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# Inequality and absolute value questions from my collection

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22 Dec 2009, 18:50
Hi Bunuel,
Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0.

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.

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22 Dec 2009, 19:09
lionslion wrote:
Hi Bunuel,
Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0.

I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them.

Hope it's clear.
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22 Dec 2009, 20:31
logan wrote:
hey Bunuel!! first i would like to thank you for posting such wonderful questions..

regarding a question that you posted above, i got a small doubt..

|x+2|=|y+2|
so lets say |x+2|=|y+2|=k (some 'k')

now |x+2|=k =====> x+2=+/- k
and x+2= +k, iff x>-2
x+2= -k, iff x<-2

also we have |y+2|=k =====> y+2=+/- k
and y+2= +k, iff y>-2
y+2= -k, iff y<-2

so x+2=y+2 ===> x=y , iff (x>-2 and y>-2) or (x<-2 and y<-2)--------eq1
and x+y=-4, iff (x<-2 and y>-2) or (x>-2 and y<-2)-------------------eq2

now coming to the options,
1) xy<0 i.e., (x=-ve and y=+ve) or (x=+ve and y=-ve)
(x=-ve and y=+ve): this also means that x and y can have values, x=-1 and y= some +ve value. so eq2 cannot be applied, x+y#-4. if x=-3 and y=some +ve value, x+y=-4. two cases. data insuff.
(x=+ve and y=-ve): this also means that x and y can have values, x=+ve value and y=-1.so eq2 cannot be applied, x+y#-4. if x=some +ve value and y=-3, x+y=-4. two cases. data insuff.

2)x>2,y>2 for this option too we cannot judge the value of x+y, with the limits of x and y being different in the question and the answer stem. so data insuff.

so i have a doubt that, why the answer cannot be E??

plz point out if i made any mistake..

Hi, Logan. Good way of thinking. Though I think that your solution is not correct.

Consider this:

We have |x+2|=|y+2|

(1) xy<0, hence x and y have opposite signs. Let's take x negative and y positive (obviously it doesn't matter which one we pick as equation is symmetric).

If y is positive RHS |y+2| will be positive as well and we can expend it as |y+2|=y+2.

Now for |x+2| we can have to cases:
A. -2<x<0 --> |x+2|=x+2=y+2 --> x=y. BUT: it's not valid solution as x and y have opposite signs and they can not be equal to each other.

B. x<=-2 --> |x+2|=-x-2=y+2 --> x+y=-4. Already clear and sufficient.

If we go one step further to see for which x and y is this solution is valid, we'll get:
As x+y=-4, x<=-2 and y>0 --> x must be <-4. If you substitute values of x<-4 you'll receive the values of y>0 and their sum will always be -4.

The same approach works for (2) as well.

Hope it's clear.
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20 Jan 2010, 03:45
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.

Quote:
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=-2.
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 positive only. Hence if xy is not positive 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.

Wonderful question, I solved it up to an extent but in reasoning I messed up.
Question for the quoted part which talks about the scenarios to reducing to two scenarios. I can think and verfy also though the validity of it, but want to drill further and understand how could we generalise this scenario if we could ?

Like what if |x+3|=|X-2| or |x+3|=|X-2| + |X+4| etc , just crude examples I have put , How could we generalise such scenarios ? instead of imagining six scenarios or more .
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20 Jan 2010, 06:15
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.

Quote:
(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;

How do you conclude A here can't make out or I am tired.; anyways, it is mental marathon .. wonderful questions Bunuel. Any more link for inequality as I need some more practise .
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21 Jan 2010, 00:41
How do you conclude A here can't make out or I am tired.; anyways, it is mental marathon .. wonderful questions Bunuel. Any more link for inequality as I need some more practise .

We have $$-s<=r<=s$$ --> $$-s<=s$$. Now if $$s$$ in negative, let's say $$-2$$, then we would have $$-(-2)<=-2$$ --> $$2<=-2$$, which is not right. Hence $$s$$ can not be negative. But $$s$$ can be zero --> $$0<=0$$, true.
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21 Jan 2010, 00:55
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.

Quote:
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=-2.
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 positive only. Hence if xy is not positive 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.

Wonderful question, I solved it up to an extent but in reasoning I messed up.
Question for the quoted part which talks about the scenarios to reducing to two scenarios. I can think and verfy also though the validity of it, but want to drill further and understand how could we generalise this scenario if we could ?

