Question

1. Car A left from a point P at 9 am at a
speed of 45 km/h. Car B left from P
in a direction same as that of Car A at
9:30 am at a speed of 'X' km/h. If
Car B travelled 90 km before it met
Car A, what is the value of X? (Both
cars travelled at constant speeds)

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Answer 1

Given:

• Car A left from a point P at 9 am at a speed of 45 km/h. Car B left from P in a direction same as that of Car A at 9:30 am at a speed of 'X' km/h.

To find:

• If Car B travelled 90 km before it met Car A, what is the value of X?

Solution:

Car B travelled 90 km before it met Car A, which means that Car B travelled 90 km at a speed 45 km/h for that time.

So speed of Car A is

= path travelled / speed of the car

= 90 / 45 h

= 2 h

Car A travelled for 2 hours and Car B met with it at (9 a.m. + 2 hours) = 11 a.m.

It implies that Car B travelled for (11 a.m. - 9.30 a.m.) = 1.30 h = 3/2 h.

Car B travelled for 3/2 hours and covered 90 km.

Thus speed of Car B is

= path travelled / time taken

= 90 / (3/2) km/h

= 90 * 2/3 km/h

= 60 km/h

Answer: The value of X is 60.