memory - Hit and miss ration in cache and average time calculation -

i'm trying solve objective type question , came in examination. don't know right answer, , don't know how , need help. thank .

question : in system main memory access time 100 ns. cache 10 time faster main memory , uses write though protocol. if hit ratio read request 0.92 , 85% of memory requests generated cpu read, remaining being write; average time consideration both read , write requests is

a) 14.62ns

b) 348.47ns

c) 29.62ns

d) 296.2ns

my work ::::

well, memory access time = 100ns

cache access time = 10 ns (10 time faster)

in order find avg time have formula  tavg = hc+(1-h)m     h = hit rate      (1-h) = miss rate        c   = time access information cache         m  = miss penalty  (time access main memory) 

write through operation : cache location , main memory location updated simultaneously.

it given 85% request generated cpu read request , 15% write request.

tavg = 0.85(avg time read request)+ 0.15(avg time write request)      = 0.85(0.92*10+0.08*100)+0.15(avg time write request) 

//* 0.92 hit ratio read request , hit ratio write request not given ??

if assume hit hit ratio write request same hit ratio read request then,

  = 0.85(0.92*10+0.08*100)+0.15(0.92*(10+100)+0.08*100)   =31 ns 

if assume hit ratio 0% write request then,

  = 0.85(0.92*10+0.08*100)+0.15(0*110+1*100)   =29.62 ns 

avg access time considering read = 0.92*10 + 0.08*100 = 17.2 ns.

avg access time considering write = 100 ns (because in write through have go memory update if hit or miss. if assume hit ratio = 0.5 , miss = 0.5 0.5*100 + 0.5*100 = 1*100)

so, total access time both read , write - 0.85*17.2 + 0.15*100 = 14.62 + 15 = 29.62 ns

**you cant assume hit ratio write same hit ratio read. write request (write through) whatever case, have write in memory. write access time equal memory access time.