Question

A train, 100 metres in length, crosses a telegraphic post in 10 seconds. another train of same length crosses a platform, 125 metres long, in 15 seconds. what is the difference of distances covered by the two trains in 3 hours?

Answer 1

$speed \: \: of \: \: 1st \: \: train = \frac{100}{10} = 10 \frac{m}{s} \\ speed \: \: of \: \: 2nd\: \: train = \frac{100 + 125}{15} = 15 \frac{m}{s} \\ \\ therefore \: \: difference \: \: of \: \: distance \\ covered \: \: by \: \: the \: \: trains \: \: in \: \: one \: \: second \\ = 15 - 10 = 5 \: m \\ \\ therefore \: \: difference \: \: of \: \: distance \\ covered \: \: by \: \: the \: \: trains \: \: in \: \: one \: \: 3 \: hours \\ = 3 \times 60 \times 60 \times 5 = 54000 \: m = 54 \: km$

Answer 2

Answer:

\begin{gathered}speed \: \: of \: \: 1st \: \: train = \frac{100}{10} = 10 \frac{m}{s} \\ speed \: \: of \: \: 2nd\: \: train = \frac{100 + 125}{15} = 15 \frac{m}{s} \\ \\ therefore \: \: difference \: \: of \: \: distance \\ covered \: \: by \: \: the \: \: trains \: \: in \: \: one \: \: second \\ = 15 - 10 = 5 \: m \\ \\ therefore \: \: difference \: \: of \: \: distance \\ covered \: \: by \: \: the \: \: trains \: \: in \: \: one \: \: 3 \: hours \\ = 3 \times 60 \times 60 \times 5 = 54000 \: m = 54 \: km\end{gathered}

speedof1sttrain=

10

100

=10

s

m

speedof2ndtrain=

15

100+125

=15

s

m

thereforedifferenceofdistance

coveredbythetrainsinonesecond

=15−10=5m

thereforedifferenceofdistance

coveredbythetrainsinone3hours

=3×60×60×5=54000m=54km