Computer A : 2 GHz = 2*109 cycle per second
Program run in 10 seconds on computer A this Program requires 10*2*109 cycle
Now for computer the number of cycle is 1.2 time of original = 1.2*2*1010 =24*109 cycle
24*109 cycle it should be done in 6 second
That means in 1 second = (24*109 )/6 = 4*109 cycle
4*109 cycle per second means 4Ghz
Hence for computer B, we need 4Ghz Clock rate.
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