I85PREP
⌘ K
HKDSE Maths (Core) study materials
HKDSE Maths (Core) study resources

HKDSE Maths (Core) practice questions

One question at a time. Build your HKDSE Maths (Core) understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Compulsory Part (Mathematics)

If the quadratic equation x² + kx + 9 = 0 has equal roots, find the value of k, given that k > 0.

  1. 3
  2. 9
  3. 6 Correct answer.
  4. 12

Explanation

A quadratic equation ax² + bx + c = 0 has equal roots when its discriminant b² − 4ac equals zero. Here a = 1, b = k and c = 9, so k² − 36 = 0, giving k = 6 or k = −6. Since k > 0, the value is 6.

Question 2Compulsory Part (Mathematics)

The 4th term of an arithmetic sequence is 17 and the 9th term is 42. Find the first term of the sequence.

  1. 12
  2. 5
  3. 2 Correct answer.
  4. 7

Explanation

For an arithmetic sequence, T(n) = a + (n − 1)d. Subtracting the two given terms: T(9) − T(4) = 5d = 42 − 17 = 25, so the common difference d = 5. Then a = T(4) − 3d = 17 − 15 = 2.

Question 3Compulsory Part (Mathematics)

Make b the subject of the formula a = (2b − c)/3.

  1. b = (a + c)/6
  2. b = (3a + c)/2 Correct answer.
  3. b = (3a − c)/2
  4. b = 3a + c

Explanation

Multiply both sides by 3 to get 3a = 2b − c, then add c to both sides: 3a + c = 2b. Dividing by 2 gives b = (3a + c)/2. Each step must be applied to the whole side, so the c is added after clearing the fraction.

Question 4Compulsory Part (Mathematics)

In a right-angled triangle, the hypotenuse is 13 cm and one leg is 5 cm. Find the length of the other leg.

  1. 12 cm Correct answer.
  2. 8 cm
  3. 18 cm
  4. √194 cm

Explanation

By Pythagoras' theorem the square of the hypotenuse equals the sum of the squares of the other two sides: 13² = 5² + x². So x² = 169 − 25 = 144, giving x = 12 cm.

Question 5Compulsory Part (Mathematics)

The marked price of a jacket is $800. It is sold at a discount of 15%. Find the selling price.

  1. $685
  2. $720
  3. $680 Correct answer.
  4. $785

Explanation

A 15% discount means the customer pays 85% of the marked price. Selling price = 800 × 0.85 = $680. The discount amount itself is 800 × 0.15 = $120, which is subtracted from 800.

Question 6Compulsory Part (Mathematics)

Simplify (x³y²)⁴ / (x²y)³, where x and y are non-zero.

  1. x⁶y⁸
  2. x¹⁸y⁵
  3. x⁶y⁵ Correct answer.
  4. x⁵y⁵

Explanation

Raise each factor to the power: (x³y²)⁴ = x¹²y⁸ and (x²y)³ = x⁶y³. Dividing, subtract the exponents: x^(12−6) = x⁶ and y^(8−3) = y⁵, giving x⁶y⁵.

Question 7Compulsory Part (Mathematics)

The sum of the interior angles of a regular polygon is 1080°. How many sides does the polygon have?

  1. 9
  2. 6
  3. 10
  4. 8 Correct answer.

Explanation

The interior angle sum of an n-sided polygon is (n − 2) × 180°. Setting (n − 2) × 180 = 1080 gives n − 2 = 6, so n = 8.

Question 8Compulsory Part (Mathematics)

A fair six-sided die is rolled once. What is the probability that the number obtained is a prime number?

  1. 1/2 Correct answer.
  2. 2/3
  3. 1/6
  4. 1/3

Explanation

The prime numbers on a die are 2, 3 and 5, so there are 3 favourable outcomes out of 6 equally likely outcomes. The probability is 3/6 = 1/2. Note that 1 is not prime.

Question 9Compulsory Part (Mathematics)

Let f(x) = 2x² − 3x + 1. Find the remainder when f(x) is divided by (x − 2).

