This topic entails solving problems of varying complexity with the help of several equations. There is only one formula and it is known to most of you: Distance = Speed * Time. However, what differentiates one problem from the other is the application of the formula. These problems are more about logic and application.

This basic formula leads to 2 special cases.

Case 1: Distance Constant

If the distance of the journey is constant, then the time taken for travel is inversely proportional to speed.

## What else you can do inside qs leap ?

=> Time α (1/Speed)

=> T1 / T2 = S2 / S1

So, if the speed of journey is tripled, the time of journey becomes one-thirds.

Case 2: Time Constant

If the time of the journey is constant, then the distance taken for travel is directly proportional to speed.

=> Distance α Speed

=> D1 / D2 = S1 /S2

So, if the speed of journey is doubled, the distance covered for the same time becomes double.

Solve the following question –

**If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/ hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon? **

**(A) 11 Km/hr **

**(B) 12 Km/hr **

**(C) 13 Km/hr **

**(D) 14 Km/hr **

**(E) 15 Km/hr**