A library has a total of 450 books, each classified as fiction, non-fiction, or reference. If one book is selected...

1. TRANSLATE the problem information

• Given information:
  • Total books: 450
  • Probability of fiction: \(\frac{2}{5}\)
  • Probability of non-fiction: \(\frac{1}{3}\)
  • Need to find: Number of reference books

2. INFER the key relationship

• Since fiction, non-fiction, and reference are the only three categories, and exactly one book is selected:
  \(\mathrm{P(fiction)} + \mathrm{P(non-fiction)} + \mathrm{P(reference)} = 1\)

• This means: \(\mathrm{P(reference)} = 1 - \mathrm{P(fiction)} - \mathrm{P(non-fiction)}\)
• Substitute: \(\mathrm{P(reference)} = 1 - \frac{2}{5} - \frac{1}{3}\)

3. SIMPLIFY the fraction calculation

• Find common denominator for 5 and 3: LCM = 15
• Convert fractions: \(\frac{2}{5} = \frac{6}{15}\) and \(\frac{1}{3} = \frac{5}{15}\)
• Add: \(\frac{6}{15} + \frac{5}{15} = \frac{11}{15}\)
• Calculate: \(\mathrm{P(reference)} = 1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15}\)

4. SIMPLIFY to find the final answer

• Number of reference books = P(reference) × Total books
• Number of reference books = \(\frac{4}{15} \times 450 = 4 \times 30 = 120\)

Answer: C. 120

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Making arithmetic errors when working with fractions, particularly when finding the common denominator or performing addition/subtraction.

For example, students might incorrectly calculate \(\frac{2}{5} + \frac{1}{3}\). Some might get \(\frac{3}{8}\) (adding numerators and denominators incorrectly), leading to \(\mathrm{P(reference)} = 1 - \frac{3}{8} = \frac{5}{8}\), which gives \(\frac{5}{8} \times 450 = 281.25\) (not a choice, causing confusion). Others might make errors converting to the common denominator, getting \(\frac{2}{15}\) instead of \(\frac{4}{15}\) for P(reference), leading them to select Choice A (60).

Second Most Common Error:

Weak SIMPLIFY execution: Calculation errors in the final multiplication step, even after correctly finding \(\mathrm{P(reference)} = \frac{4}{15}\).

Students might incorrectly compute \(\frac{4}{15} \times 450\), perhaps getting \(\frac{3}{15} \times 450 = 90\) due to a copying error, leading them to select Choice B (90).

The Bottom Line:

This problem requires solid fraction arithmetic skills combined with the key insight about probabilities summing to 1. Most errors stem from computational mistakes rather than conceptual misunderstanding.