Television Viewing Habits

Explanation

Analyze

• \(\frac16\) of the student body watches no TV daily (Given)
• \(\frac14\) of the student body watches 1 hour of TV daily. (Given)
• \(\frac{7}{15}\) of the student body watches 2 hours of TV daily. (Given)
• What fraction of the student body watches 0 to 2 hours of TV daily? (Find)

Form

We translate the words of the problem to numbers and symbols

[The fraction of the student body that watches from 0 to 2 hours of TV daily] is equal to [the fraction that watches no TV daily] plus [the fraction that watches 1 hour of TV daily] plus [the fraction that watches 1 hour of TV daily] plus [the fraction that watches 2 hours of TV daily]

The fraction of the student body that watches from 0 to 2 hours of TV daily = \(\frac16 + \frac14 + \frac{7}{15}\)

Solve

We must find the sum of three fractions with different denominators. To find the LCD, we prime factor the denominators and use each prime factor the greatest number of times it appears in any one factorization:

6 = 2 • 3

4 = 2 • 2

15 = 3 • 5

LCD = 2 • 2 • 3 • 5 = 60

Build each fraction so that its denominator is 60

$\frac16 + \frac14 + \frac{7}{15} = \frac16 \color{red}{\cdot \frac{10}{10}} + \frac14 \color{blue}{\cdot \frac{15}{15}} + \frac{7}{15}  \color{green}{\cdot \frac44}$

Multiply the numerators and the denominators

$= \frac{10}{60} + \frac{15}{60} + \frac{28}{60}$

Add the numerators and write the sum over the common denominator 60

$= \frac{10 + 15 + 28}{60}$

The fraction is in simplest form

$= \frac{53}{60}$

State

The fraction of the student body that watches 0 to 2 hours of TV daily is \(\frac{53}{60}\)

Check

We can check by estimation.  The result, \(\frac{53}{60}\), is approximately \(\frac{50}{60}\), which simplifies to \(\frac56\).  The red, yellow, and blue shaded areas appear to shade about \(\frac56\) of the pie chart the result seems reasonable.