If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:Sa/Sb = √(b/a)
i.e. Ratio of speeds is given by the square root of the inverse ratio of time taken.
Two trains A and B starting from two points and travelling in opposite directions, reach their destinations 9 hours and 4 hours respectively after meeting each other. If the train A travels at 80kmph, find the rate at which the train B runs.
Sa/Sb = √(4/9) = 2/3This gives us that the ratio of the speed of A : speed of B as 2 : 3.Since speed of A is 80 kmph, speed of B must be 80 * (3/2) = 120 kmph
A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?
Sa/Sb = sqrt(90/40) = 3/2
This gives us that the ratio of the speed of A : speed of B as 3:2. We know that time taken is inversely proportional to speed. If ratio of speed of A and B is 3:2, the time taken to travel the same distance will be in the ratio 2:3. Therefore, since B takes 90 mins to travel from the meeting point to Opladen, A must have taken 60 (= 90*2/3) mins to travel from Opladen to the meeting point(adsbygoogle = window.adsbygoogle || []).push({});
So time taken by A to travel from Opladen to Cologne must be 60 + 40 mins = 1 hr 40 mins
[Credits: Veritasprep]

Kaprekar's constant
6174 is known as Kaprekar's constant.
Take any four-digit number, using at least two different digits. (Leading zeros are allowed.)Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary.Subtract the smaller number from the bigger number.Go back to step 2 and repeat.The above process will always yield 6174, in at most 7 iterations. The only four-digit numbers for which Kaprekar's routine does not reach 6174 are repdigits such as 1111, which give the result 0000 after a single iteration.