Sohan covers a certain distance between his house and a park by cycle. At 30 kmph, he is late by 20 minutes. However, at 40 kmph, he reaches the park 10 minutes early. Find the distance between Sohanâs house and the park.

Points to Remember:

This is a problem involving speed, time, and distance.
The key is to understand the relationship between speed, time, and distance (Distance = Speed Ã Time).
We will use algebraic equations to solve for the unknown distance.

Introduction:

This question is a classic example of a time and distance problem. It tests our understanding of the inverse relationship between speed and time when covering a fixed distance. Faster speed means less time taken, and slower speed means more time taken. We will use the formula Distance = Speed Ã Time to solve this problem. The core challenge lies in setting up equations that reflect the given information about Sohan’s journey to the park.

Body:

Setting up the Equations:

Let ‘d’ represent the distance between Sohan’s house and the park (in km).
Let ‘t’ represent the scheduled time to reach the park (in hours).

Scenario 1: Speed = 30 kmph, Late by 20 minutes:
The time taken is t + (20/60) = t + (1/3) hours.
Therefore, the equation is: d = 30(t + 1/3) —(1)
Scenario 2: Speed = 40 kmph, Early by 10 minutes:
The time taken is t – (10/60) = t – (1/6) hours.
Therefore, the equation is: d = 40(t – 1/6) —(2)

Solving the Equations:

Since both equations equal ‘d’, we can equate them:

30(t + 1/3) = 40(t – 1/6)

Expanding the equations:

30t + 10 = 40t – 20/3

Solving for ‘t’:

10t = 10 + 20/3 = 50/3
t = 5/3 hours

Finding the Distance:

Substitute the value of ‘t’ (5/3 hours) into either equation (1) or (2) to find ‘d’. Let’s use equation (1):

d = 30(5/3 + 1/3) = 30(6/3) = 30(2) = 60 km

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Therefore, the distance between Sohan’s house and the park is 60 km.

Conclusion:

By setting up and solving two simultaneous equations based on the given information about Sohan’s journey at different speeds, we determined that the distance between his house and the park is 60 km. This problem highlights the practical application of algebraic equations in solving real-world problems involving speed, time, and distance. Understanding the relationship between these three variables is crucial for efficient problem-solving in various fields, including transportation planning and logistics. This approach can be applied to similar problems involving varying speeds and time constraints to accurately calculate distances or travel times. Further, incorporating factors like traffic conditions and varying terrain into such calculations would enhance the accuracy and practical applicability of these methods.

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