in this case, you have added

M of HCl, which completely ionises so
[1]this drives the
\leftrightharpoons H_3O^+ (aq) + OH^-(aq))
to the left, assuming it is 25 degrees [assuming 1 litre, no other sources of these ions and no other impurities, letting x be the number of moles consumed in the backwards reaction]:
(10^{-7}-x) = 10^{-14} \implies x=4.88\times 10^{-9}mol)
hence, the new concentration of

is

, and the pH is 6.98
so as you see...

[in the exam, we'd just say it's still 7]
normal calculations are just [1], and since we normally deal with 0.1M or concentrations of that like [a lot higher than 10^-7], we can usually ignore the subsequent calculations because it will have insignificant impact.
but in this case, there is significant impact [of pH ~0.02!!!]
if the acid was to the power of -9 [or less], then we can safely say that the pH will still be 7, the change will be too insignificant to do anything.