Question

Train X started from town A towards town B at
6:00 a.m. and reached B at 2:00 p.m. Train Y started
from B towards A at 7:00 a.m. and reached A at
3:00 p.m. At what time did both the trains cross each
other?​

Answer 1

Answer:

Answer: Option 3

Solution:

Let the speeds of the trains X and Y be ‘a’ and ‘b’ kmph respectively and the total distance between the town be ‘d’ kms

We can see that both the trains travelled 8 hours to reach from one point to other (6 : 00 am to 2 : 00 pm and 7 : 00 am to 3 : 00 pm)

⇒ a = b = (d/8) kmph

Distance covered by train X in the first one hour from 6 : 00 am to 7 : 00 am = (d/8)/1

⇒ (d/8) kms

Remaining distance travelled by both the trains till them they meet each other = d - (d/8)

⇒ (7d/8) kms

Time taken to meet for the first time when moving towards each other = (Distance)/Relative Speed

⇒ Relative Speed = a + b

⇒ (d/8) + (d/8)

⇒ (d/4)

⇒ Time taken = (7d/8)/(d/4) = 7/2 hours = 3.5 hours

⇒ Meeting time is 3.5 hours from 7 am = 10.30 am

∴ Both the trains cross each other at 10 : 30 am

Answer 2

Both the trains cross each other at 10:30 a.m

Given :

• Train X started from town A towards town B at 6:00 a.m. and reached B at 2:00 p.m.

• Train Y started from B towards A at 7:00 a.m. and reached A at 3:00 p.m

To find :

The time when both the trains cross each other

Solution :

Step 1 of 2 :

Form the equation to calculate the time when both the trains cross each other

Let distance between town A and town B is d km

Train X started from town A towards town B at 6:00 a.m. and reached B at 2:00 p.m.

Time taken by train X = 8 hour

$\displaystyle \sf \therefore \: Speed \: of \: train \: X = \frac{d}{8} \: \: km/hr$

Train Y started from B towards A at 7:00 a.m. and reached A at 3:00 p.m

Time taken by train Y = 8 hour

$\displaystyle \sf \therefore \: Speed \: of \: train \: Y = \frac{d}{8} \: \: km/hr$

Let Let both the train meet t hours after 7 a.m.

Train X travelled (t + 1) hours and Train Y travelled t hours

$\displaystyle \sf \therefore \: Distance \: travelled \: by \: train \: X = \frac{d(t + 1)}{8} \: km$

$\displaystyle \sf \therefore \: Distance \: travelled \: by \: train \: Y = \frac{dt}{8} \: km$

By the given condition

$\displaystyle \sf \frac{d(t + 1)}{8} + \frac{dt}{8} = d$

Step 2 of 2 :

Calculate time when both the trains cross each other

$\displaystyle \sf{ \implies }\frac{(t + 1)}{8} + \frac{t}{8} = 1$

$\displaystyle \sf{ \implies }\frac{(t + 1) + t}{8} = 1$

$\displaystyle \sf{ \implies }\frac{2t + 1}{8} = 1$

$\displaystyle \sf{ \implies }2t + 1 = 8$

$\displaystyle \sf{ \implies }2t = 7$

$\displaystyle \sf{ \implies }t = \frac{7}{2}$

$\displaystyle \sf{ \implies }t = 3 \frac{1}{2}$

$\displaystyle \sf Now \: \: 3 \frac{1}{2} \: hours = 3 \: hours \: 30 \: minutes$

Hence, the time when both the trains cross each other

= 7 a.m + 3 hours 30 minutes

= 10:30 a.m

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