Solving Complex Ratio Problems

This question asks us to find the value of a specific ratio, \(b : (c + a)\), given two other ratios involving the variables \(a\), \(b\), and \(c\). We are given:

• \(a : (b + c) = 1 : 3\)
• \(c : (a + b) = 5 : 7\)

We can use the property of ratios that if \(x : y = p : q\), then \(x / y = p / q\). Also, a useful property is that \(x / (x + y) = p / (p + q)\) and \(y / (x + y) = q / (p + q)\). This property helps relate parts of the ratio to the total sum of the parts.

Step-by-Step Solution using Ratio Proportions

Analyzing the First Given Ratio: \(a : (b + c) = 1 : 3\)

From the ratio \(a : (b + c) = 1 : 3\), we can express this as a fraction:

\[ \frac{a}{b + c} = \frac{1}{3} \]

To relate \(a\) to the total sum \(a + b + c\), we can add 1 to both sides of the equation:

\[ \frac{a}{b + c} + 1 = \frac{1}{3} + 1 \]

\[ \frac{a + (b + c)}{b + c} = \frac{1 + 3}{3} \]

\[ \frac{a + b + c}{b + c} = \frac{4}{3} \]

This equation tells us the ratio of the total sum to \(b+c\). From the original ratio \(a : (b + c) = 1 : 3\), we know \(a\) is 1 part and \(b + c\) is 3 parts, making the total \(a + b + c\) equal to \(1 + 3 = 4\) parts. Therefore, \(a\) is \(\frac{1}{4}\) of the total sum \(a + b + c\).

\[ a = \frac{1}{4} (a + b + c) \]

Analyzing the Second Given Ratio: \(c : (a + b) = 5 : 7\)

Similarly, from the ratio \(c : (a + b) = 5 : 7\), we write:

\[ \frac{c}{a + b} = \frac{5}{7} \]

Adding 1 to both sides to relate \(c\) to the total sum \(a + b + c\):

\[ \frac{c}{a + b} + 1 = \frac{5}{7} + 1 \]

\[ \frac{c + (a + b)}{a + b} = \frac{5 + 7}{7} \]

\[ \frac{a + b + c}{a + b} = \frac{12}{7} \]

This shows the ratio of the total sum to \(a+b\). From the original ratio \(c : (a + b) = 5 : 7\), we know \(c\) is 5 parts and \(a + b\) is 7 parts, making the total \(a + b + c\) equal to \(5 + 7 = 12\) parts. Therefore, \(c\) is \(\frac{5}{12}\) of the total sum \(a + b + c\).

\[ c = \frac{5}{12} (a + b + c) \]

Finding the Proportion of \(b\) in the Total Sum

Let \(S = a + b + c\) be the total sum. We have found that:

• \(a = \frac{1}{4} S\)
• \(c = \frac{5}{12} S\)

Since \(a + b + c = S\), we can write:

\[ a + b + c = S \]

Substitute the expressions for \(a\) and \(c\) in terms of \(S\):

\[ \frac{1}{4} S + b + \frac{5}{12} S = S \]

To find \(b\) as a proportion of \(S\), subtract the fractions for \(a\) and \(c\) from \(S\):

\[ b = S - \frac{1}{4} S - \frac{5}{12} S \]

Find a common denominator for the fractions, which is 12:

\[ b = \frac{12}{12} S - \frac{3}{12} S - \frac{5}{12} S \]

\[ b = \left(\frac{12 - 3 - 5}{12}\right) S \]

\[ b = \left(\frac{4}{12}\right) S \]

\[ b = \frac{1}{3} S \]

So, \(b\) is \(\frac{1}{3}\) of the total sum \(a + b + c\).

Calculating the Required Ratio: \(b : (c + a)\)

We need to find the ratio \(b : (c + a)\). We already have \(b = \frac{1}{3} S\).

Now, let's find \(c + a\) in terms of \(S\):

\[ c + a = \frac{5}{12} S + \frac{1}{4} S \]

Using the common denominator 12:

\[ c + a = \frac{5}{12} S + \frac{3}{12} S \]

\[ c + a = \frac{5 + 3}{12} S \]

\[ c + a = \frac{8}{12} S \]

\[ c + a = \frac{2}{3} S \]

Now, we can form the ratio \(b : (c + a)\):

\[ b : (c + a) = \left(\frac{1}{3} S\right) : \left(\frac{2}{3} S\right) \]

Since \(S\) is a common factor and is not zero (assuming \(a, b, c\) are positive quantities as suggested by ratios), we can cancel \(S\):

\[ b : (c + a) = \frac{1}{3} : \frac{2}{3} \]

To simplify the ratio of fractions, multiply both sides by the common denominator (3):

\[ \left(\frac{1}{3} \times 3\right) : \left(\frac{2}{3} \times 3\right) \]

\[ 1 : 2 \]

The value of \(b : (c + a)\) is \(1 : 2\).

Summary of Proportions relative to \(S = a+b+c\)

Check: \(\frac{1}{4} + \frac{1}{3} + \frac{5}{12} = \frac{3}{12} + \frac{4}{12} + \frac{5}{12} = \frac{12}{12} = 1\). The proportions add up to the total sum.

Calculation of the Desired Ratio \(b : (c + a)\)

Final Result

The calculated ratio \(b : (c + a)\) is \(1 : 2\).

Revision Table: Key Concepts for Ratio Problems

Additional Information: Working with Proportions

Problems involving ratios with sums or differences often benefit from expressing each component as a fraction of the total sum or difference. If you have a ratio like \(x : y = p : q\), it implies that \(x = pk\) and \(y = qk\) for some constant \(k\). While this substitution method can work, especially with simultaneous equations, relating everything to a common total (like \(a+b+c\) in this case) often simplifies the algebraic steps.

Consider the ratio \(a : (b+c) = 1 : 3\). This means \(a\) is 1 part and \(b+c\) is 3 parts of some division. The total parts are \(1+3=4\). So, \(a\) is \(1/4\) of the total amount \(a+(b+c)\). Similarly, \(b+c\) is \(3/4\) of the total amount \(a+(b+c)\).

This method allows you to find the individual fractional contribution of \(a\), \(b\), and \(c\) to the sum \(a+b+c\), making it easy to calculate any other ratio involving these terms or their sums.