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Inequality and absolute value questions from my collection

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

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New post 27 Aug 2015, 18:39
Bunuel wrote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(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.


I thought of this one graphically...

____________________-s____________s____________________________
(1) xxxxxxxxxxxxxxxxxrrrrrrrrrrrrrrrrrrrrxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
(2) rrrrrrrrrrrrrrrrrrrrrrrrrxxxxxxxxxxxxxrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr
(1)&(2) xxxxxxxxxxxxxrxxxxxxxxxxxxxrxxxxxxxxxxxxxxxxx

(1) Everything between and including \(-s\) and \(s\) INSUFF
(2) Everything outside of but still including \(-s\) and \(s\) INSUFF
(1) & (2) \(r = -s\) or \(s\)... still INSUFF
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New post 28 Aug 2015, 12:55
Bunuel wrote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(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.


Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 04 Oct 2015, 12:37
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a


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

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Answer is C (Together, they can answer the initial question).

Am I wrong, or right?
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New post 05 Oct 2015, 22:22
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.


Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.
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New post 06 Oct 2015, 05:57
Sapient 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.


Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.


Statement 2: \(y<1\)

Since we know that \(y = |x| + x\)

case 1: x>0.... In this case y = 2x and will be positive
case 2: x<0.... In this case y = 0
i.e. Y can never be Negative

This statement tells us that Y is an Integer less than 1 therefore 0 is the only possible value of y

Hence, SUFFICIENT

I hope this helps!
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 08 Oct 2015, 13:34
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.


Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

Thanks in advance!
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New post 09 Oct 2015, 21:22
petocities wrote:
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2


Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

Thanks in advance!


Hi petocities,

I think this question requires more of observation about given information

|x+2|=|y+2| will be true for two possible cases of x and y

Case 1: when x = y
Case 2: For Values like (x=1 and y=-5) or (x=2 and y=-6) or (x=3 and y=-7) ...etc.

Case 1 gives inconsistent answers because for each different value of x and y, x+ywill be different
but
Case 2 always gives a consistent value of x+y=-4 (Check all set of values mentioned above in Case 2)

Statement 1 suggests that x and y are not equal (for x and y equal, their product must be Non-Negative) i.e. Case 2 prevails which always gives us x+y=-4
i.e. SUFFICIENT

Statement 2 also rules out the scenario in which x and y may be equal i.e. Case 2 prevails again leading to a consistent value of x+y=-4
i.e. SUFFICIENT

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

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New post 18 Oct 2015, 13:18
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 Bunuel, I've solved this one correctly, but have one question. A is ok - no questions.
Can we manipulate Statement 2 and say |n|*n<1 as |n| is always positive we must be able to do this - but |n|*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.
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New post 18 Oct 2015, 13:25
BrainLab 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 Bunuel, I've solved this one correctly, but have one question. A is ok - no questions.
Can we manipulate Statement 2 and say |n|*n<1 as |n| is always positive we must be able to do this - but |n|*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.


\(\frac{1}{| n |}>n\) --> multiply by \(|n|\) (we can safely do that since |n|>0): \(n*|n| < 1\).

If \(n>0\), then we'll have \(n^2<1\) --> \(-1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\).
If \(n<0\), then we'll have \(-n^2<1\) --> \(n^2>-1\). Which is true for any n from the range we consider. So, \(n*|n| < 1\) holds true for any negative value of n.

Thus \(\frac{1}{| n |}>n\) holds true if \(n<0\) and \(0<n<1\).
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 04 Nov 2015, 19:46
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.



How did you figure out that \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.
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New post 23 Feb 2016, 00:25
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.


hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!
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New post 23 Feb 2016, 00:43
nishantdoshi 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.


hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!


An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.
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New post 23 Feb 2016, 00:53
Bunuel wrote:
nishantdoshi 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.


hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!


An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.


thanks for the reply
my understanding about this topic is that...

if x>=0 then |x|=x
and if x<0 then |x|=-x

am i wrong?
please reply!!!
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New post 14 May 2016, 00:34
In your step you have taken it as (x-3)^2 but isn't (x^2-6x+9) = (x-3)^2 or (3-x)^2?

Also, what does 'reduce the expression by y' mean?

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.

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.
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New post 14 May 2016, 01:06
ameyaprabhu wrote:
In your step you have taken it as (x-3)^2 but isn't (x^2-6x+9) = (x-3)^2 or (3-x)^2?

Also, what does 'reduce the expression by y' mean?

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.

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.


Both (x-3)^2 and (3-x)^2 are the same.
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New post 02 Sep 2016, 23:52
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.



Hi Bunuel,

Can you please explain why have you not considered (for option1) the other case. I mean, x(x-2)<=0 can lead to two possiblities, one that you have mentioned, and the other one could be just the opposite x<=0 and x>=2. I need to understand this, please reply.

Thanks.
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New post 04 Sep 2016, 04:50
rpradhan25 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.



Hi Bunuel,

Can you please explain why have you not considered (for option1) the other case. I mean, x(x-2)<=0 can lead to two possiblities, one that you have mentioned, and the other one could be just the opposite x<=0 and x>=2. I need to understand this, please reply.

Thanks.


x(x-2)<=0 is true for 0<=x<=2 and not true for any other range.

Check the links below:
Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

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

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New post 27 Dec 2016, 05:11
[quote="Bunuel"]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.

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

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here.
From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3?
That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

Let me know if I am correct.

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

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New post 27 Dec 2016, 06:52
naren01 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.

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

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here.
From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3?
That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

Let me know if I am correct.

Thanks!


No, you are not correct.

From \(y*(x-3)^2=0\) it follows that either \(x=3\) or/and \(y=0\). (2) says that \(x^3<0\), thus x is not 3, therefore y must be 0 --> xy = 0.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 30 Dec 2016, 07:50
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 Bunuel, thank you so much for such an amazing post, so so so helpful.

A quick query, regarding statement 2 here:
(2) \(\frac{1}{|n|} > n\), Shouldn't it be true for all values of n such that n<1 (n#0) ?

Eg: n =1/2: \(\frac{1}{|(1/2)|} > \frac{1}{2}\) : \(2 > \frac{1}{2}\)

PS: It doesn't alter the final answer though.
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Re: Inequality and absolute value questions from my collection   [#permalink] 30 Dec 2016, 07:50

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