10. (10 points) A school provides students with three types of fruits: apples, bananas, and pears, to be used as snacks during breaks. Each student chooses at least one type, and can choose more than one. The statistics show: $70 \%$ of the students chose apples, $40 \%$ chose bananas, and $30 \%$ chose pears. What is the maximum percentage of students who chose all three types of fruits?
【Analysis】Divide all students into four categories, that is, the number of people who choose all three fruits $a$, the number of people who choose both apples and bananas $b$, the number of people who choose both pears and apples $c$, and the number of people who choose both bananas and pears $d$. Then, based on the number of people who choose each fruit, list the relationship: $2 a+b+c+d=70+40+30-100=40$, and then use the range of each value to find the maximum number of people who choose all three fruits.

【Solution】Solution: According to the analysis, let the total number of students be 100. Therefore, 70 students choose apples, 40 students choose bananas, and 30 students choose pears.

The number of students who choose all three fruits is $a$, the number of students who choose both apples and bananas is $b$, the number of students who choose both pears and apples is $c$,

and the number of students who choose both bananas and pears is $d$. Then: $2 a+b+c+d=70+40+30-100=40 \Rightarrow a=$ $\frac{40-(\mathrm{b}+\mathrm{c}+\mathrm{d})}{2}$, and since $\because b+c+d \geqslant 0, \therefore a \leqslant \frac{40-0}{2}=20$, so when $b+c+d=0$, $a$ reaches its maximum value of 20, which is 20% of the total. Therefore, the answer is $20 \%$.