If \(\displaystyle\frac{a}{3}=\frac{b}{5}=\frac{c}{8}\), then find the value of \(\displaystyle\frac{2a+b}{a+b+c}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{11}{16}\)
When several ratios are equal, each letter is its denominator times the same number. Calling that common value \(k\), every term of the fraction carries \(k\), which then cancels.
\(a=3k\), \(b=5k\), \(c=8k\)
\(2a+b=6k+5k=11k\)
\(a+b+c=3k+5k+8k=16k\)
\(\displaystyle\frac{2a+b}{a+b+c}=\frac{11k}{16k}=\frac{11}{16}\)