Digital SAT Math · 18 question styles
DSAT Math Question Styles
The 18 question styles that make up most of the Digital SAT Math section. For each one: how to spot it, a step-by-step method, the tips and traps that decide the point, and worked pattern examples.
By Engin Savaş · MATHIGH · mathematics educator with 20+ years of AP and IB teaching
AlgebraAdvanced MathProblem-Solving and DataGeometry and Trigonometry
Rewriting and solving linear equations with parentheses, fractions or a parameter, and deciding when an equation has one solution, no solution or infinitely many.
Spot it
- Parentheses, fractions or coefficients on both sides
- A parameter such as k or a: “Which value of k makes…”
- “No solution” or “infinitely many solutions”
- Both sides linear in x, for example 5(x − 2) = 3x + 8
Method
- Distribute every set of parentheses.
- Clear fractions by multiplying every term by the LCD.
- Combine like terms on each side.
- Move the x-terms to one side and the constants to the other.
- Solve for x, or compare coefficients if there is a parameter.
Tips
- Clear fractions first: it removes most arithmetic slips.
- For parameter questions, compare the x-coefficients and the constants instead of solving.
- Same x-coefficient, different constants → no solution.
- Same x-coefficient and same constant → infinitely many solutions.
Common traps
- Sign errors after distributing a negative
- Multiplying only some terms when clearing fractions
- Treating 2(x + 3) as 2x + 3
- Solving for x when the question asks for k
Pattern examples
Solve 5(x − 2) = 3x + 8.
5x − 10 = 3x + 8 → 2x = 18 → x = 9
Solve (x − 3)/4 + x/2 = 6.
× 4: x − 3 + 2x = 24 → x = 9
For what value of k does 3(x + k) = 3x + 12 have infinitely many solutions?
3x + 3k = 3x + 12 → k = 4
Does 4x − 8 = 4(x − 2) have a solution?
Both sides are 4x − 8: infinitely many solutions
Why it works
Every linear equation reduces to ax = b
Distributing and collecting terms always leads to “something times x equals something”. That form shows the number of solutions at once.
No solution
If the x-terms cancel and leave a false statement such as 5 = 3, no value of x works.
Infinitely many solutions
If the x-terms cancel and leave a true statement such as 0 = 0, every value of x works.
Mastery levels
LEVEL 1
Recognise the structure: parentheses, fractions, parameters.
LEVEL 2
Distribute, clear fractions and isolate x accurately.
LEVEL 3
Decide instantly whether a parameter gives no solution or an identity.
Reading slope and intercept from a graph, a table or a description, and explaining what they mean in context.
Spot it
- Rate words: every, per, each, for every increase of
- Initial-value words: starts at, initially, when t = 0
- A straight-line graph
- A table with a constant difference in y for equal steps in x
- “What does the number 3 represent?”
Method
- State what x and y represent, with units.
- Find the slope: rise/run on a graph, Δy/Δx in a table, rate words in a description.
- Find the intercept: where the graph crosses the y-axis, the value at x = 0, or the starting amount.
- Write y = mx + b.
- Interpret slope and intercept in the units of the context.
- Answer exactly what is asked.
Tips
- “Starting at” almost always means the y-intercept.
- “Every” or “per” means the slope.
- Use exact points from a table or the text rather than estimating from a graph.
- If x-values in a table are unevenly spaced, compute Δy/Δx, not the change between rows.
Common traps
- Mixing up what x and y represent
- Misreading a negative slope as positive
- Treating the first row of a table as the intercept when x ≠ 0 there
- Assuming a relationship is linear when the rate is not constant
Pattern examples
A taxi charges $4 plus $0.30 per mile.
y = 0.30x + 4
x: 0, 3, 6, 9 and y: 10, 22, 34, 46
slope = 12/3 = 4, so y = 4x + 10
You have $15 and earn $2 per hour.
y = 2x + 15
Which line shows a faster rate of change?
Compare the slopes; the larger absolute slope changes faster
Why it works
Slope is a rate
It is the change in y for each one-unit increase in x, measured in y-units per x-unit.
Intercept is a starting value
It is the value of y when x = 0: where the situation begins.
