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Inequality and absolute value questions from my collection

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New post 17 Nov 2009, 10:13
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

Statement 1:

2(1)-2(1/2)=1 , x,y are both positve

2(1/2)-2(-1/2)=1 x is positive, y is negative

INSUFFICIENT

Statement 2:

Either (x,y) are both positive or both negative

INSUFFICENT

Statement 1 and 2:

With both requirements x must be greater than y and satisfy this equation: 2x-2y=1

2(1)-2(1/2)=1 , x,y are both positve and x>y

2(1/2)-2(-1/2)=1 x is positive, y is negative and x>y

Answer: E
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New post 17 Nov 2009, 10:27
12. Is r=s?

(1) -s<=r<=s

(2) |r|>=s


E – for this - both can be true or false when 0< r < 1
For example , take r as 0.8
S = 0.86 i.e. -0.86 < = 0.8 < = 0.86
|0.8|>= 0.86 i.e. 1 >= 0.86
Combining , any values can be taken , on values > =1 , both r and s
will be same

3. Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a

(2) (x – y)^2 = a
C is the answer

Combined both and the equation will give x^2 + y^2 = 5a
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New post 17 Nov 2009, 10:43
Bunuel wrote:
6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0


Statement 1:

Nothing about y... INSUFFICIENT

Statement 2:

two equations, two unknowns... INSUFFICIENT

Statements 1 and 2:

From x +1 > 0 and the fact that x must be an integer, we know that x must be [0,1,2,3...]

Because we know that xy > 0, we know that x cannot be 0... therefore y must be a positive integer!

SUFFICIENT

ANSWER: C.
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New post 17 Nov 2009, 10:48
4)
I) 2x-2y=1 so y=x-1/2 NS
II)x/y>0 so x and y have the same sign and the modulus of x has to be larger than the modulus of y NS
Together, to satisfy both clues needs to be larger than 1/2 and x becomes larger than 0; the stem is true, therefore C
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New post 17 Nov 2009, 10:53
h2polo wrote:
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

Statement 1:

2(1)-2(1/2)=1 , x,y are both positve

2(1/2)-2(-1/2)=1 x is positive, y is negative

INSUFFICIENT

Statement 2:

Either (x,y) are both positive or both negative

INSUFFICENT

Statement 1 and 2:

With both requirements x must be greater than y and satisfy this equation: 2x-2y=1

2(1)-2(1/2)=1 , x,y are both positve and x>y

2(1/2)-2(-1/2)=1 x is positive, y is negative and x>y

Answer: E


Your last choice of numbers: x=1/2, y=-1/2 does not satisfy clue I, because 2*(1/2)-2*(-1/2)=2, not 1
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New post 17 Nov 2009, 14:01
4
Quote:
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1


Question Stem : x > 0 ; y > 0 ?

St. (1) : 2x -2y = 1
x = y + 0.5
Equation can be satisfied for both positive and negative values of x and y.
Hence Insufficient.

St. (2) : x/y > 1
Equation can be satisfied when both x and y are either positive or negative.
Hence Insufficient.

St. (1) and (2) together : (y + 0.5)/y > 1
1 + 0.5/y > 1
0.5/y > 1
For this to be true, y must be positive.
If y is positive then x will also be positive.
Hence Sufficient.

Answer : C

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New post 17 Nov 2009, 14:10
2
Quote:
5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11


Question stem : What is the exact value of y?

St. (1) : 3*|x^2 -4| = y - 2
y = 3*|x^2 -4| + 2
From this we can infer that y will be a positive value. That is, y > 0. However, we want to know the exact value of y.
Therefore, Insufficient.

St. (2) : |3 - y| = 11
(a) When (3 - y) > 0 ; 3 - y = 11 ; y = -8.
(b) When (3 - y) < 0 ; - (3 - y) = 11 ; y = 14.
Thus we can see that there are two possible values for y.
Hence Insufficient.

St. (1) and (2) together : y > 0 ; y = 14 or -8.
Obviously since y has to be greater than 0, it cannot be -8. Therefore value of y = 14.
Hence Sufficient.

Answer : C
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New post 17 Nov 2009, 14:48
Quote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|


Question stem : Neither a nor b can hold the value 0 ; |a|/|b|=a/b
For condition to be true, both a and b must hold the same sign.

St. (1) : |a*b| = a*b
This condition will be satisfied only when both a and b are either both positive or both negative.
Hence Sufficient.

St. (2) : |a|/|b| = |a/b|
This condition can be satisfied when a and b are same sign as well as opposite sign.
Hence Insufficient.

Answer : A
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New post 17 Nov 2009, 15:36
1
Quote:
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
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New post 17 Nov 2009, 15:48
2
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.

Answer : E

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New post 17 Nov 2009, 18:24
sriharimurthy wrote:
Quote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16


Question Stem : Is n negative?

St. (1) : -n = |-n|
Let -n = A ; therefore the statement becomes : A = |A|.
This can only be valid when A is positive. This in turn means that n must be negative.
Thus Sufficient.


-n=|-n| also for n=0, hence not sufficient.

Everything else is as you suggested, therefore C
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New post 18 Nov 2009, 01:01
Marco83 wrote:
sriharimurthy wrote:
Quote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16


Question Stem : Is n negative?

St. (1) : -n = |-n|
Let -n = A ; therefore the statement becomes : A = |A|.
This can only be valid when A is positive. This in turn means that n must be negative.
Thus Sufficient.


-n=|-n| also for n=0, hence not sufficient.

Everything else is as you suggested, therefore C


Yes, you are right. I overlooked that.
Thanks.
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New post 18 Nov 2009, 17:43
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.


Just curious if my thinking is correct.

on the 2nd part I get y = -8 and y =14
Then I substituted the values into the first equation:
3|x^2-4|=-10
the answer will never give -10/3

do the same for 14
3|x^2-4|=12
x = 0

using my methodology I also got C, but is my thinking correct?
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New post 18 Nov 2009, 19:45
Awesome stuff Bunuel! Hats off to you dude.
+5 from me.
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New post 01 Dec 2009, 11:58
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.


Bunuel, I tried to solve this in another way.

1) 3|x^2 -4| = y - 2
if (x^2 -4) is positive, we can rewrite above as 3(x^2 -4) = y - 2
=> 3x^2-y = 10 -> Eqn. 1
if (x^2 -4) is negative, we can rewrite above as 3(4-x^2) = y - 2
=> -3x^2-y = -14 -> Eqn. 2
Adding equation 1 and 2, we get -2y = -4 or y = 2. So (A) as the answer is tempting.

I know this is not correct and carries the assumption that y is an integer which is not the case here.

If y indeed were an integer in the question, do you think the above approach had any problems ? I am a little confused because every inequality problem appears to have a different method for solving it!

Thanks
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New post 13 Dec 2009, 19:52
lagomez wrote:
Bunuel wrote:

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


I'm getting B for this one

1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient

2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient


hi

how would mod(1-x)<1 would resolve, i mean the interval of x
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New post 14 Dec 2009, 01:21
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New post 22 Dec 2009, 13:55
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.

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.

Answer: D.


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..
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New post 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.

Answer: C.
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New post 22 Dec 2009, 19:09
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Re: Inequality and absolute value questions from my collection   [#permalink] 22 Dec 2009, 19:09

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