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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.

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)

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).

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
_________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

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!

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.

(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.

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.

(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.
_________________

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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