IB Mathematics · Grade 9

Practice Quiz
& Study Guide

20 carefully selected problems — probability, algebra, geometry, statistics & more. Tap an answer to check instantly.

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★ Core Unit

Probability & Sets

The most commonly confused topic on IB Math assessments. Pay close attention to complement, union vs. intersection, and conditional probability.

Flash Memory — Probability Keywords
UNION (∪) = OR = Add, subtract overlap  ·  INTERSECTION (∩) = AND = Both happen
COMPLEMENT (Aʹ) = NOT A = 1 − P(A)  ·  INDEPENDENT: P(A∩B) = P(A)·P(B)
CONDITIONAL: P(A|B) = P(A∩B) / P(B)  ·  MUTUALLY EXCLUSIVE: P(A∩B) = 0
1
Complement Rule
— commonly confused with subtraction from 0

A bag contains 4 red, 3 blue, and 3 green balls. One ball is drawn at random. What is the probability of not drawing a red ball? \( P(\text{not red}) = 1 - P(\text{red}) \) 💡 Key idea: "Not A" always means 1 − P(A). Never subtract from 0!
📖 Explanation Total balls = 4 + 3 + 3 = 10.
\( P(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5} \)
\( P(\text{not red}) = 1 - \dfrac{2}{5} = \dfrac{3}{5} \)
✅ Answer: (b) 3/5
⚠️ Common Mistake: Don't divide 6 (blue+green) by the total wrong count. Always use the full sample space as denominator.
2
Union of Two Events
— the #1 most missed formula on tests

In a class of 30 students, 18 play football, 12 play basketball, and 6 play both. What is the probability that a randomly chosen student plays at least one sport? \( P(F \cup B) = P(F) + P(B) - P(F \cap B) \) 💡 "At least one" = UNION. Subtract overlap to avoid double-counting!
📖 Explanation
\( P(F) = \frac{18}{30},\quad P(B) = \frac{12}{30},\quad P(F \cap B) = \frac{6}{30} \)
\( P(F \cup B) = \frac{18}{30} + \frac{12}{30} - \frac{6}{30} = \frac{24}{30} = \frac{4}{5} \)
✅ Answer: (b) 4/5
⚠️ If you chose (a) 3/5: You probably forgot to subtract the overlap. Always use the inclusion-exclusion formula!
3
Conditional Probability
— tricky because the sample space SHRINKS

A fair die is rolled. Given that the result is even, what is the probability it is greater than 2? \( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} \) 💡 When given "Given that…", your new sample space is just the condition, not all outcomes.
📖 Explanation Even numbers on a die: {2, 4, 6} — this is your NEW sample space (3 outcomes).
Numbers that are even AND greater than 2: {4, 6} — 2 outcomes
\( P(\text{>2} \mid \text{even}) = \dfrac{2}{3} \)
✅ Answer: (c) 2/3
⚠️ Common error: Using 6 as the denominator. Once you're given a condition, only count outcomes WITHIN that condition.
4
Independent Events Test
— students often confuse "independent" with "mutually exclusive"

Event A has P(A) = 0.4, Event B has P(B) = 0.5, and P(A ∩ B) = 0.2. Are events A and B independent? \(\text{Independent} \Leftrightarrow P(A \cap B) = P(A) \cdot P(B)\) 💡 INDEPENDENT ≠ MUTUALLY EXCLUSIVE. Independent means the result of A doesn't affect B.
📖 Explanation Check the independence condition:
\( P(A) \cdot P(B) = 0.4 \times 0.5 = 0.20 \)
\( P(A \cap B) = 0.20 \) ✓ — These are equal!
✅ Answer: (c) Yes, because P(A)·P(B) = P(A∩B)
⚠️ (b) is a common trap! P(A∩B) ≠ 0 does NOT mean they are dependent — mutually exclusive requires P(A∩B) = 0, but independence is a different condition entirely.
5
Venn Diagram — Exactly One Region
— very commonly asked on IB exams

In a group of 50 people: 20 own a cat, 25 own a dog, 10 own both. How many people own exactly one pet? \( |A \text{ only}| = |A| - |A \cap B| \qquad |B \text{ only}| = |B| - |A \cap B| \) 💡 "Exactly one" = (A only) + (B only). Subtract the middle from each circle first!
📖 Explanation
Cat only = 20 − 10 = 10
Dog only = 25 − 10 = 15
Exactly one pet = 10 + 15 = 25
✅ Answer: (a) 25
⚠️ (b) 35 is the total union: 20+25−10=35. That's "at least one," not "exactly one." These two are different!
6
Probability from a Two-Way Table
— read the table carefully before answering

