Solving Real-world Speed and Time Problems Using Algebraic Equations

Understanding how to solve problems involving speed, distance, and time is a fundamental skill in mathematics that has numerous practical applications. Let's explore a problem scenario where a person covers a distance of 110 kilometers between two cities in 10 hours, traveling partly on foot at 9 kilometers per hour and partly on a bicycle at 15 kilometers per hour. We will use algebraic equations to find the distance traveled on foot.

In this article, we will break down the problem step-by-step, providing a clear explanation and solution.

The Problem

A man covers a distance of 110 km between two cities in 10 hours. He traveled partly on foot at 9 km/hr and partly on a bicycle at 15 km/hr. Find the distance traveled on foot.

Step-by-Step Solution

Defining Variables

Let's define the following variables:

d_f - the distance traveled on foot (in kilometers) d_b - the distance traveled by bicycle (in kilometers)

Setting Up Equations

We know the following:

The total distance covered is 110 km: d_f d_b 110 The speed on foot is 9 km/hr, and the speed on the bicycle is 15 km/hr. The time taken for each part of the journey can be expressed as: Time on foot: (frac{d_f}{9}) Time on bicycle: (frac{d_b}{15})

The total time taken is 10 hours:

(frac{d_f}{9} frac{d_b}{15} 10)

Solving the Equations

Let's solve this system of equations step by step.

Step 1: Solve for d_b

From the first equation, we can express d_b in terms of d_f:

d_b 110 - d_f

Step 2: Substitute into the Second Equation

Substituting d_b into the second equation:

(frac{d_f}{9} frac{110 - d_f}{15} 10)

Step 3: Clear the Fractions

To eliminate the fractions, let's multiply through by the least common multiple of the denominators, which is 45:

45 left(frac{d_f}{9} frac{110 - d_f}{15}right) 45 cdot 10)

This simplifies to:

5d_f 3(110 - d_f) 450)

Step 4: Expand and Simplify

Expanding the equation:

5d_f 330 - 3d_f 450)

Combining like terms:

2d_f 330 450)

Step 5: Solve for d_f

Subtract 330 from both sides:

2d_f 120)

Dividing by 2:

d_f 60)

Conclusion: The distance traveled on foot is 60 km.

Verification

To verify, we can calculate d_b:

d_b 110 - d_f 110 - 60 50 km)

Now, let's check the time:

Time on foot: (frac{60}{9} approx 6.67) hours Time on bicycle: (frac{50}{15} approx 3.33) hours

Total time:

6.67 3.33 10 hours

The calculations are consistent, confirming that the distance traveled on foot is indeed 60 km.

Additional Insight: Consider the problem from another perspective. Let t1 and t2 be the times of travel by cycle and on foot, respectively. The time equations can be set as:

Equations:

t1 t2 9 1 9t1 4 t2 61 2

2 - 1 4:

9t1 4t2 61 4t1 4t2 36

5t1 61 - 36 25

t1 25/5 5 hours

t2 9 - 5 4 hours

So, the distance traveled by bicycle is 9 5 45 km, and the distance traveled on foot is 4 4 16 km.

Note: There are slight variations in the methods, but both solutions confirm the total distance and time constraints.