If (10 - x)/3

Question

If  \frac{10 - x}{3} < -2x , which of the following must be true?

I.  2 < x

II.  |x - 5| \geq 7

III.  \frac{|x - 1|}{|x|} > 1

A. I Only
B. II Only
C. III Only
D. II and III
E. I, II, and III

chetan2u · Answer

\frac{10 - x}{3} < -2x 10 - x< 3*-2x 6x-x<-10 5x<-10 x<-2 We are looking for MUST be true. Do not confuse it with range of x. Whatever range consists of x<-2 will be true. I. 2 < x x<-2 does not fall in the given range. So never true. II. |x - 5| \geq 7 Two cases x - 5\geq 7……….x\geq 12 5-x\geq 7……….-2\geq x x<-2 does fall in the range. -2\geq x True III. \frac{|x - 1|}{|x|} > 1…….|x-1|>|x| Square both sides x^2-2x+1>x^2……….x<\frac{1}{2} x<-2 does fall in the range. x<\frac{1}{2} True D

PyjamaScientist · Answer

chetan2u sir, I also did solve the question like this, but could not choose second option as valid.

Quote:

\(x - 5\geq 7……….x\geq 12\)
\( 5-x\geq 7……….-2\geq x\)
x<-2 does fall in the range. \( -2\geq x\)

In the above condition, x can be \(\geq 12\). So, is it enough to be good for a "must be true" question?

Also, I have a question with respect to the method you chose the third condition,
\(\frac{|x - 1|}{|x|} > 1…….|x-1|>|x|\)
I solved it by making three different ranges of absolute modulus's,
First range: x<0. Second range: 0<x<1. Third range: x>1.
I could ascertain from it that \( x<\frac{1}{2}\).
But, you instead just went ahead and squared both the sides, I read somewhere that squaring or multiplying on both sides of inequality is not a good idea as it flips the signs of inequality. Then on that principle is your "squaring" method based upon?

chetan2u · Answer

Yes, x<12 would also be in MUST be true cases.

You can safely square two sides without flipping sign when both sides are non negative. 
Here, both are MOD, so non negative.

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

Given that \(\frac{10 - x}{3} < -2x\)
=> 10 - x < 3*-2x
=> 10 - x < -6x
=> 10 < x - 6x
=> 10 < -5x
=> -5x > 10
=> x < -2

I. \(2 < x\)
Since x < -2 so it cannot be > 2
=> FALSE

II. \(|x - 5| \geq 7\)

Now, lets put values of x < -2
Ex: x = -2.5
=> | -2.5 - 5| = |-7.5| = \(7.5 \geq 7\)
And this will be true for all values of x < -2
=> TRUE

III. \(\frac{|x - 1|}{|x|} > 1\)

Let's try with the same value of x = -2.5
=> \(\frac{|-2.5 - 1|}{|-2.5|}\) = \(\frac{|-3.5|}{|-2.5|}\) = \(\frac{3.5}{2.5}\) > 1
And this will be true for all values of x < -2
=> TRUE

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

gowthamkanirajan · Answer

I also have the same doubt 
For example lets take x = 11 . It satisfy that x<12 but this does not tell me that x <-2 and also on the other side its x<= -2 so if x=-2 it fails and for 3rd option if x=0 then it fails . So please clarify, this is a must be true case

Posted from my mobile device

bumpbot · Answer

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If (10 - x)/3 < -2x, which of the following must be true?

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