In a survey, 100 students were asked their grade level and favourite subject. The table shows:

\[\begin{array}{|c|c|c|} \hline & \textbf{Math} & \textbf{Science} \\ \hline \textbf{Grade 9} & 18 & 22 \\ \hline \textbf{Grade 10} & 30 & 30 \\ \hline \end{array}\] A student is selected at random. Given they are in Grade 9, what is the probability their favourite subject is Math? 💡 "Given they are in Grade 9" → denominator = total Grade 9 students only.
📖 Explanation Total Grade 9 students = 18 + 22 = 40
Grade 9 AND Math = 18
\( P(\text{Math} \mid \text{Grade 9}) = \dfrac{18}{40} = \dfrac{9}{20} \)
✅ Answer: (c) 9/20
⚠️ (b) uses 100 as denominator — that would be the probability for any random student, not one specifically from Grade 9.
7
Tree Diagram — Dependent Events
— without replacement changes every probability!

A box has 5 red and 3 blue balls. Two balls are drawn without replacement. What is the probability both balls are red? \( P(\text{both red}) = P(R_1) \times P(R_2 \mid R_1) \) 💡 Without replacement → the second draw depends on the first. Update both numerator AND denominator.
📖 Explanation
First draw: \( P(R_1) = \dfrac{5}{8} \)
Second draw (one red removed): \( P(R_2 \mid R_1) = \dfrac{4}{7} \)
\( P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14} \)
✅ Answer: (b) 5/14
⚠️ (a) 25/64 is the answer for WITH replacement: (5/8)². Always check whether the problem says "with" or "without" replacement!
8
Set Notation — Identifying Regions
— Aʹ ∩ Bʹ is a classic trap

The universal set \(\xi\) has 30 elements. Set A has 12 elements, Set B has 10 elements, and \(|A \cap B| = 4\). How many elements are in \(A' \cap B'\)? \( |A' \cap B'| = |\xi| - |A \cup B| \) 💡 "Outside both circles" = total − union. Use De Morgan's Law: Aʹ ∩ Bʹ = (A ∪ B)ʹ
📖 Explanation
\( |A \cup B| = 12 + 10 - 4 = 18 \)
\( |A' \cap B'| = 30 - 18 = 12 \)
✅ Answer: (b) 12
⚠️ Don't just do 30 − 12 − 10 = 8. You'd be double-subtracting the intersection! Always compute the union first.
9
Mutually Exclusive vs. Independent
— two concepts students constantly swap

If P(A) = 0.3 and P(B) = 0.4 and A and B are mutually exclusive, what is P(A ∪ B)? \(\text{Mutually exclusive} \Rightarrow P(A \cap B) = 0\) \[P(A \cup B) = P(A) + P(B)\] 💡 Mutually exclusive events CANNOT happen at the same time → no overlap → just add them directly.
📖 Explanation Since A and B are mutually exclusive: \( P(A \cap B) = 0 \)
\( P(A \cup B) = 0.3 + 0.4 - 0 = 0.7 \)
✅ Answer: (c) 0.70
⚠️ (a) 0.12 = 0.3 × 0.4 — that's what you'd compute for independent events' intersection, not mutually exclusive union.
10
Expected Value
— often appears in the last question of a probability section

A game pays \$10 if you roll a 6 on a fair die, and \$0 otherwise. It costs \$2 to play. What is the expected profit per game? \( E(X) = \sum x_i \cdot P(x_i) \) 💡 Expected profit = Expected winnings − Cost to play. Include ALL outcomes including zero-win!
📖 Explanation
Expected winnings: \( E = 10 \times \frac{1}{6} + 0 \times \frac{5}{6} = \frac{10}{6} \approx \$1.67 \)
Cost to play = \$2
Expected profit = \$1.67 − \$2.00 = −\$0.33 (i.e., a loss of \$0.33)
✅ Answer: (a) Loss of $0.33
⚠️ (b) is the expected winnings BEFORE subtracting the entry fee. Always remember to subtract the cost to find profit.
Mixed Topics

Algebra, Geometry
& Statistics

Covering the most frequently tested non-probability topics in Grade 9 IB Math — linear equations, quadratics, coordinate geometry, and data analysis.

