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# If x^3y^4 = 5,000, is y = 5? (1) y is a positive integer.

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Intern
Joined: 08 Nov 2008
Posts: 29
If x^3y^4 = 5,000, is y = 5? (1) y is a positive integer.  [#permalink]

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11 Nov 2008, 20:17
If x^3y^4 = 5,000, is y = 5?
(1) y is a positive integer.
(2) x is an integer.

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.

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Director
Joined: 14 Aug 2007
Posts: 694

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11 Nov 2008, 20:36
petercao wrote:
If x^3y^4 = 5,000, is y = 5?
(1) y is a positive integer.
(2) x is an integer.

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.

C.

We need to know that x is an integer and then y is positive integer
then we have
5000 = 5*5*5*5*2*2*2 = 5^4 2^3

[but it can also be that (-5)^4 2^3 thus y can be -5]
Manager
Joined: 16 Oct 2008
Posts: 54

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12 Nov 2008, 01:25
alpha_plus_gamma wrote:
petercao wrote:
If x^3y^4 = 5,000, is y = 5?
(1) y is a positive integer.
(2) x is an integer.

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.

C.

We need to know that x is an integer and then y is positive integer
then we have
5000 = 5*5*5*5*2*2*2 = 5^4 2^3

[but it can also be that (-5)^4 2^3 thus y can be -5]

Why we need to know x is an integer? coz i cannot stimulate the case with x is not integer.
Director
Joined: 14 Aug 2007
Posts: 694

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12 Nov 2008, 01:40
Natasha0123 wrote:
Why we need to know x is an integer? coz i cannot stimulate the case with x is not integer.

simple case:
y = 1, x = cuberoot(5000)
Manager
Joined: 16 Oct 2008
Posts: 54

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12 Nov 2008, 01:42
alpha_plus_gamma wrote:
simple case:
y = 1, x = cuberoot(5000)

I see, I totally forgot cube root, thank you
Intern
Joined: 28 Jul 2008
Posts: 29

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15 Nov 2008, 04:27
The answer should be A. Even though Y can be both 5 and - 5, based on the question, Y must be 5, because Y must be positive.
VP
Joined: 05 Jul 2008
Posts: 1331

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15 Nov 2008, 10:49
Got a C

(1) x^3 Y^4 = 5000

y= +ve integer means x3 is also positive, means x is also positive.

Y can be 5 or any thing because we don't know whether x is integer or not. For example y =10 y^ 4= 10,000 and x^3 =1/2. Hence A & D out

(2) x is integer.

Again similar logic. we dont know anything about Y

Together.

X & Y are +ve integers. X=2, Y=5 Hence C

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Re: x y integer---27 &nbs [#permalink] 15 Nov 2008, 10:49
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# If x^3y^4 = 5,000, is y = 5? (1) y is a positive integer.

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