kushtayal wrote:
Can someone please post an easy to understand solution to the question!
@bunel, waiting to see your answer!
This question is easy when you practice various types of them. Lets jump into it
There are 3 scenarios : 1) When the vehicle has no Engine Malfunctions, 2) When Engine Malfunction occurs after 80 Kms, 3) When the Engine Malfunctions after 120 Kms
Let us take Scenario 1. When the engine doesn't malfunction, we can assume that the car travels a distance 'D' at a rate 'R' at time 'T'. We only have variables to work with so lets just keep it that way. We get D = RT (from the formula)
From Scenario 2, note that the distance traveled by the car at the end of the trip is same as above 'D', but the rate and the time it takes to reach the destination is different. Lets say that Scenario 2 has two trips [Trip 1 + Trip 2] [NOTE: Distance of Trip 1 + Distance of Trip 2 = D (When No Engine Malfunctions Occur]
For Trip 1, the car travels 80 Km with no engine malfunctions. In this case we get D = RT --> 80 = RT1
For Trip 2, the car travels the remaining distance (D - 80) at 4/5 the original rate. So we get D-80 = 4/5(R)T2
Note that we only have a relation between T & T1 & T2 from the question as follows -> T1 + T2 = T + 1hr 24 mins;
Now we can use the above equations to perform the following (Remember the criteria -- Distance traveled Is the same SAME regardless of when the malfunction occurred)
So we have D = 80 (Trip 1 Distance) + D-80 (Trip 2 Distance)
RT = 80 + 4/5(R)T2
RT = 80 + 4/5(R)[T + 1hr 24 mins - T1]
RT = 80 + 4/5(R)[T + 1hr 24 mins - 80/R] (The 80/R is the time from Trip 1 --> Refer to equation Trip 1) ---> 1
The same procedure is followed for Trip 2, we get the equation
RT = 120 + 4/5(R)[T + 1 - 120/R] --> 2
Subtracting 2 from 1, we can isolate variable R and end up with 25 Km/hr
Hope its clear now
_________________
Siddharth Ramachandran
B School Aspirant, 2021 Intake