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

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If x^3y^4 = 5,000, is y = 5? (1) y is a positive integer. [#permalink]  24 Mar 2009, 22:12
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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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Re: DS: Is y=5 [#permalink]  25 Mar 2009, 03:17
Answer should be C. Factors of 5000 are 2x2x2x5x5x5x5 (2^3x5^4). Only if both x and y have to be integer (statement one and two) there is no other option than x=2 and more importantly y=5. The alternative y=-5 - since in y^4 the power is even - can also be excluded by the addition positive integer in statement 1.
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Re: DS: Is y=5 [#permalink]  10 May 2009, 14:20
why it is not D?????
milo wrote:
Answer should be C. Factors of 5000 are 2x2x2x5x5x5x5 (2^3x5^4). Only if both x and y have to be integer (statement one and two) there is no other option than x=2 and more importantly y=5. The alternative y=-5 - since in y^4 the power is even - can also be excluded by the addition positive integer in statement 1.
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Re: DS: Is y=5 [#permalink]  11 May 2009, 20:36
We need Statement 2.

With Statement 1 alone, we cannot say that y = 5.
E.g. Let's say y = 4, so it satisfies Statement 1.
Then x^3 = 5000/256 = 19.53125, therefore x = 2.69 (rounded off to the last two decimal places)
This is a perfectly legitimate solution.

We need Statement 2 to find a unique solution to x^3y^4 = 5000; otherwise with just Statement 1, there will be an infinite numbers of solutions.

Only when you also specify that x is an integer, do you get a single solution (x = 2, y = 5)

[EDIT: To correct an incorrect value of x]
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Re: DS: Is y=5 [#permalink]  14 May 2009, 02:37
If x^3y^4 = 5,000, is y = 5?
(1) y is a positive integer.
(2) x is an integer.

Question: (y==5)?
Rearrange x^3y^4 = 5,000 to y=\sqrt[4]{(5000/x^3))}

Substitute y=\sqrt[4]{(5000/x^3))} into Question:(y==5)?,
Question: (\sqrt[4]{(5000/x^3))}==5)?
Rearrange:
Question: ((5000/x^3)==625)?
Question: ((x^3==5000/625)?
Question: ((x^3==8)?
Question: ((x==2)? (don't have to worry about negative numbers since we have an odd exponent)

(1) If y is an integer, we don't know whether x==2 or y==5, insufficient.

(2) If x is an integer, we don't know whether x==2 or y==5, insufficient.

(1&2) We know they're both integers, but again that's it. Let's factorize 5000 into 2^3*5^4

Rearrange fact, (x^3y^4 == 2^3*5^4) ==>> x=2,y=4 (since exponents match, and are different, and x&y are both integers therefore the bases MUST be the same).

Hence Question:(y==5)?==>NO, Sufficient.

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Re: DS: Is y=5 [#permalink]  14 Dec 2009, 20:42
So I got thoroughly confused with this question. Hope I can articulate my thoughts correctly here.
x^3 * y^4 = 5000. Is y=5.

so y^4 = 5000/x^3[/m]

y = 4\sqrt{5000/x^3}

y = 4\sqrt{625 * 8 /x^3}

y = 5 * 4\sqrt{625 * 8 /x^3}

If y has to be 5 then 4\sqrt{625 * 8 /x^3}should be 1.

Combining a and b 4\sqrt{625 * 8 /x^3} can or cannot be 1. What am I missing ?
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Re: DS: Is y=5 [#permalink]  15 Dec 2009, 04:44
pleonasm wrote:
So I got thoroughly confused with this question. Hope I can articulate my thoughts correctly here.
x^3 * y^4 = 5000. Is y=5.

so y^4 = 5000/x^3[/m]

y = 4\sqrt{5000/x^3}

y = 4\sqrt{625 * 8 /x^3}

y = 5 * 4\sqrt{625 * 8 /x^3}

If y has to be 5 then 4\sqrt{625 * 8 /x^3}should be 1.

Combining a and b 4\sqrt{625 * 8 /x^3} can or cannot be 1. What am I missing ?

on combining we get two more restriction and i.e x and y are int
now keeping that in mind the exp can be simplified further as
4th root of ([5^4 *2^3)/x^3] and for this to be an int there is only one possible value for x i.e is 2

in your approach u are getting confused because u are accepting x to be an int and are forgetting that y also has to be an int as per the statements

HTH
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Re: DS: Is y=5 [#permalink]  15 Dec 2009, 14:39
xcusemeplz2009 wrote:
pleonasm wrote:
So I got thoroughly confused with this question. Hope I can articulate my thoughts correctly here.
x^3 * y^4 = 5000. Is y=5.

so y^4 = 5000/x^3[/m]

y = 4\sqrt{5000/x^3}

y = 4\sqrt{625 * 8 /x^3}

y = 5 * 4\sqrt{625 * 8 /x^3}

If y has to be 5 then 4\sqrt{625 * 8 /x^3}should be 1.

Combining a and b 4\sqrt{625 * 8 /x^3} can or cannot be 1. What am I missing ?

on combining we get two more restriction and i.e x and y are int
now keeping that in mind the exp can be simplified further as
4th root of ([5^4 *2^3)/x^3] and for this to be an int there is only one possible value for x i.e is 2

in your approach u are getting confused because u are accepting x to be an int and are forgetting that y also has to be an int as per the statements

HTH

aah ok got it .. Thanks
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Re: DS: Is y=5   [#permalink] 15 Dec 2009, 14:39
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