Constant differences mean linear
If y changes by the same amount for each equal step in x, the relationship is linear.
Mastery levels
LEVEL 1
Identify slope, intercept and rate language.
LEVEL 2
Compute the slope and build y = mx + b.
LEVEL 3
Explain what slope and intercept mean in context.
Turning a real-world description into a linear model and using it to predict or interpret.
Spot it
- Keywords: per, each, every, starting at, initially, after t hours
- Costs, distances, money, temperature or volume that change steadily
- “How many after …?” or “When will …?”
- Two data points given in the text
Method
- Define the variables and their units.
- The rate of change is the slope m.
- The amount before any change is the intercept b.
- Write the model y = mx + b.
- Substitute, solve or interpret as the question requires.
Tips
- Check units first: wrong units give the wrong slope.
- Fixed fee + rate × quantity is the most common structure.
- With two data points, compute the slope, then the intercept.
- A decreasing quantity has a negative slope.
Common traps
- Using a later value as the initial value
- Using a total instead of a per-unit rate
- Forgetting that the slope is negative when something decreases
- Swapping the roles of x and y
Pattern examples
A gym charges a $25 sign-up fee and $12 per month.
C = 12m + 25
A car is 120 miles from a city and drives towards it at 60 mph.
d = 120 − 60t
A tank holds 800 L and drains at 12 L per minute.
V = 800 − 12t
At 2 hours the depth is 14 cm; at 6 hours it is 26 cm.
slope = 12/4 = 3, so d = 3t + 8
Why it works
Every linear story has the same shape
Start with an amount, then add or subtract the same amount for each unit of time or quantity.
Rate = slope
Anything that changes per hour, per item or per mile is the slope.
Initial value = intercept
Whatever is there before the repeated change starts is the intercept.
Mastery levels
LEVEL 1
Find the rate and starting-value language.
LEVEL 2
Build the model correctly with units.
LEVEL 3
Interpret, compare and predict with the model.
Finding where two linear relationships are both true, choosing an efficient method and recognising special cases.
Spot it
- Two linear equations in x and y
- “Intersection”, “same value”, “solution to both”
- A question asking for x + y, 2x − y or another expression
- Parallel lines (no solution) or the same line (infinitely many)
Method
- Compare slopes: same slope with different intercepts means no solution; the same line means infinitely many.
- Choose a method: substitution if a variable is isolated, elimination if coefficients line up.
- Solve, or combine the equations to get the requested expression directly.
- In context, the solution is where two quantities are equal.
Tips
- If the question asks for x + y or 2x, try adding or subtracting the equations first.
- Line the equations up vertically before eliminating.
- Multiply an equation only when it creates a clean cancellation.
- Desmos: graph both lines and read the intersection when the numbers are messy.
Common traps
- Solving for x and y when only an expression is needed
- Sign errors when subtracting equations
- Swapping the x and y values when substituting
- Missing a no-solution case
Pattern examples
3x + 2y = 16 and −3x + y = −1
add: 3y = 15 → y = 5, x = 2
y = 4x − 3 and 2x + y = 15
2x + 4x − 3 = 15 → x = 3, y = 9
y = 2x + 1 and y = 2x − 5
same slope, different intercepts: no solution
Plan A: $20 + $5 per visit. Plan B: $50 + $2 per visit. When do they cost the same?
20 + 5v = 50 + 2v → v = 10 visits
Why it works
A solution is an intersection
Solving a system finds the point (x, y) where both lines meet.
Substitution or elimination?
Substitution suits an isolated variable; elimination suits aligned coefficients such as 2x and −2x.
Special cases
Parallel lines never meet (no solution). Identical lines meet everywhere (infinitely many).
Mastery levels
LEVEL 1
Recognise the structure and the best method.
LEVEL 2
Solve accurately by elimination or substitution.
LEVEL 3
Spot special cases and evaluate expressions without solving fully.
Solving equations with absolute value by treating |A| as the distance of A from zero.
Spot it
- Vertical bars: |x − a| or |expression|
- Words such as distance, how far, away from, difference
- An isolated absolute value equal to a number: |A| = B
Method
- Isolate the absolute value: |A| = B.
