A village has 650 youths. 250 of them are girls. 425 youths like music and 90 girls do not like music. How many boys do not like music?

Total youths = 650
Number of girls = 250
Number of boys = Total youths - Number of girls

$$\text{Number of boys} = 650 - 250 = 400$$

Total youths = 650
Youths who like music = 425
Youths who do not like music = Total youths - Youths who like music

$$\text{Youths who do not like music} = 650 - 425 = 225$$

Youths who do not like music = 225
Girls who do not like music = 90
Boys who do not like music = Youths who do not like music - Girls who do not like music

$$\text{Boys who do not like music} = 225 - 90 = 135$$

The correct option is A.
The number of boys who do not like music is $\boxed{\text{A. } 135}$.

:

Total boys = 20
Boys not allowed to play games = 2
Boys who play games = Total boys - Boys not allowed to play games

$$n(S \cup H) = 20 - 2 = 18$$

Let $S$ be the set of boys who play soccer and $H$ be the set of boys who play hockey.
We are given:
$n(S) = 16$
$n(H) = 12$
$n(S \cup H) = 18$
The formula is $n(S \cup H) = n(S) + n(H) - n(S \cap H)$.
We need to find $n(S \cap H)$.

$$18 = 16 + 12 - n(S \cap H)$$ $$18 = 28 - n(S \cap H)$$ $$n(S \cap H) = 28 - 18$$ $$n(S \cap H) = 10$$

The correct option is C.
The number of students who play soccer and hockey is $\boxed{\text{C. } 10}$.

:
The set difference $P - Q$ represents all elements that are in set $P$ but not in set $Q$.
This is equivalent to the intersection of set $P$ and the complement of set $Q$, denoted as $Q'$.
Therefore, $P - Q = P \cap Q'$.
The correct option is C.
$P - Q$ is same as $\boxed{\text{C. } (P \cap Q')}$.

:
We are given:
$n(A - B) = 18$
$n(A \cup B) = 70$
$n(A \cap B) = 25$
We need to find $n(B)$.

$n(A - B) = n(A) - n(A \cap B)$

$$18 = n(A) - 25$$ $$n(A) = 18 + 25$$ $$n(A) = 43$$

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

$$70 = 43 + n(B) - 25$$ $$70 = 18 + n(B)$$ $$n(B) = 70 - 18$$ $$n(B) = 52$$

The correct option is D.
$n(B)$ is $\boxed{\text{D. } 52}$.

:
Let $B$ be the set of people who buy and $S$ be the set of people who sell.
We are given:
Percentage of people who buy, $n(B) = 70\%$
Percentage of people who sell, $n(S) = 90\%$
All people buy or sell or do both, which means $n(B \cup S) = 100\%$.
We need to find the percentage of people who do both, $n(B \cap S)$.

Use the Principle of Inclusion-Exclusion for percentages:
$n(B \cup S) = n(B) + n(S) - n(B \cap S)$

$$100\% = 70\% + 90\% - n(B \cap S)$$ $$100\% = 160\% - n(B \cap S)$$ $$n(B \cap S) = 160\% - 100\%$$ $$n(B \cap S) = 60\%$$

The correct option is A.
The percentage of people who do both is $\boxed{\text{A. } 60\%}$.