workout wrote:
In a school outing with only adults and young children, the average weight of the adults is 60 kg and the average weight of the children is 24 kg. If the average weight of everyone on the school outing is 30 kg, what is the ratio of adults to children on this outing?
A) \(\frac{1}{12}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{5}\)
D) \(\frac{1}{4}\)
E) \(\frac{1}{3}\)
Let's solve this question using
weighted averagesWeighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...Let \(A\) = the number of adults
Let \(C\) = the number of children
This means \(A+C\) = the total number of people
Also \(\frac{A}{A+C}\) = the proportion of adults and the group
And \(\frac{C}{A+C}\) = the proportion of children and the group
Substitute values into our
equation to get:
\(30 = (\frac{A}{A+C})(60) + (\frac{C}{A+C})(24)\) Multiply both sides of the equation by \((A+C)\) to get: \(30(A+C) = 60A + 24C\)
Expand to get: \(30A+30C = 60A + 24C\)
Subtract \(30A\) from both sides: \(30C = 30A + 24C\)
Subtract \(24C\) from both sides: \(6C = 30A\)
Divide both sides by \(C\) to get: \(6 = \frac{30A}{C}\)
Divide both sides by \(30\) to get: \(\frac{6}{30} = \frac{A}{C}\)
Simplify: \(\frac{1}{5} = \frac{A}{C}\)
Answer: C
Cheers,
Brent