- If B < 0, there is no solution.
- If B = 0, solve A = 0.
- If B > 0, solve both A = B and A = −B.
- Check the answers if the equation had other restrictions.
Tips
- Translate words directly: “x is 3 units from 6” → |x − 6| = 3.
- Solutions of |x − a| = d are a ± d, symmetric around a.
- Check B first: a negative right side means no solution.
- Questions often ask for the sum of the solutions or the greater one.
Common traps
- Solving only one case
- Missing that B must be at least 0
- Forgetting to isolate the absolute value before splitting
- Thinking direction matters for distance
Pattern examples
|x − 5| = 3
x = 8 or x = 2
|2x + 1| = −4
no solution: an absolute value cannot be negative
x is 4 units from 10.
|x − 10| = 4 → x = 6 or x = 14
|x − 8| = |x − 2|
x is equally far from 2 and 8 → x = 5
Why it works
Absolute value is distance
|x − a| = d means x is d units from a, so the solutions sit symmetrically at a − d and a + d.
Graph view
y = |x − a| is a V-shape with its vertex at (a, 0). Solving |x − a| = d finds where the V reaches height d.
Mastery levels
LEVEL 1
Recognise the structure and check the sign of B.
LEVEL 2
Solve both cases and verify.
LEVEL 3
Use the distance meaning to answer quickly.
Solving and interpreting inequalities in one or two variables, including compound inequalities and contexts.
Spot it
- Symbols <, ≤, >, ≥
- Words: at least, no more than, fewer than, at most
- Compound statements: between … and …
- A shaded region on a coordinate graph
Method
- Simplify both sides.
- Collect the variable terms on one side.
- Isolate the variable; if you multiply or divide by a negative number, reverse the inequality sign.
- Write the answer in the form the question uses.
Tips
- Rewrite 8 > 2x + 4 as 2x + 4 < 8 before solving.
- “At least” → ≥ and “no more than” → ≤.
- For two-variable inequalities, test a point such as (0, 0) to see which side is shaded.
- Solid boundary line → ≤ or ≥; dashed line → < or >.
Common traps
- Forgetting to reverse the sign after dividing by a negative
- Mixing up open and closed circles
- Reading “and” as “or”
- Translating word phrases the wrong way round
Pattern examples
3x − 7 ≤ 11
3x ≤ 18 → x ≤ 6
−2(x − 3) > 8
−2x + 6 > 8 → −2x > 2 → x < −1 (sign reversed)
4 < 3x + 1 ≤ 13
3 < 3x ≤ 12 → 1 < x ≤ 4
x is at least 5 and no more than 12.
5 ≤ x ≤ 12
Why it works
Why the sign reverses
Multiplying by a negative reflects the number line about 0, which reverses the order of numbers.
Compound inequalities
a < x < b means x lies strictly between a and b; a ≤ x ≤ b includes the endpoints.
Regions
A linear inequality in x and y describes a half-plane on one side of a boundary line.
Mastery levels
LEVEL 1
Read the symbols and the word phrases correctly.
LEVEL 2
Solve accurately, reversing the sign when needed.
LEVEL 3
Move between words, algebra and graphs.
Solving quadratic equations efficiently and using the discriminant and the forms of a quadratic to answer questions without full working.
Spot it
- ax² + bx + c = 0
- Words: roots, zeros, solutions, x-intercepts, maximum, minimum
- “One solution”, “two solutions”, “no real solutions”
- Factored or vertex form given
Method
- Look at the form: factored → read the roots; vertex → read the maximum or minimum; standard → factor, use the formula or the discriminant.
- Factor when it is quick; otherwise use the quadratic formula.
- For the vertex in standard form, x = −b/(2a).
- For the number of solutions, use the discriminant b² − 4ac.
- Interpret roots and the vertex in context.
Tips
- DSAT quadratics often factor cleanly: try factoring first.
- “How many solutions?” → go straight to the discriminant.
- The sum of the roots is −b/a: useful when the question asks for it.
- In context, roots are where the quantity is zero (height, profit).
