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by sanju09 » Thu Apr 16, 2009 3:17 am
Is x > y?

(1) √x > y

(2) x^3 > y



OA C
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Source: — Data Sufficiency |
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by sands_of_time » Thu Apr 16, 2009 11:18 pm
Hi sanju09,

Isn't the [spoiler]condition (1)[/spoiler] sufficient here ?
Am i missing something ?
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by sanju09 » Fri Apr 17, 2009 12:18 am
sands_of_time wrote:Hi sanju09,

Isn't the [spoiler]condition (1)[/spoiler] sufficient here ?
Am i missing something ?
If √x > y, then x could be both more or less than y; see this in the following examples:

&#8730;(1/4) > 1/3 but 1/4 < 1/3, and

&#8730;16 > 3 and also, 16 > 3.
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by cubicle_bound_misfit » Fri Apr 17, 2009 1:14 pm
I am not convinced with my own solution !!!!

(1) & (2) together tell us

x has to be >0 and x can not be a fraction then both x^3 and X^1/2 can not be greater than y

hence if x is a positive integer and x^1/2 > y, x > y

any better explanation?
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by sanju09 » Sat Apr 18, 2009 1:00 am
cubicle_bound_misfit wrote:I am not convinced with my own solution !!!!

(1) & (2) together tell us

x has to be >0 and x can not be a fraction then both x^3 and X^1/2 can not be greater than y

hence if x is a positive integer and x^1/2 > y, x > y

any better explanation?
Your interpretation is correct, though x can be a fraction also, provided it is a mixed fraction to the right of 0 on the real number line. We know the square root of such numbers is always less than the number and remains greater than 1. So if the square root of a number as such is greater than a given number, then the number in question is definitely greater than the given number; read x and y, respectively, in this case. The amalgamation was needed, so C.
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by ketkoag » Mon Apr 20, 2009 10:47 am
take the values less between 0 and 1 as well coz then x^2< x.
and thus check some possibilities by assuming values less than 1 as well as greater than 1. u'll get answer to be C. :)
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by Uri » Tue Apr 21, 2009 4:05 am
(1) Is insufficient.
&#8730;(1/4) > 1/3
and ¼ < 1/3
Again, &#8730;16 > 3
And, 4 > 3
Since both are possible, (1) is insufficient
(2) Is also insufficient
3^3>4 and 3<4
Again, 3^3>2 and 3>2
Since both are possible, (2) is insufficient
(1) And (2) both are together sufficient. No proper fraction can satisfy both the conditions. Negative numbers are also ruled out, since (1) talks about taking square root. So, we are left with only positive integers and improper fractions for x. And if x satisfies both (1) and (2), then x>y.
Ans. (C)

Can anyone please provide a better explanation? My logic took me more than 10 minutes, perhaps. Is there any general way to attack this type of problems?
Last edited by Uri on Tue Apr 28, 2009 9:53 pm, edited 1 time in total.
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by sanju09 » Tue Apr 21, 2009 4:51 am
Uri wrote:(1) Is insufficient.
&#8730;(1/4) > 1/3
and ¼ < 1/3
Again, &#8730;16 > 3
And, 4 < 3
Since both are possible, (1) is insufficient
(2) Is also insufficient
3^3>4 and 3<4
Again, 3^3>2 and 3>2
Since both are possible, (2) is insufficient
(1) And (2) both are together sufficient. No proper fraction can satisfy both the conditions. Negative numbers are also ruled out, since (1) talks about taking square root. So, we are left with only positive integers and improper fractions for x. And if x satisfies both (1) and (2), then x>y.
Ans. (C)

Can anyone please provide a better explanation? My logic took me more than 10 minutes, perhaps. Is there any general way to attack this type of problems?
This should not have taken you more than 10 minutes; please take it seriously. How is 4 < 3, by the way?
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by Uri » Tue Apr 28, 2009 9:57 pm
sanju09 wrote:
Uri wrote:(1) Is insufficient.
&#8730;(1/4) > 1/3
and ¼ < 1/3
Again, &#8730;16 > 3
And, 4 < 3
Since both are possible, (1) is insufficient
How is 4 < 3, by the way?
Thanks sanju for pointing at the typo. I have rectified it now.
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