Is x 0 ? (1) |3 - x|

Question

Competition Mode Question

Is x > 0 ?

(1) |3 - x| < |x + 5|

(2) |3 - 2x| < x - 1

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ArunSharma12 · Accepted Answer

statement 1: critical points for x are -5 and 3. case A: x < -5 3 - x < -x - 5, x cancels out and 3 is not less than -5 so this is not a valid range. case B: -5< x < 3 3 - x < x + 5 2x > -2; x > -1 valid range for x: -1< x < 3 case c: x > 3 -3 + x < x + 5, x cancels out and -3 is less than 5 so this range is valid for all values of x. valid range for x: x > 3. -1 < x < infinity not sufficient statement 2: |3 - 2x| < x - 1 critical points are x = 3/2 case A: x < 3/2 3 - 2x < x - 1 3x > 4; x > 4/3 valid range for x: 1.3 < x < 1.5 case B: x > 3/2 -3 + 2x < x -1 x < 2 valid range for x: 1.5< x < 2 1.3 < x < 2 sufficient Ans: B

nick1816 · Answer

Statement 1-  The distance between x and -5 is greater than that between x and 3.

Hence,  x>\frac{3+(-5)}{2}  or x >-1

Insufficient

Statement 2-  since LHS (|3x-2|) is always non-negative, solution of this inequality must be greater than 1.

Sufficient

QuantMadeEasy · Answer

Step 1: Understanding statement 1 alone 
Using value of x as positive and negative
Taking positive value as x = 0.1, |3-0.1|< |0.1+5| which is true. Answer to the question "yes"
Taking negative value as x = -0.1, |3+0.1|< |-0.1+5| which is true. Answer to the question "no"
Insufficient

Step 2: Understanding statement 2 alone 
Manipulating the inequality
|3 - 2x| + 1 < x 
|3 - 2x| is always positive as it is an absolute quantity, and as x is greater than |3 - 2x| +1, hence x greater than 0.
Sufficient

IMO B

minustark · Answer

1) | 3-x | < |x + 5 |. We can notice the inequality holds true for x = 0 and x = 2. As for negative number, it will hold true until x is greater than -1. No unique values ascertained. Not sufficient.

2) | 3 - 2x| < x -1. The left hand side is never negative. So, x - 1 >= 0. So, x >= 1. Actually x can take any value from 1 to 2 to hold the inequality true. Sufficient.

B is the answer.

exc4libur · Answer

(1) insufic x=0: 3<5 true x=1: 2<6 true (2) sufic pos: 3-2x>0, 3>2x, x<1.5 pos: 3-2x1.333, so 1.3333, x>1.5 neg: -3+2x 0 Ans (B)

monikakumar · Answer

Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 1) 3-x < x+5 -2<2x -1-1 Insufficient 2)|3-2x|0 or +ve x>1 Ans B Posted from my mobile device

NCC · Answer

Hello,
Here you mean to say in Statement 1 that all 3 regions are valid right? x> -1, x<-1 and -3<5??

NCC · Answer

Hello folks,
Do share your solution for this.
I approached it like this, please tell me if it is correct.

1} | 3-x| < | x+5|

roots = 3,-5
plotting on number line
-5 3
reg1 reg2 reg3

region1:
3-x >0 & x+5 <0
3-x < -x-5
3<-5
not possible so this region is rejected.

region2:
both >0
3-x<x+5
x>-1

so accepted region is -1 to 3

region3:
3-x <0 & x+5>0
-3+x<x+5
-3<5
accepted

therefore, -1 < x < infinity
insuff

2} | 3-2x| < x-1
When x>3/2 --> 3-2x<x-1 -----> x>4/3

When x<3/2 ---> -3+2x<x-1 -----> x<2

So, 4/3<x<2
sufficient

Ans B

chetan2u · Answer

Quick solution (1) |3 - x| < |x + 5| a) Clearly x=0, gives a NO. b) x as any positive value 0.1, 1, 100 etc will give a YES. Insuff (2) |3 - 2x| < x - 1 Surely x-1>0 as x-1 is greater than an absolute value, which always nonnegative. Sufficient B Algebraic solution for statement 1 (1) |3 - x| < |x + 5| Square two sides 9+x^2-6x-16..............x>-1 Insufficient

gmatophobia · Answer

Hi  nick1816 ,

Can you explain the logic / formula used in the highlighted portion.

I obtained the result, x > -1, by squaring both sides of the equation. Your approach is shorter, however I am unable to figure out the logic behind the equation.

My thought process so far

We know that "The distance between x and -5 is greater than the distance between x and 3. "

---------- -5 ----------------- 0 -----------3 ----------------

x can be at the following locations in the number line

---------- -5 ------------- -1 -- X -- 0 ----- X ------3 ---- X -----------

Not sure how you arrived at  x>\frac{3+(-5)}{2}

Thanks in advance !

chetan2u · Answer

Hi

We can say that x is closer to 3 than to -5. So, if you take the mid point of 3 and -5,  \frac{3+(-5)}{2} , x should lie between this midpoint and 3.
Hence, x> \frac{3+(-5)}{2}  or x>-1

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Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1

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