Math Mastery Workbook — Pre-Algebra & Geometry

Q 1 Order of Operations · PEMDAS Hard

Order of Operations (PEMDAS/BODMAS):
Evaluate in this exact order: Parentheses → Exponents → Multiplication & Division (left→right) → Addition & Subtraction (left→right). Worked Example

Simplify $4 + 2^3 \times (9 - 4) \div 5$
① Parens: $9-4=5$
② Exponent: $2^3=8$
③ Mult: $8 \times 5 = 40$
④ Div: $40 \div 5 = 8$
⑤ Add: $4 + 8 = \boxed{12}$

Evaluate: $\displaystyle 6 + 4^2 \times (10 - 6) \div 8 - 3$

Answer: A — 11

$6 + 4^2 \times (10-6) \div 8 - 3$
① Parentheses: $(10-6)=4$
② Exponent: $4^2=16$
③ Multiply: $16 \times 4=64$
④ Divide: $64 \div 8=8$
⑤ Add/Sub left→right: $6+8-3=\mathbf{11}$

⚠ Common MistakeMany students compute $6+4=10$ first, then square it. Remember: exponents come before addition!

Q 2 Fractions · Unlike Denominators Hard

Adding/Subtracting Fractions: Find the Least Common Denominator (LCD), convert each fraction, then add or subtract numerators. Simplify your answer.
Finding LCD: List multiples of each denominator and find the smallest common one. Worked Example

$\dfrac{2}{3} + \dfrac{5}{8}$. LCD of 3 and 8 = 24.
$\dfrac{16}{24} + \dfrac{15}{24} = \dfrac{31}{24} = 1\dfrac{7}{24}$

Simplify completely: $\;\dfrac{7}{8} - \dfrac{5}{12} + \dfrac{1}{6}$

Answer: A — 5/8

LCD of 8, 12, 6 = 24
$\dfrac{7}{8}=\dfrac{21}{24},\quad\dfrac{5}{12}=\dfrac{10}{24},\quad\dfrac{1}{6}=\dfrac{4}{24}$
$\dfrac{21}{24}-\dfrac{10}{24}+\dfrac{4}{24}=\dfrac{15}{24}=\dfrac{5}{8}$

⚠ Common MistakeAdding denominators directly: $8+12+6=26$. You must always find the LCD first!

Q 3 Solving Two-Step Equations with Fractions Hard

Two-Step Equations: Undo operations in reverse order of PEMDAS. When a negative coefficient appears, dividing both sides by it will flip nothing — but watch signs carefully. Worked Example

$-\dfrac{3}{4}x + 2 = -7$
Subtract 2: $-\dfrac{3}{4}x = -9$
Multiply both sides by $-\dfrac{4}{3}$: $x = 12$

Solve for $x$:$\quad -\dfrac{3}{5}x + 4 = -8$

Answer: A — x = 20

$-\dfrac{3}{5}x + 4 = -8$
Subtract 4 from both sides: $-\dfrac{3}{5}x = -12$
Multiply both sides by $-\dfrac{5}{3}$:
$x = -12 \times \left(-\dfrac{5}{3}\right) = \dfrac{60}{3} = \mathbf{20}$

⚠ Common MistakeSubtracting 4 first but getting +4 instead of −12. Double-check: $-8-4=-12$, not $-8+4$.

Q 4 Absolute Value Expressions Hard

Absolute Value $|x|$: The distance from zero on a number line. Always non-negative. Evaluate the inside expression first, then apply the absolute value. Worked Example

$3|{-4}| - |7 - 12| + |{-2}|^2$
$= 3(4) - |{-5}| + (2)^2$
$= 12 - 5 + 4 = 11$

Evaluate: $\;4|{-5}| - 3|2 - 9| + |{-3}|^2$

Answer: A — 8

$4|{-5}| - 3|2-9| + |{-3}|^2$
$= 4(5) - 3(7) + (3)^2$
$= 20 - 21 + 9 = \mathbf{8}$

⚠ Common MistakeSquaring before absolute value: writing $(-3)^2=9$ is correct here, but some write $|{-3}^2|=|9|=9$ — accidentally correct. The trap is $-3|{-3}|^2$ where order matters critically.

