1/x, 2x, x^2,
Number plugging, and since x is positive, test with x = 1/2 and x = 3
With
x = 1/2, can find that I is correct.
With
x = 3, I find that x^2 > 2x > 1/x.
Therefore for me, I only found I to be correct.
How is that II is also correct ?
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Stuart@KaplanGMAT
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II is correct for large fractions.jymloke wrote:1/x, 2x, x^2,
Number plugging, and since x is positive, test with x = 1/2 and x = 3
With
x = 1/2, can find that I is correct.
With
x = 3, I find that x^2 > 2x > 1/x.
Therefore for me, I only found I to be correct.
How is that II is also correct ?
For example, if x = 9/10, then we get:
81/100 < 10/9 < 18/10
which is true.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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logitech
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Stuart,Stuart Kovinsky wrote:
II is correct for large fractions.
For example, if x = 9/10, then we get:
81/100 < 10/9 < 18/10
which is true.
How did you know that ? I just tried
three values:
0, 1/2 and 2
What is the algebra behind number 2 ?
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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jymloke
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Hi Stuart,
I like to understand your thought process when answering this sort of questions.
My initial thought process went like this,
try with x > 1 (integer) , x > 1 (fractions), x < 1(fractions)
I never thought of trying it with bigger fractions. The fractions I tested were
1/2 and 10/9 , integer that I tried was 3, since trying 2 would give 2x == x^2
I like to understand your thought process when answering this sort of questions.
My initial thought process went like this,
try with x > 1 (integer) , x > 1 (fractions), x < 1(fractions)
I never thought of trying it with bigger fractions. The fractions I tested were
1/2 and 10/9 , integer that I tried was 3, since trying 2 would give 2x == x^2
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logitech
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1/x < 2x
2x - 1/x > 0
2x^2-1 > 0
2x^2 > 1
x^2 > 1/2
x > 1/1.4
x > 10/14
x > 5/7
NICE!!!
lets say x=6/7
X^2 = 36/49
1/X = 7/6
2X = 12/7
and this satisfies the evil statement II :twisted:
2x - 1/x > 0
2x^2-1 > 0
2x^2 > 1
x^2 > 1/2
x > 1/1.4
x > 10/14
x > 5/7
NICE!!!
lets say x=6/7
X^2 = 36/49
1/X = 7/6
2X = 12/7
and this satisfies the evil statement II :twisted:
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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Stuart@KaplanGMAT
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Hi!jymloke wrote:Hi Stuart,
I like to understand your thought process when answering this sort of questions.
My initial thought process went like this,
try with x > 1 (integer) , x > 1 (fractions), x < 1(fractions)
I never thought of trying it with bigger fractions. The fractions I tested were
1/2 and 10/9 , integer that I tried was 3, since trying 2 would give 2x == x^2
We know when comparing 1/x and x^2 that we don't need to try different fractions, since all fractions follow the same rule in this regard. However, when comparing 1/x and 2x, we can't be sure that different fractions will all give the same result.
A good general rule for picking numbers is to think about what types of numbers could give different answers. Going to extremes is often a good way to test things out. So, since I wasn't sure if all fractions would work out the same, I tried both a small one and a big one.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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cramya
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Stuart,A good general rule for picking numbers is to think about what types of numbers could give different answers. Going to extremes is often a good way to test things out. So, since I wasn't sure if all fractions would work out the same, I tried both a small one and a big one.
I am sure that's why u r a 800 scorer on GMAT
Thanks again for your valuable inputs!
Regards,
Cramya
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logitech
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Cramya, I guess you have a value for Stuart's inputs, but I do not..they are really INVALUABLE!cramya wrote:
Stuart,
I am sure that's why u r a 800 scorer on GMAT
Thanks again for your valuable inputs!
Got ya!
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
I used another approach:
in my mind, plot the three functions on a plane and see where they intersect by solving for x:
1/x = 2x
1/x = x^2
2x = x^2
1st: ~ 1/1.4 = 0.7
2nd: 1
3rd: 2
so if 0<x<1, then x^2 is smallest and 1/x intersects 2x at approximately 0.7
and if x > 1, then 1/x is smallest and 2x intersects x^2 at 2
in my mind, plot the three functions on a plane and see where they intersect by solving for x:
1/x = 2x
1/x = x^2
2x = x^2
1st: ~ 1/1.4 = 0.7
2nd: 1
3rd: 2
so if 0<x<1, then x^2 is smallest and 1/x intersects 2x at approximately 0.7
and if x > 1, then 1/x is smallest and 2x intersects x^2 at 2
