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# Inequality and absolute value questions from my collection

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

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16 Nov 2009, 10:33
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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: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#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: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
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21 May 2010, 01:37
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.

I am a little confused on this one . Can the answer be E??

From A:
2x-2y=1
=> x-y= 0.5 INSF

From B
x/y > 1
=> x > y INSF

From A & B
x-y =0.5 and x > y

If x = -0.5 and y = -1 then
x > y
and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5
Hence both x and y can be negative

If x= 1 and y = 0.5 then
x > y
and x- y = 1 -0.5 = 0.5
Hence both x and y can be positive

Ans = E ??
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21 May 2010, 02:21
Bunuel wrote:
ManishS wrote:
I am a little confused on this one . Can the answer be E??

From A:
2x-2y=1
=> x-y= 0.5 INSF

From B
x/y > 1
=> x > y INSF

From A & B
x-y =0.5 and x > y

If x = -0.5 and y = -1 then
x > y
and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5
Hence both x and y can be negative

If x= 1 and y = 0.5 then
x > y
and x- y = 1 -0.5 = 0.5
Hence both x and y can be positive

Ans = E ??

Problem with your solution is that the red part is not correct.

$$\frac{x}{y}>1$$ does not mean that $$x>y$$. If both x and y are positive, then $$x>y$$, BUT if both are negative, then $$x<y$$.

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.

Hope it's clear.

Thank you so much. This really was helpful.
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14 Jun 2010, 21:47
h2polo wrote:
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Another way of looking at the problem is to ask, is x<0? Because if it is, then we know that y is zero. The only case in which y will not be zero is if x is positive.

Statement 1:

SUFFICIENT

Statement 2:

y<1

Because y is an integer, it must be one of the following values: 0, -1, -2, -3...

BUT |x| + x can never be a negative value. The lowest value that it can be is 0.

Hence, y can never be negative and the only possible value it can be then is 0.

SUFFICIENT

----------------------------
hear it is not give that X is also a integer,

S1: if X=1/2 or -1/2 then also Y is integer ,

in this case Ans: B

Is i am missing somthing?
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15 Jun 2010, 12:54
varun2410 wrote:
----------------------------
hear it is not give that X is also a integer,

S1: if X=1/2 or -1/2 then also Y is integer ,

in this case Ans: B

Is i am missing somthing?

OA' s and solutions for all the problems are given in my posts on pages 2 and 3.

OA for this question is D. Below is solution for it.

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

Note: $$y=|x|+x$$, this expression 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.

Hope it's clear.
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16 Jun 2010, 14:07
3. is X^2+Y^2 > 4a?

1) (X+Y)^2= 9a

x^2+2xy+Y^2=9a

Insufficient

2) (X-y)^2= a

x^2-2XY+Y^2 = a

Insufficient

Combining 1 & 2, 2(X^2+Y^2)=10a X^2+Y^2 = 5a, thus C
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05 Jul 2010, 14:23
this thread has helped a lot.. thanks!
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05 Jul 2010, 14:31
Whoa, inequalities and AV's did me in for my first GMAT! Great thread!
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Re: Inequality and absolute value questions from my collection [#permalink]

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06 Jul 2010, 23:47
Ive got C for this question..

when both together yields x^2 + y^2 = 5a
why it is E?

Also I dont understand the explanation of below,
St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true.
When x and y are both 0, question stem is false.

Can somebody clarify how to solve this please...

sriharimurthy wrote:
Quote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

St. (1) : (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
Insufficient.

St. (2) : (x - y)^2 = a
x^2 + y^2 - 2xy = a
Insufficient.

St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true.
When x and y are both 0, question stem is false.

Hence insufficient.

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07 Jul 2010, 02:38
gmatJP wrote:
Ive got C for this question..

when both together yields x^2 + y^2 = 5a
why it is E?

Also I dont understand the explanation of below,
St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true.
When x and y are both 0, question stem is false.

Can somebody clarify how to solve this please...

sriharimurthy wrote:
Quote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

St. (1) : (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
Insufficient.

St. (2) : (x - y)^2 = a
x^2 + y^2 - 2xy = a
Insufficient.

St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true.
When x and y are both 0, question stem is false.

Hence insufficient.

OA' s and solutions for all the problems are given in my posts on pages 2 and 3.

OA for this question is E.

When we consider statement together we'll have: $$x^2+y^2=5a$$. Now, if $$x$$, $$y$$ and $$a$$ are different from zero (for example: $$x=3$$, $$y=4$$ and $$a=5$$) then $$x^2+y^2=5a>4a$$ and the answer to the question is YES, but if $$x=y=a=0$$, then $$x^2+y^2=5a=4a=0$$ and the answer to the question is NO. Two different answers, thus not sufficient.

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

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16 Jul 2010, 07:03
Hi Bunuel,
If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

y*(x-3)^2 = 0 means x=3 or/and y=0.

I have been thinking that ONLY either of the variables can be zero. i.e x=3 or y=0.
So should i always consider the possibility of both X=3 and Y=0 ?