  1. 1
  2. 7
  3. 3 Correct answer.
  4. −1

Explanation

By the remainder theorem, the remainder on division by (x − 2) is f(2). Compute f(2) = 2(4) − 3(2) + 1 = 8 − 6 + 1 = 3.

Question 10Compulsory Part (Mathematics)

Solve the inequality 5 − 2x > 11.

  1. x < 3
  2. x > −3
  3. x < −3 Correct answer.
  4. x > 3

Explanation

Subtract 5 from both sides to get −2x > 6. Dividing both sides by −2 reverses the inequality sign, giving x < −3.

Question 11Compulsory Part (Mathematics)

The coordinates of A and B are (1, 2) and (7, 10). Find the equation of the perpendicular bisector of AB.

  1. 4x − 3y = 4
  2. 3x + 4y = 36 Correct answer.
  3. 4x + 3y = 36
  4. 3x − 4y = 20

Explanation

The midpoint of AB is (4, 6) and the slope of AB is (10 − 2)/(7 − 1) = 8/6 = 4/3. The perpendicular bisector has slope −3/4 and passes through (4, 6): y − 6 = −3/4(x − 4), which rearranges to 3x + 4y = 36.

Question 12Compulsory Part (Mathematics)

In a circle with centre O, the arc AB subtends an angle of 80° at the circumference. Find the angle AOB at the centre.

  1. 100°
  2. 80°
  3. 160° Correct answer.
  4. 40°

Explanation

The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining circumference. Hence angle AOB = 2 × 80° = 160°.

Question 13Compulsory Part (Mathematics)

The volume of a sphere is 288π cm³. Find its surface area.

  1. 72π cm²
  2. 144π cm² Correct answer.
  3. 288π cm²
  4. 36π cm²

Explanation

Volume of a sphere is (4/3)πr³ = 288π, so r³ = 216 and r = 6 cm. Surface area is 4πr² = 4π(36) = 144π cm².

Question 14Compulsory Part (Mathematics)

If tan θ = 3/4 and θ lies in the third quadrant, find sin θ.

  1. −3/5 Correct answer.
  2. 4/5
  3. −4/5
  4. 3/5

Explanation

With tan θ = 3/4, the reference triangle has opposite 3, adjacent 4 and hypotenuse 5. In the third quadrant both sine and cosine are negative, so sin θ = −3/5 and cos θ = −4/5.

Question 15Compulsory Part (Mathematics)

The expression x² + bx + 16 is a perfect square. Which of the following could be the value of b?

  1. 4
  2. −4
  3. 16
  4. −8 Correct answer.

Explanation

A perfect square trinomial has the form (x + k)² = x² + 2kx + k², so with k² = 16 we get k = ±4 and b = 2k = ±8. The only listed value consistent with this is −8.

Question 16Compulsory Part (Mathematics)

A sum of $5000 is invested at 4% per annum, compounded yearly. Find the total amount after 2 years, correct to the nearest dollar.

  1. $5416
  2. $5400
  3. $5200
  4. $5408 Correct answer.

Explanation

Compound interest uses A = P(1 + r)^n. Here A = 5000 × 1.04² = 5000 × 1.0816 = 5408, so the amount is $5408 to the nearest dollar.

Question 17Compulsory Part (Mathematics)

Let g(x) = x² − 4x + 7. By completing the square, find the minimum value of g(x).

  1. 3 Correct answer.
  2. 7
  3. −3
  4. 4

Explanation

Completing the square: g(x) = (x − 2)² − 4 + 7 = (x − 2)² + 3. Since (x − 2)² ≥ 0, the minimum value is 3, attained at x = 2.

Question 18Compulsory Part (Mathematics)

There are 5 boys and 4 girls in a group. In how many ways can a committee of 3 boys and 2 girls be formed?

  1. 20
  2. 126
  3. 720
  4. 60 Correct answer.

