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consider a cache memory which is 30 times faster than the main memory and uses 90% of the total time.

what is the speedup gain by the cache memory

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Let, access time without cache = $M$  (Main Memory access time)

Cache is $30$ times faster than MM. So, $T_c = \frac{M}{30}$
Access time with cache $= hT_c$ + (1-h)M = $0.9 \frac{M}{30} + 0.1M$

Speed up $=\frac{Speed_{old}}{Speed_{new}}\\ =\frac{M}{ 0.9 \frac{M}{30} + 0.1M}\\ =\frac{1}{0.13}\\ =7.69$
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