Finding the Ratio of Two Numbers from Their Sum and Difference Ratio

This problem asks us to find the ratio of two numbers given the ratio of their sum and their difference. Let's break it down step by step.

Let the two unknown numbers be $x$ and $y$. We are given the ratio of their sum and difference. Assume, without loss of generality, that $x \ge y$ so that the difference $x-y$ is non-negative. The sum of the numbers is $(x+y)$ and the difference between the numbers is $(x-y)$.

According to the question, the ratio of the sum and the difference is 6 ∶ 5. We can write this ratio as a fraction:

\( \frac{\text{Sum of numbers}}{\text{Difference of numbers}} = \frac{x+y}{x-y} \)

We are given that this ratio is 6 ∶ 5, which means:

\( \frac{x+y}{x-y} = \frac{6}{5} \)

Now, we need to solve this equation to find the ratio $x : y$. We can do this by cross-multiplication.

Solving the Equation for the Ratio of Numbers

Cross-multiplying the equation \( \frac{x+y}{x-y} = \frac{6}{5} \), we get:

\( 5 \times (x+y) = 6 \times (x-y) \)

Now, distribute the numbers on both sides of the equation:

\( 5x + 5y = 6x - 6y \)

Our goal is to find the ratio $x : y$, which is equivalent to finding the value of \( \frac{x}{y} \). To do this, let's rearrange the equation to gather terms involving $x$ on one side and terms involving $y$ on the other side.

Move the $5x$ term from the left side to the right side by subtracting $5x$ from both sides:

\( 5y = 6x - 6y - 5x \)

\( 5y = (6x - 5x) - 6y \)

\( 5y = x - 6y \)

Now, move the $-6y$ term from the right side to the left side by adding $6y$ to both sides:

\( 5y + 6y = x \)

\( 11y = x \)

We have found a relationship between $x$ and $y$. To find the ratio $x : y$, we can divide both sides by $y$ (assuming $y \neq 0$, which must be true if the difference is non-zero as implied by the ratio 6:5) and rearrange:

\( \frac{x}{y} = \frac{11}{1} \)

This means the ratio $x : y$ is 11 : 1.

Therefore, the ratio of the two numbers is 11 ∶ 1.

Verification

Let the numbers be $11k$ and $1k$ (where $k$ is a constant). Sum = \( 11k + 1k = 12k \) Difference = \( 11k - 1k = 10k \) Ratio of sum to difference = \( \frac{12k}{10k} = \frac{12}{10} = \frac{6}{5} \). This matches the given ratio 6 ∶ 5, so our finding that the ratio of the numbers is 11 ∶ 1 is correct.

Revision Table: Ratio of Numbers Problem

Additional Information: Componendo and Dividendo

This type of problem can also be solved quickly using a property called Componendo and Dividendo. If \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \).

In our problem, we have \( \frac{x+y}{x-y} = \frac{6}{5} \). Let $a = x+y$ and $b = x-y$. Applying Componendo and Dividendo to \( \frac{a}{b} = \frac{6}{5} \), we get:

\( \frac{a+b}{a-b} = \frac{6+5}{6-5} \)

Substitute back $a = x+y$ and $b = x-y$:

\( \frac{(x+y) + (x-y)}{(x+y) - (x-y)} = \frac{11}{1} \)

\( \frac{x+y+x-y}{x+y-x+y} = \frac{11}{1} \)

\( \frac{2x}{2y} = \frac{11}{1} \)

\( \frac{x}{y} = \frac{11}{1} \)

Thus, the ratio $x : y$ is 11 : 1. This method provides a more direct way to arrive at the ratio of the numbers when the ratio of their sum and difference is given.

Understanding ratios and how to manipulate algebraic equations is fundamental to solving problems involving numbers and their relationships. This problem demonstrates a classic application of these concepts.