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

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Name: Ronak Amin
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If x^3y^4 = 5,000, is y = 5?  [#permalink]

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Updated on: 24 Oct 2013, 01:14
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If $$x^3y^4 = 5,000$$, is y = 5?

(1) y is a positive integer.
(2) x is an integer.

Originally posted by Economist on 24 Mar 2009, 21:12.
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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Joined: 02 Sep 2009
Posts: 52231

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24 Oct 2013, 00:35
2
7
sidvish 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.

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.

Hope it's clear.
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25 Mar 2009, 02:17
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.
Manager
Joined: 31 Mar 2008
Posts: 143
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11 May 2009, 19:36
1
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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14 May 2009, 01:37
2
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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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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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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23 Oct 2013, 21:31
2
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.

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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Posts: 52231

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24 Oct 2013, 00:36
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.

Hope it's clear.

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tough-and-tricky-exponents-and-roots-questions-125967-20.html

Hope it helps.
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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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30 May 2014, 22:36
Bunuel wrote:
sidvish 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.

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.

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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Posts: 52231

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31 May 2014, 03:37
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.

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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Re: If x^3y^4 = 5,000, is y = 5?  [#permalink]

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25 Aug 2016, 21:16
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Economist wrote:
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

(1) y is a positive integer.
(2) x is an integer.

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Re: If x^3y^4 = 5,000, is y = 5?  [#permalink]

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13 Dec 2016, 18:11
I have solved this in another way. Experts please comment if this is correct.

x^3*y^4 = 5000.
taking square roots on both sides.

x*y^2 *\sqrt{x} = 2*25*\sqrt{2}

Since we are already given that odd power of X when combined with y^4 which is obviously a positive number gives us positive number 5000. So X has to be positive, no need to consider negative root of x.

just by comparing powers we can say that X = 2 and y = 5.

But question stem does not say that both X and Y are integers. So both statements are needed to confirm the above logic.
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Re: If x^3y^4 = 5,000, is y = 5?  [#permalink]

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14 May 2018, 00:12
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Re: If x^3y^4 = 5,000, is y = 5? &nbs [#permalink] 14 May 2018, 00:12
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