Solving Ratio and Proportion Problems

The question asks us to find the value of a specific expression involving squares of variables, given a relationship between ratios of these variables. We are given that the ratio a:b is equal to the ratio c:d, and both are equal to the ratio 1:6. This can be written as:

\(\frac{a}{b} = \frac{c}{d} = \frac{1}{6}\)

From this information, we can work towards finding the value of the expression \(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}}\).

Method 1: Using Substitution

Since \(\frac{a}{b} = \frac{1}{6}\), we can express 'a' in terms of 'b':

\(a = \frac{1}{6}b\)

Similarly, since \(\frac{c}{d} = \frac{1}{6}\), we can express 'c' in terms of 'd':

\(c = \frac{1}{6}d\)

Now, we substitute these expressions for 'a' and 'c' into the expression we need to evaluate:

\(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}}\)

Substitute \(a = \frac{b}{6}\) and \(c = \frac{d}{6}\):

\(\frac{{\left(\frac{b}{6}\right)^2}\; + \;{\left(\frac{d}{6}\right)^2}}{{{b^2}\; + \;{d^2}}}\)

Now, square the terms in the numerator:

\(\frac{{\frac{b^2}{36}}\; + \;{\frac{d^2}{36}}}{{{b^2}\; + \;{d^2}}}\)

Combine the terms in the numerator by finding a common denominator:

\(\frac{{\frac{b^2\; + \;d^2}{36}}}{{{b^2}\; + \;{d^2}}}\)

This is a fraction divided by an expression. We can rewrite this as multiplying the numerator by the reciprocal of the denominator:

\(\left(\frac{b^2\; + \;d^2}{36}\right) \times \left(\frac{1}{b^2\; + \;d^2}\right)\)

Assuming that \({b^2\; + \;d^2}\) is not zero, we can cancel out the term \({b^2\; + \;d^2}\) from the numerator and the denominator:

\(\frac{1}{36}\)

Method 2: Using Ratio Property

There is a property of ratios that can be applied here. If we have a ratio \(a:b = k\), and another ratio \(c:d = k\), then it is also true that:

• \(\frac{a+c}{b+d} = k\) (Property of Addition)
• \(\frac{a-c}{b-d} = k\) (Property of Subtraction)
• \(\frac{a^2+c^2}{b^2+d^2} = k^2\) (Property of Squares)

In our given problem, we have \(\frac{a}{b} = \frac{c}{d} = \frac{1}{6}\). Here, the common ratio 'k' is \(\frac{1}{6}\).

We are asked to find the value of \(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}}\). Using the property of squares, where k is the common ratio:

\(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}} = k^2\)

Substitute the value of k = \(\frac{1}{6}\):

\(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}} = \left(\frac{1}{6}\right)^2\)

\(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}} = \frac{1^2}{6^2}\)

\(\frac{{{a^2}\; + \;{c^2}}}{{{b^2}\; + \;{d^2}}} = \frac{1}{36}\)

Both methods lead to the same result.

Revision Table: Key Concepts

Additional Information: More on Ratio Properties

When you have a series of equal ratios, like \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = \dots = \frac{a_n}{b_n} = k\), several useful properties hold:

• Componendo: If \(\frac{a}{b} = k\), then \(\frac{a+b}{b} = k+1\).
• Dividendo: If \(\frac{a}{b} = k\) (and \(b \ne 0\)), then \(\frac{a-b}{b} = k-1\).
• Componendo and Dividendo: If \(\frac{a}{b} = \frac{c}{d}\) (and \(b \ne 0, d \ne 0, b+d \ne 0, b-d \ne 0\)), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\).
• Sum/Difference of Numerators and Denominators: If \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n} = k\), then \(\frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n} = k\) and \(\frac{a_1-a_2}{b_1-b_2} = k\) (if \(b_1 \ne b_2\)), and more generally, \(\frac{m_1 a_1 + m_2 a_2 + \dots + m_n a_n}{m_1 b_1 + m_2 b_2 + \dots + m_n b_n} = k\) for any constants \(m_i\) (provided the denominator is not zero).
• Property of Powers: If \(\frac{a}{b} = k\), then \(\frac{a^n}{b^n} = k^n\) for any power n. Combining this with the sum property gives the property used in Method 2: if \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = k\), then \(\frac{a_1^2+a_2^2}{b_1^2+b_2^2} = \frac{k^2 b_1^2 + k^2 b_2^2}{b_1^2+b_2^2} = \frac{k^2(b_1^2+b_2^2)}{b_1^2+b_2^2} = k^2\).

Understanding these properties can simplify solving complex problems involving ratios and proportions, especially in competitive exams.