Question 85

A train started at 9AM from a station A with a speed of 72 km/hr. Another train after tvvo hours started from the station B towards A with a speed of 90 km/ph. The two trains are expected to cross each other at 1.30 PM. At 12 noon because of the signals both the trains reduced their speeds by the same quantity and they crossed each other at 4.30 PM. The speed of the train, after 12 noon, that started from the station A, is

Solution

The train which started from station A traveled 144 km in first 2 hours.

The train from station A and station B had speed of 72 km/h and 90 km/h respectively.

And they expected to be meet in 1:30 PM.

But from 11:00 AM they started to travel towards eachother.It means that they expected to meet eachother by 2.5 hours from 11:00AM.

Their relative velocity is=(72+90)=162 km/h.

So, in 2.5 hours they should travel=2.5×162=405 km. So they were 405 km apart from eachother at 11:00 AM.

Let say, due to signal,they reduced their speed by x km/h.

So,train from station A and station B will have (72-x) km/h and (90-x) km/h as their respective speed.

But,they shifted their speed to this reduced speed at 12 noon, which means that they travel for more 1 hour from 11:00AM at their previous speeds.

So,in that 1 hour they travel for more 72 km and 90 km further towards eachother.

Which in turn reduced their total journey of 405 km by (72+90)=162 km by this 1 hour.

So,they traveled only (405-162)=243 km in (12 noon-4:30 PM)= 4.5 hours.

But their new relative velocity is (162-2x) km/h.

So, they will take (243/(162-2x)) hours to cross eachother.

So,

$$(243/(162-2x))=4.5$$

or,$$54=162-2x$$

or,$$2x=108$$

or,$$x=54$$.

So, the train ,which left from station A,have reduced its speed to (72-54)=18 km/h.

C is correct choice.


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