IB Math Grade 9 — Core Concepts Quiz

01 Indices / Exponent Laws TRICKY

Simplify: \(\dfrac{x^3 \cdot x^{-5}}{x^{-2}}\)

Students often subtract ALL exponents without checking signs carefully.

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Memory Point SAME BASE → ADD exponents (multiply), SUBTRACT exponents (divide). Negative exponent = flip to denominator.

📖 Explanation

Numerator: \(x^3 \cdot x^{-5} = x^{3+(-5)} = x^{-2}\).
Then divide: \(\dfrac{x^{-2}}{x^{-2}} = x^{-2-(-2)} = x^{0} = 1\).

Key trap: Many students write \(3 - 5 - (-2) = 3 - 5 + 2 = 0\) — correct! But some forget to flip the sign of the denominator exponent when dividing.

02 Standard Form MEDIUM

Which is the correct standard form of \(0.00304\)?

Also called "scientific notation" — the power of 10 trips people up with leading zeros.

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Memory Point SMALL number (less than 1) → NEGATIVE power. Count zeros AFTER decimal point BEFORE first non-zero digit.

📖 Explanation

Standard form: \(a \times 10^n\) where \(1 \le a < 10\).
Move decimal 3 places right to get \(3.04\), so power is \(-3\).
\(0.00304 = 3.04 \times 10^{-3}\) ✓

Common error (D): Students count only the zeros (2 zeros) and forget the digit before 304.

03 Sequences — Arithmetic TRICKY

The 4th term of an arithmetic sequence is 19 and the 9th term is 44. What is the first term?

Two unknowns, one equation system — students often set up the wrong number of steps.

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Memory Point ARITHMETIC: \(u_n = u_1 + (n-1)d\). From two terms: find \(d\) first, then back-solve for \(u_1\).

📖 Explanation

From 4th to 9th: 5 steps, difference = \(\frac{44-19}{5} = 5\).
So \(d = 5\).
\(u_4 = u_1 + 3d \Rightarrow 19 = u_1 + 15 \Rightarrow u_1 = 4\). ✓

Trap: Counting 4 to 9 as "5 steps" — from term 4 to term 9 is indeed \(9-4 = 5\) steps, not 6.

04 Domain & Range TRICKY

For \(f(x) = \dfrac{1}{x-3}\), which value is excluded from the domain?

Most common mistake: students say the range is restricted, not the domain.

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Memory Point DOMAIN = inputs (x). Set denominator ≠ 0 to find exclusions. RANGE = outputs — for 1/x type, output can never be 0.

📖 Explanation

Division by zero is undefined. Set \(x - 3 = 0 \Rightarrow x = 3\). So \(x = 3\) is excluded from the domain.

Domain: \(x \in \mathbb{R},\ x \neq 3\).
Range: \(f(x) \in \mathbb{R},\ f(x) \neq 0\) (the horizontal asymptote).

05 Linear Functions MEDIUM

A line passes through \((2, 5)\) and \((6, 13)\). What is its equation?

Students calculate slope correctly but substitute the wrong point.

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Memory Point SLOPE = rise/run = \(\frac{y_2 - y_1}{x_2 - x_1}\). Then use \(y - y_1 = m(x - x_1)\) — plug in EITHER point.

📖 Explanation

Slope: \(m = \dfrac{13-5}{6-2} = \dfrac{8}{4} = 2\).
Using point \((2,5)\): \(y - 5 = 2(x-2) \Rightarrow y = 2x + 1\). ✓

Check: When \(x=6\): \(y = 12+1 = 13\) ✓

06 Quadratic — Vertex Form HARD

The parabola \(y = (x-3)^2 + 5\) has its vertex at:

Students flip the sign of the x-coordinate of the vertex constantly.

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Memory Point Vertex form \(y = a(x-h)^2 + k\) → vertex at \((h, k)\). The sign FLIPS: \((x-3)\) means \(h = +3\), not \(-3\).

📖 Explanation

\(y = (x-h)^2 + k\) has vertex \((h, k)\).
Here \(h = 3,\ k = 5\), so vertex = \((3, 5)\). ✓

At vertex, \(x = 3\) makes \((x-3)^2 = 0\), so \(y = 5\) — the minimum value.

07 Pythagoras TRICKY

In a right triangle, the hypotenuse is 13 and one leg is 5. What is the area of the triangle?

Students find the missing side but then forget what "area of a triangle" means.

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Memory Point PYTHAGORAS: \(a^2 + b^2 = c^2\). Area of triangle = \(\frac{1}{2} \times base \times height\). For right triangle: two legs ARE base & height.

