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Inequality and absolute value questions from my collection

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Bunuel
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Inequality and absolute value questions from my collection [#permalink] 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: https://gmatclub.com/forum/inequality-a ... ml#p653690

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

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

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

Solution: https://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: https://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: https://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: https://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: https://gmatclub.com/forum/inequality-an ... ml#p653789

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

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

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

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

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

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

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

Solution: https://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] New post   20 May 2014, 02:57
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JusTLucK04 wrote:
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.


Is this also a right way to solve this:
x^2+2xy+y^2=9a --> If we have to find x^2 + y^2 > 4a we should know whether 2xy>5a or 2xy<5a...No info on this.Hence Insuff
x^2-2xy+y^2=a---> If we have to find x^2 + y^2 > 4a we should know whether 2xy>3a or 2xy<3a...No info..Hence Insuff

Combining both..we still dont know as both the cases are possible and hence E..

Bunuel Please Confirm..


Yes, the question from (1) becomes: is 2xy < 5a and from (2) it becomes: is 2xy > 3a. So, when we combine the question becomes is 3a < 2xy < 5a. It's unclear how you deduced that we cannot answer that question without plugging values.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   22 May 2014, 18:55
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 Bunnel,

I don't quite get the part on "substitute x". Does it mean \(\frac{1}{y}(x-y)>0\) and so (x-y) is zero?

Thanks in advance!
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Re: Inequality and absolute value questions from my collection [#permalink] New post   23 May 2014, 01:50
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pretzel 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 Bunnel,

I don't quite get the part on "substitute x". Does it mean \(\frac{1}{y}(x-y)>0\) and so (x-y) is zero?

Thanks in advance!


No, we are substituting \(x=y+\frac{1}{2}\) into \(\frac{x-y}{y}>0\).

From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).

When we consider two statement together:

From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.

OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.

Graphic approach for this question: inequality-and-absolute-value-questions-from-my-collection-86939-260.html#p1269802

Hope this helps.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   23 Oct 2014, 01:21
Bunuel wrote:
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.




hi
m a little confused here. How is A sufficient

isn't n^2>16 ----> n^2-16>0 ---> (N^2-4^2) >0 ----> n+4>0 or n-4>0 ----> n> -4 or n>4.....how did u get n<-4....

kindly correct me if m wrong
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Re: Inequality and absolute value questions from my collection [#permalink] New post   23 Oct 2014, 01:23
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sugand wrote:
Bunuel wrote:
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.




hi
m a little confused here. How is A sufficient

isn't n^2>16 ----> n^2-16>0 ---> (N^2-4^2) >0 ----> n+4>0 or n-4>0 ----> n> -4 or n>4.....how did u get n<-4....

kindly correct me if m wrong


n^2 > 16 means |n| > 4, which means that n > 4 or n < -4.

Notice that n > -4 or n > 4 does not make any sense. What are the possible value of n in this case?
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Re: Inequality and absolute value questions from my collection [#permalink] New post   09 Feb 2015, 08:14
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

Answer: C.


Great question.
I looked at S1 and as usual thought of special numbers, but then proceeding with -0=|-0| did not make sense to me because I thought "0 is neither +ve nor -ve so why negate it and then find absolute value". Is there some logic i'm missing or is it just a 'possible calculation'?
Thank you
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Re: Inequality and absolute value questions from my collection [#permalink] New post   09 Feb 2015, 08:17
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deeuk wrote:
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

Answer: C.


Great question.
But -0=|-0| did not make sense to me because I thought "0 is neither +ve nor -ve so why negate it and then find absolute value". I did not know that was a possible direction. Is there some logic i'm missing or is it just a 'possible calculation'?
Thank you


Yes, 0 is neither negative nor positive but there is nothing wrong in writing -0, or |0|, because -0=0 and |0| = 0.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   09 Feb 2015, 10:15
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.


About S1: can i say |x-1|<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   10 Feb 2015, 06:00
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deeuk 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.


About S1: can i say |x-1|<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me.


Yes, that;s correct. We CAN take the square root from both sides of (x-1)^2 <= 1 because both sides are non-negative.

For more on inequalities check here: inequalities-tips-and-hints-175001.html

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   29 May 2015, 09:51
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.


Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 May 2015, 04:01
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sinhap07 wrote:
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.


Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?


First of all |-n| = |n|, so we are given that -n = |n|, which implies that n <= 0.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 May 2015, 04:15
Bunuel wrote:
sinhap07 wrote:
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.


Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?


First of all |-n| = |n|, so we are given that -n = |n|, which implies that n <= 0.


How did we arrive at |-n| = |n|?
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 May 2015, 04:17
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sinhap07 wrote:
How did we arrive at |-n| = |n|?


It's true for ANY x that |-x| = |x|: |-1| = |1|.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   01 Jun 2015, 01:35
lagomez 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

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.

(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.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or 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.


would you think this is a 700+ question?


Hi Bunuel,

Can you please comment if my approach is correct?

Given: 6*x*y = x^2*y + 9*y, which is reduced to y*(x-3)^2 = 0.......(1)

Statement [1] y-x=3 or y = x+3
sub'ing y in (1)
(x+3)(x-3)(x-3) = 0
(x^2-9)(x-3)=0
so, x^2 = 9 or x = +/-3
and x=3
therefore, y = 6 or 0 for x = 3 and -3 respectively..........Not sufficient

Statement [2]....says x < 0, so (x-3)^2 will never be = 0, y has to be it.....therefore xy=0 so sufficient!

So my question to you is whether I handled [1] properly? and whether substituting y into the original problem equation and arriving at x = +/- 3 is correct?
So far, I've learnt that Sqrt(x) will only have +x as solution as GMAT doesn;t consider -ve solution, but x^2 =25 will have +/-5 as solution so we should consider both solutions for x^2 = 25 in our analysis.....Did I get this right?
Please opine.
Thanks!
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Re: Inequality and absolute value questions from my collection [#permalink] New post   01 Jun 2015, 01:59
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rohitd80 wrote:
lagomez 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

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.

(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.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or 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.


would you think this is a 700+ question?


Hi Bunuel,

Can you please comment if my approach is correct?

Given: 6*x*y = x^2*y + 9*y, which is reduced to y*(x-3)^2 = 0.......(1)

Statement [1] y-x=3 or y = x+3
sub'ing y in (1)
(x+3)(x-3)(x-3) = 0
(x^2-9)(x-3)=0
so, x^2 = 9 or x = +/-3
and x=3
therefore, y = 6 or 0 for x = 3 and -3 respectively..........Not sufficient

Statement [2]....says x < 0, so (x-3)^2 will never be = 0, y has to be it.....therefore xy=0 so sufficient!

So my question to you is whether I handled [1] properly? and whether substituting y into the original problem equation and arriving at x = +/- 3 is correct?
So far, I've learnt that Sqrt(x) will only have +x as solution as GMAT doesn;t consider -ve solution, but x^2 =25 will have +/-5 as solution so we should consider both solutions for x^2 = 25 in our analysis.....Did I get this right?
Please opine.
Thanks!


Yes, everything tis correct.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root.

That is:
\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   02 Jun 2015, 03:20
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camlan1990 wrote:
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder
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Re: Inequality and absolute value questions from my collection [#permalink] New post   02 Jun 2015, 04:14
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camlan1990 wrote:
EgmatQuantExpert wrote:
camlan1990 wrote:
Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder


Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)


Oops camlan1990, that was my bad! I didn't notice the '<' sign in the "<=" :-D

Now I do see how you got to x^2+y^2 >= 4.5a

But please note that x = 0, y = 0 and a = 0 is one set of values that satisfies this inequality. For these values of x, y and a, the answer to the question 'Is x^2 + y^2 > 4a' is NO

For all other values of x, y and a, the answer will be YES.

Since we are not able to rule out x, y, a = 0, we cannot infer a unique answer to the posed question using St. 1 alone. So, St. 1 is not sufficient.

Hope this helped.

Japinder
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Re: Inequality and absolute value questions from my collection [#permalink] New post   03 Jun 2015, 03:17
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adityanukala wrote:
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.


Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

Thanks.


The question asks whether x^2 + y^2 > 4a. If x = y = a = 0, then the answer is NO but if this is not so then the answer is YES. Please read the whole thread. This was discussed many, many, many times before.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   03 Jun 2015, 07:34
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NavenRk 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 tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance


x = -1 and y = -1.5 does not satisfy x/y > 1.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   05 Jun 2015, 19:38
Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

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