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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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27 Jun 2013, 20:01
Is x^2 + y^2 > 4a?

I ran into trouble when I instinctively got the absolute value of x^2 + y^2 > 4a ===> |x| + |y| > 4a and did the same for number one and two. We need to find the values of x, y and a to solve for this problem. if x = 10 and y = 20 and a = 2 then the inequality would hold true but if x = 0 and y = 0 and a = 0 then it wouldn't. I'm still having trouble figuring out why this problem is solved the way it is, as opposed to the way I outline below:

|x| + |y| > √4a
x + y > 4a
OR
x + -y > 4a

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

|x + y| = √9a
x + y = √9a
y - x = √9a

However, this method is taking too long so I will solve the way Bunuel did (though I still don't know how I would know to solve it this way on the test!

(1) (x + y)^2 = 9a
x^2 + 2xy + y^2 = 9a
This tells us nothing about x, y and a
INSUFFICIENT

(2) (x – y)^2 = a
x^2 - 2xy + y^2 = a
Again, this tells us nothing about x, y and a
INSUFFICIENT

1+2) Both statements can be added up

x^2 + 2xy + y^2 = 9a
+ x^2 - 2xy + y^2 = a
= 2x^2 + 2y^2 = 10a

(x^2+y^2) = 5a
Again, we know nothing of x, y and a.
INSUFFICIENT

(E)
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28 Jun 2013, 06:54
4. Are x and y both positive?

(1) 2x-2y=1

2(x - y) = 1
(x-y) = 1/2
(1-.5) = 1/2
(-.5 - [-1]) = 1/2
INSUFFICIENT

(2) x/y>1
3/1 > 1
-3/-1 > 1
INSUFFICIENT

1+2) combined, both statements have a valid answers where x and y are both negative. No other valid answers are shared.
SUFFICIENT

(C)

OR

2(x - y) = 1
x/y>1

2x - 2y = 1
x - y = 1/2
x = 1/2 + y

x/y > 1
[ (1/2) + y ] / y > 1
(1/2) / y + (y/y) > 1
(1/2) / y + 1 > 1
(1/2) / y > 0
(1/2y) > 0
y > 0

If y is greater than zero, then (x = 1/2 + y) must be greater than zero as well
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28 Jun 2013, 07:08
6. If x and y are integer, is y > 0?

(1) x +1 > 0

Tells us nothing about the value of y.
INSUFFICIENT

(2) xy > 0

x and y must both be positive or negative. We don't know if y is positive or negative.
INSUFFICIENT

x + 1 > 0
xy > 0

x + 1 must be > 0. x cannot be negative as -x + 1 wouldn't be greater than zero. Therefore, x is an integer ≥ 0. xy must be > 0 which means xy = (+)*(+) or (-)*(-) but neither x or y can equal zero. So, if x and y cannot = zero then x must be an integer > 0. If x is positive then y also must be positive for xy > 0.

SUFFICIENT
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28 Jun 2013, 13:58
7. |x+2|=|y+2| what is the value of x+y?

I. (x+2) = (y+2) ===> x-y = 0 (x = y)
OR
II. (x+2) = -y-2 ===> x + y = -4

(1) xy<0
if xy < 0 then x,y ≠ 0 and either x OR y is negative.

This is a VERY tricky problem!

To start, let's look at just |x+2|=|y+2|. #1) tells us that either x OR y is negative and that neither = 0. Of course, |x+2| must = |y+2| so we can try a few different values for x and y.

|x+2|=|y+2|
x= -8, y= 4
|-8+2| = |4+2|
|-6|=|6|
6=6 Valid

x=2, y=-6
|2+2| = |-6+2|
|4| = |-4|
4=4 Valid

If we were told that xy>0, we wouldn't be able to solve. Now, back to the stem:

(x+2) = (y+2) ===> x-y = 0 (x = y)
OR
(x+2) = -y-2 ===> x + y = -4

x and y don't = 0. In scenario 1 we have x-y=0 or (x=y) #1) says that xy<0 meaning either x or y is negative and therefore x ≠ y.