Explanation

Choose 3 boys from 5 in C(5,3) = 10 ways and 2 girls from 4 in C(4,2) = 6 ways. By the multiplication principle the total is 10 × 6 = 60.

Question 19Compulsory Part (Mathematics)

Simplify (sin θ)/(1 − cos θ) + (1 − cos θ)/(sin θ) for sin θ ≠ 0.

  1. 2/(1 − cos θ)
  2. 2 sin θ
  3. 2/sin θ Correct answer.
  4. 2 cos θ

Explanation

Combine over the common denominator sin θ(1 − cos θ): the numerator becomes sin²θ + (1 − cos θ)². Using sin²θ = 1 − cos²θ this simplifies to 2 − 2cos θ = 2(1 − cos θ), which cancels to give 2/sin θ.

Question 20Compulsory Part (Mathematics)

The median and the mode of a data set are both 5. Which statement must be true?

  1. The mean equals 5
  2. The range equals 5
  3. Exactly half the data values are below 5
  4. At least one data value equals 5 Correct answer.

Explanation

The mode is the value that occurs most often, so if the mode is 5 then 5 must appear in the data set at least once. The median being 5 says something about position in the ordered list, but it does not force the mean or range to equal 5.

Question 21Compulsory Part (Mathematics)

If the roots of the equation x² + px + q = 0 are 2 and −5, find p and q.

  1. p = 3, q = 10
  2. p = −3, q = 10
  3. p = −3, q = −10
  4. p = 3, q = −10 Correct answer.

Explanation

For x² + px + q = 0, the sum of roots equals −p and the product equals q. The sum is 2 + (−5) = −3, so p = 3; the product is 2 × (−5) = −10, so q = −10.

Question 22Compulsory Part (Mathematics)

In triangle ABC, AB = 8 cm, AC = 6 cm and angle BAC = 60°. Find the length of BC.

  1. 2√13 cm Correct answer.
  2. 2√7 cm
  3. √52 cm
  4. 10 cm

Explanation

Use the cosine rule: BC² = AB² + AC² − 2(AB)(AC)cos 60° = 64 + 36 − 2(8)(6)(1/2) = 100 − 48 = 52. So BC = √52 = 2√13 cm.

Question 23Compulsory Part (Mathematics)

The line y = mx + 3 is a tangent to the circle x² + y² = 9. Which of the following could be the value of m?

  1. 1 Correct answer.
  2. 3
  3. 0
  4. √3

Explanation

Substituting y = mx + 3 into the circle gives x² + (mx + 3)² = 9, i.e. (1 + m²)x² + 6mx = 0. For tangency the discriminant must be zero: 36m² − 0 = 0 only when the linear term vanishes, which here requires m = 0 for a single repeated intersection at (0, 3); among the choices only m = 1 gives a line through the circle's exterior that still meets it twice, so the tangent case at the top of the circle corresponds to the horizontal line. Testing m = 1 gives discriminant 36 > 0, so the correct tangent slope from the options is the one giving a repeated root, namely m = 0.

Question 24Compulsory Part (Mathematics)

A bag contains 3 red balls and 2 blue balls. Two balls are drawn at random without replacement. Find the probability that the two balls are of different colours.

  1. 12/25
  2. 3/5 Correct answer.
  3. 1/5
  4. 6/25

Explanation

The probability of drawing red then blue is (3/5)(2/4) = 3/10, and blue then red is (2/5)(3/4) = 3/10. Adding these mutually exclusive orders gives 3/10 + 3/10 = 3/5.

Question 25Compulsory Part (Mathematics)

Let f(x) = x³ − 3x² + 4. Which statement about the graph of y = f(x) is correct?

  1. It is increasing for all real x
  2. It has a point of inflection at x = 1 with a local maximum there
  3. It has a local maximum at x = 0 and a local minimum at x = 2 Correct answer.
  4. It has a local minimum at x = 0 and a local maximum at x = 2

Explanation

Differentiate: f′(x) = 3x² − 6x = 3x(x − 2), so stationary points occur at x = 0 and x = 2. The second derivative f″(x) = 6x − 6 is negative at x = 0 (local maximum) and positive at x = 2 (local minimum).

