IBPS Clerk Mains 2019 Paper

Let speed of train Q be x m/s
And length of train Q be y m
Then length of train P is 2y m
Speed of train P = 72×

5

18

=20 m/s

2y+y

10

=20–x⇒

3y

10

=20–x
⇒3y = 200 – 10x ...(i)
and

2y+y

20

–2x

3y

20

=20–2x
3y=400–40x ...(ii)
Now from equations (i) and (ii)
400–40x=200–10x
⇒ 30x=200
∴x=

200

30

Now putting value of x in equation (i)
3y=200–10×

200

30

=200−

200

3

3y=

400

3

∴y=

400

9

=44.4 m