(2) x>2 y<2

|x+2|=|y+2| tells us that x+2 must equal y+2. #2.) rules out the possibility of scenario #1) which means the only other possibility (|x+2|=|y+2| is stated as fact so one of the scenarios must hold true) is that x+y=-4. This can be confirmed by plugging in x values greater than 2 and y values less than -2 to get 0x+y=-4.
SUFFICIENT

(D)
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28 Jun 2013, 16:18
8. a*b # 0. Is |a|/|b| = a/b?

a,b ≠ 0
|a|/|b| will always be positive
are a,b BOTH positive or BOTH negative?

(1) |a*b|=a*b

|a*b|=a*b

a*b MUST be positive as it is equal to an absolute value. The only way for a*b to be positive is if a and b are BOTH positive or BOTH negative.
SUFFICIENT.

(2) |a|/|b|=|a/b|
We could have positive or negative values for a and b and get correct results:

|2|/|1| = |2/1|
a=2, b=1
2/1 = 2/1
2=2
OR
|-2|/|1| = |-2/1|
a=-2, b=1
2/1 = |-2|
2=2

INSUFFICIENT

(A)
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28 Jun 2013, 16:23
9. Is n<0?

(1) -n=|-n|
if -n = |-n| then -n must be positive as it is equal to an absolute value. Therefore, -n = |-n| ===> -(-n) = |-n|
N CAN BE ≥0 AS -(-0) = |-0| ===> 0=0!
INSUFFICIENT

(2) n^2=16
|n| = 4
n=4 OR n=-4
N could be less than zero or greater than zero.
INSUFFICIENT

1+2) #1 tells us that N is ≥0. #2 tells us that n could be 4 or -4. The only value of n that is ≥0 is 4.
SUFFICIENT

(C)
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28 Jun 2013, 16:41
10. If n is not equal to 0, is |n| < 4 ?

(1) n^2 > 16
|n| > 4

n>4
OR
n<-4

If n is greater than four or less than negative four the absolute value will NOT be less than four.
SUFFICIENT

(2) 1/|n| > n
1/|n| > n
Example:

1/|-2| > -2 ===> 1/2 > -2

n is negative. However, we don't know whether it's absolute value is less than or more than four.
INSUFFICIENT

(A)
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28 Jun 2013, 17:36
11. Is |x+y|>|x-y|?

(x+y)>(x-y)
x+y>x-y
2y>0
y>0

OR

(x+y)>-(x-y)
x+y>-x+y
2x>0
x>0

(1) |x| > |y|

|x+y| will be greater than |x-y| only when x and y have the same sign (+) (+) or (-) (-)

|x| > |y| tells us nothing about the signs. For example:

|x| > |y|
|4| > |2|
4 > 2
OR
|-4| > |2|
4 > 2

INSUFFICIENT

(2) |x-y| < |x|

|x-y| < |x|
|3-1| < |3| ===> |2|< |3| VALID
|-3-1| < |-3| ===> |4| < |3| INVALID
|3-(-1)| < |3| ===> |4| < |3| INVALID
|-3-(-1)| < |-3| ===> |2| < |3| VALID

For |x-y| < |x| to hold true then x and y have to have the same signs. |x+y|>|x-y| is true only when x and y have the same signs.
SUFFICIENT

(B)
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29 Jun 2013, 10:45
12. Is r=s?

