Confusion

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dkumar.83: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Thu Jan 14, 2010 5:42 am; Thanked: 2 times

by dkumar.83 » Tue May 18, 2010 4:59 am

If x and y are positive, is x3 > y?
(1) x^1/2 > y
(2) x > y

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

The answer as per solution sheet is E, however i feel it should be D. Kindly check if its correct?

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Source: Beat The GMAT — Data Sufficiency | Tue May 18, 2010 4:59 am

sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Tue May 18, 2010 5:44 am

dkumar.83 wrote:If x and y are positive, is x3 > y?
(1) x^1/2 > y
(2) x > y

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

The answer as per solution sheet is E, however i feel it should be D. Kindly check if its correct?

Here, x and y are only positive, any of the two intervals are fine with x and y here, 0 < x < 1, or x â‰¥ 1; 0 < y < 1, or y â‰¥ 1. I guess the typo and hence the question as saying, is x^3 > y?

Fine

(1) If x^1/2 > y, then (x^1/2) ^6 > y^6, when x > 1, y > 1, but it's not so when 0 < x < 1, 0 < y < 1. We can take suitable numbers in the said intervals for a quick check. Hence, insufficient

(2) Same explanation as in (1). If x > 1, y > 1, and x > y, then here's a YES for the question, but when 0 < x < 1, 0 < y < 1, here comes my NO!

Taken as one

When x^1/2 > y with x > y, then it at first glance, seems like x > 1, y > 1 case. But remember, our intention is to deny the grounds for x^3 > y or not, instead of supporting it. So, let's look for suitable values for x and y in the ranges 0 < x < 1, 0 < y < 1 which could still disprove our hasty deem.

How if x = 1/9, y = 1/10, here x > y and also x^1/2 > y as 1/3 > 1/10, but still (1/9) ^3 is not greater than 1/10. [spoiler]Still insufficient

E[/spoiler]

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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dkumar.83: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Thu Jan 14, 2010 5:42 am; Thanked: 2 times

by dkumar.83 » Tue May 18, 2010 7:53 am

Thanks for the detailed explanation.

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