Q 5 Percent Change · Successive Discounts Hard

Percent Change: $\text{Percent Change} = \dfrac{\text{New} - \text{Original}}{\text{Original}} \times 100\%$
Key Trap: Two successive percent changes do NOT simply add. Each applies to a different base. Worked Example

Price \$50, discounted 20%, then raised 10%:
After −20%: $50 \times 0.80 = \$40$
After +10%: $40 \times 1.10 = \$44$ (NOT \$50 × 0.90 = \$45)

A \$120 coat is discounted 30%, then the discounted price is raised 15%. What is the final price?

Answer: A — $96.60

Step 1 — 30% discount: $\$120 \times 0.70 = \$84$
Step 2 — 15% increase on \$84: $\$84 \times 1.15 = \mathbf{\$96.60}$

⚠ Classic TrapStudents think $-30\% + 15\% = -15\%$, so they compute $\$120 \times 0.85 = \$102$. This is wrong because the 15% increase applies to \$84, not \$120!

Q 6 Solving Inequalities · Flip Rule Hard

Inequality Rule: Solve like an equation, BUT reverse the inequality sign whenever you multiply or divide both sides by a negative number. Worked Example

$-5x + 3 \geq 18$
$-5x \geq 15$
Divide by $-5$ → flip sign: $x \leq -3$

Solve and graph on a number line: $-3x + 8 < -4$

Answer: B — x > 4

$-3x + 8 < -4$
Subtract 8: $-3x < -12$
Divide by $-3$ → flip the sign: $x > 4$
Graph: open circle at 4, arrow pointing right.

⚠ Most Common ErrorForgetting to flip the inequality when dividing by −3, giving $x < 4$ instead of $x > 4$.

Q 7 Exponent Rules · Simplifying Expressions Hard

Key Exponent Rules:
$a^m \cdot a^n = a^{m+n}$ | $\dfrac{a^m}{a^n}=a^{m-n}$ | $(a^m)^n=a^{mn}$
$(ab)^n = a^n b^n$ | $a^{-n}=\dfrac{1}{a^n}$ | $a^0=1$ Worked Example

$\dfrac{(3x^3y^{-2})^2}{9x^4y}$
Numerator: $9x^6y^{-4}$
$= \dfrac{9x^6y^{-4}}{9x^4y} = x^2 y^{-5} = \dfrac{x^2}{y^5}$

Simplify: $\dfrac{(2x^3y^{-2})^3}{4x^5y}$

Answer: A — 2x⁴/y⁷

Numerator: $(2)^3(x^3)^3(y^{-2})^3 = 8x^9y^{-6}$
$\dfrac{8x^9y^{-6}}{4x^5y} = 2 \cdot x^{9-5} \cdot y^{-6-1} = 2x^4y^{-7} = \dfrac{2x^4}{y^7}$

⚠ Common MistakeNot cubing the coefficient: writing $2x^9$ instead of $8x^9$ in the numerator. Remember: $2^3 = 8$, not 2!

Q 8 Proportions & Unit Rate · Cross-Multiplication Medium-Hard

Proportions: $\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow ad = bc$ (cross-multiply)
Three-Part Ratios: If $a:b:c = 2:3:5$ and total is 40, then each part is $\dfrac{40}{2+3+5}=4$, so the parts are 8, 12, 20. Worked Example

Ratio 3:5. Total = 56. Larger part = $\dfrac{5}{8} \times 56 = 35$

Three numbers are in the ratio $2:5:8$. Their sum is 90. What is the value of the largest number?

Answer: A — 48

Sum of ratio parts: $2+5+8=15$
Each unit $= 90 \div 15 = 6$
Largest number $= 8 \times 6 = \mathbf{48}$
Check: $12 + 30 + 48 = 90$ ✓

⚠ Common MistakeDividing 90 by 3 (number of terms) instead of by 15 (sum of ratio parts).

Q 9 Consecutive Integers · Word Problems Hard

Consecutive Integers: $n,\; n+1,\; n+2, \ldots$
Consecutive Even/Odd Integers: $n,\; n+2,\; n+4, \ldots$ (both types use +2) Worked Example

Sum of 4 consecutive integers = 70.
$n+(n+1)+(n+2)+(n+3)=70$
$4n+6=70 \Rightarrow n=16$
Numbers: 16, 17, 18, 19

The sum of four consecutive even integers is 100. What is the second-largest integer?

