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What is the value of xy?

(1) \(3^x*5^y=75\)

(2) \(3^{(x-1)(y-2)}=1\)

M27-22

What is the value of \(xy\)?

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.

Answer: C.

P.S. Please read carefully and follow: https://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank youl.


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
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Bunuel
swati007
What is the value of xy?

(1) \(3^x*5^y=75\)

(2) \(3^{(x-1)(y-2)}=1\)

M27-22

What is the value of \(xy\)?

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.

Answer: C.

P.S. Please read carefully and follow: https://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank youl.


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.
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P.S. Please read carefully and follow: https://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank youl.


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
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(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.

Dear Bunuel

I know that 3^0 = 1. But why did you limited \((x-1)(y-2)=0\). Why did not you consider \((x-1)(y-2)=10\) or any value. 1^ (any value) = 1.

Thanks
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(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.

Dear Bunuel

I know that 3^0 = 1. But why did you limited \((x-1)(y-2)=0\). Why did not you consider \((x-1)(y-2)=10\) or any value. 1^ (any value) = 1.

Thanks

Hello, let me try to explain here:

Given is - \(3^{(x-1)(y-2)}=1\)

This is only possible when the power of 3 is ZERO.

Hence, we have equate the power of three, in this case


\((x-1)(y-2)\) as ZERO and it cannot be equal to any other number.

Hope this is clear.

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Sorry guys, I still have a problem with this:
In statement 2, why we assumed that (x-1)(y-2) is an integer (zero)? it might be any ir/rational number that when powered to 3 equals 1.
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Bunuel ydmuley
Sorry guys, I still have a problem with this:
In statement 2, why we assumed that (x-1)(y-2) is an integer (zero)? it might be any ir/rational number that when powered to 3 equals 1.

No. Only 3 in the power of 0 equals to 1.
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Hi can someone explain why is the answer not D as we get x 1 and y 2 from both the equations how does combining them help us ?
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Hi can someone explain why is the answer not D as we get x 1 and y 2 from both the equations how does combining them help us ?

Your doubt is addressed above. Here it is again:

The issue is that neither statement alone gives both values.

From (1), x = 1 and y = 2 only if x and y are integers. Since that is not given, many other values are possible.

From (2), you get x = 1 OR y = 2, not both.

Only when you combine the two statements does x = 1 force y = 2, or y = 2 force x = 1.

So the answer is C.
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