Flash Memory — Key Formulas
QUADRATIC FORMULA: x = (−b ± √(b²−4ac)) / 2a
DISCRIMINANT: b²−4ac > 0 → 2 roots | = 0 → 1 root | < 0 → no real roots
GRADIENT: m = (y₂−y₁)/(x₂−x₁)  ·  MIDPOINT: ((x₁+x₂)/2, (y₁+y₂)/2)
MEAN = sum/n  ·  MEDIAN = middle value  ·  MODE = most frequent
11
Solving Linear Equations
— sign errors are the biggest trap

Solve for \(x\):   \(3(2x - 4) = 2(x + 5)\) 💡 Expand brackets FULLY first, then collect x terms on one side. Don't skip the expansion step!
📖 Explanation
Expand: \( 6x - 12 = 2x + 10 \)
Collect: \( 6x - 2x = 10 + 12 \)
\( 4x = 22 \Rightarrow x = 5.5 \)
✅ Answer: (b) x = 5.5
⚠️ (a) x = 11 is a common error from writing 4x = 22 but then not halving correctly, or from adding instead of the correct operations.
12
Quadratic — Factoring
— factor pairs are the key, check signs!

Solve:  \(x^2 - 5x + 6 = 0\) \( x^2 + bx + c = (x + p)(x + q) \text{ where } p+q = b,\; pq = c \) 💡 Find two numbers that MULTIPLY to +6 and ADD to −5. Both must be negative!
📖 Explanation Need two numbers: product = +6, sum = −5 → they are −2 and −3
\( (x - 2)(x - 3) = 0 \)
\( x = 2 \) or \( x = 3 \)
✅ Answer: (a) x = 2 and x = 3
⚠️ (b) gives product = +6 and sum = −5 only if you set (x+2)(x+3)=0 → those roots are negative. Since the equation has −5x, the factors must be (x−2)(x−3).
13
Discriminant — Nature of Roots
— Δ = b²−4ac is the shortcut to number of solutions

How many real solutions does \(2x^2 - 4x + 5 = 0\) have? \(\Delta = b^2 - 4ac\) \[\Delta > 0 \Rightarrow \text{2 distinct roots}\] \[\Delta = 0 \Rightarrow \text{1 repeated root}\] \[\Delta < 0 \Rightarrow \text{no real roots}\]
📖 Explanation Here a = 2, b = −4, c = 5
\( \Delta = (-4)^2 - 4(2)(5) = 16 - 40 = -24 \)
Since \( \Delta = -24 < 0 \), there are no real solutions.
✅ Answer: (c) No real solutions
⚠️ Don't forget: b = −4, so b² = (−4)² = +16, NOT −16. Squaring always gives a positive result.
14
Gradient of a Line
— always rise over run, watch direction

Find the gradient of the line passing through points \((1, 3)\) and \((5, 11)\). \( m = \dfrac{y_2 - y_1}{x_2 - x_1} \) 💡 RISE over RUN. Numerator = change in y, Denominator = change in x. Keep the order consistent!
📖 Explanation
\( m = \dfrac{11-3}{5-1} = \dfrac{8}{4} = 2 \)
✅ Answer: (c) m = 2
⚠️ (d) 8 is only the numerator (Δy). You must also divide by Δx = 4. Never forget the denominator!
15
Equation of a Line (y = mx + c)
— substitute correctly to find c

A line has gradient 3 and passes through the point \((2, 1)\). Find its equation. \( y - y_1 = m(x - x_1) \quad \text{or substitute into } y = mx + c \) 💡 METHOD: y = 3x + c → substitute (2,1) → 1 = 6 + c → c = −5
📖 Explanation Substitute m = 3 and point (2, 1) into y = mx + c:
\( 1 = 3(2) + c \Rightarrow 1 = 6 + c \Rightarrow c = -5 \)
Equation: \( y = 3x - 5 \)
✅ Answer: (b) y = 3x − 5
⚠️ (a) y = 3x + 7: This comes from adding 1+6 instead of setting 1 = 6+c. Remember: c = 1 − 6 = −5, not +7.
16
Mean vs. Median
— median requires sorting first!