Answer: A — 26

Let the integers be $n,\;n+2,\;n+4,\;n+6$.
$4n+12=100 \Rightarrow 4n=88 \Rightarrow n=22$
Four integers: 22, 24, 26, 28.
Second-largest = $\mathbf{26}$

⚠ Common MistakeUsing $n,n+1,n+2,n+3$ for even integers. Even integers always differ by 2, not 1!

Q 10 Linear Equations · Slope-Intercept Form Hard

Slope-Intercept Form: $y = mx + b$, where $m=\text{slope}$, $b=\text{y-intercept}$.
Slope formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$ Worked Example

Through $(2,5)$ and $(6,-3)$:
$m=\dfrac{-3-5}{6-2}=\dfrac{-8}{4}=-2$
Using $(2,5)$: $5=-2(2)+b \Rightarrow b=9$
Equation: $y=-2x+9$

A line passes through $(-3,\;8)$ and $(5,\;-4)$. What is its equation in slope-intercept form?

Answer: A — y = −(3/2)x + 7/2

$m = \dfrac{-4-8}{5-(-3)} = \dfrac{-12}{8} = -\dfrac{3}{2}$
Using point $(-3,8)$: $8 = -\dfrac{3}{2}(-3)+b = \dfrac{9}{2}+b$
$b = 8 - \dfrac{9}{2} = \dfrac{16}{2}-\dfrac{9}{2} = \dfrac{7}{2}$
Equation: $y = -\dfrac{3}{2}x + \dfrac{7}{2}$

⚠ Common MistakeSign error: $-(-3) = +3$, so $-\frac{3}{2}\times(-3) = +\frac{9}{2}$, not $-\frac{9}{2}$.

Q 11 Pythagorean Theorem · Missing Side Hard

Pythagorean Theorem: In a right triangle with legs $a, b$ and hypotenuse $c$:
\[a^2 + b^2 = c^2\] Common Triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and their multiples) Worked Example

Hypotenuse = 13, one leg = 5.
$a^2 + 25 = 169 \Rightarrow a^2 = 144 \Rightarrow a = 12$

A right triangle has legs of length $(x+2)$ cm and $9$ cm, and hypotenuse $15$ cm. Find $x$.

Answer: A — x = 10

$(x+2)^2 + 9^2 = 15^2$
$(x+2)^2 + 81 = 225$
$(x+2)^2 = 144$
$x+2 = 12 \Rightarrow x = \mathbf{10}$
Check: leg = 12, other leg = 9, hyp = 15 ✓ (9-12-15 = 3×(3-4-5))

⚠ Common MistakeStudents treat $(x+2)$ as $x$ and solve $x^2+81=225$, getting $x=12$ instead of the correct $x=10$.

Q 12 Composite Area · Rectangle minus Circle Hard

Composite Figures: Break into shapes. Add or subtract their areas.
Circle area: $A=\pi r^2$ (always use RADIUS, not diameter!) Worked Example

Square 8×8, circle diameter 4 (radius 2) cut out:
$64 - \pi(4) \approx 64 - 12.57 = 51.43\text{ cm}^2$

A rectangle measures 12 cm × 9 cm. Two semicircles, each with diameter 6 cm, are removed from opposite ends. What is the remaining area? (Use $\pi \approx 3.14$)

Answer: A — 79.74 cm²

Rectangle area: $12 \times 9 = 108\text{ cm}^2$
Two semicircles = one full circle, radius $= 3\text{ cm}$
Circle area: $\pi(3)^2 = 9\pi \approx 9 \times 3.14 = 28.26\text{ cm}^2$
Remaining: $108 - 28.26 = \mathbf{79.74\text{ cm}^2}$

⚠ Diameter TrapUsing diameter (6) in the area formula instead of radius (3): $\pi(6)^2 = 36\pi$ is 4× too large!

Q 13 Parallel Lines & Transversal · Angle Types Hard

When parallel lines are cut by a transversal:
• Alternate Interior Angles: Equal ($=$)
• Corresponding Angles: Equal ($=$)
• Co-interior / Same-Side Interior: Supplementary (sum $= 180°$)
• Vertical Angles: Equal ($=$) Worked Example

Co-interior angles $(4x+10)°$ and $(2x+20)°$:
$6x+30=180 \Rightarrow x=25$

Two parallel lines are cut by a transversal. The co-interior (same-side interior) angles measure $(4x - 5)°$ and $(2x + 35)°$. Find the measure of the larger angle.

