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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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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: http://gmatclub.com/forum/inequality-an ... ml#p653690

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653695

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653697

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653709

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653731

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653740

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653783 AND http://gmatclub.com/forum/inequality-an ... l#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: http://gmatclub.com/forum/inequality-an ... ml#p653789

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653792

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653796

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653853

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653870

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.


PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 02 Jun 2015, 02:20
camlan1990 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.

Answer: E.

Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

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

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New post 02 Jun 2015, 02:30
EgmatQuantExpert wrote:
camlan1990 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.

Answer: E.

Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder


Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 02 Jun 2015, 03:14
camlan1990 wrote:
EgmatQuantExpert wrote:
camlan1990 wrote:
Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder


Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)


Oops camlan1990, that was my bad! I didn't notice the '<' sign in the "<=" :-D

Now I do see how you got to x^2+y^2 >= 4.5a

But please note that x = 0, y = 0 and a = 0 is one set of values that satisfies this inequality. For these values of x, y and a, the answer to the question 'Is x^2 + y^2 > 4a' is NO

For all other values of x, y and a, the answer will be YES.

Since we are not able to rule out x, y, a = 0, we cannot infer a unique answer to the posed question using St. 1 alone. So, St. 1 is not sufficient.

Hope this helped.

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

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New post 03 Jun 2015, 02:13
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.

Answer: E.


Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

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

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New post 03 Jun 2015, 02:17
adityanukala 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.

Answer: E.


Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

Thanks.


The question asks whether x^2 + y^2 > 4a. If x = y = a = 0, then the answer is NO but if this is not so then the answer is YES. Please read the whole thread. This was discussed many, many, many times before.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 03 Jun 2015, 06:14
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 Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 03 Jun 2015, 06:34
NavenRk 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 Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance


x = -1 and y = -1.5 does not satisfy x/y > 1.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 03 Jun 2015, 06:37
NavenRk 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 Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance


Hi NavinRK,

The Flaw is

You can NOT take Values x=-1 and y=-1.5 after combining the two statements because in that case second statement x/y>1 will NOT be satisfied.

Hence the only set of values that all allowed to be taken are the one in which
Absolute value of x is greater than Absolute value of y

and

the Sign of both x and y must be same

so only Positive values are acceptable now


I hope it answers your query.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 05 Jun 2015, 15:16
Bunuel wrote:
Bunuel wrote:
sriharimurthy wrote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

Answer : C



hi sriharimurthi,

If we have the statements -s <= r < = s and -s >= r >= s, there could be 2 possible answers; r=s or r=-s. Both the possible answers satisfy the 2 equations/inequalities. So isn't the answer E and not C?

Am I missing something. Could you please help?
Thanks v much.


Sriharimurthi solution is not correct. OA for this question is E, not C. The links to the OA's and solutions are given in the initial post (if the links does not work, then switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p652806 will lead you to the solutions).

As for this question:
12. Is r=s?

(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.

Hope it helps.


Bunuel I have seen the solution & found it intriguing. As advised, I have seen the entire thread & I am amazed to see this topics alive after 6 years of its release.
I could not find the graphical solution for the above problem except on Page 14 madn800 has tried to solve it but has considered many possibilities.

Hence, I have solved it using the graphical approach as indicated below.

IN essense, even if you combine I & II you have two points at which statements one & two intersect ( precisely r=s & r=-s) . Hence we cannot answer with C also Hence E.

I am interested in knowing what benefit we derived by stating the highlighted part & how it is relevant to derive the answer using your approach.

Also, as you being a Mahaguru of GMAT Land, can cite certain examples where this particular conclusion that ( s cannot be negative given -s<= r < s) is emphasized & utilised & tested on.

Thanks & Best regards,
Ankush.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 05 Jun 2015, 18:38
Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

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

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New post 05 Jun 2015, 21:45
lipsi18 wrote:
Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

Thanks


Hi Lipsi18,

The Explanations is as mentioned below:

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


Statement 2: 1/|n| > n

Multiplying |n| which is a positive value on both sides of the inequation, we get,

1> n*|n|
i.e. n*|n| < 1

Case 1: n is positive:
i.e. n^2 <1
i.e. 0<n<1

Case 2: n is negative
n*|n| <1 is true for all negative values of n

But in context of question this statement is NOT SUFFICIENT as it doesn't provide us any specific value of n
and for n=-5, |n| >4
and for n=-3, |n| <4
INCONSISTENT answer
Hence, NOT SUFFICIENT

I hope it answers your query!!!
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 06 Jun 2015, 11:19
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.

Answer: E.


The above question will be sufficient only if a) Either x or y is zero & a > 0 or b) x,y & a are non-zeroes.

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

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New post 16 Jul 2015, 20:20
May I know what is the difficulty of the questions? I only got 7 out from the 13 questions right...perhaps I should practice more in this area?
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 16 Jul 2015, 20:41
Hi,

for Q7, can I interpret the 2 scenario as following?
a. x=y, meaning X and Y must be have the same sign
b. x+y=-4, meaning X and Y are different sign

may I know why do we need to set the criteria for scenario to x and y both >=- 2 or x and Y both <= -2?

Thanks in advance.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 18 Jul 2015, 03:13
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

Can it be done without 2 statements as below ?
\((x+2)^2\)=\((y+2)^2\)

\(x^2-y^2\)=\(4(y-x)\)

\((x+y)(x-y)=4(y-x)\)

\(x+y=-4\).

Can someone please correct me if i am wrong.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 03 Aug 2015, 01:48
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.

Answer: C.


Could you help me understand this. When we combine (1) and (2)

We should have two cases:

a) y>=2 AND y<3 ---> y=-8

b) y>=2 AND y>3 ---> y=-14

Insuff.

So answer should be E

I'm confused :|
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 16 Aug 2015, 09:55
ferash 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.

Answer: C.


Could you help me understand this. When we combine (1) and (2)

We should have two cases:

a) y>=2 AND y<3 ---> y=-8

b) y>=2 AND y>3 ---> y=-14

Insuff.

So answer should be E

I'm confused :|


If y = -14, then |3 - y| = |3 - (-14)| = |3 + 14| = 17, not 11.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 25 Aug 2015, 02:59
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.


Also - Viewing graphically
A. 2x - 2y = 1
Put x = 1, you get y = 1.
Put x = 2, you get y = 1.5.
Make a rough plot. This is a line which agrees to negative values of both x and y as well. Insufficient.

B. x/y > 1
Both x an y can be negative. Insufficient

A + B

which means for all values x > y, which 2x - 2y = 1. This is possible only in the first quadrant whenever the value of x > 1. Sufficient.
Answer C

sorry for the flipped image :)
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 25 Aug 2015, 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.

Answer: C.


I found it easy to rule out (1) and (2) as individually being insufficient. but the conclusion I drew from (2) was obviously that the absolute value of \(x\) has to be bigger than \(y\) (and of course that they are the same size), so regardless of the sign \(2x\) had to be of greater magnitude than \(2y\), and the only way for \(2x-2y=1\) was if they were both positive... I know this isn't ground breaking but it's the very simple way I arrived at the correct answer without getting too "mathsy"
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 27 Aug 2015, 17:39
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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Re: Inequality and absolute value questions from my collection   [#permalink] 27 Aug 2015, 17:39

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