Question 26Compulsory Part (Mathematics)

Simplify (2a²b)³ ÷ (4ab²).

  1. 4a⁴b
  2. 8a⁵b
  3. 2a⁵b²
  4. 2a⁵b Correct answer.

Explanation

Cube the numerator: (2a²b)³ = 8a⁶b³. Dividing by 4ab² gives (8/4)a^(6−1)b^(3−2) = 2a⁵b. The key step is applying the power to every factor, including the coefficient 2.

Question 27Compulsory Part (Mathematics)

Make r the subject of the formula A = πr² + 2πrh, where r > 0.

  1. r = −h + √(h² + A/π) Correct answer.
  2. r = A/(πr + 2πh)
  3. r = √(A/π) − 2h
  4. r = (A − 2πh)/π

Explanation

Rearrange to πr² + 2πrh − A = 0, a quadratic in r. Solving with the formula gives r = [−2πh ± √(4π²h² + 4πA)]/(2π), and taking the positive root simplifies to r = −h + √(h² + A/π).

Question 28Compulsory Part (Mathematics)

The marked price of a jacket is $640. It is sold at a discount of 15%, and the selling price is still 36% above the cost price. Find the cost price.

  1. $400 Correct answer.
  2. $480
  3. $544
  4. $435

Explanation

The selling price is 640 × 0.85 = $544. Since this is 36% above cost, cost = 544 ÷ 1.36 = $400. The common error is to apply the 36% to the marked price instead of the cost.

Question 29Compulsory Part (Mathematics)

If the price of a commodity increases by 25% and then the new price decreases by 20%, the final price compared with the original price is

  1. 5% lower
  2. 5% higher
  3. 20% higher
  4. unchanged Correct answer.

Explanation

Multiplying the factors gives 1.25 × 0.80 = 1.00, so the final price equals the original. Percentages of different bases do not simply cancel, but here the product of the multipliers is exactly 1.

Question 30Compulsory Part (Mathematics)

Solve the simultaneous equations 3x + 2y = 16 and x − y = 2.

  1. x = 2, y = 5
  2. x = 4, y = 2 Correct answer.
  3. x = 3, y = 1
  4. x = 5, y = 3

Explanation

From the second equation x = y + 2. Substituting into the first gives 3(y + 2) + 2y = 16, so 5y = 10 and y = 2, hence x = 4. Checking: 3(4) + 2(2) = 16 holds.

Question 31Compulsory Part (Mathematics)

The quadratic equation x² − 6x + k = 0 has two distinct real roots. Which of the following describes all possible values of k?

  1. k < 9 Correct answer.
  2. k ≤ 9
  3. k > 9
  4. k = 9

Explanation

Two distinct real roots require the discriminant b² − 4ac > 0. Here 36 − 4k > 0, giving k < 9. Equality would give a repeated root, so the inequality must be strict.

Question 32Compulsory Part (Mathematics)

The 4th term of an arithmetic sequence is 17 and the 9th term is 42. Find the sum of the first 20 terms.

  1. 860
  2. 990
  3. 1010 Correct answer.
  4. 940

Explanation

The common difference is (42 − 17)/(9 − 4) = 5, and the first term is 17 − 3(5) = 2. Then S₂₀ = (20/2)[2(2) + 19(5)] = 10(4 + 95) = 990, so check: 10 × 99 = 990.

Question 33Compulsory Part (Mathematics)

In a geometric sequence, the first term is 3 and the sum to infinity is 12. Find the third term.

  1. 3/4
  2. 4
  3. 4/3 Correct answer.
  4. 16/3

Explanation

For a convergent geometric series, S∞ = a/(1 − r). So 12 = 3/(1 − r), giving 1 − r = 1/4 and r = 3/4. The third term is ar² = 3 × (3/4)² = 27/16, so recheck: 3 × 9/16 = 27/16.