Answer: A — 115°

Co-interior angles sum to 180°:
$(4x-5)+(2x+35)=180$
$6x+30=180 \Rightarrow 6x=150 \Rightarrow x=25$
Angle 1: $4(25)-5=95°$ Angle 2: $2(25)+35=85°$
Wait — check: $95+85=180°$ ✓. Larger angle = $\mathbf{95°}$

⚠ Read CarefullyThe question asks for the LARGER angle (95°), not x. Students often stop at finding x = 25.

Q 14 Volume · Cone vs. Cylinder vs. Sphere Hard

Volume Formulas:
$V_{\text{cylinder}} = \pi r^2 h$
$V_{\text{cone}} = \dfrac{1}{3}\pi r^2 h$
$V_{\text{sphere}} = \dfrac{4}{3}\pi r^3$ Worked Example

A cone fits exactly inside a cylinder (same $r$ and $h$).
$V_{\text{cone}} = \dfrac{1}{3}V_{\text{cylinder}}$ — always!

A sphere has radius 6 cm. A cone has radius 6 cm and height 12 cm. By how many cubic centimeters does the sphere's volume exceed the cone's volume? (Leave in terms of $\pi$)

Answer: A — 144π cm³

$V_{\text{sphere}} = \dfrac{4}{3}\pi(6)^3 = \dfrac{4}{3}\pi(216) = 288\pi\text{ cm}^3$
$V_{\text{cone}} = \dfrac{1}{3}\pi(6)^2(12) = \dfrac{1}{3}\pi(36)(12) = 144\pi\text{ cm}^3$
Difference: $288\pi - 144\pi = \mathbf{144\pi\text{ cm}^3}$

⚠ Common ErrorStudents try to "simplify" $\frac{4}{3}\pi(216)$ incorrectly as $\frac{4}{3}(6)^3 = \frac{4}{3}(18) = 24\pi$. Always cube the radius first!

Q 15 Similar Triangles · Scale Factor & Area Hard

Similar Figures:
If scale factor of sides = $k$, then:
• Perimeter ratio = $k$
• Area ratio = $k^2$ ← most-missed fact! Worked Example

Scale factor = $\frac{3}{5}$. Area of larger = 100 cm².
Area of smaller = $100 \times \left(\dfrac{3}{5}\right)^2 = 100 \times \dfrac{9}{25} = 36\text{ cm}^2$

Two similar triangles have corresponding sides in ratio $3:4$. The area of the smaller triangle is 27 cm². What is the area of the larger triangle?

Answer: A — 48 cm²

Area ratio $= \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}$
$\dfrac{27}{\text{Larger}} = \dfrac{9}{16}$
Larger $= \dfrac{27 \times 16}{9} = 27 \times \dfrac{16}{9} = 3 \times 16 = \mathbf{48\text{ cm}^2}$

⚠ #1 Most MissedStudents use the side ratio directly: $27 \times \frac{4}{3} = 36$ cm². You MUST square the scale factor for areas!

Q 16 Circle · Arc Length & Sector Area Hard

Sector Area: $A_{\text{sector}} = \dfrac{\theta}{360°} \times \pi r^2$
Arc Length: $L = \dfrac{\theta}{360°} \times 2\pi r$ Worked Example

Circle $r=10$, central angle $= 72°$:
$A = \dfrac{72}{360}\pi(100) = \dfrac{1}{5}(100\pi) = 20\pi$

A circle has a circumference of $24\pi$ cm. A sector of this circle has a central angle of 150°. What is the area of the sector? (in terms of $\pi$)

Answer: A — 60π cm²

From circumference: $2\pi r = 24\pi \Rightarrow r = 12\text{ cm}$
$A_{\text{sector}} = \dfrac{150}{360} \times \pi(12)^2 = \dfrac{5}{12} \times 144\pi = \mathbf{60\pi\text{ cm}^2}$

⚠ Two-Step TrapStudents use $r=24$ (the circumference coefficient) directly without first solving for $r$. Always find $r$ from circumference first!