Common traps
- Using the quadratic formula when factoring is faster
- Dropping the ± in the formula
- Assuming the vertex is always a maximum
- Mixing up roots and the y-intercept
Pattern examples
(x − 3)(x + 5) = 0
x = 3 or x = −5
2x² − 7x + 3 = 0
(2x − 1)(x − 3) = 0 → x = 1/2 or x = 3
For what c does x² + 6x + c = 0 have exactly one real solution?
36 − 4c = 0 → c = 9
h(t) = −5t² + 20t + 1. When is the height greatest?
t = −20/(2 × −5) = 2
Why it works
Discriminant
b² − 4ac > 0: two real solutions; = 0: one; < 0: none.
Forms
Standard form shows the y-intercept c; factored form shows the roots; vertex form shows the maximum or minimum.
Symmetry
The vertex lies halfway between the roots, on the axis x = −b/(2a).
Mastery levels
LEVEL 1
Identify the form and the key features.
LEVEL 2
Solve by factoring, the formula or structure.
LEVEL 3
Use the discriminant and vertex to answer in context.
Predicting the shape, direction, vertex and number of x-intercepts of a parabola from its equation, with little or no calculation.
Spot it
- Vertex form a(x − h)² + k
- Factored form a(x − p)(x − q)
- Words: opens upward or downward, vertex, axis of symmetry, shift
- “Which graph matches …?”
Method
- Identify the form.
- Sign of a: a > 0 opens up (minimum), a < 0 opens down (maximum).
- Size of a: |a| > 1 narrower, |a| < 1 wider.
- Vertex: (h, k) from vertex form, or x = −b/(2a) from standard form.
- Number of x-intercepts: read from factored form, from the vertex position, or from the discriminant.
Tips
- Check the sign of a first: it tells you half the story.
- Vertex above the x-axis and opening up → no x-intercepts.
- Vertex below the x-axis and opening down → no x-intercepts.
- The axis of symmetry is halfway between the roots.
Common traps
- (x − 3) means a shift right 3, not left
- A negative in front reflects vertically, not horizontally
- Forgetting that symmetric points share the same y-value
- Mixing up roots with the y-intercept
Pattern examples
f(x) = 2(x + 4)² − 3
vertex (−4, −3), opens up, narrower than x²
g(x) = −(x − 2)(x − 8)
opens down, roots 2 and 8, vertex (5, 9)
y = (x − 1)² + 4: how many x-intercepts?
vertex above the axis and opens up → none
Which is wider: y = 0.5x² or y = 3x²?
y = 0.5x², because |0.5| < |3|
Why it works
Vertex form
a(x − h)² + k gives the vertex (h, k), the direction from the sign of a and the width from |a|.
Factored form
a(x − p)(x − q) gives the roots p and q and the axis x = (p + q)/2.
Standard form
ax² + bx + c gives the y-intercept c, the axis x = −b/(2a) and, through the discriminant, the number of roots.
Mastery levels
LEVEL 1
Read direction, vertex and roots from the form.
LEVEL 2
Find the axis and predict shifts and stretches.
LEVEL 3
Match equations to graphs and tables quickly.
Recognising constant-percentage change, building y = a · bˣ and interpreting growth and decay in tables, graphs and contexts.
Spot it
- Words: grows by, decreases by, each year, per month, doubles, half-life
- A table with a constant ratio, not a constant difference
- Models such as y = abˣ or y = a(1 ± r)ᵗ
- A curved graph rising or falling ever faster or slower
Method
- Find the multiplier b: growth r% → 1 + r/100, decay r% → 1 − r/100, table → next ÷ previous.
- Find the initial value a: the value at x = 0.
- Write y = a · bˣ.
- Interpret: b > 1 is growth, 0 < b < 1 is decay.
- Convert back: the percent change is (b − 1) × 100%.
Tips
- Constant difference → linear; constant ratio → exponential.
- Doubling or halving is always exponential.
- If x goes up in steps of 2 in a table, the row-to-row ratio is b².
- Look for the row where x = 0 to find a.
