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Source: — Data Sufficiency |
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by jimmiejaz » Fri Nov 28, 2008 5:10 am
IMO C

From 1 we have x^1/2 > y
if x = 4, y=1 x^1/2 > y but if we have x=1/4, y=1/3
we have x^1/2>y but x<y hence insuff.
eliminate A,D

From 2 we have x^3>y
x=2, y=1 x^3>y and x>y but if x=2, y=3 x^3>y and x<y
Hence insuff.
Eliminate B

Combining both
we have x^1/2 > y and x^3>y, which is only possible if x>y.
Hence suff.
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by vittalgmat » Fri Nov 28, 2008 11:33 pm
Hi Jimmiejaz,
can u pls explain how to arrived that C is sufficient. I cant picturize it.

thanks
-V
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by pbanavara » Mon Dec 01, 2008 10:56 pm
Lemme try this:

From I:

x^1/2 >y => x> y^2. Try positive negative and fraction values and the equation falls apart. - Not sufficient. - Eliminate A, D

From II:

x^1/3 >y => x> y^3 , Try positive and negative values and the equation falls apart. Eliminate B

Together : x> y^2 and x > y^3 => x> y Sufficient.

Hope I haven't gone wrong anywhere.
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by bluementor » Tue Dec 02, 2008 1:22 am
jimmiejaz wrote:IMO C

From 1 we have x^1/2 > y
if x = 4, y=1 x^1/2 > y but if we have x=1/4, y=1/3
we have x^1/2>y but x<y hence insuff.
eliminate A,D

From 2 we have x^3>y
x=2, y=1 x^3>y and x>y but if x=2, y=3 x^3>y and x<y
Hence insuff.
Eliminate B

Combining both
we have x^1/2 > y and x^3>y, which is only possible if x>y.
Hence suff.
Correct me if I'm wrong here:

If you add both statements, you will get:
x^1/2 + x^3 > 2y

If x = 2, y = 1, (2)^1/2 + (2)^3 > 2(1) and x > y
If x = 2, y = 3, (2)^1/2 + (2)^3 > 2(3) but x < y. (So insufficient)

Answer E?

-BM-
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by parallel_chase » Tue Dec 02, 2008 2:58 am
Statement I & II insufficient alone.

Combining I & II

if x is a positive integer
x^3 > x > sqrtx

if sqrt x is greater than y, x will also be greater than y.


if x is negative, we cannot have this case because sqrt x cannot be negative.


if x is less than 1 and greater than 0
sqrtx > x > x^3

if x^3 is greater than y, then x will certainly be greater than y.

Sufficient.

Hence C.

Hope this helps. Let me know if you still have any doubts.
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by bluementor » Tue Dec 02, 2008 3:12 am
parallel_chase wrote:Statement I & II insufficient alone.

Combining I & II

if x is a positive integer
x^3 > x > sqrtx

if sqrt x is greater than y, x will also be greater than y.


if x is negative, we cannot have this case because sqrt x cannot be negative.


if x is less than 1 and greater than 0
sqrtx > x > x^3

if x^3 is greater than y, then x will certainly be greater than y.

Sufficient.

Hence C.

Hope this helps. Let me know if you still have any doubts.
Chase, thats one great solution!! Thanks.

But could you tell me why can't I see that with my approach (i.e. by adding both statements)?

-BM-
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by parallel_chase » Tue Dec 02, 2008 4:08 am
bluementor wrote: If you add both statements, you will get:
x^1/2 + x^3 > 2y

If x = 2, y = 1, (2)^1/2 + (2)^3 > 2(1) and x > y
If x = 2, y = 3, (2)^1/2 + (2)^3 > 2(3) but x < y. (So insufficient)

Answer E?

-BM-
I dont know why you added both the statements, you treat them as seperate
statements.

The first case is perfect. i.e. x = 2, y = 1

The second case, x = 2, y = 3 , actually contradicts one of the statement

x^3 > y => 2^3 > 3

sqrtx > y => sqrt2 > 3 [this case contradicts the first statement]

Therefore, there shouldnt be any such case.

Hope this helps
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