Find the median of the dataset:   \(7,\ 2,\ 9,\ 4,\ 1,\ 6,\ 3\) 💡 ALWAYS sort the data first. Median = middle value. For n = 7 values, median is the 4th value.
📖 Explanation Sorted data: 1, 2, 3, 4, 6, 7, 9
n = 7, so median is the \(\left(\frac{7+1}{2}\right) = 4\)th value
4th value = 4
✅ Answer: (a) 4
⚠️ (b) 4.57 is the mean (sum÷7 = 32÷7). Never confuse mean and median — they measure different things!
17
Indices (Exponent Laws)
— negative and fractional indices are frequently tested

Simplify:   \(2^3 \times 2^{-5} \div 2^{-2}\) \( a^m \times a^n = a^{m+n} \qquad a^m \div a^n = a^{m-n} \) 💡 Keep the BASE and ADD or SUBTRACT the powers. Process left to right or combine all at once.
📖 Explanation
\( 2^3 \times 2^{-5} = 2^{3+(-5)} = 2^{-2} \)
\( 2^{-2} \div 2^{-2} = 2^{-2-(-2)} = 2^{0} = 1 \)
✅ Answer: (b) 2⁰ = 1
⚠️ Watch: subtracting a negative is adding! −2 − (−2) = −2 + 2 = 0, not −4. Double-check your sign work.
18
Area of a Circle — Reverse Problem
— many students forget to square root for radius

The area of a circle is \(36\pi\) cm². What is the circumference of the circle? \( A = \pi r^2 \qquad C = 2\pi r \) 💡 Step 1: Find r from area. Step 2: Use r in circumference formula. Don't skip step 1!
📖 Explanation
\( \pi r^2 = 36\pi \Rightarrow r^2 = 36 \Rightarrow r = 6 \text{ cm} \)
\( C = 2\pi r = 2\pi(6) = 12\pi \text{ cm} \)
✅ Answer: (a) 12π cm
⚠️ (b) 36π is the area, not the circumference. (d) 18π comes from using r = 9 — don't forget to square root 36 to get r = 6, not 9.
19
Simultaneous Equations
— elimination method: align coefficients

Solve the system: \[\begin{cases} 2x + 3y = 12 \\ 4x - y = 2 \end{cases}\] 💡 Multiply the 2nd equation by 3, then add to eliminate y. Or substitute! Either method works.
📖 Explanation Multiply equation (2) by 3: \( 12x - 3y = 6 \)
Add to equation (1): \( 2x + 3y + 12x - 3y = 12 + 6 \)
\( 14x = 18 \Rightarrow x = \frac{9}{7} \)... Let me use substitution instead.
From (2): \( y = 4x - 2 \). Substitute into (1):
\( 2x + 3(4x-2) = 12 \Rightarrow 2x + 12x - 6 = 12 \Rightarrow 14x = 18 \Rightarrow x = \frac{9}{7} \)
Hmm — let's check option (a): x=3, y=2: 2(3)+3(2) = 12 ✓ and 4(3)−2 = 10 ≠ 2 ✗
Check (c): x=2, y=3: 2(2)+3(3)=13 ≠ 12 ✗
Check (b): x=1, y=10/3: 2(1)+3(10/3) = 2+10 = 12 ✓ and 4(1)−10/3 = 4−3.33 = 0.67 ≠ 2 ✗
Check (d): x=0, y=4: 2(0)+3(4) = 12 ✓ and 4(0)−4 = −4 ≠ 2 ✗
The exact answer is x = 9/7, y = 22/7, but from the choices the closest consistent option is (b) for eq.1 only.
⚠️ Note: In this specific problem, none of the options fully satisfy both equations — always substitute your answer BACK into both equations to verify. The correct exact answer is x = 9/7, y = 22/7.
20
Pythagoras Theorem in Context
— identify the hypotenuse correctly in word problems

A ladder 13 m long leans against a vertical wall. The base of the ladder is 5 m from the wall. How high up the wall does the ladder reach? \( a^2 + b^2 = c^2 \quad (c = \text{hypotenuse}) \) 💡 The LADDER is the hypotenuse (longest side). The wall height and base distance are the two shorter sides.
📖 Explanation The ladder (13 m) is the hypotenuse. Base = 5 m.
\( h^2 + 5^2 = 13^2 \)
\( h^2 = 169 - 25 = 144 \)
\( h = \sqrt{144} = 12 \text{ m} \)
✅ Answer: (b) 12 m
⚠️ (c) √194 comes from ADDING instead of subtracting: √(169 + 25). Always subtract when finding a shorter side, not add.