Like what if |x+3|=|X-2| or |x+3|=|X-2| + |X+4| etc , just crude examples I have put , How could we generalise such scenarios ? instead of imagining six scenarios or more .

We can count the # of scenarios for the above examples as well, but as in these examples there is only one variable we can directly solve them using the ranges, so no need to use the approach we used in solving the original question.

|x+3|=|x-2|, two check points -3 and 2, thus we'll have three ranges in which we should expand the absolute values:

A. x<-3 --> -(x+3)=-(x-2) --> -3=2, which is not true, hence in this range we have no solution;
B. -3<=x<=2 --> x+3=-(x-2) --> x=-1/2, which IS in the range we are expanding so it's a valid solution;
C. x>2 --> x+3=x+2 --> 3=2, which is not true, hence in this range we have no solution.

The equation |x+3|=|x-2| has only one solution x=-1/2. If you look at the scenarios A,B and C you'll see that |x+3|=|x-2| can take also only TWO forms x+3=x+2 or x+3=-(x-2), but the ranges are important here and we should consider these forms in their respective ranges.

Hope it's clear.
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15 Feb 2010, 02:23
logan wrote:
xyztroy 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.

Hi Bunuel,

I kind of disagree with your conclusion when you combined both the stmts. If x,y, and a all are 0 then the actual question (x^2+y^2 > 4a) itself will become whether 0 > 0 ?....so I would say that the answer should be C.

hi xyztroy,

i think i can answer ur question.

in question, no limits for x and y are given, like x&y are integers or x&y are real numbers. so x and y can assume any values, including 0. but we have to conclusively show that (x^2+y^2 > 4a).
as you see, 1&2 are individually insufficient.
combining 1&2 we have (x^2+y^2 = 5a), which is definitely greater than 4a. when you substitute values for x,y and a, all values of x,y and a which satisfy (x^2+y^2 = 5a) also satisfies (x^2+y^2 > 4a), except the values x=y=a=0. so two cases arise.
hence insufficient.

I think, when statements are given we have to answer whether based on statements given question can be answered, we should not question the validity of statements or a statements derived form the statements.
after deriving from statement 1 and 2 we deduce that x^2+y^2 =5a, this is enough to show that the question asked x^2+y^2>4a is satisfied. we are not supposed to validate whether statement x^2+y^2 =5a holds.
if u try to put x,y as 0 then just ask your self what question is trying to ask?
is 0>0.
so answer has to be C
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15 Feb 2010, 06:15
sandeep25398 wrote:
logan wrote:
xyztroy wrote:

Hi Bunuel,

I kind of disagree with your conclusion when you combined both the stmts. If x,y, and a all are 0 then the actual question (x^2+y^2 > 4a) itself will become whether 0 > 0 ?....so I would say that the answer should be C.

hi xyztroy,

i think i can answer ur question.

in question, no limits for x and y are given, like x&y are integers or x&y are real numbers. so x and y can assume any values, including 0. but we have to conclusively show that (x^2+y^2 > 4a).
as you see, 1&2 are individually insufficient.
combining 1&2 we have (x^2+y^2 = 5a), which is definitely greater than 4a. when you substitute values for x,y and a, all values of x,y and a which satisfy (x^2+y^2 = 5a) also satisfies (x^2+y^2 > 4a), except the values x=y=a=0. so two cases arise.
hence insufficient.

I think, when statements are given we have to answer whether based on statements given question can be answered, we should not question the validity of statements or a statements derived form the statements.
after deriving from statement 1 and 2 we deduce that x^2+y^2 =5a, this is enough to show that the question asked x^2+y^2>4a is satisfied. we are not supposed to validate whether statement x^2+y^2 =5a holds.
if u try to put x,y as 0 then just ask your self what question is trying to ask?
is 0>0.
so answer has to be C

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

Consider two cases:

1. $$x=2$$, $$y=1$$ and $$a=1$$. These values satisfy both statements and the answer to the question "is $$x^2+y^2>4a$$ true" is YES, as $$5>4$$ is true. (Note here that these values naturally satisfy $$x^2+y^2=5a$$ too)

2. $$x=0$$, $$y=0$$ and $$a=0$$. These values ALSO satisfy both statements and the answer to the question "is $$x^2+y^2>4a$$ true" is NO, as LHS is $$0$$, RHS is also $$0$$ and $$0>0$$ is NOT TRUE. (Note here that these values naturally satisfy $$x^2+y^2=5a$$ too)

So question is not asking whether $$0>0$$. Question is asking whether "$$x^2+y^2>4a$$ true". We got that for smoe values YES and for some values NO.

