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

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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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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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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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11 Dec 2010, 22:35
Hi
for question#12, the answer should be E. Lets take some examples. Let r = -5, s=5. Then both the statements 1 and 2 are valid. But r is not equal to s. when r = 3 and s = 3, then also both the statements are valid, but r = s.
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19 Dec 2010, 02:12
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.

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.

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19 Dec 2010, 02:27
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.

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.

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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19 Dec 2010, 02:48
Oh yeah..my mistake again...Thanks again for your prompt reply...I guess more concentration is required from my end..Thanks for your help...
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05 Feb 2011, 08:26
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.

Bunuel...here why cant we use squaring on both sides of the statements.
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05 Feb 2011, 08:35
ajit257 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.

Bunuel...here why cant we use squaring on both sides of the statements.

I done't understand what you mean. Can you please SHOW what you mean.
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06 Feb 2011, 14:11
nnnnnnnniiiiiiiceeeeeee!
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07 Feb 2011, 16:26
Bunuel wrote:
ajit257 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.

Bunuel...here why cant we use squaring on both sides of the statements.

I done't understand what you mean. Can you please SHOW what you mean.

well i was just wanted to know why cant we do this

1. x + y = 3sqrt a
2. x - y = sqrt a

then solve the 2 equations to get x and y.
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07 Feb 2011, 16:37
ajit257 wrote:

well i was just wanted to know why cant we do this

1. x + y = 3sqrt a
2. x - y = sqrt a

then solve the 2 equations to get x and y.

2 equations 3 unknowns.
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07 Feb 2011, 16:47
oh yes ....yup got it ...thanks
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07 Feb 2011, 18:06
lagomez wrote:
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.

So here in statement 1 we are not at all concerned with |X^2 -4|
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08 Feb 2011, 03:33
ajit257 wrote:
lagomez wrote:
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.

So here in statement 1 we are not at all concerned with |X^2 -4|

Yes. We just have that left hand side is some absolute value and as absolute value is always non-negative then right hand side must also be non-negative.
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22 Feb 2011, 05:24
This is amazing. Thanks!
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22 Feb 2011, 05:30
Those are very tough. Couldn't finish many of them.
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02 Mar 2011, 09:13
This was good Thanks alot for the explanations
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18 Jul 2011, 13:41
absolute brilliance from Bunuel!
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02 Aug 2011, 02:11
Nice questions. Thanks for sharing
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08 Aug 2011, 10:20
Quote:
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

Thanks ImJun for the above solution.

I just could not understand how option B was sufficient. I had left out on an important bit, (x-3)^2 will have both positive and negative roots.
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