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# Inequality and absolute value questions from my collection

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23 Dec 2012, 05:37
mridulparashar1 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.

Hi Bunuel,

Thanks for the explanation to the above Q.

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

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

Absolute value properties:

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$;

So, if $$x<0$$, then $$|x|=-x$$ and $$y=|x|+x=-x+x=0$$.

For more check here: math-absolute-value-modulus-86462.html
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23 Dec 2012, 05:40
mridulparashar1 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.

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.

First of all: $$|-n|=|n|$$, so $$-n=|-n|$$ is the same as $$-n=|n|$$, which means that $$n\leq{0}$$.
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21 Feb 2013, 21:23
Bunuel wrote:
5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8
y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

Bunuel, I think I need some conceptual help. Why should we not solve statement 1 by rewriting the two statements and then adding them together? (Besides the fact that it's time consuming....) I rewrote them and found 3x^2 -10 = y for the positive absolute vlaue, and -3x^2+14=y for the negative abs value. From this, I added them together and got y=4..

Can you please explain what I'm getting wrong conceptually? Thanks so much!!!! I appreciate your kindness.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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24 Apr 2013, 05:04
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.

hi Bunuel,

Thank you very much for all the explanations. I have a query on this one

Combining both we get x^2+y^2=5a or x,y,a = 0

aren't those sufficient to answer the question is x^2+y^2>4a

Is the first case where x^2+y^2=5a, the answer is yes

Second case where x,y,a=0, the answer is no

Kindly do elaborate. Thanks
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24 Apr 2013, 05:26
Transcendentalist 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.

hi Bunuel,

Thank you very much for all the explanations. I have a query on this one

Combining both we get x^2+y^2=5a or x,y,a = 0

aren't those sufficient to answer the question is x^2+y^2>4a

Is the first case where x^2+y^2=5a, the answer is yes

Second case where x,y,a=0, the answer is no

Kindly do elaborate. Thanks

First of all when we combine we get that x^2+y^2=5a. If $$xya\neq{0}$$, then the answer is YES but if $$xya={0}$$, then the answer is NO.

Next, it's a YES/NO DS question. In a Yes/No Data Sufficiency question, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

Hope it's clear.
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21 Sep 2013, 22:21
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.

Bunuel , Can you please show how we can reach to C using graphical approach ?
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21 Sep 2013, 23:30
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.

Hi Bunuel ,

I know saying (1/|n|) < n will be true for all n<0 is quite clear logically. Still I want to reach this conclusion mathematically.

I got swayed solving for n|n| < 1 .
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22 Sep 2013, 05:27
StormedBrain 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.

Bunuel , Can you please show how we can reach to C using graphical approach ?

4. Are x and y both positive?

The question asks whether point (x, y) is in the first quadrant.

(1) 2x-2y=1 --> draw line y=x-1/2:
Attachment:

graph.png [ 6.13 KiB | Viewed 2367 times ]
Not sufficient.

(2) x/y>1 --> Draf line x/y=1. The solutions is the green region:
Attachment:

graph (2).png [ 5.95 KiB | Viewed 2346 times ]
Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Hope it helps.
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13 Oct 2013, 23:13
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.

Also insufficient as x,y, and a could be 0
Why did you assume that ALL COULD be zero.
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14 Oct 2013, 00:02
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.

Also insufficient as x,y, and a could be 0
Why did you assume that ALL COULD be zero.

Because they could be 0, why not? And IF x=y=a=0, then the answer would be NO but if they take some other values then the answer would be YES.
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19 Feb 2014, 05:00
I get super confused in solving the below kind of inequalities.
Can some expert help?

Are the below working's correct?

1/| n | >n
• Case 1: n > 0… n 2 < 1… - 1 < n < 1
• Case 2: n < 0… - n 2 < 1… n 2 > -1…

x/| x | < x
• Case 1: x > 0… x^2 > x… as x > 0, divide by x without changing signs… x > 1
• Case 2: x < 0… - x^2 > x… as x < 0, divide and only flip sign… - x < 1… x > -1
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Re: Inequality and absolute value questions from my collection  [#permalink]

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21 Feb 2014, 09:00
1
Bunuel wrote:
11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2
OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Hi,

I didnt quite understand this part "(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin."

what in the second equation tells us that they both have to be the same sign?

Thanks
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21 Feb 2014, 09:37
anishasjkaul wrote:
Bunuel wrote:
11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2
OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Hi,

I didnt quite understand this part "(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin."

what in the second equation tells us that they both have to be the same sign?

Thanks

If x and y have the opposite signs: ---x---0---y---- the distance between the origin and x will always be less than x and y.

Hope it helps.
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04 Mar 2014, 08:41
NS13983 wrote:
I get super confused in solving the below kind of inequalities.
Can some expert help?

Are the below working's correct?

1/| n | >n
• Case 1: n > 0… n 2 < 1… - 1 < n < 1
• Case 2: n < 0… - n 2 < 1… n 2 > -1…

x/| x | < x
• Case 1: x > 0… x^2 > x… as x > 0, divide by x without changing signs… x > 1
• Case 2: x < 0… - x^2 > x… as x < 0, divide and only flip sign… - x < 1… x > -1

1. $$\frac{1}{| n |}>n$$ --> 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 $$0<n<1$$.

2. $$\frac{x}{|x|} < x$$.

If $$x>0$$, then we'll have $$\frac{x}{x} < x$$ --> $$1<x$$.
If $$x<0$$, then we'll have $$\frac{x}{-x} < x$$ --> $$-1<x$$ and as we considering the range for which $$x<0$$ then completer range would be: $$-1<x<0$$.

Thus $$\frac{x}{|x|} < x$$ holds true if $$-1<x<0$$ or $$x>1$$.

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

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20 May 2014, 02:47
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.

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..

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20 May 2014, 02:57
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.

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..

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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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.

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?

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23 May 2014, 01:50
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.

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?

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]

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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.

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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Joined: 02 Sep 2009
Posts: 59740
Re: Inequality and absolute value questions from my collection  [#permalink]

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23 Oct 2014, 01:23
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.

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] 23 Oct 2014, 01:23

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