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If x^3y^4 = 5,000, is y = 5?

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If x^3y^4 = 5,000, is y = 5? [#permalink] New post 24 Mar 2009, 21:12
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If x^3y^4 = 5,000, is y = 5?

(1) y is a positive integer.
(2) x is an integer.
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Oct 2013, 01:14, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: DS: Is y=5 [#permalink] New post 25 Mar 2009, 02:17
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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] New post 11 May 2009, 19:36
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)

So the answer is C.

[EDIT: To correct an incorrect value of x]
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Re: DS: Is y=5 [#permalink] New post 14 May 2009, 01:37
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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.

Final Answer, C.
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Re: DS: Is y=5 [#permalink] New post 14 Dec 2009, 19:42
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] New post 15 Dec 2009, 03:44
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] New post 23 Oct 2013, 21:31
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Hades wrote:
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.

Final Answer, C.


The key distinguishing factor here is that in statement 1 it says y is a POSITIVE integer.

1. Statement one says y is a positive integer. Therefore y^4 is also a positive integer. There are two possible values for y -> 1 or 5 (we dont know whether x is an integer, so x could be an irrational number and y could be 1, similarly y could also be 5). Not sufficient

2. Statement 2 says x is an integer. Again multiple possible values here -> x could be 1 and y would be the 4th root of 5000, x could be 2 and y could be 5 OR -5 (we dont even know if it's positive or negative)

Combined: x and y are both integers, therefore both x^3 and y^4 have to be integers. The only option for x is 2 (it cannot be -2 since it's raised to an odd power). The important part is that y is a POSITIVE integer, therefore it has to be 5. If this was not specified then y could be 5 or -5 and the answer would be E. As it stands the correct answer is C.

Takeaway- never assume anything not given and always take a contrarian view in DS! Watch out for even powers - they hide the sign while odd powers don't. Great question!
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Re: DS: Is y=5 [#permalink] New post 24 Oct 2013, 00:35
Expert's post
sidvish wrote:
Hades wrote:
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.

Final Answer, C.


The key distinguishing factor here is that in statement 1 it says y is a POSITIVE integer.

1. Statement one says y is a positive integer. Therefore y^4 is also a positive integer. There are two possible values for y -> 1 or 5 (we dont know whether x is an integer, so x could be an irrational number and y could be 1, similarly y could also be 5). Not sufficient

2. Statement 2 says x is an integer. Again multiple possible values here -> x could be 1 and y would be the 4th root of 5000, x could be 2 and y could be 5 OR -5 (we dont even know if it's positive or negative)

Combined: x and y are both integers, therefore both x^3 and y^4 have to be integers. The only option for x is 2 (it cannot be -2 since it's raised to an odd power). The important part is that y is a POSITIVE integer, therefore it has to be 5. If this was not specified then y could be 5 or -5 and the answer would be E. As it stands the correct answer is C.

Takeaway- never assume anything not given and always take a contrarian view in DS! Watch out for even powers - they hide the sign while odd powers don't. Great question!


This is not 100% correct.

If x^3y^4 = 5,000, is y = 5?

(1) y is a positive integer. For ANY positive integer value of y, there exists some x satisfying x^3y^4 = 5,000. For example:
y=1 --> x=\sqrt[3]{5,000};
y=2 --> x=\sqrt[3]{\frac{5,000}{16}};
y=3 --> x=\sqrt[3]{\frac{5,000}{81}};
...
y=5 --> x=2;
...
y=10 --> x=\sqrt[3]{\frac{5,000}{10,000}};
...

Not sufficient.

(2) x is an integer. The same logic applies here. Not sufficient.

(1)+(2) Since x is an integer and y is a positive integer, then from x^3y^4 = 2^35^4, it follows that y=5. Sufficient.

Answer: C.

Hope it's clear.
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Re: DS: Is y=5 [#permalink] New post 24 Oct 2013, 00:36
Expert's post
Bunuel wrote:
If x^3y^4 = 5,000, is y = 5?

(1) y is a positive integer. For ANY positive integer value of y, there exists some x satisfying x^3y^4 = 5,000. For example:
y=1 --> x=\sqrt[3]{5,000};
y=2 --> x=\sqrt[3]{\frac{5,000}{16}};
y=3 --> x=\sqrt[3]{\frac{5,000}{81}};
...
y=5 --> x=2;
...
y=10 --> x=\sqrt[3]{\frac{5,000}{10,000}};
...

Not sufficient.