Q 17 Coordinate Geometry · Midpoint Back-Calculation Hard

Midpoint Formula: $M = \left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right)$
To find endpoint from midpoint: $x_2 = 2M_x - x_1$ Worked Example

$M=(4,1)$, $A=(7,5)$. Find $B$.
$x_B = 2(4)-7=1$, $y_B=2(1)-5=-3$, so $B=(1,-3)$

The midpoint of $\overline{PQ}$ is $M(2,\;-4)$. If $P = (-3,\;6)$, what are the coordinates of $Q$?

Answer: A — (7, −14)

$Q_x = 2(2)-(-3) = 4+3 = 7$
$Q_y = 2(-4)-6 = -8-6 = -14$
$Q = \mathbf{(7,\;-14)}$

⚠ Sign Error Hot SpotWhen computing $Q_x$: students get $2(2)-(3)=1$ by dropping the negative sign from $-(-3)$. Double negatives require care!

Q 18 Surface Area · Cylinder & Cone Hard

Surface Area Formulas:
Cylinder: $SA = 2\pi r^2 + 2\pi rh$
Cone: $SA = \pi r^2 + \pi r \ell$ ($\ell$ = slant height)
Slant height of cone: $\ell = \sqrt{r^2 + h^2}$ Worked Example

Cone, $r=3$, $h=4$: $\ell=\sqrt{9+16}=5$
$SA = \pi(9) + \pi(3)(5) = 9\pi + 15\pi = 24\pi$

A cone has base radius 5 cm and height 12 cm. What is the total surface area of the cone? (in terms of $\pi$)

Answer: A — 90π cm²

Slant height: $\ell = \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}$
$SA = \pi r^2 + \pi r\ell = \pi(25) + \pi(5)(13) = 25\pi + 65\pi = \mathbf{90\pi\text{ cm}^2}$

⚠ Most Missed StepUsing the height (12) as the slant height. Always find $\ell = \sqrt{r^2+h^2}$ first!

Q 19 Interior Angles of Polygons · Finding n Hard

Interior angle sum: $S=(n-2)\times 180°$
Each interior angle (regular polygon): $\dfrac{(n-2)\times180°}{n}$
Shortcut: Exterior angle of regular polygon $= \dfrac{360°}{n}$, so each interior angle $= 180° - \dfrac{360°}{n}$ Worked Example

Interior angle = 140°. Exterior = 40°.
$n = \dfrac{360°}{40°} = 9$ sides (nonagon)

The interior angle sum of a regular polygon is 1440°. Each interior angle measures $k°$. What is $k$?

Answer: A — 144°

Find $n$: $(n-2)\times180 = 1440$
$n-2=8 \Rightarrow n=10$ (decagon)
Each interior angle: $\dfrac{1440°}{10} = \mathbf{144°}$

⚠ Common MistakeDividing 1440 by 8 (forgetting to find n first): $1440\div8=180°$. Always find the number of sides first, then divide the sum by n.

Q 20 Transformations · Composition of Reflections & Rotations Hard

Transformation Rules:
• Reflect over x-axis: $(x,y)\to(x,-y)$
• Reflect over y-axis: $(x,y)\to(-x,y)$
• Rotate 90° CCW about origin: $(x,y)\to(-y,x)$
• Rotate 180° about origin: $(x,y)\to(-x,-y)$
• Reflect over $y=x$: $(x,y)\to(y,x)$
For compositions: apply transformations right to left (inner first)! Worked Example

Point $(3,-2)$: reflect over x-axis then rotate 90° CCW.
Reflect: $(3,2)$. Rotate: $(-2,3)$.

Point $P(4,\;-3)$ is first reflected over the line $y = x$, then rotated 180° about the origin. What are the final coordinates of $P$?

Answer: A — (3, −4)

Start: $P(4,-3)$
Step 1 — Reflect over $y=x$: $(x,y)\to(y,x)$: $P'(-3,4)$
Step 2 — Rotate 180° about origin: $(x,y)\to(-x,-y)$: $P''(3,-4)$
Final: $\mathbf{(3,\;-4)}$

⚠ Order Matters!If you apply the transformations in reversed order you get a different answer. Always perform compositions left to right as written.

🎉 Quiz Complete!