(1)
u^3 < v (we can multiply without changing sign safely here since we know u and v are positive)
for this to be true, v is always greater than u (since u and v are both positive and real)
Sufficient
(2)
u^(1/3) < v
u = 8, v = 4 satisfies this
u = 8, v = 16 also satisfies this
Not Sufficient
Hence, (A)
OA?
DS real numbers
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Frankenstein
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Hi,GmatKiss wrote:If u and v are positive real numbers, is u > v?
(1) u^3/v < 1
(2) u^(1/3) / v < 1
u and v are positive real numbers. We shouldn't take for granted that they are integers.
From(1):
u=0.2, v= 0.1 -> u^3/v < 1, u>v
u=2, v=9 -> u^3/v < 1, u<v
Not sufficient
From(2):
Not sufficient as shown in prev. post
Both(1) and (2):
u^3/v < 1
and u/v^3 < 1
So, (u^3/v)*(u/v^3) < 1
=> (u/v)^4<1
So, u < v. As all are positive, we can multiply and take roots this way.
Sufficient
Hence, C
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
