Math Master

Question 1 · Order of Operations

Common Error

📖 Concept — PEMDAS / Order of Operations Evaluate expressions in this order: Parentheses → Exponents → Multiplication & Division (left to right) → Addition & Subtraction (left to right).

✏ Worked Example Evaluate $3 + 4 \times 2^2 - (6 \div 3)$
Step 1 — Parentheses: $(6 \div 3) = 2$
Step 2 — Exponents: $2^2 = 4$
Step 3 — Multiply: $4 \times 4 = 16$
Step 4 — Add/Subtract: $3 + 16 - 2 = \mathbf{17}$

Evaluate: $\displaystyle 18 \div 3 + 5 \times 2^3 - (4 + 6)$

Solution Step 1 — Parentheses: $(4+6)=10$
Step 2 — Exponents: $2^3=8$
Step 3 — Division: $18 \div 3=6$
Step 4 — Multiplication: $5 \times 8=40$
Step 5 — Left to right: $6+40-10=\mathbf{36}$
⚠ Common mistake: students add 18 ÷ 3 + 5 first before multiplying.

Question 2 · Integer Operations

Common Error

📖 Concept — Multiplying & Dividing Integers Same signs → Positive: $(-3)(-4)=+12$ | Different signs → Negative: $(-3)(+4)=-12$

✏ Worked Example Simplify $(-2)^3 \times (-5)$
$(-2)^3 = -8$ → $-8 \times (-5) = +40$

Simplify: $\displaystyle \frac{(-4)^2 \times (-3)}{-6 + (-2)}$

Solution Numerator: $(-4)^2 = 16$, then $16 \times (-3) = -48$
Denominator: $-6 + (-2) = -8$
Result: $\dfrac{-48}{-8} = \mathbf{6}$
⚠ Watch out: $(-4)^2 = +16$, NOT $-16$. The exponent applies to the entire $-4$.

Question 3 · Solving One-Step Equations

Core Skill

📖 Concept — Solving Equations Use inverse operations to isolate the variable. Whatever you do to one side, do to the other.

✏ Worked Example Solve $3x - 7 = 14$:
Add 7: $3x = 21$ → Divide by 3: $x = 7$

Solve for $x$: $\displaystyle \frac{2x + 4}{3} = 6$

Solution Multiply both sides by 3: $2x + 4 = 18$
Subtract 4: $2x = 14$
Divide by 2: $x = \mathbf{7}$

Question 4 · Ratios & Proportions

Common Error

📖 Concept — Cross-Multiplication If $\dfrac{a}{b} = \dfrac{c}{d}$, then $ad = bc$. Use this to solve for an unknown.

✏ Worked Example Solve $\dfrac{3}{4} = \dfrac{x}{20}$:
Cross-multiply: $3 \times 20 = 4x$ → $60 = 4x$ → $x = 15$

A recipe uses 2.5 cups of flour for 12 cookies. How many cups of flour are needed to make 30 cookies?

Solution Set up proportion: $\dfrac{2.5}{12} = \dfrac{x}{30}$
Cross-multiply: $12x = 2.5 \times 30 = 75$
Divide: $x = \dfrac{75}{12} = \mathbf{6.25}\text{ cups}$

Question 5 · Percentages

Common Error

📖 Concept — Percent Change $\text{Percent Change} = \dfrac{\text{New} - \text{Original}}{\text{Original}} \times 100\%$

✏ Worked Example A shirt costs $40 and is now $32. Find the percent decrease:
$\dfrac{32-40}{40} \times 100 = \dfrac{-8}{40} \times 100 = -20\%$ (20% decrease)

A store increases a price from $\$80$ to $\$92$. What is the percent increase? Round to the nearest tenth.

Solution $\dfrac{92 - 80}{80} \times 100 = \dfrac{12}{80} \times 100 = \mathbf{15.0\%}$
⚠ Common error: dividing by the NEW value (92) instead of the original (80).

Question 6 · Combining Like Terms

Core Skill

📖 Concept — Like Terms Like terms have the same variable raised to the same power. Only their coefficients are combined: $3x^2 + 5x^2 = 8x^2$.

✏ Worked Example Simplify $4a - 3b + 2a + 7b - a$:
$a$-terms: $4a + 2a - a = 5a$ | $b$-terms: $-3b + 7b = 4b$ → $\mathbf{5a + 4b}$

Simplify: $3x^2 - 2x + 5 + 4x^2 + 7x - 3$

Solution $x^2$ terms: $3x^2 + 4x^2 = 7x^2$
$x$ terms: $-2x + 7x = +5x$
Constants: $5 - 3 = 2$
Result: $\mathbf{7x^2 + 5x + 2}$

Question 7 · Distributive Property

Common Error

📖 Concept — Distributive Property $a(b + c) = ab + ac$. Multiply the term outside by EVERY term inside the parentheses.

