If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ?
I.x^2<2x<1/x
II.x^2<1/x<2x
III.2x<x^2<1/x
None
I only
III only
I and II only
I II and III
Ans D
I picked a fraction and proved I right but as I started picking values I realised I am using more than 2 mins for this question..Any easier way to solve this problem??
Exponents
This topic has expert replies
-
- GMAT Instructor
- Posts: 1223
- Joined: Thu May 01, 2008 3:29 pm
- Location: Los Angeles, CA
- Thanked: 185 times
- Followed by:15 members
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!
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!
Jim S. | GMAT Instructor | Veritas Prep
-
- Senior | Next Rank: 100 Posts
- Posts: 77
- Joined: Thu Apr 10, 2008 10:13 pm
- Thanked: 4 times
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
II.x^2<1/x<2x
how is this true if x=1/2 ?
pls suggest some value for which this is true..
Thanks
Aks
-
- Legendary Member
- Posts: 631
- Joined: Mon Feb 18, 2008 11:57 pm
- Thanked: 29 times
- Followed by:3 members
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.
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
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
-
- Senior | Next Rank: 100 Posts
- Posts: 47
- Joined: Wed Oct 08, 2008 11:29 am
- Location: Toronto, Ontario
- Thanked: 5 times
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).
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).
-
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
- Followed by:11 members
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)
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)