Inequality and absolute value questions from my collection

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 (or equal to 0). This in turn means that n must be negative (or equal to 0).
-n=|-n| also for n=0, hence not sufficient.

St. (2) : n^2 = 16
n = ±4.
Thus Insufficient.



Answer : C

Question Stem : n # 0 ; -4 > n > 4 (excluding 0)

St. (1) : n^2 > 16
(n - 4)*(n +4) = 0 Therefore boundary conditions are 4 and - 4. Thus we can write it as : n < -4 and n > 4.
Sufficient.

St. (2) : 1/|n| > n
This condition will be valid for all n < 1 excluding 0.
Thus it will be impossible to tell whether |n| < 4.
Hence Insufficient.

Answer : A

Question Stem : Is the distance between X and -Y greater than the distance between X and Y?
Note : Using number line approach.

St. (1) : |x| > |y|
OR, The distance between X and the origin is greater than the distance between Y and the origin.

Now we can have two cases :

(a) When X is positive : In this case, X > Y for the above condition to be true.

_________|_________|_________|_________|_________|__________
_________________-Y______Origin______Y........ Region of X.........

Thus we can see that the distance between X and - Y will always be greater than the distance between X and Y. Hence question stem is true.

(b) When X is negative : In this case, X < -Y for the statement to be true.

_________|_________|_________|_________|_________|__________
.... Region of X.....-Y_____Origin_______Y___________

Thus we can see that the distance between X and -Y will always be less than the distance between X and Y. Hence, question stem becomes false.

Due to conflicting statements, St (1) becomes Insufficient.

St. (2) : |x-y| < |x|
OR, the distance between X and Y is less than the distance between X and the origin.

Now again let us consider the different cases :

(a) When X is positive : For the statement to be true and for X to be positive, X must be greater than Y/2. For any value of X less than Y/2 the statement will become false. The statement will be true for any value greater than Y/2.

Thus we see that only one case comes into the picture. Now let us see how it relates to the question stem.

_________|_________|____;____|_________|________
________-Y____Origin__y/2...Y...... Region of X....

Thus we can see that the distance between X and - Y will be greater than the distance between X and Y for all values of X > Y/2. Thus question stem is true.
St. (2) is sufficient.

Answer : B

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

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

-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.

just to chime in here your thanks for all this..it's really useful

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?

Well you can even not calculate for x. Statement 1 says that y must be greater than or equal to 2. Statement 2 gives 2 values of y -8 OR 14. Combining we get that y=14.

Remember we are asked to determine the value of y not x.

Awesome stuff Bunuel! Hats off to you dude.
+5 from me.

Awesome, not only have u put the question, but solution to all the problems.
I am learning a lot. Thanks to Bunuel.
Bunuel, more questions please.

Thank you very much for this catch. +1. There was a typo. So you are right with the first one:
|x-1| < 1 means 0<x<2. Already edited the post.

As for the second one:
x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.

x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.

Hope it helps.

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

This approach is not correct not because we are not told that y is an integer, but because you can not add inequalities like you did.

3(x^2 -4) = y - 2 OR 3(4-x^2) = y - 2, in fact these equation are derived from one and from them only one is right. It's not that we have 3(x^2 -4) = y - 2 AND 3(4-x^2) = y - 2 and we are asked to solve fro unknowns. If it were then your solution would be right.

Hope it's clear.

hi

how would mod(1-x)<1 would resolve, i mean the interval of x

|x-1|<1:

|x-1| switch sign at x=1, which means that we should check for the two ranges:

A. x<1 --> -x+1<1 --> x>0;

And

B. x>=1 --> x-1<1 --> x<2;

Hence |x-1|<1 can be rewritten as 0<x<2.

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..