try with value x= 0.25, then Y would be |0.25|+0.25 = 0.5, which is less than 1. It is not mentioned that x is also an integer.

And also as you mentioned for negative value of x, it will be equal to zero. So it doesnt give definite ans whether y=0. So option (2) not sufficient.
Re: Inequality and absolute value questions from my collection   [#permalink] 02 Apr 2012, 19:09

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