CAT 2021 Quant Question [Slot 1] with Solution 01

Option: 3
Solution:
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, \(x = 190\;m\)

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