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

Mathematics

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Step-by-step solution

4 steps
  1. Determine the ratio of pens in Bag A to Bag B and the total units.

    Bag A has twice as many pens as Bag B.
    If Bag B has 1 unit of pens, then Bag A has 2 units of pens.
    Total pens = 2 units (Bag A) + 1 unit (Bag B) = 3 units.

  2. Calculate the number of red pens in each bag in terms of units.

    Red pens in Bag A = 15\frac{1}{5} of 2 units = 25\frac{2}{5} units.
    Red pens in Bag B = 13\frac{1}{3} of 1 unit = 13\frac{1}{3} units.

  3. Find the total number of red pens.

    To add the fractions, find a common denominator for 5 and 3, which is 15.

    Total red pens=25+13=2×35×3+1×53×5=615+515=1115 units\text{Total red pens} = \frac{2}{5} + \frac{1}{3} = \frac{2 \times 3}{5 \times 3} + \frac{1 \times 5}{3 \times 5} = \frac{6}{15} + \frac{5}{15} = \frac{11}{15} \text{ units}
  4. Calculate the fraction of Sam's total pens that are red pens.

    Fraction of red pens=Total red pensTotal pens=1115 units3 units\text{Fraction of red pens} = \frac{\text{Total red pens}}{\text{Total pens}} = \frac{\frac{11}{15} \text{ units}}{3 \text{ units}} Fraction of red pens=1115÷3=1115×13=1145\text{Fraction of red pens} = \frac{11}{15} \div 3 = \frac{11}{15} \times \frac{1}{3} = \frac{11}{45}

    The fraction of Sam's pens that were red pens is 1145\boxed{\frac{11}{45}}.

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