Is |x/x^1/2|1

Question

Is  |\frac{x}{\sqrt{x}}| > 1

(1) x^2 -5x + 6 > 0
(2) |x|^2 - 5|x| + 6 > 0

Zarrolou · Answer

Is |\frac{x}{\sqrt{x}}| > 1 the function is not defined for x<0 because in \sqrt{x} x cannot be <0 so the question is : Is \frac{x}{\sqrt{x}}> 1 or \sqrt{x}>1 , x>1 ? (1) x^2 -5x + 6 > 0 x>3 or x<2. Not sufficient to say that x>1 (2) |x|^2 - 5|x| + 6 > 0 Because x cannot be negative this becomes x^2 -5x + 6 > 0, exactly the same as (1). Bad question IMO Of course is not sufficient Because 1 and 2 are the same, together they add no new info. E What is the source?

Archit143 · Answer

Great question..
Solve the the stem...the equation boils down to Mod(Sqr root (x))>1
 Using the statement 1 u can get x<2 or x>3 ......So substitute values and evaluate a yes no condition...u can get both the answers...

Using statement 2 ...u end up getting values for x that are +ve values less than 2 and + values greater than 3.... So here is a catch...u have X = 1 hence square root of 1 is 1 itself.....combining both the situation remains the same as in 2nd statement and hence the answer is E.

Consider kudos if my post helps!!!!!!!!

Archit

blueseas · Answer

sorry Zarrolou,

as you dint liked it..
this question was framed by me.

SKM

KarishmaB · Answer

First thing to note is that since root x is in the denominator, x must be positive.

When is  |\frac{x}{\sqrt{x}}| > 1 ?
When x is greater than  \sqrt{x} . When does that happen? When x is greater than 1. 
So we basically need to answer whether x is greater than 1 or not.

(1) x^2 -5x + 6 > 0
(x - 2)(x - 3) > 0
This tells us that either x < 2 or x > 3. Not sufficient alone.

(2) |x|^2 - 5|x| + 6 > 0
Since x must be positive, this boils down to x^2 - 5x + 6 > 0
This is the same as statement 1. Hence not sufficient alone.

Both together are essentially the same statement so they are not sufficient together.

Answer (E)

WholeLottaLove · Answer

Is (x/√x)>1

First things first, x cannot be negative or zero, as you cannot take the square root of a negative number, nor can x be = ) as that would leave ) in the denominator which isn't valid.

(1) x^2 -5x + 6 > 0
(x - 2)*(x - 3) > 0
x>3 OR x<2
If x>3 then (x-2)*(x-3) is (+)*(+) which is greater than zero.
if x<2 then (x-2)*(x-3) is (-)*(-) which is positive thus greater than zero.

So
x>3
(x/√x)>1
(3/√3)>1 TRUE

x<1
(1/√1)>1 FALSE (1/√1 =1)
INSUFFICIENT

(2) |x|^2 - 5|x| + 6 > 0
As we determined in the stem, x cannot be negative nor can it be zero. Therefore, x must be positive.
|x|^2 - 5|x| + 6 > 0
(x)^2 - 5(x) + 6 > 0
x^2-5x +6 > 0
(x-2)*(x-3) > 0
x>3 OR x<2
If x>3 then (x-2)*(x-3) is (+)*(+) which is greater than zero.
if x<2 then (x-2)*(x-3) is (-)*(-) which is positive thus greater than zero.

We're left with the same information we determined in #1)
INSUFFICIENT

(E)

bumpbot · Answer

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Is |x/x^1/2|>1

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