Common traps
- Adding a percentage instead of multiplying
- Using 0.30 for “decreases by 30%” (the factor is 0.70)
- Thinking a constant difference means exponential
- Reading b = 0.85 as an 85% decrease
Pattern examples
A population grows by 12% each year.
y = a(1.12)ᵗ
A value decreases by 18% each year.
y = a(0.82)ᵗ
x = 0, 1, 2, 3 and y = 50, 75, 112.5, 168.75
ratio 1.5 → y = 50(1.5)ˣ
b = 0.72: what is the percent change?
0.72 − 1 = −0.28 → a 28% decrease
Why it works
Percent change is multiplication
A 20% increase multiplies by 1.20; it does not add 20.
Constant ratio
A table is exponential when value₂ ÷ value₁ = value₃ ÷ value₂ = …
Reading b
b = 2 doubles each step; b = 1.05 grows 5% each step; b = 0.5 halves each step.
Mastery levels
LEVEL 1
Recognise exponential behaviour.
LEVEL 2
Build y = a · bˣ from the multiplier and initial value.
LEVEL 3
Convert between multipliers and percentages and compare models.
Following an input through two functions in order, using formulas, tables or graphs.
Spot it
- f(g(x)), g(f(x)) or f(g(3))
- Two functions given as formulas, tables or graphs
- “Which expression is equivalent to f(g(x))?”
Method
- Start inside: for f(g(x)), find g(x) first.
- Use that output as the input of f.
- With tables or graphs: read g at the given input, then read f at that value.
- With formulas: replace every x in f with the whole expression g(x).
- Simplify if the answer choices are simplified.
Tips
- Work inside out: g first, then f.
- Write the intermediate value down before the second step.
- f(g(3)) is a number; f(g(x)) is an expression.
- f(g(x)) and g(f(x)) are usually different.
Common traps
- Applying f first
- Reading the wrong row or column of a table
- Mixing up inputs and outputs
- Substituting x into f instead of g(x)
Pattern examples
g(3) = 7 and f(7) = −2. Find f(g(3)).
f(g(3)) = f(7) = −2
f(x) = 3x − 1, g(x) = x² + 2. Find f(g(x)).
3(x² + 2) − 1 = 3x² + 5
f(x) = 2x + 1, g(x) = x² − 3. Find f(g(x)).
2(x² − 3) + 1 = 2x² − 5
g doubles a number; f adds 5. Find f(g(x)).
2x + 5
Why it works
A pipeline
f(g(x)) is a two-step machine: apply g, then apply f. Changing the order usually changes the result.
Graphs
For f(g(2)), read g(2) on g’s graph, then use that value as the input on f’s graph.
Mastery levels
LEVEL 1
Recognise composition and its order.
LEVEL 2
Evaluate compositions from formulas, tables and graphs.
LEVEL 3
Build and compare composite expressions.
Solving equations with a variable in a denominator or under a root, and rejecting solutions that do not work in the original equation.
Spot it
- A variable in a denominator
- A square root containing a variable
- Answer choices that include a value that makes a denominator zero
Method
- Write the restrictions: denominators ≠ 0, and the expression under a square root ≥ 0.
- Clear the structure: multiply by the LCD, or isolate the root and square both sides.
- Solve the resulting linear or quadratic equation.
- Check every solution in the original equation and reject any that fail.
Tips
- Write the restrictions before solving.
- Isolate the root before squaring.
- Always substitute back: extraneous answers often appear among the choices.
- A square root is never negative, so √(…) = negative number has no solution.
Common traps
- Forgetting that a denominator cannot be zero
- Keeping an extraneous solution after squaring
- Squaring before isolating the root
- Multiplying by the wrong common denominator
Pattern examples
2/(x − 3) = 5
2 = 5x − 15 → x = 17/5 (x ≠ 3)
√(x + 7) = x − 5
x² − 11x + 18 = 0 → x = 9 or 2; only x = 9 checks
√(x + 4) = −3
no solution: a square root cannot be negative
For which x is 5/(x² − 9) undefined?
x = 3 or x = −3
Why it works
Why squaring creates false solutions
Squaring loses the sign: √(x + 4) = −3 has no solution, but squaring gives x + 4 = 9, so x = 5, which does not check.
Why denominators matter
Multiplying by an expression that can be zero can introduce a value that makes the original equation undefined.
