harsha1034 wrote:
Question is asking if \(\frac{x}{y}\) is an integer.
We could pick numbers
S1)
\(x=2\). Hence \(x^{2}=4\). Factors of \(x^{2}\) are 1, 2, and 4.
Since S1 says that \(y\) is prime, \(y\) could only be 2.
Hence \(\frac{x}{y}\)=\(\frac{2}{2}=integer\) YES.
Lets pick another number to see if we can get a NO.
\(x=-6.\) Hence \(x^{2}=36\). Factors of \(x^{2}\) are 1,2,3,4,6,9,12,18,and 36.
Since S1 says that \(y\) is prime, \(y\) could now be 2 or 3 only.
Hence in both the cases \(\frac{x}{y}=integer\). YES.
So we have tried a prime number and a composite number for \(x\) and in both the cases the answer is YES. So Statement 1 is sufficient. Answer Choices could be A or D.
S2) Since we are at A or D, we could verify if statement 2 is definitely true using information from Statement 1, and if proved answer would be D.(Because Statement 1 returned a definite YES and it cannot be contradicted)
Statement 2 says \(z=y^{2}\).
If we look into the first plug-in that we did in S1, \(z\) could only be 4 meeting the condition of S1 that \(y\)=Prime=2. There was no other possible value. And it can be seen that \(z =4\) is a factor of \(x^{2}= 4\) meeting the condition in question stem also. So statement 2 is proved using statement 1.
We could try for the second plug-in we did in statement 1 also. Here \(z\) could only be 4 or 9 meeting the condition of S1 that\(y=prime=2\) or \(3\). There was no other possible value. And it can be seen that \(z=4\) or \(9\) is a factor of \(x^{2}=36\), meeting the condition in question stem also. So statement 2 is proved using statement 1.
Hence answer is D.
Hi Harsha
What if the value of x is sqt(2)? i.e x is not integer.
Would it still be true for statement 1?
x=sqt(2)--> x^2=2--> Factore 1,2
y could be 2 as y is prime and z could be 1
but in this case x wouldn't be divisible by y.
hence, statement one is not suffice.
Please correct me if I'm missing something.