(1) -s<=r<=s

r could be positive or negative. For Example:

-s≤r≤s

I.) -4≤4≤4
II.) -4≤-1≤4

INSUFFICIENT

(2) |r|>=s

r=5, s=1
|5|≥1 Valid
r=-3, s=-5
|-3|≥-5 Valid
r=5, s=-1
|5|≥-1 Valid

There are many different ways |r|>=s holds true with different combinations of +/- r,s.
INSUFFICIENT

-s≤r≤s ===> this means that s≥0. For example, s couldn't be negative for the following reason:

-(-s)≤r≤(-s)
s≤r≤-s (not valid)

~ Obviously a positive number cannot be less than a negative number.
~ r can be equal to -s or s or it can be in between the range of -s and s.

|r|≥s

The absolute value of r may be greater than or equal to s, but r could be negative. If s was 5 r = -5 in which case:
|r|≥s
|-5|≥5
5≥5
In this case, r = -5 and s = 5, on the other hand
|5|≥5
5≥5
In this case r = 5 and s = -5

r is somewhere between - s and s or it could = s or negative s. The absolute value of r is greater than or equal to s (positive s) so let's say r = -5 and s = 5. The absolute value or r would equal s and would satisfy the inequality however, r wouldn't = s as -5 ≠ 5. On the other hand, r could =5 and s could = 5. |5|≥5 is valid and r=s ans 5=5.

INSUFFICIENT

(E)
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29 Jun 2013, 11:35
13. Is |x-1| < 1?

Is 0 < x < 2

(1) (x-1)^2 <= 1

|x-1| ≤ 1
x-1 ≤ 1 ===> x≤2
OR
-x+1 ≤ 1
x ≥ 0

0≤x≤2

The stem asks if |x-1| is less than one. #1 says it could be less than OR equal to 1. Furthermore, #1 says that 0≤x≤2 but the question asks if 0<x<2.
INSUFFICIENT

(2) x^2 - 1 > 0
x^2 > 1
|x| > 1
x>1 OR x<-1

If x = 1.5 then |x-1| < 1 = |.5| < 1 Valid
OR
if x =1 4 then |x-1| < 1 = |4| < 1 Invalid
INSUFFICIENT

1+2) let's say x =2, then: |2-1| ≤ 1 ==> |1|<1 but we're looking to see if |x-1| < 1 and in that case it is not, or x = 1.1 in which case |1.1-1| < 1 ===> .1 < 1

Another way to look at it is to examine the intersection of the ranges. From #1 the range is 0≤x≤2 and from #2 the range is x>1 or x<-1 so the combined range is 1<x≤2. In the stem x must be between 0 and 2 but 1+2 combined allows for the possibility of x = 2.
INSUFFICIENT

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03 Jul 2013, 21:02
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=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.

Option 2 would mean that |x+2 | is == x + 2
but for -2<=y<2 |y+2| is == y + 2

==> x + 2 = y + 2 ==> x = y
Only for y < -2 would |y+2| be == -y - 2
which would give
x + y = -4
i.e. INSUFFICIENT
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03 Jul 2013, 22:37
ankagl63 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=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.

Option 2 would mean that |x+2 | is == x + 2
but for -2<=y<2 |y+2| is == y + 2

==> x + 2 = y + 2 ==> x = y
Only for y < -2 would |y+2| be == -y - 2
which would give
x + y = -4
i.e. INSUFFICIENT

Notice that x>2 and y<2 means that $$x\neq{y}$$. Now, in case of $$-2\leq{x}<2$$ you get x=y which is contradicts $$x\neq{y}$$, thus $$-2\leq{x}<2$$ case is not possible (valid).

Alternative solution: inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747

Hope it helps.
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06 Jul 2013, 07:09
Hi Bunuel,

Thanks for the excellent collection of questions and good to see this thread is active for more than 3 years

Please can you help me out with the below:

For (1), I started like this:

6*x*y = x^2*y + 9*y
as the question is asking for xy, I converted the above eqn to ==> x*y = (x^2*y + 9*y)/6
and then tried finding the value of xy from the two statements, and ended up selecting (e) as my answer choice.
If I had converted the equation to (x-3)^2*Y = 0, I am sure I would have done it right.

For (5) as the question is asking for y,
I converted the equation in question to y=3x^2-10 or y=14-3x^2.
then tried finding the value of Y and chose (e) as my answer as I was getting two diff answers.
Again, if I had thought that Y>=2 as 3|x^2-4| will always be >=0 from the question stem, I would have got it right.

