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A 280 meter long train crosses a platform thrice its length in 50 seconds. What is the speed of the train in km/hr?

160.48
264.86
380.64
482.38

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Length of train \\ =  280 m \\ Length of platform \\ =  (3 \times 280) m  =  840m \\ Speed of train \\ = (\frac{280 + 840}{50}) m / \sec \\ \frac{1120}{50} m / \sec \\ = (\frac{1120}{50} \times \frac{18}{5}) k m / h r \\ = 80.64 k m / h r \end{aligned}

$\eqalign{ & {\text{Length of train}} \cr & {\text{ = 280 m }} \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Length of train \\ =  280 m \\ Length of platform \\ =  (3 \times 280) m  =  840m \\ Speed of train \\ = (\frac{280 + 840}{50}) m / \sec \\ \frac{1120}{50} m / \sec \\ = (\frac{1120}{50} \times \frac{18}{5}) k m / h r \\ = 80.64 k m / h r \end{aligned}

$\eqalign{ & {\text{Length of train}} \cr & {\text{ = 280 m }} \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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