if x is positive is x<1?

conomav wrote:This is question is from kaplan cd...
if x is positive number is x<1?
1. x > sqrt(x)
2. - sqrt(x) >-x

how to approach this one?
OA:D

for 1 If i take x=1/4 then sqrt(x) can be -1/2/1/2 which mean x can be <1
but if x=4 sqrt(x) is -2/2 so can go eitherways..

Hi conomav,

For 1, if x=1/4, sqrt(x) = 1/2. it can't be -1/2 as x is a positive number.
so x cant be 1/4.
Anyways, according to 1.) x>1 and it gives us a definite no. hence, insuff.

For 2,
- sqrt(x) >-x which is just a rephrase of 1 choice.
Multiply with -1 and we get
sqrt(x) < x which is same as option 1.

hence the ans is D.
hope it clears the confusion....

if x is positive number is x<1?
1.x > sqrt(x)
2.- sqrt(x) >-x
from 1

square both sides

x^2>x..........suff

from 2

* by -1

gives you the same statement as1

D

conomav wrote:This is question is from kaplan cd...
if x is positive number is x<1?
1.x > sqrt(x)
2.- sqrt(x) >-x

how to approach this one?
OA:D

for 1 If i take x=1/4 then sqrt(x) can be -1/2/1/2 which mean x can be <1
but if x=4 sqrt(x) is -2/2 so can go eitherways..

You need to remember the information provided in the question itself. In this case, we know that x must be greater than 0, so we can ignore any non-positive solutions.

(1) x > sqrt(x)

As with most number property yes/no questions, there are two approaches we can take: pick numbers or use our knowledge of number properties rules.

When we pick numbers, we MUST do so in two stages:

First, we pick numbers that follow all the rules we've been given.

Second, we plug those numbers back into the original question, with the goal of getting both a "yes" and a "no" answer.

Here, we have two rules to follow: x must be positive and x must be greater than the square root of x.

We want to see if we can get both a "yes" and a "no" answer to x being less than 1, so let's try a number bigger than 1 and a number smaller than 1.

x = 16 (we're allowed to pick this number, since it's positive and greater than it's square root).

Is 16 < 1? NO

x = 1/2 (we're not allowed to pick this number, since it's NOT greater than it's square root).

x = 1/4 (we're not allowed to pick this number, since it's NOT greater than it's square root).

x = any number between 0 and 1 (we're not allowed to pick this number, since it's NOT greater than it's square root).

So, the only answer we can get to the question is "NO"; therefore, statement (1) is sufficient.

(2) -sqrt(x) > -x

Let's simplify by multiplying both sides by (-1). Remember, when you multiply both sides of an inequality by a negative number, the inequality swaps direction. So:

(-1)(-sqrt(x)) > (-1)(-x)
sqrt(x) < x

Hey, that's the exact same as (1)! Well, if (1) was sufficient, (2) certainly is as well.

Each of (1) and (2) is sufficient alone: choose (D).

* * *

As an aside, any time the two statements are identical, the only possible answers to the question are (D) and (E). So, even if you didn't understand this question, if you saw that the statements were exactly the same you had a 50/50 chance at getting the question correct.

yezz wrote:if x is positive number is x<1?
1.x > sqrt(x)
2.- sqrt(x) >-x
from 1

square both sides

x^2>x..........suff

from 2

* by -1

gives you the same statement as1

D

You CANNOT safely square both sides of an inequality unless you know the signs of each side.

For example, if you square both sides of the correct statement:

2 > -2

you get:

4 > 4

which is clearly not true.

If you do know the signs of both sides, then:

(1) if both sides are positive and you square them, nothing funny happens.

(e.g. if you square both sides of 3 > 2, you end up with 9 > 4)

(2) if both sides are negative and you square them, then the inequality swaps direction.

(e.g. if you square both sides of -2 > -3, you end up with 4 < 9)

(3) if the sides have different signs, then you're not actually multiplying both sides by the same number, so you should NOT square both sides.

(note: in this particular question, since we know that x is positive, it is in fact safe to square both sides of statement (1), but if you square both sides of statement (2) without eliminating the negatives first, you need to swap the direction of the inequality after squaring.)

thanks Stuart.