Calculation of Average of Two Numbers
The given problem is to find the value of \(\frac{a + b}{2}\).
First, we'll calculate the value of \(a\) and \(b\) separately and then find \(\frac{a + b}{2}\).
\(a = 4\frac{1}{2} + 1\frac{1}{2}\)
\(a = \frac{9}{2} + \frac{3}{2}\) (converting mixed fractions to improper fractions)
\(a = \frac{9 + 3}{2}\)
\(a = \frac{12}{2}\)
\(a = 6\) (simplifying the fraction)
\(b = 6\frac{1}{2} + 1\frac{3}{4}\)
\(b = \frac{13}{2} + \frac{7}{4}\) (converting mixed fractions to improper fractions)
To add these fractions, they must have a common denominator. The least common denominator for 2 and 4 is 4.
\(b = \frac{26}{4} + \frac{7}{4}\) (making the denominators the same)
\(b = \frac{26 + 7}{4}\)
\(b = \frac{33}{4}\) (simplifying the fraction)
Now, let's calculate \(\frac{a + b}{2}\).
\(\frac{a + b}{2} = \frac{6 + \frac{33}{4}}{2}\)
First, express 6 as a fraction with a denominator of 4 to be able to add it to \(\frac{33}{4}\).
\(\frac{6}{1} = \frac{24}{4}\)
Now, add the fractions:
\(\frac{a + b}{2} = \frac{\frac{24}{4} + \frac{33}{4}}{2}\)
\(\frac{a + b}{2} = \frac{24 + 33}{4 \cdot 2}\)
\(\frac{a + b}{2} = \frac{57}{8}\)
So the value of \(\frac{a + b}{2}\) is \(\frac{57}{8}\).
