a) 1 : 16
b) 4 : 1
c) 2 : 1
d) 1 : 4
Answer: b
Explanation: Let the original speed be S1 and time t1 and distance be D.
$$\eqalign{
& \frac{{ {\frac{D}{2}} }}{{2{t_1}}} = {S_2} \cr
& {S_2} = \frac{D}{{4{t_1}}}\,{\text{and}}\,{S_1} = \frac{D}{{{t_1}}} \cr
& \frac{{{S_1}}}{{{S_2}}} = \frac{4}{1} = 4:1 \cr} $$
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