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by sanju09 » Wed Apr 06, 2011 12:54 am
Is the integer a less than the integer b?

(1) a^3 < b^3.

(2) a^2 < b^2.



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Source: — Data Sufficiency |
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by vineeshp » Wed Apr 06, 2011 5:07 am
OA A?

1) cubes dont change the sign of a and b.

a ^ 3 = 1
b ^ 3 = 8
a = 1 , b = 2

a ^ 3 = -1
b ^ 3 = -1/8
a = -1 , b = -1/2

It holds for all values.

2) not sufficient,
squares can change the sign. So a and b can be positive or negative.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by sanju09 » Wed Apr 06, 2011 5:38 am
vineeshp wrote:OA A?

1) cubes dont change the sign of a and b.

a ^ 3 = 1
b ^ 3 = 8
a = 1 , b = 2

a ^ 3 = -1
b ^ 3 = -1/8
a = -1 , b = -1/2

It holds for all values.

2) not sufficient,
squares can change the sign. So a and b can be positive or negative.
why did you consider that example in bold?
The mind is everything. What you think you become. -Lord Buddha



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by umesh321 » Wed Apr 06, 2011 6:13 am
sanju09 wrote:Is the integer a less than the integer b?

(1) a^3 < b^3.

(2) a^2 < b^2.



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{a^3-b^3}< 0
(a-b)(a^2+b^2-ab)<0
(a-b)(a^2+b^2-ab)<0

since [a^2+b^2]/2>ab
thus a<b
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by vineeshp » Wed Apr 06, 2011 6:18 am
sanju09 wrote:
vineeshp wrote:OA A?

1) cubes dont change the sign of a and b.

a ^ 3 = 1
b ^ 3 = 8
a = 1 , b = 2

a ^ 3 = -1
b ^ 3 = -1/8
a = -1 , b = -1/2

It holds for all values.

2) not sufficient,
squares can change the sign. So a and b can be positive or negative.
why did you consider that example in bold?
Actually, I was just verifying that for negative values greater than -1, there are no loopholes.
Have I made a mistake somewhere?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by sanju09 » Thu Apr 07, 2011 1:15 am
vineeshp wrote:
sanju09 wrote:
vineeshp wrote:OA A?

1) cubes dont change the sign of a and b.

a ^ 3 = 1
b ^ 3 = 8
a = 1 , b = 2

a ^ 3 = -1
b ^ 3 = -1/8
a = -1 , b = -1/2

It holds for all values.

2) not sufficient,
squares can change the sign. So a and b can be positive or negative.
why did you consider that example in bold?
Actually, I was just verifying that for negative values greater than -1, there are no loopholes.
Have I made a mistake somewhere?
If the question calls a and b as integers, how can we possibly consider b = -1/2 in any case?
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by arjunshn » Thu Apr 07, 2011 1:22 am
hey can some one help with the previous topic
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by vineeshp » Thu Apr 07, 2011 1:32 am
Oops! I dint read the word integer there! :S
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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