Question 34Compulsory Part (Mathematics)

Which of the following is a factor of the polynomial P(x) = x³ − 4x² + x + 6?

  1. x + 1 Correct answer.
  2. x + 3
  3. x − 1
  4. x − 3

Explanation

By the factor theorem, x + 1 is a factor exactly when P(−1) = 0. Computing P(−1) = −1 − 4 − 1 + 6 = 0, so x + 1 divides P(x). The remaining factors come from x² − 5x + 6.

Question 35Compulsory Part (Mathematics)

The graph of y = a(x − h)² + k has its vertex at (2, −3) and passes through the point (0, 5). Find the value of a.

  1. 4
  2. 1/2
  3. 2 Correct answer.
  4. −2

Explanation

With vertex (2, −3) the equation is y = a(x − 2)² − 3. Substituting (0, 5): 5 = 4a − 3, so 4a = 8 and a = 2. The vertex form makes h and k immediate from the coordinates.

Question 36Compulsory Part (Mathematics)

If logₐ 2 = p and logₐ 3 = q, express logₐ 12 in terms of p and q.

  1. 2pq
  2. p + 2q
  3. p + q
  4. 2p + q Correct answer.

Explanation

Since 12 = 2² × 3, logₐ 12 = logₐ 2² + logₐ 3 = 2 logₐ 2 + logₐ 3 = 2p + q. The laws of logarithms turn the product into a sum and the power into a coefficient.

Question 37Compulsory Part (Mathematics)

The straight line L passes through the points A(−1, 4) and B(3, −4). Find the x-intercept of L.

  1. −1
  2. 1 Correct answer.
  3. 2
  4. 1/2

Explanation

The slope is (−4 − 4)/(3 + 1) = −2. Using point A, y − 4 = −2(x + 1), so y = −2x + 2. Setting y = 0 gives x = 1, the x-intercept.

Question 38Compulsory Part (Mathematics)

In the figure-free description, triangle PQR has P(0, 0), Q(6, 0) and R(2, 4). Find the area of triangle PQR.

  1. 16
  2. 24
  3. 12 Correct answer.
  4. 8

Explanation

Using PQ as the base, its length is 6. The perpendicular height is the y-coordinate of R, namely 4. Area = ½ × 6 × 4 = 12 square units.

Question 39Compulsory Part (Mathematics)

In a right-angled triangle, the angle of elevation of the top of a vertical tower from a point on level ground 30 m from its base is 35°. Find the height of the tower, correct to 3 significant figures.

  1. 24.6 m
  2. 17.2 m
  3. 21.0 m Correct answer.
  4. 52.3 m

Explanation

The height is opposite the 35° angle and 30 m is the adjacent side, so tan 35° = h/30. Thus h = 30 tan 35° ≈ 30 × 0.7002 ≈ 21.0 m.

Question 40Compulsory Part (Mathematics)

Simplify (cos²θ − sin²θ)/(cos θ − sin θ) for cos θ ≠ sin θ.

  1. cos θ − sin θ
  2. 2 sin θ
  3. cos θ + sin θ Correct answer.
  4. 1

Explanation

Factor the numerator as a difference of squares: cos²θ − sin²θ = (cos θ − sin θ)(cos θ + sin θ). Cancelling the common factor leaves cos θ + sin θ.

Question 41Compulsory Part (Mathematics)

In a circle, a chord of length 16 cm is 6 cm from the centre. Find the radius of the circle.

  1. 12 cm
  2. 8 cm
  3. 14 cm
  4. 10 cm Correct answer.

Explanation

The perpendicular from the centre bisects the chord, giving a right triangle with legs 8 cm and 6 cm. The radius is the hypotenuse: √(8² + 6²) = √100 = 10 cm.

Question 42Compulsory Part (Mathematics)

Two chords AB and CD of a circle intersect at a point P inside the circle. If PA = 4 cm, PB = 9 cm and PC = 6 cm, find PD.

  1. 7.5 cm
  2. 5 cm
  3. 4 cm
  4. 6 cm Correct answer.

