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:

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Last edited by Bunuel on 08 Oct 2012, 04:18, edited 2 times in total.

3 * 109 * 3 * 17 * 2 * 5 * z = 2 * 29 * 2 * 457 * x * y (all the numbers are prime)

x = (3^2 * 5 * 17 * 109 * z) / (29 * 2 * 457 * y) (cancelling 2 on numerator and denominator)

1) y is prime Not sufficient y = 3, z = 29*2*457 * n (where n is an intereger ) and values of x could vary depending on n or depending on y(3,5,109, 457 etc) (2) x is prime Not sufficient y = 9 * 5 * 17,z = 29*2*457 , x = 109 y = 9 * 5 * 109,z = 29*2*457 , x = 17

combining .. there is no value of x that will satify the equation as if y is prime, then y could be any of the values of 3,5, 17, 109 or a different number that is a factor of z. and if z is an interger, x cannot be prime.

E ( not sure if E is the correct answer as there is no valid value of x that satisfies the conditions)

Here's the solution for the re-posted problem If x, y, and z are integers greater than 1, and 3^27 * 5^10 * z =5^8 * 9^14 * x * y , then what is the value of x? 3^27 * 5^10 * z =5^8 * 9^14 * x * y 3^27 * 5^ 10 * z = 5^ 8 * 3^28 * x * y 5^2 * z = 3 * x * y x = (5^2 * z) / 3y

(1) y is prime y can be a factor of z, and z can be a factor of y and there could be a lot of possibilities of x y = 3, z = 27, x = 75 y = 5, z = 3, x = 5 Not sufficient (2) x is prime z has to be a multiple of 3 and y has to be a multiple of 5 z= 3*a (where a is prime not equal to 5), y = 25, then x=a(many values for a) z=3, y = 5, x=5 Not sufficient

combining .. x and y both are prime .. and as 5^2 is in the numerator, y has to be 5 and z has be to 3 or else x cannot be prime. y=5, z=3, x=5

After simplifying the given question, we end up with 5*5*z/(3*y) = x.

Stmt 1 - y can take a whole lot of prime numbers from 2 to infinity (not the car!!!) and until we know z, we can't comment on value of x.

Stmt 2 - x itself can take a whole lot of prime numbers from 2 to infinity (again, not the car!!!) and we would have to struggle to get appropriate values for z and y, and there can very many that match the criteria.

Combining both stmts - as long as z is some composite, that can be expressed as a multiple of 3, x (any prime number choice) and y (prime number choice), it is good, and there are whole lot of possibilities for x as well.

I am not very sure, how do we get to B as the OA. In my opinion, it must be E. Math experts and Math God Bunuel might wish to throw some light!!!! _________________

I had this question in the test today too. I went with E but QA is B. I didnt understand the explanation. I think QA is wrong here, 2 is a prime too.

---

The best way to answer this question is to use the rules of exponents to simplify the question stem, then analyze each statement based on the simplified equation.

(327)(510)(z) = (58)(914)(xy) Simplify the 914 (327)(510)(z) = (58)(328)(xy) Divide both sides by common terms 58, 327 (52)(z) = 3xy

(1) INSUFFICIENT: Analyzing the simplified equation above, we can conclude that z must have a factor of 3 to balance the 3 on the right side of the equation. Similarly, x must have at least one factor of 5. Statement (1) says that y is prime, which does no tell us how many fives are contained in x and z.

For example, it is possible that x = 5, y = 2, and z = 3: 52 · 3 = 3 · 52

It is also possible that x = 25, y = 2, and z = 75: 52 · 75 = 3 · 252 52 · 52 · 3 = 3 · 252

(2) SUFFICIENT: Analyzing the simplified equation above, we can conclude that x must have a factor of 5 to balance out the 52 on the left side. Since statement (2) says that x is prime, x cannot have any other factors, so x = 5. Therefore statement (2) is sufficient.

If the answer is really B, then I think question should be: \(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\). If I'm right, then it's B indeed.

In it's current form the answer is C as explained.

Bunuel, can you please explain the answer for choice B _________________

_______________________________________________________________________________________________________________________________ If you like my solution kindly reward me with Kudos.

If the answer is really B, then I think question should be: \(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\). If I'm right, then it's B indeed.

In it's current form the answer is C as explained.

Bunuel, can you please explain the answer for choice B

You mean what would be the solution if it were x^y instead of xy?

If x, y, and z are integers greater than 1, and \(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\), then what is the value of x?

