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by vijaynarayanan » Thu Jul 15, 2010 10:47 pm
If x and y are positive, is x^3 > y?
(1) Sq.root(x) > 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.

Correct Answer: E
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Source: — Data Sufficiency |
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by selango » Fri Jul 16, 2010 12:31 am
x^3>y?

From stmt1,

sqrt(x)>y

sqrt(9)>2,x^3>y

sqrt(1/16)>1/5,x^3<y

Insufficient.

From stmt2,

x>y

2>1,x^3>y

1/2>1/4,x^3<y

Insufficient.

Combining 1 and 2

sqrt(9)>2 and 9>2,x^3>y

sqrt(1/4)>1/5,1/4>1/5,x^3<y

Insufficient.

Hence E
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by ryantherocket » Fri Jul 16, 2010 5:49 am
vijaynarayanan wrote:If x and y are positive, is x^3 > y?
(1) Sq.root(x) > y
(2) x > y
(1) if we're dealing with positive fractions, x could be smaller than y even though its square root is bigger than y. In that case, x^3 will be smaller than y. If we're dealing with positive integers, then x will be bigger than y and x^3 will be bigger than y. INSUFFICIENT

(2) if we're dealing with positive fractions, even though x >y, x^3 could be smaller than y. If we're dealing with positive integers, x>y --> x^3 is bigger than y. INSUFFICIENT

The same examples are used to dismiss both statements, so putting the statements together will not be any more helpful. the answer is E
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by Rahul@gurome » Fri Jul 16, 2010 5:49 am
vijaynarayanan wrote:If x and y are positive, is x^3 > y?
(1) Sq.root(x) > 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.

Correct Answer: E
(1) If x = 9 and y = 2, then √x = 3 > 2. Here x^3 > y.
If x = 1/9 and y = 1/4, then √x = 1/3 > 1/4. Here x^3 < y.
We don't get a unique answer.
So, (1) is NOT SUFFICIENT.

(2) If x = 9 and y = 2, then √x = 3 > 2. Here x^3 > y.
If x = 1/3 and y = 1/4, then x^3 < y
We don't get a unique answer.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), If x = 16/25 and y = 1/2, then x^3 < y.
If x = 9 and y = 2, then √x = 3 > 2. Here x^3 > y. Again no unique answer.

The correct answer is (E).
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