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
