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.
ROOTS and CUBES
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Source: Beat The GMAT — Data Sufficiency |
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vittalgmat
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Hi Jimmiejaz,
can u pls explain how to arrived that C is sufficient. I cant picturize it.
thanks
-V
can u pls explain how to arrived that C is sufficient. I cant picturize it.
thanks
-V
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pbanavara
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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.
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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bluementor
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Correct me if I'm wrong here: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.
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?
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parallel_chase
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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.
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.
No rest for the Wicked....
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bluementor
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Chase, thats one great solution!! Thanks.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.
But could you tell me why can't I see that with my approach (i.e. by adding both statements)?
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parallel_chase
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I dont know why you added both the statements, you treat them as seperatebluementor 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?
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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
No rest for the Wicked....
