Determining the Distance Between Home and School: A Comprehensive Guide

Solving the problem of finding the distance between home and school involves understanding the relationship between speed, time, and distance. This guide walks you through the process using a step-by-step approach, highlighting algebraic equations that are commonly used in school mathematics. We'll explore various methods and provide a clear answer using algebra.

Solving the Problem Using Algebra

The problem states that walking at 5 km/hr takes 5 minutes more to reach school than walking at 6 km/hr. Conversely, walking at 6 km/hr makes you 5 minutes early. Let's denote the distance between home and school as ( d ) kilometers.

Step 1: Setting up the Equations

When walking at 5 km/hr:

Time taken ( frac{d}{5} ) hours This takes 5 minutes longer than the time taken at 6 km/hr.

When walking at 6 km/hr:

Time taken ( frac{d}{6} ) hours This is 5 minutes less than the time taken at 5 km/hr.

Step 2: Converting Time Differences to Hours

Since 5 minutes is ( frac{5}{60} frac{1}{12} ) hours, we can express the time relationships as:

( frac{d}{5} frac{d}{6} frac{1}{12} )

Step 3: Solving the Equation

Multiply every term by 60 to eliminate the denominators:

( 60 left( frac{d}{5} right) 60 left( frac{d}{6} right) frac{60}{12} )

( 12d 10d 5 )

Rearranging the equation gives:

( 12d - 10d 5 )

( 2d 5 )

( d frac{5}{2} 2.5 ) kilometers

Conclusion

The distance between home and school is ( 2.5 ) kilometers.

Alternative Methods

Method 1: Using a Direct Distance Formula

Let's denote the distance as ( d ) kilometers.

For walking at 5 km/hr:

Time taken ( frac{d}{5} ) hours

For walking at 6 km/hr:

Time taken ( frac{d}{6} ) hours

The time difference is 10 minutes, which is ( frac{10}{60} frac{1}{6} ) hours.

Therefore,

( frac{d}{5} - frac{d}{6} frac{1}{6} )

( frac{6d - 5d}{30} frac{1}{6} )

( frac{d}{30} frac{1}{6} )

( d 5 ) kilometers

Method 2: Algebraic Approach with T1 and T2

Let ( T1 ) and ( T2 ) be the times to school moving at 5 km/hr and 6 km/hr respectively.

We know:

( 5T1 d ) and ( 6T2 d ) ( T1 T2 frac{10}{60} T2 frac{1}{6} ) ( T2 T1 - frac{10}{60} T1 - frac{1}{6} )

Substituting ( T1 T2 frac{1}{6} ) into ( 5T1 d ), we get:

( 5 left( T2 frac{1}{6} right) d )

( 5T2 frac{5}{6} d )

( 6T2 5 6d )

( 6T2 5 6 left( frac{6T2}{5} right) )

( 6T2 5 frac{36T2}{5} )

( 30T2 25 36T2 )

( 25 6T2 )

( T2 frac{25}{6} ) hours

( d 6T2 6 times frac{25}{6} 25/6 4.17 ) kilometers

Verification

Using the simpler method, the answer is ( 5 ) kilometers, which is more accurate.

Conclusion

The distance between home and school is ( 5 ) kilometers. This method involves understanding and applying basic algebraic equations to solve real-world problems effectively.

Keywords

distance calculation speed and time algebraic equations