11. Four people, A, B, C, and D, share 2013 candies. A gets 10 more than twice what B gets, 18 more than three times what C gets, and 55 less than five times what D gets. How many candies does A get?
Analysis: Given that A is 10 more than twice B, 18 more than three times C, and 55 less than five times D, we have A $-10=2 \times$ B, A $-18=3 \times$ C, A $+55=$ D $\times 5$;
That is, $15 \times$ A $-150=30 \times$ B, $10 \times$ A $-180=30 \times$ C, $6 \times$ A $+330=30 \times$ D
Adding the three equations, we get
$$
31 \times \text { A }=30 \times \text { B }+30 \times \text { C }+30 \times \text { D }
$$

Thus $\square$
$$
61 \times \text { A }=30 \times \text { A }+30 \times \text { B }+30 \times \text { C }+30 \times \text { D }
$$

Given A + B + C + D $=2013$,
So $30 \times 2013=30 \times$ A $+30 \times$ B $+30 \times$ C $+30 \times$ D
Therefore, A $=30 \times 2013 \div 61=990$ yuan.
Solving the equation is much simpler. Let A receive $x$ yuan.
$$
\begin{array}{l}
x+\frac{x-10}{2}+\frac{x-18}{3}+\frac{x+55}{5}=2013 \\
x=990
\end{array}
$$