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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 01 Nov 2010, 00:55
Dear Bunuel,

I don't quite understand your explanation on question 4, could you please explain to me again?.
When I solved both statement 1 and 2, the answer can be + or -, therefore, how the answer comes up with only 1 sign?.

Thanks
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New post 01 Nov 2010, 01:04
sumitra wrote:
Dear Bunuel,

I don't quite understand your explanation on question 4, could you please explain to me again?.
When I solved both statement 1 and 2, the answer can be + or -, therefore, how the answer comes up with only 1 sign?.

Thanks
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Welcome to GMATClub !


4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1


(1) 2(x-y)=1 OR (x-y)=0.5
Does not need both to be positive or not, EG. x=5, y=4.5 ... x=-4.5, y=-5

(2) x/y>1
Again not sufficient. EG. x=5, y=2 ... x=-5,y=-2

(1+2) x/y>1 does mean that the sign of x & y has to be the same since their ratio is greater than 0
Now (x-y)=0.5 is also true which means x is greater than y
Either x&y are both negative or both positive
If x and y are both negative, and we know x is greater than y, then (x/y) will be less than 1 (x=-3, y=-4 .. x/y=3/4<1) which contradicts statement 1
Hence x & y must both be positive
Sufficient

Answer is (C) : Both statements together but neither alone is sufficient
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New post 01 Nov 2010, 07:46
sumitra wrote:
Dear Bunuel,

I don't quite understand your explanation on question 4, could you please explain to me again?.
When I solved both statement 1 and 2, the answer can be + or -, therefore, how the answer comes up with only 1 sign?.

Thanks
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Hi, and welcome to Gmat Club.

I guess you understand why each statement alone is not sufficient. As for (1)+(2):

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.

Hope it helps.
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New post 19 Dec 2010, 03:12
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

I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient.

Consider putting the values as 1) x=1 y=1/2 ,now x>y and x-y=1/2.In this case both x and y are positive
Now2) x=-1 and y=-1/2,now again x>y and x-y=-1/2.In this case both are negative.

Hence insufficient to deduce whther x and y are both positive.

Please share your inputs on the same.
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New post 19 Dec 2010, 03:27
Eshika 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

I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient.

Consider putting the values as 1) x=1 y=1/2 ,now x>y and x-y=1/2.In this case both x and y are positive
Now2) x=-1 and y=-1/2,now again x>y and x-y=-1/2.In this case both are negative.

Hence insufficient to deduce whther x and y are both positive.

Please share your inputs on the same.


OA for this question is C, not E.

The red part in your reasoning is not correct: if x=-1 and y=-1/2 then y>x.

Here is the logic for C:

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

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.

Does it make sense now?
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New post 24 Jan 2012, 09:01
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=-2.
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.
(...)


Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!
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New post 24 Jan 2012, 13:47
metallicafan 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=-2.
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.
(...)


Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!


We have \(|x+2|=|y+2|\).
If both absolute values expand with + or - sign we'll get: \(x+2=y+2\) (notice that \(-(x+2)=-(y+2)\) is exactly the same);
If they will expand with different signs we'll get: \(-(x+2)=y+2\) (notice that \(x+2=-(y+2)\) is exactly the same).

So only two forms.

Hope it's clear.
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New post 23 Feb 2012, 11:41
Quote:
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.




In the first statement from 2x-2y=1 --> we can sat x-y=1/2

So it cud be 8.5-8 or 0.25 - (-0.25)
HOw can we say both x and y are positive?


similarily statement 2

x/y>1
=>x>y

how can we be sure x and y have the same sign
we can have 8>7
or
8>-8

Bunel can you pls xplain..or am i missing sumtin fundamental?
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New post 25 Feb 2012, 07:51
Quote:
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.



when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

Thanks in advance.
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New post 25 Feb 2012, 09:06
shankar245 wrote:
Quote:
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.



when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

Thanks in advance.


