If x operations cost O(x) amortized then how much xy operations cost? True or False?

Say some data structure can perform $x$ operations in amortized $O(x)$ time.
Then for a big enough $y$ it can perform $xy$ operations in worst case $O(xy)$ time.

My attempt:

$x$ operations in $O(x)$ amortized means $O(1)$ expected time for $1$ operation.
Then for $xy$ operations it'd be $O(xy)$ amortized (and I think $O(x^2y)$ worst case). Therefore, the statement is incorrect.

But the answers sheet says i'm wrong. Why?
Amortized is not just probabilistic, it means that for some big enough $y$,  $xy$ operations can't take a long time and will guaranteed to be $O(x)$ in average in worst case (and therefore $O(xy)$ for all $xy$ operations), even through some of operations may take even $O(xy)$ time itself.  

https://stackoverflow.com/questions/200384/constant-amortized-time