sidvish wrote:
Hades wrote:
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.
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.