There are various methods to approach questions from the topic Speed, Time and Distance.
This article demonstrates the usage of Ratios in solving various types of questions:
Basics: Distance = Speed ×Time
1. If Time is fixed, then Distance α Speed, i.e., D1/D2 = S1/S2.
So for example, if two trains start from Delhi and Chandigarh towards each other at 8 am, and meet somewhere in between at 10 am, this means both the trains have travelled for 2 hours each. So, their distances travelled will be in the ratio of their speeds.
2. If Speed is fixed, then Distance α Time i.e., D1/D2 = T1/T2.
This basically means that if a person is moving at a fixed speed, then he will take more time in covering a greater distance as compared to the time he would take to cover a shorter distance.
3. If Distance is fixed, then Speed α 1/Time i.e., S1/S2 = T2/T1.
This means that a body at a higher speed takes less time to cover the distance as compared to the time it would take to cover the same distance at a lower speed.
So if a body takes time T to cover a distance D at a certain speed, then it would take time T/2 to cover the same distance in case its speed is doubled.
You will understand the concept better with the help of the following examples:
Example 1:
A thief steals a car at 9 am and drives it at a speed of 60 kmph. The police start chasing him at 10 am at the speed of 90 km/hr. How much more distance would be covered by the thief before he is caught.
Solution:
Distance covered by thief from 9 am to 10 am = 60 km.
Now, after the police starts chasing him, let the distance covered by thief be Dt and that of police be Dp. So Dt/Dp = 60/90 = 2/3. So distances will be 2x and 3x.
As per question, 3x – 2x = 60 i.e. x = 60. So Dt = 120 km and Dp = 180 km.
Extra practice:
The same question can also be used to find the time when the thief will be caught. Now, the thief has to run a further distance of 120 km as found above, so that means 120/60 = 2 hours. The chase started at 10am, so he will be caught at 12 Noon.
Example 2:
A thief steals a car at 2:30 pm and drives away at 60 kmph. The theft is discovered at 3 pm and the owner sets off in another car at 75 kmph. At what time will the thief be caught?
Solution:
Distance covered by thief from 2:30 pm to 3 pm = 60 × ½ = 30 km.
After the chase starts, let the distance covered by thief is Dt and that of owner is Do. So Dt/Do = 60/75 = 4/5 i.e. 4x and 5x.
Now, as per question 5x – 4x = 30 i.e. x = 30 km. So Dt = 120 km and Do = 150 km.
Thief will be caught after travelling adistance of 120kms i.e. 120/60 = 2 hours from the time the chase began i.e. 5 pm.
Example 3:
A train after travelling 50 km meets with an accident and then proceeds at ¾ of its former speed and arrives at its destination 35 mins late. Had the accident occurred 24 km further, it would have reached only 25 mins late. Find the speed of the train.
Solution:
In the second case, the train travelled an additional 24 km with its normal speed, hence saving 10 mins. So instead of making equations, we will focus only on these 24 km.
Distance Speed Time
i. scenario: 24 kms 3/4 S 4/3 T
ii. Scenario: 24 kms S T
Clearly, in the first case the train takes 1/3 T extra mins i.e. 1/3 T = 10 mins.
So T = 30 mins. This means that the train takes 30 mins to cover 24 kmat its normal speed, hence the normal speed should be 48 kmph.
