Children Crawling Puzzle

#196

Two children were sitting on the floor. 1st child was at one corner of the floor(Point A) and other child is on the opposite corner of the floor(Point B). Both of them started crawling to approach each other but by different speeds. In between, they meet at point X and pass each other. After their meeting at point X, 1st child took 40 seconds to reach at point B and the other child took only 10 seconds to reach at point A.
Can you find out the total time each of them took to change their positions?

  Difficulty Level - 3/5

   
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  Solution :

Assume that "a" is the speed of 1st Child  and "b" is the speed of other child.
Let "t" be the number of seconds it takes them to meet at point "X". 
Using Distance = Speed * Time
>>> For 1st Child
Distance = a * t before meeting
And 
Distance = a * 40 after meeting

>>> For other child
Distance = b * t before meeting
And
Distance = b * 10 after meeting
Now the distance travelled by 1st child before meeting is equal to the distance travelled by the other child after meeting and vice versa.
Thus
a * t = b * 10 And b * t = a * 40 
>>> a * t = (a * 40)/t * 10
>>> t^2 = 400
>>> t = 20
Thus 1st Child requires 20 + 40 = 60 seconds,
Whereas the other Child requires 20 + 10 = 30 seconds to change the position..