Question 21.

Speed of the faster train = $\frac{160}{12} = \frac{40}{3}$ m/s

Speed of the slower train = $\frac{40}{3} - \left(6 \times \frac{5}{18}\right) = \frac{35}{3}$ m/s

Sum of speeds (when the trains travel towards each other) = $\frac{40}{3} + \frac{35}{3} = 25$ m/s

Let the slower train be $x$ metres long; then: $\frac{160 + x}{25} = 14$

On solving, $x = 190\text{ m}$