5 Sam kept pens in bags \( A \) and \( B \). Bag \( A \) contained twice as many pens as bag \( B \). In bag \( A, \frac{1}{5} \) of the pens were red pens. In bag \( B, \frac{1}{3} \) of the pens were red pens. What fraction of Sam's pens were red pens?

Let's denote the number of pens in bag \( B \) as \( x \). Given that bag \( A \) contained twice as many pens as bag \( B \), the number of pens in bag \( A \) is \( 2x \). In bag \( A \), \( \frac{1}{5} \) of the pens were red pens, so the number of red pens in bag \( A \) is \( \frac{1}{5} \times 2x = \frac{2x}{5} \). In bag \( B \), \( \frac{1}{3} \) of the pens were red pens, so the number of red pens in bag \( B \) is \( \frac{1}{3} \times x = \frac{x}{3} \). The total number of pens Sam kept is \( 2x + x = 3x \). The total number of red pens Sam kept is \( \frac{2x}{5} + \frac{x}{3} \). To find the fraction of Sam's pens that were red pens, we need to calculate \( \frac{\frac{2x}{5} + \frac{x}{3}}{3x} \). Let's simplify this expression. Simplify the expression by following steps: - step0: Solution: \(\frac{\left(\frac{2x}{5}+\frac{x}{3}\right)}{3x}\) - step1: Remove the parentheses: \(\frac{\frac{2x}{5}+\frac{x}{3}}{3x}\) - step2: Add the terms: \(\frac{\frac{11x}{15}}{3x}\) - step3: Multiply by the reciprocal: \(\frac{11x}{15}\times \frac{1}{3x}\) - step4: Reduce the fraction: \(\frac{11}{15}\times \frac{1}{3}\) - step5: Multiply the terms: \(\frac{11}{15\times 3}\) - step6: Multiply the terms: \(\frac{11}{45}\) The fraction of Sam's pens that were red pens is \( \frac{11}{45} \) or approximately 0.2444.