Question

The final grade in one of your friend's courses is determined by a term paper, exams, quizzes, and class participation. The term paper is worth \(\frac{1}{5}\), the exams \(\frac{1}{3}\), and the quizzes \(\frac{1}{4}\) of the final grade. a. If \(x\) represents the fractional part of the final grade for class participation, write an equation describing the situation. b. Solve the equation for \(x\)

Short answer

Answer: The fractional part of the final grade for class participation is $\frac{13}{60}$.

Step-by-step solution

Equation Formulation

To represent the situation, devise an equation adding up the fractional parts of each component to the final grade that equals 1. Use the given fractions and the variable \(x\) for the class participation fraction. The final grade consists of the term paper, exams, quizzes, and class participation fractions. Thus, the equation is: $$(\frac{1}{5}) + (\frac{1}{3}) + (\frac{1}{4}) + x = 1$$ #b. Solve the equation for x.#

Combine Common Fractions

First, combine the fractions in the equation to simplify it: $$(\frac{1}{5}+\frac{1}{3}+\frac{1}{4})+x=1$$ Find the least common denominator (LCD) of the three denominators: LCD = \(5 \times 3 \times 4 = 60\) Transform the fractions individually: \(\frac{1}{5} \rightarrow\frac{12}{60}\), \(\frac{1}{3} \rightarrow\frac{20}{60}\), and \(\frac{1}{4} \rightarrow\frac{15}{60}\) Add the fractions: $$(\frac{12}{60}) + (\frac{20}{60}) + (\frac{15}{60}) + x = 1$$ $$(\frac{47}{60}) + x = 1$$

Solve for x

Now, subtract the fraction from both sides of the equation to isolate \(x\), which will solely represent the class participation fraction of the final grade: $$ x = 1 - (\frac{47}{60}) $$ Then, rewrite \(1\) as a fraction with a denominator of \(60\): $$ x = \frac{60}{60} - \frac{47}{60} $$ Now, subtract the fractions: $$ x = \frac{60 - 47}{60} $$ Finally, simplify the fraction: $$ x = \frac{13}{60} $$ Thus, the fractional part of the final grade for class participation is \(\frac{13}{60}\).