5. (10 points) The amount of rice in warehouses A and B is the same. Dad, Mom, and Yangyang can each transport one warehouse of rice alone in 10 days, 12 days, and 15 days, respectively. Dad and Mom start transporting the rice from warehouses A and B simultaneously, and Yangyang first helps Mom, then helps Dad, and they finish transporting the rice from both warehouses at the same time. How many days did Yangyang help Mom? 
A. 3
B. 4
C. 5
D. 6
【Analysis】Assuming the amount of rice in a warehouse is "1", the efficiency of Dad, Mom, and Yangyang are $\frac{1}{10}$, $\frac{1}{12}$, and $\frac{1}{15}$, respectively. The time required for the three to move two warehouses together: $(1+1) \div\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)=8$ (days); The total amount Mom moved in 8 days: $8 \times \frac{1}{12}=\frac{2}{3}$ (warehouse); The remaining amount for Mom is what Yangyang helped Mom move, so Yangyang helped Mom move for (1- $\frac{2}{3}$ ) $\div \frac{1}{15}=5$ (days).
【Solution】Solution: The total time for the three to move:
$$
\begin{aligned}
& (1+1) \div\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right), \\
= & 2 \div \frac{1}{4}, \\
= & 8 \text { (days); }
\end{aligned}
$$

The number of days Yangyang helped Mom move:
$$
\begin{aligned}
& \left(1-\frac{1}{12} \times 8\right) \div \frac{1}{15}, \\
= & \frac{1}{3} \times 15, \\
= & 5 \text { (days); }
\end{aligned}
$$

Answer: Yangyang helped Mom move for 5 days.
Therefore, the answer is: $C$.