Hi moneyman,

First off, picking fractions or any number, really (as long as it's positive) is the right way to approach this problem, so great job!

A quicker way to look at this is that you don't really need to do the calculations and solve everything. If you practice just looking at the problems and figuring out how to plug in numbers and do some estimating in your head (without actually having to write them out and solving for the numbers), you can save a lot of time.

Hope this helps!

Just a quick question,

II.x^2<1/x<2x

how is this true if x=1/2 ?

pls suggest some value for which this is true..

Thanks

Ya akshat I had the same doubt..need help with the second choice guys!!

Ak/Maxx,

The relation x^2<1/x will only work for x <1> 1 this is not going to work.

As per your assumption of x = 1/2

If you deduce the value as
x^2 = 1/4 = 0.25
1/x = 2

and 0.25< 2

Hope this helps.

DK ...

Well dk you have missed out on 2x and have only calculated for 1/x and x^2..If x=1/2 then 2x=1 which is not greater than 1/x(according to stat 2)

The number 0.9 works with Statement II. It took me a few tries to get that

Figuring out why eq 2 holds is tricky.

Important point to note is that we have to prove 1/x<2x for a fraction between 0 and 1

as we already know that x^2<2x and x^2<1/x holds for all fractions between 0 and 1 (from eq 1)

1/x <2x or x^2 > 1/2, which equates to x>1/sqrt(2) ~ 0.71

so all numbers between 0.71 and 1 will satisfy this condition.

akshatsingh wrote:Just a quick question,

II.x^2<1/x<2x

how is this true if x=1/2 ?

pls suggest some value for which this is true..

Thanks

I was just looking through this problem. I agree that the correct approach is to pick up fractions and work. But if we consider the value of x to 1 that the equation might not hold true.

Or is it that the questions does not require the consideration of x=1 because of "could" used in the question.

Thanks in advance.
-Abhi

Yes
As the question says "could be"
so even if the result holds for one value
it would be considered correct.

This is clearly not a terribly difficult question, but it's tricky!

So what's the takeaway to get this thing done in under 2 mins?

The way I looked at it:

I replaced x with 1/2, 1, and 3, and tried the equations. The only equation that worked was (I). (II) is tricky!!!

What's the trick to "quickly" recognizing that we also need to replace x with 0.71 - 0.9999? Under that assumption, we could sit there for hours, and try and also prove that (III) is right (to no avail obviously).

The key to these problems are picking extreme values and testing.Since its a could be true if we prove any statement to be true for one set of values then thats sufficient

Eg: Pick 1/10, 1/2 and 9/10

Its a tricky problem for sure where picking number is the only viable option.

I) x=1/2

1/4<1<2

TRUE

II) x=9/10

81/100<10/9<1*9/10

TRUE

III) for x=1/10,1/2 and 9/10 it could never be true

D)