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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 05 Apr 2018, 08:37
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.

Answer: C.


Hi Bunuel, I solved it as below
(1) 2x-2y=1 -> x-y= 1/2. This means x>y by 1/2 but x can be 1/2 and y=0 so this is insufficient
(2) x/y=1 -> x>y which is no different from (1) so insufficient again

But I didn't follow your approach on combining the 2 statements, how did you get x to be substituted by 1/y? The fact that x is positive is already proved by (1) isn't it?
Please help out.
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New post 05 Apr 2018, 12:24
Aditi10 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 solved it as below
(1) 2x-2y=1 -> x-y= 1/2. This means x>y by 1/2 but x can be 1/2 and y=0 so this is insufficient
(2) x/y=1 -> x>y which is no different from (1) so insufficient again

But I didn't follow your approach on combining the 2 statements, how did you get x to be substituted by 1/y? The fact that x is positive is already proved by (1) isn't it?
Please help out.


The red parts are 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 \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So from (2) \(\frac{x}{y}>1\), we can only deduce that \(x\) and \(y\) have the same sigh (either both positive or both negative).

Also, from (1) we can only say that x > y (because x = y + 1/2 = y + positive number) but we cannot say that x is positive. For example, consider x = -2 and y = -2.5.

Finally, when we substitute \(x = y + \frac{1}{2}\) into \(\frac{x}{y}>1\) we'll get:

\(\frac{y + \frac{1}{2}}{y}>1\);

\(1 + \frac{1}{2y}>1\);

\(\frac{1}{2y}>0\);

\(y > 0\).

If \(y > 0\), then \(x > y > 0\).

Hope it helps.
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New post 17 Apr 2018, 09:29
Hi
This is really an interesting question,
I want to share an alternative approach.
Concept |x-a|=b, it means distance of x from a = b
it can be logically deducted that , if |x-a|=|y-a|, it means either x= y or the a and y are equidistant from a, hence average of x & y = a , or x+y = 2a
it can be proved but if visualization can save time.

7. |x+2|=|y+2| what is the value of x+y?

|x+2|=|y+2|, it means either x and y are equal or are equidistant from -2.

(1) xy<0
it means one is positive and one is negative, hence they are not equal. so x and y are equidistant from -2, hence sum of x and y = -4. SUFFICIENT.

(2) x>2 y<2
x not equal to y, hence their sum is -4. SUFFICIENT.

Hence Answer D

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.

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New post 18 Apr 2018, 01:15
Is |x-1| <1?
Since Absolute value function is always non negative, we can square both sides,

We get is (x-1)^2<1?

Statement 1 is (x-1)^2< = 1

If (x-1)^2<1: answer to question is yes

(x-1)^2= 1: hence answer is No,

So Statement 1 is NOT SUFFICIENT.



Statement2: Question stem (x-1)^2<1?

Can be reduced to is x(x-2)<0

Or is 0<x<2

Now St2: x^2-1>0 gives x >1 or x <-1

Now x can be -2, answer to question stem is No

Or 1.5 answer to question stem is yes.

Hence NOT SUFFICIENT.



Combining both Statement 1 & 2, we get

X can be 2, answer to question stem- NO

Or 1.5 answer to question stem is yes.



Hence Answer E


Buttercup3 wrote:
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


Still not clear on this one.
Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient?
Thanks in Advance

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New post 31 Jul 2018, 14:56
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Answer: A.


Can someone clarify whether |a/b|=|a|\|b| exists as a property?
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New post 31 Jul 2018, 20:49
arunjohn43 wrote:
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Answer: A.


Can someone clarify whether |a/b|=|a|\|b| exists as a property?


Yes, \(|\frac{a}{b}|=\frac{|a|}{|b|}\) and \(|ab|=|a|*|b|\) are generally true.
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New post 14 Sep 2018, 11:36
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?

Notice that from \(y=|x|+x\) it follows that y cannot be negative:
If\(x>0\), then \(y=x+x=2x=2*positive=positive\);
If \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.


Answer: D.


HiBunuel,

Shouldn't it be like this

if |x|=x if \(x\geq 0\)
then y=x+x which means y=2x

and if f |x|= - x if x<0
then
y=-x+x which means y=0

Probus.
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New post 15 Sep 2018, 03:53
Probus wrote:
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?