When we consider ranges to expand an absolute value we should put equal sign (=) in either of the range. For our question we put equal sign for the first range (case A) when we are analyzing the case when x and y are both \(\geq{-2}\) than -2 OR both \(\leq{-2}\). So the scenario when \(x=y=-2\) (\(x+y=-2-2=-4\)) is included in case A.

Hope it's clear.
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New post 09 Aug 2012, 17:06
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
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New post 26 Aug 2012, 01:23
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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New post 03 Oct 2012, 05:39
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!
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New post 03 Oct 2012, 07:13
liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.
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New post 03 Oct 2012, 12:26
Quote:
liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.


Great.. I get it now.. Thanks Bunuel.

I am also stuck at Q10 :

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

We need to see if |n|<4 (this means -4<n<4)

1) n^2>16 => n<-4 and n>4
So from n<-4, |n|<|-4| = |n|<4 (works)
But n>4 does not work.
Doesn't that make 1) Insufficient?

Could you please tell me what I am doing wrong here ??
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New post 04 Oct 2012, 03:39
liarish wrote:
Quote:
liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.


Great.. I get it now.. Thanks Bunuel.

I am also stuck at Q10 :

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

We need to see if |n|<4 (this means -4<n<4)

1) n^2>16 => n<-4 and n>4
So from n<-4, |n|<|-4| = |n|<4 (works)
But n>4 does not work.
Doesn't that make 1) Insufficient?

Could you please tell me what I am doing wrong here ??


If n<-4, then n, for example can be -4.5 --> |-4.5|=4.5>4, so |n|<4 doesn't hold true.

If n is not equal to 0, is |n| < 4 ?

Question basically asks whether \(-4<n<4\), so whether \(n\) is some number from this range.

(1) n^2>16. This implies that either \(n>4\) or \(n<-4\). No number from these ranges is between -4 and 4, thus the answer to the question whether \(-4<n<4\) is NO. Since we have a definite answer then this statement is sufficient.

Hope it's clear.
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New post Updated on: 04 Oct 2012, 04:03

Originally posted by carcass on 04 Oct 2012, 03:53.
Last edited by carcass on 04 Oct 2012, 04:03, edited 2 times in total.
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New post 04 Oct 2012, 04:13
Correct me if I'm wrong because the concepts are always the same but the gmat blonds them and as consequence blow your mind.

in other words, you are saying

1/|n| > n 2 cases

1/-n > n ----> n^2 > -1 this implies that any squared number is positive and therefore greater than -1, all negative n values work as solutions. your n < 0

1/n > n -----> 1 > n^2 ----> n^ 2 < 1 --------> -1 < n < 1 . your second range. so in the end we have all those information and we are not sure of course of - 4 < n < 4.

Correct ???

Thanks
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New post 23 Dec 2012, 04:40
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,

Thanks for the explanation to the above Q.

Regarding st 1 i.e X less than zero then [m]y=|x|+x = -x+x=0,

1. we know any value in modulus is positive then ideally the above should be interpreted as [m]y=|x|+x--> [m]y=x-x=0.
2.Also if from St 1 if we x<0 then [m]y=|x|+x= -x-x=-2x

3. Where as we also know that |x|= -x for X<0 and |x|= x for X>/ 0

So can you please tell me where am I going wrong with the concept.

Thanks
Mridul
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New post 23 Dec 2012, 05:29
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.



Hello Bunuel,

I got A as the answer to the Q.

From St1, we have -n=|-n|---> -n=n (As Mod value is +ve)---> we have 2n=0 or -2n=0. In both case we can say that n=0 and hence Ans should be A.

From your explanation, it is very clear that either n<0 or n=0. Could you tell me what was your approach to this Question. I mean did you assume values of
1. n as less than zero,
2. ngreater than zero and
3. n equal to zero

and check under which condition the St1 holds true.

If so, would this be a standard way of doing a modulus Question because clearly I just considered only 1 of the above conditions here.

Your inputs please

Thanks
Mridul
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