CSCA Math Questions: 10 Practice Questions With Worked Answers

Q1. Let M = {x | x² − 4x + 3 < 0} and N = {x | 2 ≤ x ≤ 5}. Then M ∩ N = ? Options: A. {x | 1 < x ≤ 5}; B. {x | 2 ≤ x < 3}; C. {x | 1 < x < 2}; D. {x | 3 < x ≤ 5}

Answer: B. Factor the quadratic: x² − 4x + 3 = (x − 1)(x − 3), which is negative between the roots, so M = {x | 1 < x < 3}. Now overlap with N, which runs from 2 to 5 inclusive. The overlap starts at 2 (included, because N includes 2) and stops just before 3 (excluded, because M excludes 3). So M ∩ N = {x | 2 ≤ x < 3}. Option A is M ∪ N, the answer you get if you mix up intersection and union.

Q2. The solution set of the inequality (x − 2)/(x + 1) ≤ 0 is: Options: A. −1 < x ≤ 2; B. −1 ≤ x ≤ 2; C. x < −1 or x ≥ 2; D. −1 < x < 2

Answer: A. A fraction is at most 0 when the top and bottom have opposite signs, or when the top equals 0. The critical points are x = 2 (top is 0) and x = −1 (bottom is 0). Test one value in each interval: x = 0 gives −2, which works; x = 3 gives 1/4, which fails; x = −2 gives 4, which fails. Include x = 2, because the fraction equals 0 there, but exclude x = −1, because you cannot divide by zero. So −1 < x ≤ 2. Option B is the most common wrong answer.

Q3. The domain of f(x) = √(x − 1) + log₂(3 − x) is: Options: A. (1, 3); B. [1, 3]; C. (1, 3]; D. [1, 3)

Answer: D. Each part of the function sets its own condition. The square root needs x − 1 ≥ 0, so x ≥ 1, and 1 itself is allowed because the square root of 0 is 0. The logarithm needs a strictly positive input, so 3 − x > 0, which gives x < 3. Together: 1 ≤ x < 3, written [1, 3). Remember the rule: square roots allow zero, logarithms do not.

Q4. What is the value of 27^(1/3) + log₂ 16 − log₁₀ 100? Options: A. 3; B. 4; C. 5; D. 7

Answer: C. Work out each term separately. 27^(1/3) is the cube root of 27, which is 3 because 3³ = 27. log₂ 16 = 4 because 2⁴ = 16. log₁₀ 100 = 2 because 10² = 100. So 3 + 4 − 2 = 5. Practice spotting exact powers of 2, 3, 5 and 10 quickly, because you will be working without a calculator.

Q5. If sin α = 3/5 and α is in the second quadrant (90° < α < 180°), then tan α = ? Options: A. 3/4; B. −3/4; C. 4/3; D. −4/3

Answer: B. Use sin²α + cos²α = 1: cos²α = 1 − 9/25 = 16/25, so cos α = 4/5 or −4/5. In the second quadrant cosine is negative, so cos α = −4/5. Then tan α = sin α ÷ cos α = (3/5) ÷ (−4/5) = −3/4. Option A is what you get if you forget the sign. Option D comes from dividing the wrong way round, cos α ÷ sin α.

Q6. In an arithmetic sequence {aₙ}, a₂ = 5 and a₅ = 14. What is the sum of the first 10 terms, S₁₀? Options: A. 145; B. 150; C. 155; D. 310

Answer: C. In an arithmetic sequence a₅ − a₂ = 3d, so 3d = 14 − 5 = 9 and d = 3. Then a₁ = a₂ − d = 2. Use Sₙ = n(2a₁ + (n − 1)d)/2: S₁₀ = 10 × (2 × 2 + 9 × 3)/2 = 10 × 31/2 = 155. Check by listing the terms: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, which add up to 155. Option D, 310, is what you get if you forget to divide by 2.

Q7. What is the equation of the tangent line to the curve y = x³ − 2x at the point where x = 1? Options: A. y = x − 2; B. y = 3x − 4; C. y = x + 2; D. y = −x − 2

Answer: A. First find the point: when x = 1, y = 1 − 2 = −1, so the point is (1, −1). Then find the slope: the derivative is y′ = 3x² − 2, so at x = 1 the slope is 3 − 2 = 1. The line through (1, −1) with slope 1 is y + 1 = x − 1, which simplifies to y = x − 2. Option B also passes through (1, −1) but uses slope 3, the result of differentiating x³ and forgetting the −2x term.

Q8. What are the coordinates of the foci of the ellipse x²/25 + y²/16 = 1? Options: A. (±5, 0); B. (0, ±3); C. (±√41, 0); D. (±3, 0)

Answer: D. For an ellipse x²/a² + y²/b² = 1 with a > b, the foci lie on the x-axis at (±c, 0), where c² = a² − b². Here a² = 25 and b² = 16, so c² = 9 and c = 3. The foci are (±3, 0). Option A gives the ends of the major axis (the vertices), not the foci. Option C comes from adding a² + b², which is the rule for a hyperbola, not an ellipse.

Q9. If i is the imaginary unit (i² = −1), then (1 + i)/(1 − i) = ? Options: A. i; B. −i; C. 1; D. 2i

Answer: A. Multiply the top and the bottom by the conjugate of the bottom, 1 + i. The new top is (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i. The new bottom is (1 − i)(1 + i) = 1 − i² = 1 + 1 = 2. So the result is 2i ÷ 2 = i. Quick check: i × (1 − i) = i − i² = i + 1, which matches the original top.

Q10. Two fair six-sided dice are rolled. What is the probability that the sum of the two numbers is 7? Options: A. 1/12; B. 1/6; C. 7/36; D. 1/9

Answer: B. Two dice give 6 × 6 = 36 equally likely outcomes. The ordered pairs that sum to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is six outcomes. The probability is 6/36 = 1/6. Option A, 1/12, comes from counting only the three unordered pairs and dividing by 36. Option C is a guess based on the number 7.

Are these official CSCA past paper questions?

No. They are original CSCA-style practice questions written by CSCA.ai to match the topics in the official 2025 mathematics syllabus. None of them is taken from a real exam.

How many questions are in the CSCA math exam?

There are 48 single-choice questions in 60 minutes, scored from 0 to 100. You can take it in Chinese or English.

Can I use a calculator for CSCA math?

Plan to work without one. A calculator is not on the list of permitted desk items in the official home-based test rules.

What topics are in CSCA math?

The official syllabus has four modules: sets and inequalities, functions (including trigonometry, sequences and basic derivatives), geometry and algebra (including analytic geometry, vectors and complex numbers), and probability and statistics.

Is CSCA math hard?

The topics match senior high school mathematics in most countries. The main difficulty is speed, about 75 seconds per question, and working without a calculator.

What is a good CSCA math score?

There is no official pass mark. Each university sets its own requirement, and competitive programs and scholarships generally expect higher scores.