(2) x is an integer. The same logic applies here. Not sufficient.

(1)+(2) Since x is an integer and y is a positive integer, then from x^3y^4 = 2^35^4, it follows that y=5. Sufficient.

Answer: C.

Hope it's clear.


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Re: If x^3y^4 = 5,000, is y = 5? [#permalink] New post 24 Oct 2013, 01:10
I wish the stem question would've been written a bit clearer cause I read it as X to the power of all (3y^4) double power !!
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Re: DS: Is y=5 [#permalink] New post 24 Oct 2013, 01:12
Super helpful. I hear you. I should've said AT LEAST 2 values of y exist.
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Re: If x^3y^4 = 5,000, is y = 5? [#permalink] New post 24 Oct 2013, 01:15
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Re: DS: Is y=5 [#permalink] New post 30 May 2014, 22:36
Bunuel wrote:
sidvish wrote:
Hades wrote:
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.

Final Answer, C.


The key distinguishing factor here is that in statement 1 it says y is a POSITIVE integer.

1. Statement one says y is a positive integer. Therefore y^4 is also a positive integer. There are two possible values for y -> 1 or 5 (we dont know whether x is an integer, so x could be an irrational number and y could be 1, similarly y could also be 5). Not sufficient

2. Statement 2 says x is an integer. Again multiple possible values here -> x could be 1 and y would be the 4th root of 5000, x could be 2 and y could be 5 OR -5 (we dont even know if it's positive or negative)

Combined: x and y are both integers, therefore both x^3 and y^4 have to be integers. The only option for x is 2 (it cannot be -2 since it's raised to an odd power). The important part is that y is a POSITIVE integer, therefore it has to be 5. If this was not specified then y could be 5 or -5 and the answer would be E. As it stands the correct answer is C.

Takeaway- never assume anything not given and always take a contrarian view in DS! Watch out for even powers - they hide the sign while odd powers don't. Great question!


This is not 100% correct.

If x^3y^4 = 5,000, is y = 5?

(1) y is a positive integer. For ANY positive integer value of y, there exists some x satisfying x^3y^4 = 5,000. For example:
y=1 --> x=\sqrt[3]{5,000};
y=2 --> x=\sqrt[3]{\frac{5,000}{16}};
y=3 --> x=\sqrt[3]{\frac{5,000}{81}};
...
y=5 --> x=2;
...
y=10 --> x=\sqrt[3]{\frac{5,000}{10,000}};
...

Not sufficient.

(2) x is an integer. The same logic applies here. Not sufficient.

(1)+(2) Since x is an integer and y is a positive integer, then from x^3y^4 = 2^35^4, it follows that y=5. Sufficient.

Answer: C.

Hope it's clear.



Hi Bunuel,

From (1) and (2) we come to conclusion that X is an integer and Y is positive but we cant say Y is integer . Y can be 5 when x = 2 or Y can be 5*(8)^1/4 when x=1.

So not Sufficient. I think OA should be E.

Am I right ?

Regards,
Ammu
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Re: DS: Is y=5 [#permalink] New post 31 May 2014, 03:37
Expert's post
ammuseeru wrote:
Bunuel wrote:
sidvish wrote:

This is not 100% correct.

If x^3y^4 = 5,000, is y = 5?

(1) y is a positive integer. For ANY positive integer value of y, there exists some x satisfying x^3y^4 = 5,000. For example:
y=1 --> x=\sqrt[3]{5,000};
y=2 --> x=\sqrt[3]{\frac{5,000}{16}};
y=3 --> x=\sqrt[3]{\frac{5,000}{81}};
...
y=5 --> x=2;
...
y=10 --> x=\sqrt[3]{\frac{5,000}{10,000}};
...

Not sufficient.

(2) x is an integer. The same logic applies here. Not sufficient.

(1)+(2) Since x is an integer and y is a positive integer, then from x^3y^4 = 2^35^4, it follows that y=5. Sufficient.

Answer: C.

Hope it's clear.



Hi Bunuel,

From (1) and (2) we come to conclusion that X is an integer and Y is positive but we cant say Y is integer . Y can be 5 when x = 2 or Y can be 5*(8)^1/4 when x=1.

So not Sufficient. I think OA should be E.

Am I right ?

Regards,
Ammu


No, you are not right. The OA is given under the spoiler in the original post and it's C, not E.

(1) directly says that y is a positive integer, while in your example, y is NOT an integer.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: DS: Is y=5   [#permalink] 31 May 2014, 03:37
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