Problem 3 - A History of Mathematics: An Introduction

Question

A hare is 150 paces ahead of a hound that is pursuing him. If the hound covers 10 paces each time the hare cover 6 , in how many paces will the hound overtake the hare?

Step-by-Step Solution

Answer

Answer: It will take the hound 375 paces to catch up to the hare.

Step 1: Set up a proportion between the distance traveled by the hare and the hound

To set up a proportion, we need to examine the relationship between the distance traveled by the hound and the distance traveled by the hare. We know that the hound covers 10 paces for every 6 paces the hare travels. This relationship can be represented by the ratio: Hound paces / Hare paces = 10 / 6

Step 2: Include the head start of the hare

Since the hare is 150 paces ahead of the hound, we need to include this head start in our proportion. Let x be the number of paces the hound needs to travel to catch up to the hare. Then, the hare will travel x - 150 paces. Our proportion now becomes: Hound paces / (Hare paces + 150) = 10 / 6

Step 3: Solve for x

Now, we can solve for x to find the number of paces it takes for the hound to catch up to the hare. Substitute x for "Hound paces" in the proportion: x / (x - 150) = 10 / 6 Since we have a proportion, we can solve for x using cross-multiplication: 6x = 10(x - 150) Now, distribute the 10 on the right-hand side: 6x = 10x - 1500 Subtract 10x from both sides: -4x = -1500 Divide by -4: x = 375

Step 4: Check the solution

The hound will catch up to the hare after 375 paces. To verify our solution, we can see if the ratio of the hound paces to hare paces still holds true. The hare travels 6 paces for every 10 paces the hound travels, so when the hound has traveled 375 paces: Hare paces = (6 / 10) * 375 = 225 Since the hare had a 150-pace head start, the hare will have traveled 225 - 150 = 75 paces when the hound catches up. The ratio of the hound paces to hare paces is: 375 / 75 = 10 / 6 This confirms our solution, and the hound will catch the hare after traveling 375 paces.

Key Concepts

Ratio and Proportion

To understand the problem of the hare and the hound, we first need to delve into the concept of ratio and proportion. Ratios are a way of comparing two or more quantities to show the relationship between them. Here, it tells us that for every 6 paces the hare travels, the hound travels 10. This relationship can be expressed as the ratio 10/6, meaning the hound moves faster compared to the hare.
Proportions take this comparison further by setting two ratios equal to each other. In this problem, we set up a proportion to determine how long it will take the hound to catch up to the hare. By incorporating the hare's head start of 150 paces, the equation becomes \[\frac{x}{x - 150} = \frac{10}{6}.\] This expression helps us judge when the two will have traveled far enough for the hound to overtake the hare. Proportions are like a bridge connecting different parts of an equation to solve for an unknown value.
When working with ratios and proportions:

• Identify the relationship between the quantities involved.
• Set up an equation involving those quantities set in proportion.
• Solve the equation by working with equivalent fractions or using techniques like cross-multiplication.

Algebraic Equations

Algebraic equations are mathematical statements where two expressions are set equal to each other, often containing variables representing unknown values. Solving these equations reveals the unknown quantities we're interested in.
In the given problem, after setting up the proportion, we transition to an algebraic equation: \[ 6x = 10(x - 150). \] This equation arises from using the mathematical technique known as cross-multiplication. By multiplying both sides of the equation by the denominators, we eliminate the fractions, simplifying the process of solving for x.
Here's a step-by-step approach when dealing with algebraic equations:

• First, simplify both sides of the equation separately if possible. Distribute and combine like terms where necessary.
• Move all terms involving the variable to one side and constant terms to the other.
• Isolate the variable to one side of the equation typically by dividing or multiplying both sides of the equation by a constant.
• Review and verify your solution by substituting back into the original equation.

When solved correctly, our equation tells us the hound will travel 375 paces to catch the hare. By knowing how to form and solve algebraic equations, we tackle real-world problems using mathematical language.

Distance and Speed Problems

Distance and speed problems frequently appear in both everyday life and mathematics. They involve understanding the relationship between distance, speed, and time. The fundamental equation connecting these quantities is: \[ ext{Distance} = ext{Speed} imes ext{Time}. \] This formula helps us figure out how far something will travel based on its speed over a given time period.
In the hare and hound scenario, speed is symbolized as how many paces are covered in a given interval. The hound travels 10 paces while the hare travels 6, meaning the hound's speed is greater concerning the pace per unit time.
To solve such problems:

• First, clearly identify the speeds of the various entities involved.
• Identify any head starts or initial leads, as in this case, with the hare.
• Use proportions or algebra to balance the distance equation if more than one variable is present.
• Once you have enough information, plug values into the core distance formula to find unknowns like time or extra distance.

Understanding these problems means piecing together these relationships through logical reasoning and descriptive equations.