is x > y?
1) sqrt(x) > y
2) cube(x) > y
Here is what i think:-
1) sqrt(x) > y
Case 1: 0<x<1
Let x = 0.36 and y = 0.50
sqrt(0.36) > 0.50
=> 0.6 > 0.5 But x=0.36 and y = 0.50 --- And 0.36 > 0.50 is false
Case 2: Let x = 16 and y=3
=> and sqrt(16) > 3 -- which is true
So Case 1 is "False" and Case 2 is "true"
Hence Statement 1 is insufficient.
Statement 2:-
cube(x) > y
Case 1:
cube(2) > 7 but 2 > 7 is "false"
Case 2:
I am NOT able to find a value of x such that 0<x<1
Probably because if cube(x) > y then x>1
Is my line of reasoning correct to decide that
Answer is (B)
Is there a way to think about this problem without substituting numnbers?
Thanks a lot.
is x>y?
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I guess ans-E.
Statement1
sqx >y, or x>y^2, now choose y =1/2, 2, in first case y^2 <y, and in second case y^2>y and hence x can be > or < y
hence insufficient
Statement 2
x^3 > y, and now x >y^1/3 and choose, y=8, x > 2or y^1/3 but less than y, now choose y=0, x=0
Insufficient
Combining 2 , X>y^2, x>y^1/3 still doesnt help, hence E
Statement1
sqx >y, or x>y^2, now choose y =1/2, 2, in first case y^2 <y, and in second case y^2>y and hence x can be > or < y
hence insufficient
Statement 2
x^3 > y, and now x >y^1/3 and choose, y=8, x > 2or y^1/3 but less than y, now choose y=0, x=0
Insufficient
Combining 2 , X>y^2, x>y^1/3 still doesnt help, hence E
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we can not do like you because, maybe y<0kg wrote:I guess ans-E.
Statement1
sqx >y, or x>y^2, now choose y =1/2, 2, in first case y^2 <y, and in second case y^2>y and hence x can be > or < y
hence insufficient
Statement 2
x^3 > y, and now x >y^1/3 and choose, y=8, x > 2or y^1/3 but less than y, now choose y=0, x=0
Insufficient
Combining 2 , X>y^2, x>y^1/3 still doesnt help, hence E
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- thephoenix
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ans can't be D as s1) is def not suffdmitriyaleyev wrote:I still don't get why the answer is not D...
Please help.
to prove
we are given
sqrt(x)>y
we are asked if x>y or not
let us try by pluging no's bth the cases
case1) when x>y ; prove sqrtx>y
e.g:- x=4 and y=1
x>y true
sqrtx> y--->2>1 true
case 2)
let x=0.5 and y=0.3
x>y true
but sqrtx > y...false
hencde sqrt X > Y ; but for different values results are diff
hence s1) is not suff and threfore ans can't be D
s2)
x^3>y
similarly
if x=3 and y=2
X>Y
and X^3>y
but when X=3 and Y=9
X^3>Y holds true but
X>Y does not holds true
hence insuff
Hence the ans shud be E
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the phoenix,
Thanks for your explanation.
I see I overlooked s2.
However,
sqrt of 0.5 --->>> appx 0.7
Therefore,
"case 2)
let x=0.5 and y=0.3
x>y true
but sqrtx > y...false "
if x=0.5 and y=0.3
sqrtx will still be more than 0.3...
or am i going crazy ? ))
Thanks for your explanation.
I see I overlooked s2.
However,
sqrt of 0.5 --->>> appx 0.7
Therefore,
"case 2)
let x=0.5 and y=0.3
x>y true
but sqrtx > y...false "
if x=0.5 and y=0.3
sqrtx will still be more than 0.3...
or am i going crazy ? ))