Notice that from \(y=|x|+x\) it follows that y cannot be negative:
If\(x>0\), then \(y=x+x=2x=2*positive=positive\);
If \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.


Answer: D.


HiBunuel,

Shouldn't it be like this

if |x|=x if \(x\geq 0\)
then y=x+x which means y=2x

and if f |x|= - x if x<0
then
y=-x+x which means y=0

Probus.


Both are correct. If x = 0, then |0| = 0 as well |0| = -0 = 0.
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New post 16 Nov 2018, 11:02
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.


Is (1) even a correct clue? How can an absolute value be negative? Please help me understand.
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New post 18 Nov 2018, 01:45
PB1712989 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.

Answer: C.


Is (1) even a correct clue? How can an absolute value be negative? Please help me understand.


We have -n=|-n|. Now, if n is negative or 0 then -n is positive or 0, so the absolute value, which is on the right hand side, does not equal to negative number. All is correct there.
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New post 25 Mar 2019, 05:30
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.

Answer: A.


also , if n lies between 0 to 1 then (2) is possible Ex: n=0.1

n=( - infinity, 1) -> not sufficient
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New post 13 Oct 2019, 17:40
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 question doesn't ask us to calculate the value of x^2+y^2, then why we need to consider the values of x,y, and a here?, could you please explain.
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New post 14 Oct 2019, 22:00
Hi Bunuel,

Can you please explain your analysis for statement (a). I don't think I understood it well

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.
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New post 14 Oct 2019, 22:05
crack1991 wrote:
Hi Bunuel,

Can you please explain your analysis for statement (a). I don't think I understood it well

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.


You might find the following solution easier: https://gmatclub.com/forum/inequality-a ... l#p1111747

Hope it helps.
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New post 14 Oct 2019, 22:20
Do I need to memorize this ?
"(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.)"

or is there some other way to remember this?

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.)

Answer: B.
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New post 23 Oct 2019, 16:23
I am just extending your explanation.
St. 1 tells Mod (R) <= S
St. 2 tells Mod (R) >= S

Combining these two
Mod(R)= S
Now here can be two conditions
either R= S
or -R =S

So I think E is correct.


lagomez wrote:
Bunuel wrote:


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



I'm getting c

1. s can be 3 and r can be 3 which makes question yes
s can be 3 and r can be 2 which makes question no
insufficient

2. r can be 3 and s can be 3 makes question yes
r can be 3 s can be 2 makes question no
insufficient

combining:
|r|>=s means
r>=s or r<=-s

and -s<=r<=s means
-s<=r and r<=s

now we have -s<=r and -s>=r so -s = r or s = r
r>=s and r<=s so s = r
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New post 23 Oct 2019, 16:56
8. a∗b≠0a∗b≠0. Is |a||b|=ab|a||b|=ab?
(1) |a∗b|=a∗b|a∗b|=a∗b
(2) |a||b|=|ab|

St-1
Here can be two cases, either both are positive or both are negative. The both cases satisfies the expression given in question stem.
Sufficient.

St.-1
This st. is also satisfied for all negative as well as positive values of both x and y.
And for all respective values of x and y, the question stem doesn't satisfy.

For ex. x= 2, y = -3
st. 2 is satisfying with these two values, but question stem not.
Another set of x, y= 2, 3
St. 2 as well as question stem is satisfying with this stem.

A is correct answer.
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New post 24 Oct 2019, 22:43
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.


If I were to substitute values instead, to check if C is correct

The first equation can be written as x-y = 0.5

Using second equation, I know x is greater than y.

If I substitute x = 2.5 and y = 2, both my values are positive and x-y = 0.5

However, if I use x = -2 and y = -2.5, then x - y = -2 - (-2.5) = 0.5 is also true.

In this case, both x and y are negative. So I thought the answer should be E.

Please explain. Thanks
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New post 21 Nov 2019, 05:31
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.



(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.:
its x^2-1^2>0
(x-1)*(x+1)>0
x>1;x>-1 ???

I will questioning everything i know if this turns out to be wrong
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 22 Nov 2019, 11:18
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

If the combination range was 1<x<2, then will correct answer be option C??

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Re: Inequality and absolute value questions from my collection   [#permalink] 22 Nov 2019, 11:18

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