Full solution

Q. 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?

Define Bag Pens: Let's say bag $B$ has $x$ pens. Then bag $A$ has $2x$ pens because it's twice as many.
Calculate Red Pens in Bag A: In bag A, $(1)/(5)$ of the pens are red. So, the number of red pens in bag A is $(1)/(5) \times 2x$ .
Calculate Red Pens in Bag B: In bag B, $(1)/(3)$ of the pens are red. So, the number of red pens in bag B is $(1)/(3) \times x$ .
Find Total Red Pens: Now, we add the red pens from both bags to find the total number of red pens: $(\frac{1}{5} \times 2x + \frac{1}{3} \times x$ .
Convert Fractions to Common Denominator: To add the fractions, we need a common denominator. The common denominator for $5$ and $3$ is $15$ . So we convert the fractions: $\frac{3}{15} \times 2x + \frac{5}{15} \times x$ .
Add Fractions: Now we add the fractions: (\frac{$3$}{$15$}) \cdot \(2x + (\frac{ $5$ }{ $15$ }) \cdot x = (\frac{ $6$ }{ $15$ })x + (\frac{ $5$ }{ $15$ })x\.
Combine Terms: Combine the terms: (\frac{$6$}{$15$})x + (\frac{$5$}{$15$})x = (\frac{$11$}{$15$})x\.
Calculate Total Pens: The total number of pens is the sum of pens in both bags, which is \(x + 2x = 3x.
Find Fraction of Red Pens: To find the fraction of red pens, we divide the number of red pens by the total number of pens: $(\frac{11}{15}x) \div 3x$ .
Divide Fractions: When we divide fractions, we multiply by the reciprocal. So, $(\frac{11}{15}x) \times (\frac{1}{3x})$ .
Multiply Fractions: Multiplying the fractions, we get $(\frac{11}{15}) \times (\frac{1}{3})$ .
Final Calculation: Now, we multiply the numerators and the denominators: $11 \times 1 = 11$ and $15 \times 3 = 45$ .