✏ Worked Example Expand $-3(2x - 5)$:
$-3 \cdot 2x = -6x$ and $-3 \cdot (-5) = +15$ → $\mathbf{-6x + 15}$

Expand and simplify: $4(3x - 2) - 2(x + 5)$

Solution $4(3x-2) = 12x - 8$
$-2(x+5) = -2x - 10$
Combine: $12x - 8 - 2x - 10 = \mathbf{10x - 18}$
⚠ Mistake: forgetting to distribute the negative sign in $-2(x+5)$.

Question 8 · Inequalities

Flip the Sign!

📖 Concept — Solving Inequalities Solve like equations, but: when you multiply or divide both sides by a negative number, FLIP the inequality sign.

✏ Worked Example Solve $-3x < 12$:
Divide by $-3$ and flip: $x > -4$

Solve for $x$: $-2x + 7 \geq 15$

Solution Subtract 7: $-2x \geq 8$
Divide by $-2$ (flip!): $x \leq \mathbf{-4}$
⚠ The most common error: forgetting to flip the inequality when dividing by a negative.

Question 9 · Prime Factorization & GCF

Core Skill

📖 Concept — GCF The Greatest Common Factor is the largest number that divides two numbers. Use prime factorization: find shared prime factors and multiply them.

✏ Worked Example GCF(24, 36): $24=2^3\cdot3$, $36=2^2\cdot3^2$ → Shared: $2^2\cdot3=12$

Find the GCF of $48$ and $72$.

Solution $48 = 2^4 \times 3$
$72 = 2^3 \times 3^2$
GCF $= 2^3 \times 3 = 8 \times 3 = \mathbf{24}$

Question 10 · Two-Variable Equations — Word Problem

Common Error

📖 Concept — Setting Up Systems Assign variables to unknowns, write two equations, and solve by substitution or elimination.

✏ Worked Example Tickets: adult $5, child $3; 10 tickets for $42. Find number of each type.
Let $a + c = 10$ and $5a + 3c = 42$. Substitution: $a=6, c=4$.

Marcus has 18 coins consisting of dimes and quarters. The total value is $\$3.00$. How many quarters does he have?

Solution Let $d$ = dimes, $q$ = quarters.
Eq 1: $d + q = 18$ → $d = 18 - q$
Eq 2: $0.10d + 0.25q = 3.00$
Substitute: $0.10(18-q) + 0.25q = 3.00$
$1.8 - 0.10q + 0.25q = 3.0$ → $0.15q = 1.2$ → $q = \mathbf{4}$

Question 11 · Pythagorean Theorem

Common Error

📖 Concept — Pythagorean Theorem In a right triangle: $a^2 + b^2 = c^2$, where $c$ is the hypotenuse (the longest side, opposite the right angle).

✏ Worked Example Legs: 6 and 8. Find hypotenuse:
$c = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$

A right triangle has legs of length $9$ and $12$. What is the length of the hypotenuse?

Solution $c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = \mathbf{15}$
⚠ Common error: adding the legs directly ($9+12=21$) without squaring.

Question 12 · Area of a Triangle

Core Skill

📖 Concept — Area Formulas Triangle: $A = \dfrac{1}{2}bh$ | Rectangle: $A = lw$ | Parallelogram: $A = bh$

✏ Worked Example Triangle with base 10 and height 6: $A = \frac{1}{2}(10)(6) = 30\text{ sq units}$

A triangle has a base of $14$ cm and an area of $63$ cm². What is its height?

Solution $63 = \dfrac{1}{2}(14)(h)$ → $63 = 7h$ → $h = \mathbf{9}\text{ cm}$
⚠ Check: $\frac{1}{2} \times 14 \times 9 = 63$ ✓

Question 13 · Circle — Circumference & Area

Radius vs Diameter

📖 Concept — Circle Formulas Circumference: $C = 2\pi r = \pi d$ Area: $A = \pi r^2$
Note: $r = \text{radius}$, $d = \text{diameter} = 2r$

✏ Worked Example Circle with diameter 10: $r=5$, $C = 10\pi \approx 31.4$, $A = 25\pi \approx 78.5$

A circle has a diameter of $18$ cm. What is its area? Leave answer in terms of $\pi$.

Solution Diameter = 18 → Radius $r = 9$
$A = \pi r^2 = \pi (9)^2 = \mathbf{81\pi}\text{ cm}^2$
⚠ Most common error: using diameter in $\pi d^2$ instead of radius.

Question 14 · Supplementary & Complementary Angles

Core Skill

📖 Concept — Angle Pairs Complementary: two angles that sum to $90°$
Supplementary: two angles that sum to $180°$

✏ Worked Example Two supplementary angles: one is $3x$, other is $x + 20$.
$3x + x + 20 = 180$ → $4x = 160$ → $x = 40$ → angles: $120°$ and $60°$.

Two supplementary angles are in the ratio $2:7$. Find the measure of the smaller angle.

