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

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New post 17 Apr 2018, 23:32
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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

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New post 27 Jun 2018, 07:14
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.


I forgot the instruction by the time, I started assessing option B. Y is an integer.
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New post 27 Jun 2018, 07:33
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.


I have done a silly mistake while solving this problem, in my mind I assumed that the statement B is an inequality rather than equal sign.
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New post 12 Jul 2018, 00:47
sriharimurthy wrote:
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


\(6*x*y = x^2*y + 9*y\)
\(6*x = x^2 + 9 = 0\)
\(x^2 - 6*x + 9 = 0\)
\((x-3)^2 = 0\)
\(x = 3\)

St. (1) : y - x = 3
y = 6
Sufficient.

St. (2) : x^3 < 0
Invalid statement. Does not give us value of y.
Insufficient.

Answer : A


solving the equation gives either y=0 or x=3
(1) y=x+3
so if y=0, x=-3, xy=0
if x=3 y =6, xy=18
insufficient
(2) x^3<0
means x<0, so y=0
and xy=0
sufficient (B)
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New post 15 Jul 2018, 03:00
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.



Hi Bunnuel, I didnt quite understand the part where you said "B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less than -2 and the other is more than -2."

why can't x & y be -2 & -2 here?
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New post 15 Jul 2018, 06:56
blueviper wrote:
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.



Hi Bunnuel, I didnt quite understand the part where you said "B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less than -2 and the other is more than -2."

why can't x & y be -2 & -2 here?


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

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

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

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New post 17 Aug 2018, 23:00
In case anybody struggles with #4.
I found it easier to think about (1+2) as 2 cases:
Stmt 1:
Step 1) 2x-2y =1
Step 2) 2(x-y) = 1
Step 3) x-y = 1/2
At this point its easy to think of cases when either x or y are negative or both are positive, for example x = 0 y = (-1/2) or x = 1 and y = 1/2, Hence, NS - BCE

Stmt 2: x/y > 1
From this statement it should be clear that both x and y are either negative or positive. NS - CE

Now Stmt 1 and 2 combined:
consider 2 cases from stmt 2 split:
case 1 - both positive:
x/y > 1 => (multiply both sides by "y", since "y" > 0 we keep the sign)
x > y, we can still satisfy the condition from stmt 1. Example x = 1 and y = 1/2

case 2 - both negative:
x/y > 1 => (multiply both sides by "y", since "y" < 0 we flip the sign)
x < y, no we can no longer satisfy the condition from stmt 1. x - y = 1/2.
(x - y) has to be positive. We cannot subtract something bigger (y) from something smaller (x) and get positive number. Hence, case 2 does not work. (1+2) combined Sufficient - Answer is C
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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

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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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Re: Inequality and absolute value questions from my collection &nbs [#permalink] 15 Sep 2018, 03:53

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