Problem 11.2. At the zoo, oranges, bananas, and coconuts were brought for feeding three monkeys, and there were equal amounts of all three types of fruit. The first monkey was fed only oranges and bananas, with bananas being $40 \%$ more than oranges. The second monkey was fed only bananas and coconuts, with coconuts being $25 \%$ more than bananas. The third monkey was fed only coconuts and oranges, with oranges being twice as many as coconuts. The monkeys ate all the fruits brought.

Let the first monkey eat $a$ oranges, and the third monkey eat $b$ oranges. Find $a / b$.
Answer: $1 / 2$.

Solution. Let the first monkey eat $x$ oranges, the second monkey eat $y$ coconuts, and the third monkey eat $z$ coconuts. Then the first monkey ate $1.4 x$ bananas; the second monkey ate $1.25 y$ coconuts; the third monkey ate $2 z$ oranges. Since all the fruits were eaten equally, we have

$$
x+2 z=1.4 x+y=1.25 y+z
$$

Rewrite this as a system:

$$
\left\{\begin{array}{l}
x+2 z=1.25 y+z \\
x+2 z=1.4 x+y
\end{array}\right.
$$

Grouping like terms, we get

$$
\left\{\begin{array}{l}
x+z=1.25 y \\
2 z=0.4 x+y
\end{array}\right.
$$

Multiplying the first equation by 2 and subtracting the second from it, we get $2 x=1.5 y-0.4 x$, hence $2.4 x=1.5 y$ and $y=1.6 x$. Substituting this into the equation $x+z=1.25 y$, we get $x+z=2 x$, i.e., $z=x$.

Now we can answer the question of the problem. The first monkey ate $x$ oranges, and the third monkey ate $2 z=2 x$. Therefore, the desired ratio is $1 / 2$.