Hope it's clear.
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17 Feb 2010, 23:19
Bunuel wrote:
How do you conclude A here can't make out or I am tired.; anyways, it is mental marathon .. wonderful questions Bunuel. Any more link for inequality as I need some more practise .

We have $$-s<=r<=s$$ --> $$-s<=s$$. Now if $$s$$ in negative, let's say $$-2$$, then we would have $$-(-2)<=-2$$ --> $$2<=-2$$, which is not right. Hence $$s$$ can not be negative. But $$s$$ can be zero --> $$0<=0$$, true.

You can't write it this way. -s<=r<=s means that either r lies between s & -s or r is equal to -s or r can be equal to s. But this certainly doesn't mean that you can equate -s<=s.

If I say that -4<=x<=4. This doesn't mean that I can write -4<=4.

But in any case the answer you've arrived at is correct. We can't derive anything out of the first statement as it says that -s<=r<=s. r can be anything -s or s or between -s & s.

Second says that lrl>=s ---> r>=s or r<=-s. This again is not giving any specific value.

If we combine the two we get that either r=s or r=-s. This is closer but still ambiguous. So we don't know whether r=s. Therefore, answer is E.
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18 Feb 2010, 08:22
honeyrai wrote:
Bunuel wrote:
How do you conclude A here can't make out or I am tired.; anyways, it is mental marathon .. wonderful questions Bunuel. Any more link for inequality as I need some more practise .

We have $$-s<=r<=s$$ --> $$-s<=s$$. Now if $$s$$ in negative, let's say $$-2$$, then we would have $$-(-2)<=-2$$ --> $$2<=-2$$, which is not right. Hence $$s$$ can not be negative. But $$s$$ can be zero --> $$0<=0$$, true.

You can't write it this way. -s<=r<=s means that either r lies between s & -s or r is equal to -s or r can be equal to s. But this certainly doesn't mean that you can equate -s<=s.

If I say that -4<=x<=4. This doesn't mean that I can write -4<=4.

But in any case the answer you've arrived at is correct. We can't derive anything out of the first statement as it says that -s<=r<=s. r can be anything -s or s or between -s & s.

First of all: you are excluding the possibility when -s=r=s=0.

-s<=r<=s, so -s can be equal to s, when s=0. In all other cases -s will be less than s, so there is nothing wrong in writing this as -s<=s.

Second of all: we CAN conclude one more thing from this statement: s is either positive or zero.

Here is what I wrote in my solution from page 3:

(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.

Hope it's clear.
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20 Feb 2010, 08:10
sriharimurthy wrote:
Quote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Question Stem : Is |x-1| < 1 ?
When x > 1 ; x - 1 < 1 ; x < 2.
When x < 1 ; -x + 1 < 1 ; x > 0.
Thus it can be written as : 0 < x < 2.

St. (1) : (x-1)^2 <= 1
x^2 + 1 - 2x <= 1
x^2 - 2x <= 0
x(x - 2) <= 0 ; Thus boundary values are 0 and 2.
Therefore statement can be written as : 0 <= x <= 2.
Since the values are inclusive of 0 and 2, it cannot give us the answer.
Insufficient.

St. (2) : x^2 - 1 > 0
(x + 1)*(x - 1) > 0
Statement can be written as x > 1 and x < -1.
Thus it is possible for x to hold values which make the question stem true as well as false.
Insufficient.

St. (1) and (2) : 0 <= x <= 2 ; x > 1 and x < -1
Thus combined, the statements become : 1 < x <= 2.
Since it is inclusive of 2, it will give us conflicting solutions for the question stem.
Hence Insufficient.

