Question

Car leaves point a for point b every 10 min. The distance between a and b is 60 km. The car travels at a speed of 60 km/h. Determine the number of cars that a man driving from b to a will meet in route, if he starts from b simultaneously with one of the cars leaving
a. The car from b travels with a speed of 60 km/h.

Answer 1

Answer:

Total number of cars that will meet on route is 7 cars.

Explanation:

Given as :

Car leaves point a for point b every 10 min

The distance between point a and b = 60 km

The car from point a to b travel at speed = 60 km/h

The car from point b to a travel at speed = 60 km/h

Let The number of cars that meet in route = n cars

According to question

Time = $\dfrac{distance}{speed}$

Or, time = $\dfrac{60 km}{60 km/h}$

i.e time = 1 hour = 60 min

Now,

At every 10 min , one car move from a to b

For first 10 min , car move = 1

For second 10 min , car move = 1

Similarly for sixth 10 min ,car move= 1

For total time of 60 min, number of cars moves from a = 1 + 1+ 1+1 + 1 +1  = 6

But at last of 10 min, the car from b already reach near by point 1 , and just before car from point a move, therefor they meet somewhere

So, one more car meet

So, The total number of car now = 6 + 1 = 7

Hence, Total number of cars that will meet on route is 7 cars. Answer