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 –

**Ralph and Steve run a race between points X and Y, 5 km apart. Ralph starts at 9 a.m. from X at a speed of 5 km/hr, reaches Y, and returns to X at the same speed. Steve starts at 9:45 a.m. from X at a speed of 10 km/hr, reaches Y and comes back to X at the same speed. ****At what time does Steve overtake Ralph?**

**(1) 10:20 a.m.**

**(2) 10:30 a.m.**

**(3) 10:40 a.m.**

**(4) 10:50 a.m.**

**(5) 11:`00 a.m.**

*img credit: Flickr*