Inequality and absolute value questions from my collection

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

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

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

Everything else is as you suggested, therefore 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?

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

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

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..

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.

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 .

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.

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.

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)

Two different answers. Not sufficient.

Answer: E.

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.

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.