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Notice that we are not told that the \(x\) and \(y\) are integers.

(1) \(3^x*5^y=75\) --> if \(x\) and \(y\) are integers then as \(75=3^1*5^2\) then \(x=1\) and \(y=2\) BUT if they are not, then for any value of \(x\) there will exist some non-integer \(y\) to satisfy given expression and vise-versa (for example if \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). Not sufficient.

(2) \(3^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) \(x=1\) then from (1) \(3^x*5^y=3*5^y=75\) --> \(y=2\) and if from (2) \(y=2\) then from (1) \(3^x*5^y=3^x*25=75\) --> \(x=1\). Thus \(x=1\) and \(y=2\). Sufficient.

Since we do not know that x and y are integers hence statement I and II will be insufficient as we can have multiple answers. Combining will be sufficient here as we will have two equations and two answers.
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Since we do not know that x and y are integers hence statement I and II will be insufficient as we can have multiple answers. Combining will be sufficient here as we will have two equations and two answers.

Notice that we are not told that the \(x\) and \(y\) are integers.

(1) \(3^x*5^y=75\) --> if \(x\) and \(y\) are integers then as \(75=3^1*5^2\) then \(x=1\) and \(y=2\) BUT if they are not, then for any value of \(x\) there will exist some non-integer \(y\) to satisfy given expression and vise-versa (for example if \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). Not sufficient.

(2) \(5^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) \(x=1\) then from (1) \(3^x*5^y=3*5^y=75\) --> \(y=2\) and if from (2) \(y=2\) then from (1) \(3^x*5^y=3^x*25=75\) --> \(x=1\). Thus \(x=1\) and \(y=2\). Sufficient.

Hello Bunuel - I have a question on this solution, I remember reading somewhere on ur post that GMAT always prefers rational numbers, in that case will this solution still exist?

Not sure if I am getting it wrong?

Posted from my mobile device _________________

"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Worried About IDIOMS?Here is a Daily Practice List: https://gmatclub.com/forum/idiom-s-ydmuley-s-daily-practice-list-250731.html#p1937393

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475

Notice that we are not told that the \(x\) and \(y\) are integers.

(1) \(3^x*5^y=75\) --> if \(x\) and \(y\) are integers then as \(75=3^1*5^2\) then \(x=1\) and \(y=2\) BUT if they are not, then for any value of \(x\) there will exist some non-integer \(y\) to satisfy given expression and vise-versa (for example if \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). Not sufficient.

(2) \(5^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) \(x=1\) then from (1) \(3^x*5^y=3*5^y=75\) --> \(y=2\) and if from (2) \(y=2\) then from (1) \(3^x*5^y=3^x*25=75\) --> \(x=1\). Thus \(x=1\) and \(y=2\). Sufficient.

Hello Bunuel - I have a question on this solution, I remember reading somewhere on ur post that GMAT always prefers rational numbers, in that case will this solution still exist?

Not sure if I am getting it wrong?

Posted from my mobile device

It's not a matter of preferences, If a solution exits, it exists and if it does not then well it does not. Here, for (1) x and y could be irrational, so the statement is not sufficient.
_________________

Hello Bunuel - I have a question on this solution, I remember reading somewhere on ur post that GMAT always prefers rational numbers, in that case will this solution still exist?

Not sure if I am getting it wrong?

Posted from my mobile device[/quote]

It's not a matter of preferences, If a solution exits, it exists and if it does not then well it does not. Here, for (1) x and y could be irrational, so the statement is not sufficient.[/quote]

Got it thanks Bunuel

Posted from my mobile device

Posted from my mobile device _________________

"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Worried About IDIOMS?Here is a Daily Practice List: https://gmatclub.com/forum/idiom-s-ydmuley-s-daily-practice-list-250731.html#p1937393

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475

(2) \(3^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

(2) \(3^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

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