DS real numbers

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GmatKiss: Legendary Member; Posts: 2789; Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members; Location: Chennai, India; Thanked: 206 times; GMAT Score:640

by GmatKiss » Mon Aug 15, 2011 4:08 am

If u and v are positive real numbers, is u > v?

(1) u^3/v < 1

(2) u^(1/3) / v < 1

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Source: Beat The GMAT — Data Sufficiency | Mon Aug 15, 2011 4:08 am

kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Mon Aug 15, 2011 5:03 am

(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?

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Mon Aug 15, 2011 5:15 am

GmatKiss wrote:If u and v are positive real numbers, is u > v?

(1) u^3/v < 1

(2) u^(1/3) / v < 1

Hi,
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!

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