Understanding which numbers to substitute for unknowns

The clearest case is a percentage question with no given total: just assume the total is $100$.
Suppose a population grows by $10\%$ one year and then by $20\%$ the next — by what percent has it grown overall?
Starting from $100$, it becomes $110$, then $110 \times 1.2 = 132$, so it has grown by $32\%$, and assuming $100$ turned the percentages into easy arithmetic.

When an expression is full of $x$-terms, substitute $x = 0$ to wipe them all out and leave only the constant.
To check whether $(x + 5)(x - 2)$ equals $x^2 + 3x - 10$, put $x = 0$: the product is $(5)(-2) = -10$, and $x^2 + 3x - 10$ is also $-10$, so they agree.
Substituting $0$ is fast because every $x$-term disappears and only the constant is left to compare.

When $0$ would break a denominator, or the constant alone is not enough, use $x = 1$ — or an even smarter value.
For the identity $A(x - 1) + B(x + 1) = 3x + 5$, substituting $x = 1$ makes the $A$-term vanish, giving $2B = 8$, so $B = 4$.
One well-chosen value can do the work of solving a whole system of equations.

For a ratio or fraction question, assume the product of the denominators so every count stays a whole number.
If $\frac{1}{4}$ of a class studies French and $\frac{1}{3}$ studies Spanish with no overlap, assume the class has $4 \times 3 = 12$ students: $3$ study French, $4$ study Spanish, and $5$ study neither, so $\frac{5}{12}$ study neither.
Just avoid any value that makes a denominator $0$, and give different unknowns different values.

A jacket's price is first raised by $25\%$ and then reduced by $20\%$. The final price is what percent of the original?
Assume the original price is $100$.
After the $25\%$ rise it is $125$, and after the $20\%$ reduction it is $125 \times 0.8 = 100$.
So the final price is $100\%$ of the original — assuming $100$ made this a two-step calculation.

Which of the following is equivalent to $(x - 4)(x + 2)$?
(A) $x^2 - 2x - 8$
(B) $x^2 + 2x - 8$
(C) $x^2 - 2x + 8$
Substitute $x = 0$: the product is $(-4)(2) = -8$, so choice (C), which gives $+8$, is eliminated.
Now try $x = 1$: the product is $(-3)(3) = -9$, and choice (A) gives $1 - 2 - 8 = -9$ while (B) gives $-5$, so (A) is correct.

At a conference, $\frac{1}{3}$ of the attendees are students and $\frac{1}{6}$ are teachers, with no overlap. What fraction are neither?
Assume there are $3 \times 6 = 18$ attendees, the product of the denominators, so the counts stay whole.
Then $6$ are students and $3$ are teachers, leaving $18 - 6 - 3 = 9$ who are neither.
So the fraction that are neither is $\frac{9}{18} = \frac{1}{2}$.