Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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. _________________

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...

I may have spoken to over 50+ Said applicants over the course of my year, through various channels. I’ve been assigned as mentor to two incoming students. A...