Calculate the effective access time This question seems to be causing a lot of debate and I'm wondering whether my working is correct.

A computer with a single cache (access time 20ns) and main memory (access
time 500ns) also uses the hard disk (average access time 0.01ms) for virtual
memory using paging. If it is found that the cache hit rate is 95% and the page
fault rate is 1%
Calculate the effective (average) access time (EAT) of this system for a
sequential access system.  (a) 73.5ns. (b) 55ns. (c) 50ns. (d) 74.75ns.

My working was as follows: (0.95x20) + 0.05 (0.01(10000+500+20) + 0.99(500+20))
Leading me to an answer of 50 being C), however others have seem to have chosen D) as their answer
Any help would be appreciated.
Seems correct , cache hit gives you 0.95*20 ns if it misses then 0.05 or 5% of the time you consult memory that is 0.05*( no page fault hence 0.99*( 500 from mem to cache + 20 from cache to cpu) else if page fault occurs then 1% or 0.01 *( 10000 ns from disk to mem +  500 ns from mem to cache + 20 ns from cache to cpu))
Should be 45.5 ns which I would give as C