Mastery levels
LEVEL 1
Recognise the structure and its restrictions.
LEVEL 2
Clear denominators or roots and solve.
LEVEL 3
Check and reject extraneous solutions.
Using the link between roots and factors: r is a root of p(x) exactly when (x − r) is a factor.
Spot it
- “Which of the following is a factor of p(x)?”
- “If p(3) = 0, which must be true?”
- Roots given and a polynomial to match
- Graphs that cross or touch the x-axis
Method
- If p(r) = 0, then r is a root.
- Root r → factor (x − r); for example, root −4 → factor (x + 4).
- Several roots → p(x) = a(x − r₁)(x − r₂)…
- Check the sign of each factor, then the degree and the leading coefficient.
Tips
- Given a root, write the factor immediately, and the other way round.
- Avoid expanding unless the choices are expanded.
- A root with even multiplicity touches the x-axis; odd multiplicity crosses it.
- The remainder when p(x) is divided by (x − a) is p(a).
Common traps
- Wrong sign: root 3 → (x − 3), not (x + 3)
- Forgetting the leading coefficient a
- Choosing a polynomial whose degree is lower than the number of distinct roots
- Expanding needlessly and running out of time
Pattern examples
4 is a root of p(x).
(x − 4) is a factor
(x + 7) is a factor of p(x).
p(−7) = 0, so −7 is a root
A polynomial has roots 1 and −2.
p(x) = a(x − 1)(x + 2)
p(x) divided by (x − 2) leaves remainder 5.
p(2) = 5
Why it works
Factor theorem
(x − r) is a factor of p(x) if and only if p(r) = 0. Evaluating and factoring are two views of the same fact.
Factored form stores information
It shows the zeros, where the graph crosses or touches the axis, and the multiplicity of each root.
Mastery levels
LEVEL 1
Convert between roots and factors with the right sign.
LEVEL 2
Build or identify polynomials from roots.
LEVEL 3
Use multiplicity and remainders to reason about graphs.
Describing how changes to a function’s equation move, stretch or reflect its graph.
Spot it
- g(x) = f(x − h) + k
- “Which transformation maps f to g?”
- Moved 3 units right, reflected across the x-axis
- Vertically stretched by a factor of 2
Method
- Use the template g(x) = a · f(b(x − h)) + k.
- h: horizontal shift (x − h moves right h; x + h moves left h).
- k: vertical shift (up if k > 0, down if k < 0).
- a: vertical stretch if |a| > 1, compression if |a| < 1, reflection across the x-axis if a < 0.
- b: horizontal compression if |b| > 1, stretch if |b| < 1, reflection across the y-axis if b < 0.
Tips
- Find the shifts first: they are the most common question.
- Changes inside the brackets act on x and work “backwards”.
- Track one landmark point, such as a vertex or an intercept.
- Do not expand: read the transformation from the structure.
Common traps
- Mixing up horizontal and vertical shifts
- Moving in the wrong direction for (x − h)
- Missing a reflection
- Confusing stretch with compression
Pattern examples
g(x) = f(x − 5)
shift right 5
g(x) = f(x) + 4
shift up 4
g(x) = −f(x)
reflect across the x-axis
g(x) = 2f(3(x + 2)) − 1
left 2, horizontal compression by a factor of 1/3, vertical stretch by 2, down 1
Why it works
Inside versus outside
Changes inside the brackets affect x-values (horizontal); changes outside affect y-values (vertical).
Why x − 3 moves right
g(x) = f(x − 3) reaches the value f(0) when x = 3, so every point moves 3 units right.
Mastery levels
LEVEL 1
Identify shift, stretch and reflection signals.
LEVEL 2
Determine direction and size correctly.
LEVEL 3
Map graphs mentally from the equation.
Working with ratios, percentages and percent change quickly by turning them into fractions and multipliers.
Spot it
- “Increase by r%”, “decrease by r%”
- “x is y% of z”
- “The ratio of A to B is …”
- Two percent changes in a row
Method
- Turn each percentage into a multiplier: increase r% → × (1 + r/100), decrease r% → × (1 − r/100).
- Identify the whole: whole × percent = part.
