Inequality and absolute value questions from my collection

Hi Bunuel, I've solved this one correctly, but have one question. A is ok - no questions.
Can we manipulate Statement 2 and say |n|*n<1 as |n| is always positive we must be able to do this - but |n|*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.

\(\frac{1}{| n |}>n\) --> multiply by \(|n|\) (we can safely do that since |n|>0): \(n*|n| < 1\).

If \(n>0\), then we'll have \(n^2<1\) --> \(-1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\).
If \(n<0\), then we'll have \(-n^2<1\) --> \(n^2>-1\). Which is true for any n from the range we consider. So, \(n*|n| < 1\) holds true for any negative value of n.

Thus \(\frac{1}{| n |}>n\) holds true if \(n<0\) and \(0<n<1\).

How did you figure out that \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages:
\(\frac{x}{y}>1\)

\(\frac{x}{y}-1>0\)

\(\frac{x}{y}-\frac{y}{y}>0\)

\(\frac{x-y}{y}>0\).

Hope it's clear.

hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!

An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.

thanks for the reply
my understanding about this topic is that...

if x>=0 then |x|=x
and if x<0 then |x|=-x

am i wrong?
please reply!!!

Yes, you are wrong.

We can say that when x<=0, then |x| is also equal to -x:

|0| = -0.

In your step you have taken it as (x-3)^2 but isn't (x^2-6x+9) = (x-3)^2 or (3-x)^2?

Also, what does 'reduce the expression by y' mean?

Both (x-3)^2 and (3-x)^2 are the same.

Hi Bunuel,

Can you please explain why have you not considered (for option1) the other case. I mean, x(x-2)<=0 can lead to two possiblities, one that you have mentioned, and the other one could be just the opposite x<=0 and x>=2. I need to understand this, please reply.

Thanks.

x(x-2)<=0 is true for 0<=x<=2 and not true for any other range.

Check the links below:
Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.

[quote="Bunuel"]SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

OR:

\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.
This one was quite tricky and was solved incorrectly by all of you.

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here.
From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3?
That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

Let me know if I am correct.

Thanks!

No, you are not correct.

From \(y*(x-3)^2=0\) it follows that either \(x=3\) or/and \(y=0\). (2) says that \(x^3<0\), thus x is not 3, therefore y must be 0 --> xy = 0.

Hi Bunuel, thank you so much for such an amazing post, so so so helpful.

A quick query, regarding statement 2 here:
(2) \(\frac{1}{|n|} > n\), Shouldn't it be true for all values of n such that n<1 (n#0) ?

Eg: n =1/2: \(\frac{1}{|(1/2)|} > \frac{1}{2}\) : \(2 > \frac{1}{2}\)

PS: It doesn't alter the final answer though.

Yes, but the fact that it's true for all negative n's was enough to discard this statement. So, we did not need to find the actual range. Still if interested here it is:

\(\frac{1}{| n |}>n\) --> multiply by \(|n|\) (we can safely do that since |n|>0): \(n*|n| < 1\).

If \(n>0\), then we'll have \(n^2<1\) --> \(-1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\).
If \(n<0\), then we'll have \(-n^2<1\) --> \(n^2>-1\). Which is true for any n from the range we consider. So, \(n*|n| < 1\) holds true for any negative value of n.

Thus \(\frac{1}{| n |}>n\) holds true if \(n<0\) and \(0<n<1\).

Hi Bunuel,
From 2, can't we say, |a|/|b|>=0, in this case, both a & B will be either +ve or -ve. Hence, sufficient? Pls pardon my ignorance.

Absolute value of a number is non-negative, so both |a| and |b| are >=0. So, naturally |a|/|b|>=0. But we cannot know whether a and b are positive or negative? Consider a = 1 and b = 2 OR a = -1 and b = 2.