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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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New post 16 Nov 2009, 11:33
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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: http://gmatclub.com/forum/inequality-an ... ml#p653690

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653695

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653697

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653709
Graphic approach: https://gmatclub.com/forum/inequality-a ... l#p1269802

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653731

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653740

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653783 AND https://gmatclub.com/forum/inequality-a ... l#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: http://gmatclub.com/forum/inequality-an ... ml#p653789

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653792

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653796

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653853

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653870

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.


PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 05 Apr 2018, 12:24
Expert's post
Aditi10 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.


Hi Bunuel, I solved it as below
(1) 2x-2y=1 -> x-y= 1/2. This means x>y by 1/2 but x can be 1/2 and y=0 so this is insufficient
(2) x/y=1 -> x>y which is no different from (1) so insufficient again

But I didn't follow your approach on combining the 2 statements, how did you get x to be substituted by 1/y? The fact that x is positive is already proved by (1) isn't it?
Please help out.


The red parts are not correct.

\(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So from (2) \(\frac{x}{y}>1\), we can only deduce that \(x\) and \(y\) have the same sigh (either both positive or both negative).

Also, from (1) we can only say that x > y (because x = y + 1/2 = y + positive number) but we cannot say that x is positive. For example, consider x = -2 and y = -2.5.

Finally, when we substitute \(x = y + \frac{1}{2}\) into \(\frac{x}{y}>1\) we'll get:

\(\frac{y + \frac{1}{2}}{y}>1\);

\(1 + \frac{1}{2y}>1\);

\(\frac{1}{2y}>0\);

\(y > 0\).

If \(y > 0\), then \(x > y > 0\).

Hope it helps.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 17 Apr 2018, 09:29
Hi
This is really an interesting question,
I want to share an alternative approach.
Concept |x-a|=b, it means distance of x from a = b
it can be logically deducted that , if |x-a|=|y-a|, it means either x= y or the a and y are equidistant from a, hence average of x & y = a , or x+y = 2a
it can be proved but if visualization can save time.

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

|x+2|=|y+2|, it means either x and y are equal or are equidistant from -2.

(1) xy<0
it means one is positive and one is negative, hence they are not equal. so x and y are equidistant from -2, hence sum of x and y = -4. SUFFICIENT.

(2) x>2 y<2
x not equal to y, hence their sum is -4. SUFFICIENT.

Hence Answer D

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.

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

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New post 17 Apr 2018, 23:32
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 cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


Still not clear on this one.
Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient?
Thanks in Advance
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 18 Apr 2018, 01:15
Is |x-1| <1?
Since Absolute value function is always non negative, we can square both sides,

We get is (x-1)^2<1?

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

If (x-1)^2<1: answer to question is yes

(x-1)^2= 1: hence answer is No,

So Statement 1 is NOT SUFFICIENT.



Statement2: Question stem (x-1)^2<1?

Can be reduced to is x(x-2)<0

Or is 0<x<2

Now St2: x^2-1>0 gives x >1 or x <-1

Now x can be -2, answer to question stem is No

Or 1.5 answer to question stem is yes.

Hence NOT SUFFICIENT.



Combining both Statement 1 & 2, we get

X can be 2, answer to question stem- NO

Or 1.5 answer to question stem is yes.



Hence Answer E


Buttercup3 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 cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


Still not clear on this one.
Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient?
Thanks in Advance

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Re: Inequality and absolute value questions from my collection   [#permalink] 18 Apr 2018, 01:15
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