- Turn ratios into fractions: A : B = 2 : 3 → A = (2/3)B, and A is 2/5 of the total.
- For successive changes, multiply the multipliers.
- Percent change = (new − old) ÷ old × 100%.
Tips
- Always convert to a multiplier first.
- Find the whole: most percent errors use the wrong base.
- Never add successive percentages: multiply.
- Keep the order of a ratio: A : B is not B : A.
Common traps
- Treating percent change as additive
- Confusing “percent of” with “percent change”
- Using the wrong whole
- Reversing a ratio
Pattern examples
Increase by 20%, then by 30%.
1.20 × 1.30 = 1.56 → a 56% increase, not 50%
Increase by 20%, then decrease by 10%.
1.20 × 0.90 = 1.08 → an 8% increase
Increase by 25%, then decrease by 25%.
1.25 × 0.75 = 0.9375 → 6.25% below the start
A : B = 3 : 5 and A + B = 64.
A = 3/8 × 64 = 24, B = 40
Why it works
Multiplication, not addition
A 20% increase multiplies by 1.20, so changes stack by multiplying.
Increases and decreases are not symmetric
+25% is undone by −20% (1.25 × 0.80 = 1), not by −25%.
Ratios are fixed relationships
If A : B = 2 : 3, then A = (2/3)B in every case, and A makes up 2/5 of A + B.
Mastery levels
LEVEL 1
Identify the whole and the type of percentage.
LEVEL 2
Apply multipliers and solve proportions.
LEVEL 3
Track multi-step changes and ratio relationships.
Converting between units, including rates, by multiplying by conversion factors so that units cancel.
Spot it
- “Convert … to …” or “How many … are in …?”
- Mixed units in a word problem
- Rates such as miles per hour or dollars per pound
- A conversion factor given in the question
Method
- Write the quantity with its unit.
- Multiply by a conversion fraction with the unit to cancel in the denominator.
- Cancel units like algebraic factors.
- Chain more fractions until only the target unit is left.
- For rates, convert the numerator and the denominator separately.
Tips
- Every conversion fraction equals 1, so it never changes the quantity.
- The unit left at the end is the unit of your answer.
- Use the conversion factors given in the question: the SAT provides non-standard ones.
- Check that the size of the answer makes sense.
Common traps
- Multiplying when you should divide (no factor chain)
- Converting the numerator of a rate but not the denominator
- Using a conversion factor upside down
- Cancelling units that only look similar
Pattern examples
7 ft to inches
7 × 12 = 84 in
3 lb to ounces
3 × 16 = 48 oz
30 mi/h to ft/s
30 × 5280 ÷ 3600 = 44 ft/s
2 gal/min to L/h, with 1 gal ≈ 3.785 L
2 × 3.785 × 60 ≈ 454 L/h
Why it works
Units behave like variables
ft × (12 in / 1 ft): the ft cancels and inches remain.
Rates convert top and bottom
For mi/h → ft/s, convert miles to feet and hours to seconds.
Always multiplicative
Conversions multiply or divide; they never add.
Mastery levels
LEVEL 1
Identify the units and the factors needed.
LEVEL 2
Build the factor chain and cancel units.
LEVEL 3
Handle conversions inside multi-step problems.
Extracting values from tables and combining groups correctly, where each group counts in proportion to its size.
Spot it
- Tables with groups and counts
- “Overall average” or “combined mean”
- A missing value or group size
- Frequency tables
Method
- For each group, total = mean × count.
- Add the totals.
- Add the counts.
- Weighted mean = total of values ÷ total count.
- For a missing value, set up an equation with the overall total.
Tips
- Never average averages unless the groups are the same size.
- Rewrite every mean as total = mean × count.
- In a frequency table, multiply each value by its frequency.
- Check which column holds counts and which holds values.
Common traps
- Averaging group means directly
- Mixing up the total value and the total count
- Using the wrong units or column
- Forgetting to rebuild the group totals
Pattern examples
Group A: mean 80, 20 people. Group B: mean 70, 10 people.
(1600 + 700) ÷ 30 ≈ 76.7
10 students average 60 and n students average 80. The overall mean is 75.
