Formulas & Methods

Convert every percent to a multiplier ($30\%$ off is $\times0.7$) and chain multipliers for successive changes — never add percents of different bases. Ratio problems: assign $k$ to one part and solve for $k$ from the given total or difference.

Set up a table with columns rate, time, amount. For "how many of each" mixture problems, use one variable and the total. For averages, always work with TOTALS, not averages.

Step 1 — as multipliers: $20\%$ off is $\times0.8$ and $25\%$ off is $\times0.75$.

Step 2 — $80\cdot0.8\cdot0.75 = 80\cdot0.6 = 48$ dollars.

Step 3 — a multiplier of $0.6$ is a single discount of $40\%$, not $45\%$: the $25\%$ was taken from the smaller sale price.

Step 1 — the increase multiplied the old price by $1.15$: $1.15p = 69$.

Step 2 — divide: $p = 69/1.15 = 60$ dollars. Taking $15\%$ OFF $\$69$ gives $\$58.65$ — wrong, because $15\%$ of $69$ is not $15\%$ of $60$.

Step 1 — multipliers: $1.25\cdot0.8 = 1$.

Step 2 — the net change is $0$: the price is back where it started. The $20\%$ fall is taken from the larger, raised price, which is why it exactly cancels a $25\%$ rise.

Factor completely before anything else. Units-digit questions: find the cycle length, reduce the exponent. "Smallest number leaving remainders…": list one sequence and filter by the other condition.

Make the sample space explicit (a table for two dice, a tree for coins). Casework by a stated criterion. When "at least" appears, use the complement.

Step 1 — "remainder $1$ on division by each" means $n - 1$ is divisible by each.

Step 2 — the smallest positive common multiple is $\text{lcm}(2, 3, 4, 5, 6) = 60$.

Step 3 — $n = 61$. The product $2\cdot3\cdot4\cdot5\cdot6 = 720$ is a common multiple too, but not the least.

Step 1 — "larger is at most $4$" means both dice are at most $4$: $4\cdot4 = 16$ outcomes.

Step 2 — "larger is at most $3$": $3\cdot3 = 9$.

Step 3 — "larger is exactly $4$" is the difference: $16 - 9 = 7$ outcomes, probability $\tfrac{7}{36}$.

Draw to scale and label everything known. Look for right triangles to apply Pythagoras; for overlapping shapes, compute (whole) $-$ (parts). Coordinates: distance formula $\sqrt{\Delta x^2 + \Delta y^2}$, midpoint $\left(\tfrac{x_1+x_2}{2}, \tfrac{y_1+y_2}{2}\right)$.

For patterns, compute the first four terms and identify the rule before extrapolating; for cyclic patterns, find the period and reduce.

On the AMC 8 and MathCounts the check is the whole strategy: answer choices can be tested, units can be verified, and a quick estimate ("about $30$, so $28$ not $280$") catches most slips. Draw every geometry problem to scale. Make a table for every counting problem. Write the equation before solving it.

Step 1 — the legs are $5 - (-1) = 6$ and $10 - 2 = 8$, so $AB = \sqrt{36 + 64} = 10$.

Step 2 — the midpoint averages the coordinates: $\left(\tfrac{-1 + 5}{2}, \tfrac{2 + 10}{2}\right) = (2, 6)$.

Step 1 — every length is multiplied by $2$.

Step 2 — areas are length times length: $\times 2^2 = 4$.

Step 3 — volumes are length times length times length: $\times 2^3 = 8$. A cube of edge $3$ has surface $54$ and volume $27$; edge $6$ gives $216 = 4\cdot54$ and $216 = 8\cdot27$.

Step 1 — as an equation: $1.2x = 0.7x + 30$, so $0.5x = 30$ and $x = 60$.

Step 2 — check against the words, which is also how to solve it without algebra: $60$ up $20\%$ is $72$, $60$ down $30\%$ is $42$, and $72 - 42 = 30$. On a multiple-choice test, trying the choices in the middle first is a legitimate method.