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(B) Let the length of the goods train be $L_1$ and its speed be $S_1$.Let the length of the passenger train be $L_2$ and its speed be $S_2$.Given the

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(B) Let the length of the goods train be $L_1$ and its speed be $S_1$.Let the length of the passenger train be $L_2$ and its speed be $S_2$.Given the ratio of speeds $S_1 : S_2 = 1 : 2$,so $S_2 = 2S_1$.When the passenger train overtakes the goods train,the relative speed is $(S_2 - S_1)$. The total distance covered to cross completely is $(L_1 + L_2)$.$(S_2 - S_1) = \frac{L_1 + L_2}{60} \implies S_1 = \frac{L_1 + L_2}{60} \dots (i)$When a passenger on the passenger train crosses the goods train,the distance covered is only the length of the goods train $(L_1)$.$(S_2 - S_1) = \frac{L_1}{40} \implies S_1 = \frac{L_1}{40} \dots (ii)$Equating $(i)$ and $(ii)$:$\frac{L_1 + L_2}{60} = \frac{L_1}{40}$$\frac{L_1 + L_2}{3} = \frac{L_1}{2}$$2(L_1 + L_2) = 3L_1$$2L_1 + 2L_2 = 3L_1$$2L_2 = L_1$Therefore,the ratio of their lengths $L_1 : L_2 = 2 : 1$.

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