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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] New post 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) 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#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|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) 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<=r<=s
(2) |r|>=s

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

13. Is |x-1| < 1?
(1) (x-1)^2 <= 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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Re: Inequality and absolute value questions from my collection [#permalink] New post 29 Mar 2012, 00:03
Expert's post
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

8-)


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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Inequality and absolute value questions from my collection [#permalink] New post 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] New post 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] New post 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.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 02 Apr 2012, 23:37
Expert's post
roshan1985 wrote:
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.


Welcome to GMAT Club.

The red part in your reasoning is not correct: y cannot be 0.25 since given that y is an integer. So, answer D is correct.

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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Re: Inequality and absolute value questions from my collection [#permalink] New post 03 Apr 2012, 06:47
h2polo wrote:
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?

INSUFFICIENT

ANSWER: A.

solved the same. a doubt!
when this is a DS , cant we take statement 1 with 2 variables 'x' and 'y' will give the answer for xy?
but the statement 2 wont.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 03 Apr 2012, 09:08
Expert's post
kashishh wrote:
h2polo wrote:
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?

INSUFFICIENT

ANSWER: A.

solved the same. a doubt!
when this is a DS , cant we take statement 1 with 2 variables 'x' and 'y' will give the answer for xy?
but the statement 2 wont.


Answer to this question is B, not A.

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
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Inequality and absolute value questions from my collection [#permalink] New post 20 Jun 2012, 17:43
sriharimurthy wrote:
Quote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s


St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

Answer : C



hi sriharimurthi,

If we have the statements -s <= r < = s and -s >= r >= s, there could be 2 possible answers; r=s or r=-s. Both the possible answers satisfy the 2 equations/inequalities. So isn't the answer E and not C?

Am I missing something. Could you please help?
Thanks v much.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 21 Jun 2012, 00:13
Expert's post
shivamayam wrote:
sriharimurthy wrote:
Quote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s


St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

Answer : C



hi sriharimurthi,

If we have the statements -s <= r < = s and -s >= r >= s, there could be 2 possible answers; r=s or r=-s. Both the possible answers satisfy the 2 equations/inequalities. So isn't the answer E and not C?

Am I missing something. Could you please help?
Thanks v much.


Sriharimurthi solution is not correct. OA for this question is E, not C. The links to the OA's and solutions are given in the initial post (if the links does not work, then switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p652806 will lead you to the solutions).

As for this question:
12. Is r=s?

(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;
B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.
But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well.
Consider: s=5, r=5 --> -5<=5<=5 |5|>=5
s=5, r=-5 --> -5<=-5<=5 |-5|>=5
Both statements are true with these values. Hence insufficient.

Answer: E.

Hope it helps.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Inequality and absolute value questions from my collection [#permalink] New post 21 Jun 2012, 01:21
Got it. Thanks Bunuel.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 24 Jun 2012, 12:45
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.



I solved it this way. I hope i am assuming it correct

is x=y= +ve = ?

(A) 2x-2y=1

y=x-1/2 equation of line,

So, y=0 then x=1/2
x=0 then y=-1/2

thus, the line passes through points (1/2,0) & (0,-1/2) i.e - I, III & IV quadrant

(+,+) or (-,-) or (+,-) values of x & y

Not sufficient infromation

(B)

x/y > 1
if y>0 x>y
if y<0 x<y
not sufficient infomration

(C)


combine (A) and (B)

if y>0 i.e line passes only through Ist quadrant & x>y because y= x - 1/2

x>y ==> therefore x=y=+ve satisfies only first quadrant


condition y<0 from (B) is not true for x<y because

when line pass through IV the quadrant x is +ve and y is -ve, so we can ignore this condition as it doesnot satisfy (A) and (B) combined

Therefore left with only y>0 and x>y where line passes thr' Ist quadrant: C
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Re: Inequality and absolute value questions from my collection [#permalink] New post 24 Jun 2012, 17:41
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.

Answer: D.



Vote for D [Need help in understanding part (B) below]

I solved it this way. I really find it difficult to open mod sign, I hope i am doing it correctly

(A) xy < 0

So we have (x = +ve and y = -ve) or (x = -ve and y = +ve)

|x + 2| = |y + 2|

critical point (x > -2 & y < -2) or (x < -2 & y > -2)

if (x > -2 & y < -2)
then

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

sufficient

(B) x > 2 and y < 2

when y < -2
(x + 2) = - (y + 2)
x + y = -4

when 2 > y > -2 we get x - y = 0 - Cannot be true statement [Not sure if this is correct. Need help here]

and considered this as sufficient
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Re: Inequality and absolute value questions from my collection [#permalink] New post 24 Jun 2012, 18:48
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.



Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???

Here is what i did

Question: |x-1| < 1
critical point x>1 or x<1

when x>1 then (x-1)<1 x<2
when x<1 then -(x-1)<1 -x<0 therefore x>0
to prove 0<x<2 --? [understood this]

(A)(x-1)^2 <= 1
x^2 - 2x + 1 <= 1
x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]

my thoughts [what m i doing wrong??]
when (x = 0) then x-2<=0 therefoe x<=2
when (x-2 = 0) then x<=0 ????? i am confused here ?????


(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]

Thank you
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Re: Inequality and absolute value questions from my collection [#permalink] New post 25 Jun 2012, 01:03
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kuttingchai wrote:
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.



Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???

Here is what i did

Question: |x-1| < 1
critical point x>1 or x<1

when x>1 then (x-1)<1 x<2
when x<1 then -(x-1)<1 -x<0 therefore x>0
to prove 0<x<2 --? [understood this]

(A)(x-1)^2 <= 1
x^2 - 2x + 1 <= 1
x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]

my thoughts [what m i doing wrong??]
when (x = 0) then x-2<=0 therefoe x<=2
when (x-2 = 0) then x<=0 ????? i am confused here ?????


