If x and y are positive, is x^3>y?
1. sqroot(x)>y
2. x>y
The OA is E, but I got D. Could someone please explain. Thanks
is x^3>y>
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- Birottam Dutta
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Hey, x and y are positive but the factor here is that they are not integers. They can be fractions and that is what the question is asking.
(1) sqrt (x)>y.
Let x=1/4. and y=1/3.
sqrt (1/4) = 1/2 >1/3 or y.
but x =1/4 is less than 1/3. so x^3 = 1/64 is also < y. So, this is not sufficient.
(2) x>y.
Using same logic, let x=1/2 and y = 1/3. Again here, x^3 is less than y. So this is not sufficient.
Combining both, sqrt (x)>y and x>y.
Let x = 1/4 and y = 1/5.
here, both x and sqrt (x) are greater than y.
but x^3 = 1/64 is less than 1/5.
So, even this is not sufficient.
All of these conditions are sufficient if x and y are integers but not sufficient for fractions.
So, E is the correct answer.
(1) sqrt (x)>y.
Let x=1/4. and y=1/3.
sqrt (1/4) = 1/2 >1/3 or y.
but x =1/4 is less than 1/3. so x^3 = 1/64 is also < y. So, this is not sufficient.
(2) x>y.
Using same logic, let x=1/2 and y = 1/3. Again here, x^3 is less than y. So this is not sufficient.
Combining both, sqrt (x)>y and x>y.
Let x = 1/4 and y = 1/5.
here, both x and sqrt (x) are greater than y.
but x^3 = 1/64 is less than 1/5.
So, even this is not sufficient.
All of these conditions are sufficient if x and y are integers but not sufficient for fractions.
So, E is the correct answer.
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https://www.youtube.com/watch?v=H7p56NzAVKc
https://www.youtube.com/watch?v=H7p56NzAVKc
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The two statements together are not sufficient because of the weird behavior of numbers between 0 and 1:
If 0 < x < 1, raising it to higher power brings the result closer to zero, so sqrt(x)>y and x>y do not guarantee that x^3 (which is brought closer to zero) will remain larger than y.
If we knew that x>1, there is no weirdness. The fact that x>y quarantees that x^3=x*x*x>y because the factors x are larger than 1 and they scale up/increase the result with each multiplication.
If 0 < x < 1, raising it to higher power brings the result closer to zero, so sqrt(x)>y and x>y do not guarantee that x^3 (which is brought closer to zero) will remain larger than y.
If we knew that x>1, there is no weirdness. The fact that x>y quarantees that x^3=x*x*x>y because the factors x are larger than 1 and they scale up/increase the result with each multiplication.
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