Can you explain the concept?

Thanks
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16 Jul 2010, 07:07
Hi Bunuel,
I forgot to thank you for these great questions and solutions. You are the BEST!!!!

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

(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence . Sufficient.

In the above problem, it is very obvious that Y is not equal to 0 from the 2nd statement.
Do you think GMAT questions will have this much obvious statements?
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16 Jul 2010, 07:21
Hi Bunuel,
I forgot to thank you for these great questions and solutions. You are the BEST!!!!

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

(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence . Sufficient.

In the above problem, it is very obvious that Y is not equal to 0 from the 2nd statement.
Do you think GMAT questions will have this much obvious statements?

The trick here is to conclude that $$y$$ can not be negative, the rest is relatively easy. And yes, I think you can see such questions on GMAT.
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20 Jul 2010, 01:34
lagomez wrote:
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

because in number 2 n can be negative or a fraction

Doubt :The question is is |N|<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions
for example for n =1/2 , 1/(|1/2|)=2 which is greater than n=1/2

Also for n=-1/2,1/(|-1/2|)=2 which is greater than n=-1/2

for negative integers also its true ,say we take n=-3 ,
1/|n|=1/3 which is greater than n=-3

But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!!
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20 Jul 2010, 01:47
amlan009 wrote:
lagomez wrote:
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

because in number 2 n can be negative or a fraction

Doubt :The question is is |N|<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions
for example for n =1/2 , 1/(|1/2|)=2 which is greater than n=1/2

Also for n=-1/2,1/(|-1/2|)=2 which is greater than n=-1/2

for negative integers also its true ,say we take n=-3 ,
1/|n|=1/3 which is greater than n=-3

But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!!

$$\frac{1}{|n|}>n$$ holds true for ALL negative values of $$n$$, as if $$n<0$$ then $$LHS=positive>RHS=negative$$. Hence we don't know whether $$-4<n<4$$ is true. That's why statement (2) is not sufficient.

The complete range of $$n$$ for which $$\frac{1}{|n|}>n$$ holds true is $$n<1$$.

P.S. OA's and solutions for all question are given in my posts on pages 2 and 3.

Hope it's clear.
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22 Jul 2010, 05:36
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, devide the whole equation by y, and
6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3
(1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient
(2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient
B

2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1
(1)Sufficient
(2)Sufficient, y couldn't be negative, the least it could be is 0.
D
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x+y)^2= x^2+y^2+2x*y
9a=x^2+y^2+2x*y Still Insufficient

(2) (x-y)^2=x^2+y^2-2x*y
a=x^2+y^2-2x*y Insufficient

Sum up equations (1) and (2) , then we can get 10a=2(x^2+y^2) => 5a=x^2+y^2, and we know a is necessarily positive since (x – y)^2 = a
With both factors, we can know x^2 + y^2 >4a

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

(1)Insufficient
(2)that only tells you x is greater than y Insufficient
(1) and (2) together: x-y=0.5, and x is great than y, Insufficient
E

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

(1) we only know y must be positive, insufficient
(2)y coule be -8 or 14 insufficient
BOth (1) and (2): from statement (1) weve learned that y must be a positive number, along with statement (2) we know y is 14

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

(1)x>-1, we wouldnt know anything about y with that, insufficient
(2)xy>0, if x is less than 0, then y too has to be less than 0 Insufficient

both (1) and (2): xy>0, x>-1, x could be 0 or any positive integer, and even if x=0, y still has to be great than 0 coz xy>0 Sufficient

C

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

(1)xy<0, that means either x or y is positive and the other negative. seems like x+y= -4, coz x+2=-2-y => x+y=-4, sufficient
(2) Sufficient too, same reasons as above

D

8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

(1) a*b has to be positive, a/b has to positive too. Sufficient

(2) we already know that for sure, Insufficient

A

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

(1)since the absolute value must be positive, so we know n has to be negative, or 0 Not Sufficient
(2)ok, n could be 4 or -4, so what? Insufficient
BOth (1) and (2) together: ok, n could be 4 or -4, and it's less than or equal to zero, so it must be -4 Sufficient

C

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

(1) n>4 or n<-4, |n| must be greater than 4 sufficient
(2)we only know n<1, |n| could be greater than 4 if n < -4 insufficient

A

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

Just leave it for a while

12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

(1) Not sufficient
(2) Not sufficient
Together: Still Not sufficient

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

ill solve it next time
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16 Aug 2010, 07:10
An excellent resource. Thanks Bunuel
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21 Oct 2010, 21:54
Great stuff.

Learned a lot from this set.
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31 Oct 2010, 23:55
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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01 Nov 2010, 00:04
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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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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Re: Inequality and absolute value questions from my collection [#permalink]

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01 Nov 2010, 06:46
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
New member

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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Re: Inequality and absolute value questions from my collection   [#permalink] 01 Nov 2010, 06:46

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