Let us assume number of operation =100;
So number of floating point operation=100*2/5=40
So number of fixed point operation=60
Lets time taken by floating point operation is t1 and for fixed point operation it take t2
So t1+t2=100(assume).
since 60 floating point operation take t1 time
then only one=t1/40
and fixed point=t2/60
and it is given that::
t1/40=2*t2/60
3t1=4t2;
since t1+t2=100
t2=300/7
t1=400/7
Now time after speed up let us assume t1' and t2'
t1'=400/3*1.1=121.22(Approx)
t'2=300/7*1.2=35.72
Tt=t'1+t'2=156.94
S2=100/156.94*S1
S2=0.64*S1