&lt;?phpvar $one = 1;var $two = 2;echo $one / $two * $one / $two * $two;?&gt;
A. 1
B. 0
C. 0.5
D. Error
Answer:A
&lt;?phpvar $one = 1;var $two = 2;echo $one / $two * $one / $two * $two;?&gt;
A. 1 B. 0 C. 0.5 D. Error
Explanation: You can not use var before a variable name.