(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]

Thank you


Check the following links:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 25 Jun 2012, 06:33
Bunuel wrote:
kuttingchai wrote:
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.



Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???

Here is what i did

Question: |x-1| < 1
critical point x>1 or x<1

when x>1 then (x-1)<1 x<2
when x<1 then -(x-1)<1 -x<0 therefore x>0
to prove 0<x<2 --? [understood this]

(A)(x-1)^2 <= 1
x^2 - 2x + 1 <= 1
x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]

my thoughts [what m i doing wrong??]
when (x = 0) then x-2<=0 therefoe x<=2
when (x-2 = 0) then x<=0 ????? i am confused here ?????


(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]

Thank you


Check the following links:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535

Hope it helps.


Thank you Bunuel for the link below that helped : if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html

So for x (x-2) <= 0
first check for point of intersection on x-axis

x=0 and x=2
then as equation is <= 0 points between (0,2) are below x axis and so we have 0<=x<=2
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Re: Inequality and absolute value questions from my collection [#permalink] New post 09 Aug 2012, 16:06
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.

Answer: D.


Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below -
I used number picking.

A. xy<0,
x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value
or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks
jay
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Re: Inequality and absolute value questions from my collection [#permalink] New post 09 Aug 2012, 16:21
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jayaddula 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.

Answer: D.


Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below -
I used number picking.

A. xy<0,
x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value
or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks
jay


In your example, both pairs give the same value for x+y: 2-6=-4 and 3-7=-4.

We can solve this question in another way:

7. |x+2|=|y+2| what is the value of x+y?

Square both sides: x^2+4x+4=y^2+4y+4 --> x^2-y^2+4x-4y=0 --> (x+y)(x-y)+4(x-y)=0 --> (x-y)(x+y+4)=0 --> either x=y or x+y=-4.

(1) xy<0 --> the first case is not possible, since if x=y, then xy=x^2\geq{0}, not <0 as given in this statement, hence we have the second case: x+y=-4. Sufficient.

(2) x>2 and y<2. This statement implies that x\neq{y}, therefore x+y=-4. Sufficient.

Answer: D.

Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 10 Aug 2012, 07:26
Bunuel wrote:
jayaddula 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.

Answer: D.


Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below -
I used number picking.

A. xy<0,
x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value
or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks
jay


In your example, both pairs give the same value for x+y: 2-6=-4 and 3-7=-4.

We can solve this question in another way:

7. |x+2|=|y+2| what is the value of x+y?

Square both sides: x^2+4x+4=y^2+4y+4 --> x^2-y^2+4x-4y=0 --> (x+y)(x-y)+4(x-y)=0 --> (x-y)(x+y+4)=0 --> either x=y or x+y=-4.

(1) xy<0 --> the first case is not possible, since if x=y, then xy=x^2\geq{0}, not <0 as given in this statement, hence we have the second case: x+y=-4. Sufficient.

(2) x>2 and y<2. This statement implies that x\neq{y}, therefore x+y=-4. Sufficient.

Answer: D.

Hope it's clear.



Thanks Bunuel.
Its a shame that i did all the work and completely misread the question.

thanks
jay
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Re: Inequality and absolute value questions from my collection [#permalink] New post 25 Aug 2012, 06:26
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Bunuel wrote:
SOLUTIONS:

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 let's simplify given expression 6*x*y = x^2*y + 9*y:

y*(x^2-6x+9)=0 --> y*(x-3)^2=0. Note here that we CAN NOT reduce this expression by y, as some of you did. Remember we are asked to determine the value of xy, and when reducing by y you are assuming that y doesn't equal to 0. We don't know that.

Next: we can conclude that either x=3 or/and y=0. Which means that xy equals to 0, when y=0 and x any value (including 3), OR xy=3*y when y is not equal to zero, and x=3.

(1) y-x=3. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case xy=18. But if y=0 then x=-3 and xy=0. Two possible scenarios. Not sufficient.

OR:

y-x=3 --> x=y-3 --> y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0 --> either y=0 or y=6 --> if y=0, then x=-3 and xy=0 or if y=6, then x=3 and xy=18. Two different answers. Not sufficient.

(2) x^3<0. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.



The answer to this one is C right? B alone is not sufficient.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 26 Aug 2012, 00:23
dvinoth86 wrote:
Bunuel wrote:
SOLUTIONS:

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 let's simplify given expression 6*x*y = x^2*y + 9*y:

y*(x^2-6x+9)=0 --> y*(x-3)^2=0. Note here that we CAN NOT reduce this expression by y, as some of you did. Remember we are asked to determine the value of xy, and when reducing by y you are assuming that y doesn't equal to 0. We don't know that.

Next: we can conclude that either x=3 or/and y=0. Which means that xy equals to 0, when y=0 and x any value (including 3), OR xy=3*y when y is not equal to zero, and x=3.

(1) y-x=3. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case xy=18. But if y=0 then x=-3 and xy=0. Two possible scenarios. Not sufficient.

OR:

y-x=3 --> x=y-3 --> y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0 --> either y=0 or y=6 --> if y=0, then x=-3 and xy=0 or if y=6, then x=3 and xy=18. Two different answers. Not sufficient.

(2) x^3<0. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.



The answer to this one is C right? B alone is not sufficient.


Don't forget that y*(x-3)^2=0.
B alone is definitely sufficient.
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Re: Inequality and absolute value questions from my collection   [#permalink] 26 Aug 2012, 00:23
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