\(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\) --> \(3^{27}*5^{2}*z =3^{28}*x^y\) --> \(5^{2}*z = 3*x^y\) --> \(\frac{x^y}{z}=\frac{5^2}{3}\), so \(x^y\) is a multiple of 25 and \(z\) is a multiple of 3.

(1) y is prime. We can have that \(x=5\), \(y=2=prime\) and \(z=3\) OR \(x=10\), \(y=2=prime\) and \(z=12\)... Not sufficient.

(2) x is prime. Since \(x^y\) is a multiple of \(5^2\) and \(x\) is a prime, then \(x=5\). Sufficient.

Re: If x, y, and z are integers greater than 1, and [#permalink]

Show Tags

17 Jul 2012, 08:20

think the answer is B even as written?

Formula simplifies to 25z/3y = x or 25z/3x=y

1: y is prime; since 25z/3y yields an integer, does it not hold that 25/y = integer since 3 is not a factor of 25. If y is prime, it has to be 5 since 5^2 are the prime factors of 25. However, in order to know X, we need to know what z might be, and Z could be any number with 3 as a prime factor. Not sufficient.

2: x is prime; since 25z/3x yields an integer, 25/x is also an integer since 3 is not a factor of 25. Holding the same logic as in 1; x must be 5. Sufficient.

Please let me know if I am thinking about this wrong

Re: If x, y, and z are integers greater than 1, and [#permalink]

Show Tags

17 Jul 2012, 16:41

JohnGalt44 wrote:

think the answer is B even as written?

Formula simplifies to 25z/3y = x or 25z/3x=y

1: y is prime; since 25z/3y yields an integer, does it not hold that 25/y = integer since 3 is not a factor of 25. If y is prime, it has to be 5 since 5^2 are the prime factors of 25. However, in order to know X, we need to know what z might be, and Z could be any number with 3 as a prime factor. Not sufficient.

2: x is prime; since 25z/3x yields an integer, 25/x is also an integer since 3 is not a factor of 25. Holding the same logic as in 1; x must be 5. Sufficient.

Please let me know if I am thinking about this wrong

Yes you are thinking about it wrongly. Why does 25/x have to be an integer? 3 is not a factor of 25, but it can be a factor of z, the other term in the numerator. Hence, if you let y=5^2, then z=3x, and x=any prime you want it to be. Not sufficient

However taking (1)+(2), both x and y have to be prime and since you need to make 25 using 3*x*y, only way to do it is using x=y=5 and z=3. C.

Please give kudos if you like.

Also, can OP please change answer in spoiler to the correct one?

You mean what would be the solution if it were x^y instead of xy?

If x, y, and z are integers greater than 1, and \(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\), then what is the value of x?

\(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\) --> \(3^{27}*5^{2}*z =3^{28}*x^y\) --> \(5^{2}*z = 3*x^y\) --> \(\frac{x^y}{z}=\frac{5^2}{3}\), so \(x^y\) is a multiple of 25 and \(z\) is a multiple of 3.

(1) y is prime. We can have that \(x=5\), \(y=2=prime\) and \(z=3\) OR \(x=10\), \(y=2=prime\) and \(z=12\)... Not sufficient.

(2) x is prime. Since \(x^y\) is a multiple of \(5^2\) and \(x\) is a prime, then \(x=5\). Sufficient.

Answer: B.

Hope it's clear.

Bunuel, if the statement (1) were- Z (instead of Y) is prime then the answer should be 'D'? Thanks!

You mean what would be the solution if it were x^y instead of xy?

If x, y, and z are integers greater than 1, and \(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\), then what is the value of x?

\(3^{27}*5^{10}*z = 5^8*9^{14}*x^y\) --> \(3^{27}*5^{2}*z =3^{28}*x^y\) --> \(5^{2}*z = 3*x^y\) --> \(\frac{x^y}{z}=\frac{5^2}{3}\), so \(x^y\) is a multiple of 25 and \(z\) is a multiple of 3.

(1) y is prime. We can have that \(x=5\), \(y=2=prime\) and \(z=3\) OR \(x=10\), \(y=2=prime\) and \(z=12\)... Not sufficient.

(2) x is prime. Since \(x^y\) is a multiple of \(5^2\) and \(x\) is a prime, then \(x=5\). Sufficient.

Answer: B.

Hope it's clear.

Bunuel, if the statement (1) were- Z (instead of Y) is prime then the answer should be 'D'? Thanks!

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...