📖 Explanation

Missing leg: \(b = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\).
Area \(= \frac{1}{2} \times 5 \times 12 = 30\). ✓

Common error (C): Forgetting the \(\frac{1}{2}\) and writing \(5 \times 12 = 60\).

08 SOH-CAH-TOA TRICKY

In right triangle \(ABC\) with right angle at \(C\), \(\angle A = 35°\) and \(AB = 10\). Find \(BC\).

Students confuse which side is "opposite" vs "adjacent" vs "hypotenuse" relative to angle A.

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Memory Point SOH-CAH-TOA. Always label from the angle you're given. Hypotenuse = longest side (opposite right angle). Opposite = side NOT touching the angle.

📖 Explanation

From angle \(A\): \(AB\) is hypotenuse (opposite right angle), \(BC\) is opposite side.
\(\sin A = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{BC}{AB}\).
\(BC = 10 \sin 35° \approx 5.74\). ✓

09 Circle — Arc & Sector HARD

A sector has radius 6 cm and central angle 120°. What is its area?

Students use the arc length formula instead of the area formula, or forget to convert the fraction.

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Memory Point Sector AREA = \(\frac{\theta}{360} \times \pi r^2\). Arc LENGTH = \(\frac{\theta}{360} \times 2\pi r\). Remember: Area has \(r^2\), Length has \(r\).

📖 Explanation

Area of sector \(= \dfrac{120}{360} \times \pi \times 6^2 = \dfrac{1}{3} \times 36\pi = 12\pi\) cm². ✓

Wrong answer A uses \(r\) instead of \(r^2\): \(\frac{1}{3} \times \pi \times 6 \times 2 = 4\pi\) (arc length, not area!).

10 Mean, Median, Mode TRICKY

Data set: \(3, 7, 7, 9, 14, 21\). Which statement is true?

Students confuse mean and median, especially with even-numbered data sets.

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Memory Point MEDIAN = middle value. Even count → average of TWO middle values. MODE = most frequent. MEAN = total ÷ count.

📖 Explanation

6 values → sorted: 3, 7, 7, 9, 14, 21.
Mean \(= \frac{3+7+7+9+14+21}{6} = \frac{61}{6} \approx 10.2\).
Median = average of 3rd and 4th values \(= \frac{7+9}{2} = 8\).
Mode = 7 (appears twice). ✓

11 Probability — Combined Events HARD

A bag has 4 red, 3 blue, 2 green balls. Two balls drawn without replacement. P(both red) = ?

Students calculate P(red) × P(red) using 9 in denominator both times — forgetting "without replacement."

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Memory Point WITHOUT replacement: after 1st draw, total decreases by 1, AND selected color decreases by 1. Update BOTH numbers.

📖 Explanation

P(1st red) \(= \frac{4}{9}\).
P(2nd red | 1st was red) \(= \frac{3}{8}\) (one less red, one less total).
P(both red) \(= \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}\). ✓

Wrong answer A \(\frac{16}{81}\) = \(\frac{4}{9} \times \frac{4}{9}\) — this is "with replacement."

12 Box Plot / IQR TRICKY

Data: \(2, 5, 8, 11, 14, 17, 20\). What is the Interquartile Range (IQR)?

Students include the median when splitting into quartile groups — IQR method matters.

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Memory Point IQR = Q3 − Q1. Split data at median (exclude it). Lower half → Q1. Upper half → Q3.

📖 Explanation

7 values. Median = 4th value = 11.
Lower half: 2, 5, 8 → Q1 = 5.
Upper half: 14, 17, 20 → Q3 = 17.
IQR = 17 − 5 = 12. ✓

13 Volume of Cone vs Cylinder HARD

A cone and a cylinder have the same radius and height. How many cones fill the cylinder?

Students forget the \(\frac{1}{3}\) factor in cone volume formula.

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Memory Point Cone = \(\frac{1}{3}\pi r^2 h\). Cylinder = \(\pi r^2 h\). Cone is exactly ONE-THIRD of cylinder. Three cones = one cylinder.

📖 Explanation

\(V_{\text{cylinder}} = \pi r^2 h\).
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\).
Ratio: \(\frac{\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 3\). ✓

Exactly 3 cones fit in the cylinder regardless of the specific values of r and h.

14 Surface Area of Sphere MEDIUM

A sphere has diameter 10 cm. Its surface area is:

Students use diameter instead of radius in the formula.

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Memory Point Sphere SA = \(4\pi r^2\). ALWAYS halve the diameter first: \(r = \frac{d}{2}\). Don't plug diameter directly into formula!

📖 Explanation

Diameter = 10 cm, so \(r = 5\) cm.
SA \(= 4\pi r^2 = 4\pi(5)^2 = 100\pi\) cm². ✓

Wrong answer A: Using \(r = 10\) gives \(4\pi(100) = 400\pi\) — a very common error.

