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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, 03:18, edited 2 times in total.
Re: If x, y, and z are integers [#permalink]
22 Feb 2010, 10:19
(327)(510)(z) = (58)(914)(xy)
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)
Re: If x, y, and z are integers [#permalink]
24 Feb 2010, 15:36
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
Re: If x, y, and z are integers [#permalink]
24 Feb 2010, 15:45
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!!!! _________________
Re: If x, y, and z are integers [#permalink]
01 Aug 2010, 15:31
1
This post received KUDOS
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.
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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]
17 Jul 2012, 07:20
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]
17 Jul 2012, 15:41
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?
Re: If x, y, and z are integers [#permalink]
18 Jul 2013, 14:05
Bunuel wrote:
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!
Re: If x, y, and z are integers [#permalink]
18 Jul 2013, 14:17
Expert's post
mneeti wrote:
Bunuel wrote:
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!
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