2014/15 AHSMC Part 1 Problem 16

There are $230$ legs in the two cages in the second column. Since $230 = 4 \times 57 + 2$, they may come from as few as $58$ animals, namely, $57$ rabbits and $1$ chicken. Hence the minimum number of animals in all four cages is $70+58=128$. If we have $55$ animals in the first cage in the first row, $0$ chickens and $5$ rabbits in the second cage in the first row, $15$ chickens and $0$ rabbits in the first cage in the second row, and $1$ chicken and $52$ rabbits in the second cage in the second row, we have $128$ animals overall. The answer is (B).