Question
A train of length 420 m can cross a platform of length
‘y’ m in 30 seconds. Also, it can cross another train of same length as that of the platform, moving at a speed of 22 m/sec in opposite direction, in 15 seconds. Find the value of ‘y’.Solution
ATQ,
Let the speed of train be ‘s’ m/sec.
According to question:
(420 + y) = 30 × s -----(1)
(420 + y) = 15 × (s + 22) -----(2)
Solving (1) and (2), we get
30 × s = 15 × (s + 22)
30s – 15s = 15 × 22
15s = 330
s = 330 / 15
s = 22 m/s
Putting the value of ‘s’ in equation (1), we get
(420 + y) = 30 × 22
(420 + y) = 660
y = 660 – 420
y = 240 m
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