Solution Let angles be $2k$ and $7k$.
$2k + 7k = 180$ → $9k = 180$ → $k = 20$
Smaller angle: $2 \times 20 = \mathbf{40°}$

Question 15 · Volume of a Rectangular Prism

Core Skill

📖 Concept — Volume Formulas Rectangular prism: $V = l \times w \times h$
Cylinder: $V = \pi r^2 h$ | Cone: $V = \dfrac{1}{3}\pi r^2 h$

✏ Worked Example Box: 5 cm × 4 cm × 3 cm → $V = 5 \times 4 \times 3 = 60\text{ cm}^3$

A rectangular swimming pool is $25$ m long, $12$ m wide, and $2$ m deep. What is its volume?

Solution $V = 25 \times 12 \times 2 = \mathbf{600}\text{ m}^3$

Question 16 · Parallel Lines & Transversals

Common Error

📖 Concept — Angle Relationships When a transversal cuts parallel lines: Alternate interior angles are equal; Co-interior (same-side interior) angles are supplementary $(180°)$; Corresponding angles are equal.

✏ Worked Example If corresponding angles are $(3x+10)°$ and $70°$:
$3x+10=70$ → $x=20$

Two parallel lines are cut by a transversal. Co-interior angles measure $(3x + 15)°$ and $(2x + 25)°$. Find the value of $x$.

Solution Co-interior angles sum to $180°$:
$(3x+15) + (2x+25) = 180$
$5x + 40 = 180$ → $5x = 140$ → $x = \mathbf{28}$
⚠ Error: treating co-interior angles as equal (they're supplementary, not equal).

Question 17 · Triangle Angle Sum

Core Skill

📖 Concept — Triangle Angle Sum The interior angles of any triangle always sum to $180°$.
Exterior angle theorem: an exterior angle equals the sum of the two non-adjacent interior angles.

✏ Worked Example Triangle angles: $x°$, $(x+20)°$, $50°$. Find $x$:
$x + x + 20 + 50 = 180$ → $2x = 110$ → $x = 55$

In triangle $ABC$, angle $A = (2x+10)°$, angle $B = (3x-5)°$, and angle $C = (x+15)°$. What is the measure of angle $B$?

Solution $(2x+10)+(3x-5)+(x+15)=180$
$6x+20=180$ → $6x=160$ → $x=\frac{160}{6}\approx 26.67$
Wait — recheck: $6x = 160$ → $x = \frac{80}{3}$?
Actually solve cleanly: $6x + 20 = 180$ → $6x = 160$ → $x = \frac{160}{6}$
Angle B $= 3\left(\frac{160}{6}\right) - 5 = 80 - 5 = \mathbf{85°}$ (using $x \approx 26.67$)

Question 18 · Similar Triangles

Common Error

📖 Concept — Similar Triangles Similar triangles have equal corresponding angles and proportional corresponding sides. If scale factor = $k$, then areas scale by $k^2$.

✏ Worked Example Triangles with sides 3, 4, 5 and 6, ?, 10. Scale factor: $10/5=2$. Missing side: $4\times2=8$.

Two similar triangles have corresponding sides in the ratio $3:5$. If the area of the smaller triangle is $27$ cm², what is the area of the larger triangle?

Solution Side ratio $= \dfrac{3}{5}$ → Area ratio $= \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}$
$\dfrac{27}{A} = \dfrac{9}{25}$ → $A = \dfrac{27 \times 25}{9} = \mathbf{75}\text{ cm}^2$
⚠ Error: multiplying area by side ratio $(\times\frac{5}{3})$ instead of the square ratio.

Question 19 · Coordinate Geometry — Midpoint & Distance

Common Error

📖 Concept — Distance & Midpoint Formulas Distance: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Midpoint: $M = \left(\dfrac{x_1+x_2}{2},\, \dfrac{y_1+y_2}{2}\right)$

✏ Worked Example Distance from $(1,2)$ to $(4,6)$:
$d=\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=\sqrt{25}=5$

Point $M(3, -1)$ is the midpoint of segment $\overline{AB}$. If $A = (-1, 5)$, find the coordinates of $B$.

Solution Midpoint formula: $\dfrac{-1+x_B}{2}=3$ → $x_B=7$
$\dfrac{5+y_B}{2}=-1$ → $y_B=-7$
Answer: $B = \mathbf{(7,\,-7)}$

Question 20 · Surface Area of a Cylinder

High Difficulty

📖 Concept — Surface Area of a Cylinder $SA = 2\pi r^2 + 2\pi r h$
$2\pi r^2$ = area of two circular bases | $2\pi r h$ = lateral (side) area

✏ Worked Example Cylinder: $r=3$, $h=5$:
$SA = 2\pi(9) + 2\pi(3)(5) = 18\pi + 30\pi = 48\pi \approx 150.8\text{ units}^2$

A cylinder has a radius of $5$ cm and a height of $11$ cm. What is its total surface area? Leave in terms of $\pi$.

Solution Two bases: $2\pi r^2 = 2\pi(25) = 50\pi$
Lateral surface: $2\pi r h = 2\pi(5)(11) = 110\pi$
Total: $50\pi + 110\pi = \mathbf{160\pi}\text{ cm}^2$
⚠ Common error: forgetting to add the two circular bases.

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