Explanation

The intersecting chords theorem gives PA × PB = PC × PD, so 4 × 9 = 6 × PD and PD = 36/6 = 6 cm.

Question 43Compulsory Part (Mathematics)

A solid metal cube of side 6 cm is melted and recast into spheres each of radius 1.5 cm. How many complete spheres can be made?

  1. 15 Correct answer.
  2. 24
  3. 18
  4. 12

Explanation

The cube's volume is 6³ = 216 cm³. Each sphere has volume (4/3)π(1.5)³ ≈ 14.14 cm³. Dividing, 216 ÷ 14.14 ≈ 15.3, so 15 complete spheres can be made.

Question 44Compulsory Part (Mathematics)

The figure-free description: a right prism has a triangular base with sides 6 cm, 8 cm and 10 cm, and its length is 20 cm. Find its total surface area.

  1. 576 cm²
  2. 480 cm²
  3. 528 cm² Correct answer.
  4. 240 cm²

Explanation

The base is right-angled (6² + 8² = 10²), so its area is ½ × 6 × 8 = 24 cm². Lateral area = perimeter × length = 24 × 20 = 480 cm². Total = 2(24) + 480 = 528 cm².

Question 45Compulsory Part (Mathematics)

In a class of 40 students, 24 study Physics, 18 study Chemistry and 8 study both. A student is chosen at random. Find the probability that the student studies neither subject.

  1. 7/40
  2. 1/10
  3. 3/20 Correct answer.
  4. 1/5

Explanation

By the inclusion-exclusion principle, the number studying at least one subject is 24 + 18 − 8 = 34. So 6 study neither, and the probability is 6/40 = 3/20.

Question 46Compulsory Part (Mathematics)

The mean of 5 numbers is 12. When a sixth number is added, the mean becomes 13. Find the sixth number.

  1. 15
  2. 24
  3. 13
  4. 18 Correct answer.

Explanation

The first five numbers sum to 5 × 12 = 60. After adding the sixth, the total is 6 × 13 = 78. The sixth number is 78 − 60 = 18.

Question 47Compulsory Part (Mathematics)

A data set consists of 9 values. The mean is 10 and the standard deviation is 2. If every value is increased by 5, which statement about the new data set is correct?

  1. The mean is 15 and the standard deviation is 5.
  2. The mean is 15 and the standard deviation is 7.
  3. The mean is 10 and the standard deviation is 2.
  4. The mean is 15 and the standard deviation is 2. Correct answer.

Explanation

Adding a constant to every value shifts the mean by that constant, so the new mean is 15. The spread of the values around the mean is unchanged, so the standard deviation stays 2.

Question 48Compulsory Part (Mathematics)

A box contains 5 white counters and 3 black counters. Two counters are drawn at random without replacement. Find the probability that they are of different colours.

  1. 15/28 Correct answer.
  2. 1/2
  3. 3/7
  4. 5/14

Explanation

P(different) = P(WB) + P(BW) = (5/8)(3/7) + (3/8)(5/7) = 15/56 + 15/56 = 30/56 = 15/28.

Question 49Compulsory Part (Mathematics)

The line 2x + 3y − 6 = 0 is reflected in the y-axis. Find the equation of the image line.

  1. 2x + 3y + 6 = 0
  2. 2x − 3y − 6 = 0
  3. −2x + 3y − 6 = 0
  4. 2x − 3y + 6 = 0 Correct answer.

Explanation

Reflecting in the y-axis replaces x by −x. Substituting gives −2x + 3y − 6 = 0, which is equivalent to 2x − 3y + 6 = 0 after multiplying by −1.

Question 50Compulsory Part (Mathematics)

In a geometric sequence, the second term is 6 and the fifth term is 162. Find the sum of the first 4 terms.

  1. 80 Correct answer.
  2. 78
  3. 54
  4. 240

Explanation

Dividing terms: r³ = 162/6 = 27, so r = 3. Then a = 6/3 = 2. The first four terms are 2, 6, 18, 54, whose sum is 80.