600 + 80n = 75(10 + n) → n = 30
Values 2, 3, 4 with frequencies 5, 3, 2
(10 + 9 + 8) ÷ 10 = 2.7
Which group contributes most to the total?
Compare mean × count, not the means alone
Why it works
Means hide totals
Mean = total ÷ count, so total = mean × count. To combine groups, rebuild their totals.
Bigger groups weigh more
A larger group pulls the combined mean closer to its own mean.
Mastery levels
LEVEL 1
Read the means, counts and frequencies in a table.
LEVEL 2
Convert to totals, combine and divide.
LEVEL 3
Solve missing-value problems with an equation.
Applying area and volume formulas, breaking composite figures into parts and using scale factors.
Spot it
- Radius, diameter, sector or arc
- “Area of the shaded region”
- “If the radius doubles …”
- Composite figures such as a rectangle with a semicircle
Method
- Identify the basic shapes.
- Apply the formula: triangle ½bh, circle πr², circumference 2πr, cylinder πr²h.
- Break composite figures into parts; add or subtract areas.
- For scale factor k: lengths × k, areas × k², volumes × k³.
- For sectors, use the fraction of the circle: θ/360 of the area or circumference.
Tips
- The reference sheet in Bluebook lists the area and volume formulas.
- Check radius against diameter: questions often give the diameter.
- The height of a triangle is perpendicular to its base.
- Use scale factors instead of recalculating from scratch.
Common traps
- Using the diameter as the radius
- Using a slanted side as the height
- Adding a region that should be subtracted
- Scaling area by k instead of k²
Pattern examples
The radius of a circle doubles.
the area becomes 4 times as large
A sector of 90° in a circle of radius 6
area = 90/360 × 36π = 9π
A cylinder’s height triples and its radius halves.
volume × (1/2)² × 3 = 3/4 of the original
A square of side 10 contains a circle of radius 5. Find the area outside the circle.
100 − 25π
Why it works
Triangle area
A triangle is half of a rectangle with the same base and height.
Scaling
Area involves two lengths and volume three, so they scale by k² and k³.
Mastery levels
LEVEL 1
Identify shapes, formulas and the needed dimensions.
LEVEL 2
Compute areas and volumes, including composite figures.
LEVEL 3
Use scale factors and structure to solve quickly.
Using sine, cosine and tangent, special right triangles and the complement rule to find sides and angles.
Spot it
- A right triangle with an angle and a side given
- Angle of elevation or depression
- 30-60-90 or 45-45-90 triangles
- sin x° = cos(90° − x°)
Method
- Label the sides relative to the angle: opposite, adjacent, hypotenuse.
- Choose the ratio: sin = opp/hyp, cos = adj/hyp, tan = opp/adj.
- Write the equation with the unknown side and solve.
- Use special triangles where possible: 30-60-90 → x, x√3, 2x; 45-45-90 → x, x, x√2.
- If two sides are known, use Pythagoras first.
Tips
- Draw and label the triangle before choosing a ratio.
- sin x° = cos(90° − x°): this complement rule appears often.
- Tangent is rise over run, like slope.
- The special triangles are on the reference sheet.
Common traps
- Labelling opposite and adjacent from the wrong angle
- Using the hypotenuse in a tangent ratio
- Assuming a diagram is drawn to scale
- Mixing degrees and radians on the calculator
Pattern examples
sin x° = 0.6. Find cos(90° − x°).
0.6
A 30-60-90 triangle has hypotenuse 10.
the shorter leg is 5 and the longer leg is 5√3
From 50 m away, the angle of elevation to the top of a tower is 30°.
h = 50 tan 30° ≈ 28.9 m
A right triangle has legs 6 and 8.
hypotenuse 10, so sin of the angle opposite 6 is 0.6
Why it works
Why ratios work
All right triangles with the same acute angle are similar, so their side ratios are equal.
Complementary angles
The two acute angles add to 90°, and the side opposite one is adjacent to the other, so sin x° = cos(90° − x°).
Mastery levels
LEVEL 1
Identify the triangle and the right ratio.
LEVEL 2
Find missing sides with SOH CAH TOA or special triangles.
LEVEL 3
Model real situations with triangles.
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