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25 Jan 2022, 07:08
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
33% (01:55) correct
68%
(02:16)
wrong
based on 40
sessions
History
Date
Time
Result
Not Attempted Yet
25 Jan 2022, 22:28
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Bunuel
(1) \(x^2 > 2y\)
If x<0, we can have both possibilities.
x=-3 and y=-2.......y>x
x=-3 and y=-4.......x>y
If x>0, again both possibilities
x=1, y=1/3....x>y
x=4, y=7.....y>x
Rather 0<x<1 will give \(x>x^2\), so \(x>x^2>2y\) or x>y
Insufficient
(2) \(x^3 > y^4\)
This tells us that x is positive.
If x>1, then \(x^3 > y^4........x>y^{\frac{4}{3}}>y\)
If x<1, then \(x^3 > y^4........x^{\frac{3}{4}}>y\)..
But \(x^{\frac{3}{4}}>x\)WHY? Because value decreases with increase in power and vice versa.
So, \(x^{\frac{3}{4}}>y>x\) or \(x^{\frac{3}{4}}>x>y\)
Insufficient
Combined
We know x is positive from statement II.
Case I: 0<x<1........x>y from statement I.
Case II: x>1........x>y from statement II.
Thus, both statements are satisfied only when x>y.
Sufficient
C
18 Jul 2022, 19:31
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hi chetan2u,
I didn't understand how we deduce this highlighted part. Can you elaborate a bit, please?
Thanks
I didn't understand how we deduce this highlighted part. Can you elaborate a bit, please?
Thanks
chetan2u
