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 \( b \). According to the problem, bag A contains twice as many pens, which means bag A has \( 2b \) pens. Now, let's find the number of red pens in each bag: - In bag A, \( \frac{1}{5} \) of the pens are red. Therefore, the number of red pens in bag A is: \[ \text{Red pens in A} = \frac{1}{5} \times 2b = \frac{2b}{5} \] - In bag B, \( \frac{1}{3} \) of the pens are red. Therefore, the number of red pens in bag B is: \[ \text{Red pens in B} = \frac{1}{3} \times b = \frac{b}{3} \] Now, let's calculate the total number of pens Sam has: \[ \text{Total pens} = b + 2b = 3b \] Next, we find the total number of red pens: \[ \text{Total red pens} = \frac{2b}{5} + \frac{b}{3} \] To add these fractions, we need a common denominator, which will be 15: \[ \frac{2b}{5} = \frac{2b \times 3}{5 \times 3} = \frac{6b}{15} \] \[ \frac{b}{3} = \frac{b \times 5}{3 \times 5} = \frac{5b}{15} \] Now, we can add the red pens: \[ \text{Total red pens} = \frac{6b}{15} + \frac{5b}{15} = \frac{11b}{15} \] Now we can find the fraction of Sam's total pens that are red: \[ \text{Fraction of red pens} = \frac{\text{Total red pens}}{\text{Total pens}} = \frac{\frac{11b}{15}}{3b} \] Simplifying the fraction: \[ = \frac{11b}{15 \times 3b} = \frac{11b}{45b} = \frac{11}{45} \] Thus, the fraction of Sam's pens that were red pens is: \[ \frac{11}{45} \]