2018/19 AHSMC Part 1 Problem 1

We have $1 + a + b = 6c$, $1 + a + c = 6b$ and $1 + b + c = 6a$. Adding gives $3 + 2a + 2b + 2c = 6a + 6b + 6c$ implying $a + b + c = \dfrac{3}{4}$, hence $\dfrac{a+b+c}{3} = \dfrac{1}{4}$. The answer is (B).

Alternative solution: Using the conditions in the question, $1 + a + b + c = 7a = 7b = 7c$, hence $a = b = c = \dfrac{1}{4}$ and therefore $\dfrac{a+b+c}{3} = \dfrac{1}{4}$.