I am doing this mistake most often in inequalities and also in other topics . Please can you tell me what I am missing here.
what parameters should I consider before converting the equation in the question, so that it helps to find the answer easily.
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09 Jul 2013, 20:54
Can someone pls help me understand the flaw in my reasoning for qn 3 ? :

(1) It is given that $$(x+y)^2 =9a$$ or $$x^2+y^2+2xy=9a$$. Let us assume that $$x^2+y^2 < 4a$$, then from (1) it follows that $$2xy >5a$$ which means that $$2xy > x^2+y^2$$ which reduces to $$(x-y)^2 < 0$$ which is not possible , thus $$x^2+y^2 > 4a$$. Sufficient.

(2) Let us assume that $$x^2+y^2 < 4a$$ , unlike above , this gives us no further info on the relation betn $$x^2+y^2$$ and $$2xy$$. Insufficient.

Thanks.
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09 Jul 2013, 21:30
vs129 wrote:
Can someone pls help me understand the flaw in my reasoning for qn 3 ? :

(1) It is given that $$(x+y)^2 =9a$$ or $$x^2+y^2+2xy=9a$$. Let us assume that $$x^2+y^2 < 4a$$, then from (1) it follows that $$2xy >5a$$ which means that $$2xy > x^2+y^2$$ which reduces to $$(x-y)^2 < 0$$ which is not possible , thus $$x^2+y^2 > 4a$$. Sufficient.

(2) Let us assume that $$x^2+y^2 < 4a$$ , unlike above , this gives us no further info on the relation betn $$x^2+y^2$$ and $$2xy$$. Insufficient.

Thanks.

Check:
inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697
inequality-and-absolute-value-questions-from-my-collection-86939-80.html#p687991
inequality-and-absolute-value-questions-from-my-collection-86939-100.html#p746278

Hope it helps.
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10 Jul 2013, 05:50
Thanks, Bunuel. I did look at the explanations before, but missed the fact that a , x and y could be 0.
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Re: Inequality and absolute value questions from my collection [#permalink]

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10 Jul 2013, 17:38
Hi Bunuel,

Thanks for the excellent collection of questions and good to see this thread is active for more than 3 years

Please can you help me out with the below:

For (1), I started like this:

6*x*y = x^2*y + 9*y
as the question is asking for xy, I converted the above eqn to ==> x*y = (x^2*y + 9*y)/6
and then tried finding the value of xy from the two statements, and ended up selecting (e) as my answer choice.
If I had converted the equation to (x-3)^2*Y = 0, I am sure I would have done it right.

For (5) as the question is asking for y,
I converted the equation in question to y=3x^2-10 or y=14-3x^2.
then tried finding the value of Y and chose (e) as my answer as I was getting two diff answers.
Again, if I had thought that Y>=2 as 3|x^2-4| will always be >=0 from the question stem, I would have got it right.

I am doing this mistake most often in inequalities and also in other topics . Please can you tell me what I am missing here.
what parameters should I consider before converting the equation in the question, so that it helps to find the answer easily.
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Re: Inequality and absolute value questions from my collection [#permalink]

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

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18 Jul 2013, 21:25
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

Hi Bunuel,
Why wouldn't the answer be 'C'. I inferred from the question that we just need to prove X^2 + y^2 > 4a, which we deduced by adding the equations. Please clarify why we are considering the values x,y or a can hold?
Thanks,
Amit
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Re: Inequality and absolute value questions from my collection [#permalink]

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18 Jul 2013, 21:28
Amit0507 wrote:
Hi Bunuel,
Why wouldn't the answer be 'C'. I inferred from the question that we just need to prove X^2 + y^2 > 4a, which we deduced by adding the equations. Please clarify why we are considering the values x,y or a can hold?
Thanks,
Amit

inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697
inequality-and-absolute-value-questions-from-my-collection-86939-80.html#p687991
inequality-and-absolute-value-questions-from-my-collection-86939-100.html#p746278

Hope it helps.
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Re: Inequality and absolute value questions from my collection   [#permalink] 18 Jul 2013, 21:28

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