a) 1 : 3
b) 3 : 2
c) 3 : 4
d) None of these
Answer: b
Explanation:
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speeds}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}}\,{\text{be}}\,x\,{\text{m/sec}} \cr
& {\text{and}}\,y\,{\text{m/sec}}\,{\text{respectively}}. \cr
& {\text{Then,}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} = 27x\,{\text{metres}}, \cr
& {\text{and}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} = 17y\,{\text{metres}}. \cr
& \frac{{27x + 17y}}{{x + y}} = 23 \cr
& 27x + 17y = 23x + 23y \cr
& 4x = 6y \cr
& \frac{x}{y} = \frac{3}{2} \cr} $$
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