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Simultaneous Access for cache and main memory

42= H(30) + (1-H) (150)

So H = 0.9

If hierarchical Access

42 =H(30) + (1-H)(150+30)

H = 0.92

But in question nothing mentioned so simultaneous Access should be the ans...
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Avg memory access time= hitratio(cache access time)+missratio(cache access time+memory access time)

 Because even in case of miss, we first look into cache and then access memory.

here,

42= x(30)+(1-x)(180)

on solving, x=0.92..

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