How do u get these boundary values. Looking at the equation, I solved it as x(x-2)<=0.... which gives x<=0 or x<=2... I know this isnt correct but can u let me know how u got... x>=0 & x <=2... Thanks
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20 Feb 2010, 10:47
jeeteshsingh wrote:
sriharimurthy wrote:
Quote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Question Stem : Is |x-1| < 1 ?
When x > 1 ; x - 1 < 1 ; x < 2.
When x < 1 ; -x + 1 < 1 ; x > 0.
Thus it can be written as : 0 < x < 2.

St. (1) : (x-1)^2 <= 1
x^2 + 1 - 2x <= 1
x^2 - 2x <= 0
x(x - 2) <= 0 ; Thus boundary values are 0 and 2.
Therefore statement can be written as : 0 <= x <= 2.
Since the values are inclusive of 0 and 2, it cannot give us the answer.
Insufficient.

St. (2) : x^2 - 1 > 0
(x + 1)*(x - 1) > 0
Statement can be written as x > 1 and x < -1.
Thus it is possible for x to hold values which make the question stem true as well as false.
Insufficient.

St. (1) and (2) : 0 <= x <= 2 ; x > 1 and x < -1
Thus combined, the statements become : 1 < x <= 2.
Since it is inclusive of 2, it will give us conflicting solutions for the question stem.
Hence Insufficient.

How do u get these boundary values. Looking at the equation, I solved it as x(x-2)<=0.... which gives x<=0 or x<=2... I know this isnt correct but can u let me know how u got... x>=0 & x <=2... Thanks

"Boundary values" are the roots of the equation $$x(x-2)=0$$ --> $$x=0$$ and $$x=2$$.

I use "parabola approach" for this kind of quadratic inequalities. Basically as $$x(x-2)=0$$ is an equation of upward parabola, $$x(x-2)\leq{0}$$ ($$x^2-2x\leq{0}$$) is the part of the parabola, which does not lie above the X-axis. Parabola $$x(x-2)\leq{0}$$ does not lie above the X-axis for the values of x in the range $$0\leq{x}\leq{2}$$. If you plug the values from this range you'll see that inequality, $$x(x-2)\leq{0}$$, holds true for them.

Consider inequality $$(x-3)(x-7)>{0}$$ (or $$x^2-10x+21>0$$). This is also upward parabola (as the coefficient of x^2 is positive), with boundary values $$x=3$$ and $$x=7$$. We want to determine for which value of x, this parabola is above (as we have > sign) the X-axis. The answer would be $$x<3$$ and $$x>7$$.

OR:

$$(7+x)(3-x)>0$$ (or $$-x^2-4x+21>0$$), this would be downward parabola, with boundary values $$x=-7$$ and $$x=3$$. This parabola lies above the X-axis for $$-7<x<3$$. We can also write this as $$x^2+4x-21<0$$ (or $$(7+x)(x-3)<0$$), this would be upward parabola, with the same boundary values: $$x=-7$$ and $$x=3$$. This parabola lies below the X-axis for $$-7<x<3$$, the same range as it should be.

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

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04 May 2010, 12:44
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.

Bunuel,

Thanks so much for all these questions and explanations!

I'm having troubles to understand the quoted question. I don't understand your explanation for statement 1, it seems like a great way to solve this problem quickly but I really just don't understand how you get there.

From this 3|x^2 -4| = y - 2 how can you conclude that y must be >=2 ?

Thx for the help!
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04 May 2010, 13:30
nifoui 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.

Bunuel,

Thanks so much for all these questions and explanations!

I'm having troubles to understand the quoted question. I don't understand your explanation for statement 1, it seems like a great way to solve this problem quickly but I really just don't understand how you get there.

From this 3|x^2 -4| = y - 2 how can you conclude that y must be >=2 ?

Thx for the help!

$$3|x^2 -4| = y - 2$$ --> Left hand side (LHS) is an expression with absolute value, absolute value is never negative ($$|expression|\geq{0}$$) --> $$3|x^2-4|\geq{0}$$. So as $$LHS\geq{0}$$, then RHS also must be more than zero ($$RHS\geq{0}$$) --> $$y-2\geq{0}$$ --> $$y\geq{2}$$.

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

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21 May 2010, 02:37
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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21 May 2010, 03:06
1
1
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.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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07 Jul 2010, 00:47
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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07 Jul 2010, 03:38
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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16 Jul 2010, 08:07
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?
Re: Inequality and absolute value questions from my collection   [#permalink] 16 Jul 2010, 08:07

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