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inequality doubt..

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inequality doubt..

by rahulg83 » Mon Jun 22, 2009 11:52 am
Is x > y?

(1) x^0.5 > y

(2) x^3 > y

[spoiler]Somehow i am not convinced with the explanation. (especially when it says that x^0.5 has to be positive)
Please explain your answers.

Source MGMAT CAT...[/spoiler]
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by abhinav85 » Mon Jun 22, 2009 12:02 pm
"Somehow i am not convinced with the explanation. (especially when it says that x^0.5 has to be positive)"

let me try to explain this!!!

x^0.5 means x^1/2

x^1/2 is equal to squareroot x.

and squareroot x will always be postive.

hope this helps
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by rahulg83 » Mon Jun 22, 2009 12:08 pm
Hey but square root can be negative as well..
It's not fair to assume it to be positive
square root of 25 can be 5 or -5. Anyways can you please explain, how to solve this particular problem?
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by abhinav85 » Mon Jun 22, 2009 12:21 pm
IMO D for this problem.

I already explained the first stat. suff.

from the second statement we get,

x^3 > y

lets take some extreme numbers for this.......

x=-1/2 and y = -1/4

that will be
-1/8<-1/4.

x is not negative.

Hence x and y can only be postive inetgers.

x^3 > y.
Last edited by abhinav85 on Mon Jun 22, 2009 12:25 pm, edited 1 time in total.
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by rahulg83 » Mon Jun 22, 2009 12:24 pm
abhinav85 wrote:There is a difference.

let say
between squareroot5 and x^2 = 5.

squareroot5 could be 25 not -25.
ya if its squareroot-5 it will be -25.

and for x^2 = 5 will be -5<x<5.
it could be -5 or 5.

IMO D for this problem.

I already explained the first stat. suff.

from the second statement we get,

x^3 > y

lets take some extreme numbers for this.......

x=-1/2 and y = -1/4

that will be
-1/8<-1/4.

x is not negative.

Hence x and y can only be postive inetgers.

x^3 > y.
OA is C
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by david4431 » Tue Jun 23, 2009 7:20 am
Answer: A.

S1: x^0.5 > y. Since you are told that the square root of x is greater than y, squaring sqrt(x) would certainly give you a value greater than y.

S2: x^3 > y. Consider 2 cases:

Case 1: x^3 = 27, and y = 7
In this case, x = 3, which is less than 7.
Case 2: x^3 = 27, and y = 2
In this case, x = 3, which is greater than 2.

In case 1, x<y or 3<7. In case 2, x>y or 3>2. Insufficient since you are getting two responses.
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by deltaforce » Tue Jun 23, 2009 7:38 am
clearly the ans is c

from the first statement if _/x > y than two values will giv diff answer

take x = 4 and y = 1 if u take the sqrt the first condition is satisfied 2>1
now take x = 1/4 and y = 1/3 - once u sqrt 1/4 it becomes 1/2 which is greater than y but when u test the values 1/4> 13 is not working

2 . x^3>y again take x = 2, y= 3 than 8>3 -> x>y no
x =2 y=1 than 8>1 -> x>y yes

two answers again
combined only integer will satisfy and that which ylds a value satisfying st 1

hope this helps
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add on

by deltaforce » Tue Jun 23, 2009 7:43 am
typo
1/4> 1/3

as a rule whenever condition / ques doesn't mention type of value x can be run fractions positive and negative nos to test all values.
secondly establish a set of 4 - 6 nos which include fractions, negatives positives and always test those numbers..yr speed will increase dramatically

something like -1/2, 1/4, 1/6
positive - 2,4,8
neg - -2,-4,-8
extreme - nos like _/2 or _/3
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by mikeCoolBoy » Tue Jun 23, 2009 7:57 am
rahulg83 wrote:Hey but square root can be negative as well..
It's not fair to assume it to be positive
square root of 25 can be 5 or -5. Anyways can you please explain, how to solve this particular problem?
in this case sqrt(x) is always positive.

when you pick up numbers you want to fulfill the condition of the statement but you don't pick up the number according to the statement.

let's phrase the next question

is x > 0 ?
statement 1) sqrt(x) > 0 this statement is sufficient to assure that x > 0
just try to pick a number x < 0 with a sqrt(x) > 0


a different question: is sqrt(x) > 0

statement 1) x > 0 this statement is not sufficient because

sqrt(x) = -4 ---> x > 0
sqrt(x) = 4 ----> x > 0

Let me know this was what you were asking for.
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by david4431 » Tue Jun 23, 2009 7:58 am
deltaforce,

That was a great tip. Thanks a lot. Answer C seems correct.
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