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]
inequality doubt..
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 179
- Joined: Mon Dec 29, 2008 4:41 am
- Thanked: 2 times
"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
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
- rahulg83
- Legendary Member
- Posts: 575
- Joined: Tue Nov 04, 2008 2:58 am
- Location: India
- Thanked: 18 times
- Followed by:4 members
- GMAT Score:710
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?
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?
-
- Master | Next Rank: 500 Posts
- Posts: 179
- Joined: Mon Dec 29, 2008 4:41 am
- Thanked: 2 times
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.
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.
- rahulg83
- Legendary Member
- Posts: 575
- Joined: Tue Nov 04, 2008 2:58 am
- Location: India
- Thanked: 18 times
- Followed by:4 members
- GMAT Score:710
OA is Cabhinav85 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.
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.
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.
-
- Junior | Next Rank: 30 Posts
- Posts: 23
- Joined: Tue May 26, 2009 5:30 am
- Thanked: 2 times
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
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
-
- Junior | Next Rank: 30 Posts
- Posts: 23
- Joined: Tue May 26, 2009 5:30 am
- Thanked: 2 times
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
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
-
- Master | Next Rank: 500 Posts
- Posts: 119
- Joined: Tue Jan 20, 2009 10:16 pm
- Thanked: 9 times
- GMAT Score:730
in this case sqrt(x) is always positive.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?
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.