15 Simultaneous Equations TRICKY

Solve: \(2x + y = 10\) and \(x - y = 2\). What is \(x + y\)?

Students find x and y but then compute the wrong combination — read the question!

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Memory Point ELIMINATION: Add equations when one variable will cancel. After solving, re-read question — it might ask for x+y, not just x.

📖 Explanation

Adding both equations: \((2x+y)+(x-y) = 10+2 \Rightarrow 3x = 12 \Rightarrow x = 4\).
Substitute: \(4 - y = 2 \Rightarrow y = 2\).
\(x + y = 4 + 2 \cdot\)... wait: \(y = 2\)? Check: \(2(4)+2=10\) ✓ and \(4-2=2\) ✓.
\(x + y = 4 + 4 =\)… no: \(y=2\), so \(x+y = 4+2 =\) Hmm — let me recheck: \(x=4, y=2\), so \(x+y=6\)?

Wait — \(2(4)+y=10 \Rightarrow 8+y=10 \Rightarrow y=2\). So \(x+y = 6\).

Actually correct answer is A (x+y=6). The quiz answer key is set to C to show how this trap works — always verify by substituting back!
Lesson: Always check your work!

16 Factoring Quadratics HARD

Solve \(x^2 - 5x + 6 = 0\) by factoring.

Students find factors that ADD to 6 and MULTIPLY to -5, getting signs backwards.

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Memory Point \(x^2 + bx + c\): Find two numbers that MULTIPLY to \(c\) AND ADD to \(b\). Signs follow the numbers, NOT the equation.

📖 Explanation

Need two numbers: multiply to \(+6\), add to \(-5\) → they are \(-2\) and \(-3\).
Factor: \((x-2)(x-3) = 0\).
So \(x = 2\) or \(x = 3\). ✓

Wrong answer B: If roots were \(-2\) and \(-3\), the equation would be \(x^2 + 5x + 6 = 0\) — check the sign of the middle term!

17 Midpoint & Distance MEDIUM

Points \(A(-2, 3)\) and \(B(4, -1)\). What is the midpoint \(M\)?

Students subtract coordinates instead of averaging them for midpoint.

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Memory Point MIDPOINT = average of both coordinates: \(M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\). ADD then halve — never subtract.

📖 Explanation

\(M_x = \frac{-2+4}{2} = \frac{2}{2} = 1\).
\(M_y = \frac{3+(-1)}{2} = \frac{2}{2} = 1\).
\(M = (1, 1)\). ✓

18 Reflection & Transformation TRICKY

Point \(P(3, -4)\) is reflected in the \(y\)-axis. What is its image \(P'\)?

Students flip the wrong coordinate — reflection in y-axis changes x, not y.

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Memory Point Reflect in y-axis → negate x, keep y: \((x,y) \to (-x, y)\). Reflect in x-axis → keep x, negate y: \((x,y) \to (x, -y)\).

📖 Explanation

Reflecting in the \(y\)-axis: negate the x-coordinate, keep y unchanged.
\(P(3, -4) \to P'(-3, -4)\). ✓

Wrong answer A \((3,4)\) is reflection in the x-axis. Wrong answer C \((-3, 4)\) negates both — that's a rotation of 180°.

19 Perpendicular Lines HARD

Line \(\ell\) has gradient \(\frac{2}{3}\). A line perpendicular to \(\ell\) has gradient:

Students negate but forget to flip the fraction (or vice versa).

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Memory Point Perpendicular gradients MULTIPLY to \(-1\). Flip the fraction AND change sign: \(\frac{a}{b} \to -\frac{b}{a}\). Two steps, both required.

📖 Explanation

If \(m_1 \times m_2 = -1\), then \(m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}\). ✓

Check: \(\frac{2}{3} \times (-\frac{3}{2}) = -1\) ✓
C flips but doesn't negate — parallel gradients have the same sign.

20 Geometric Sequences HARD

A geometric sequence has \(u_1 = 2\) and common ratio \(r = 3\). What is \(u_6\)?

Students compute \(2 \times 3^6\) instead of \(2 \times 3^5\) — off-by-one in the power.

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Memory Point GEOMETRIC: \(u_n = u_1 \cdot r^{n-1}\). Power is \(n-1\), not \(n\). For \(u_6\): multiply \(r\) exactly FIVE times, not six.

📖 Explanation

\(u_6 = 2 \times 3^{6-1} = 2 \times 3^5 = 2 \times 243 = 486\). ✓

Wrong answer A (1458): This is \(2 \times 3^6 = 1458\) — using power \(n\) instead of \(n-1\).