Grade 10 — Practice Paper 3

Challenge Paper | 70 Marks | 90 Minutes | Calculator Allowed | IGCSE Extended Level

🏆 Challenge 📋 70 marks | ⏱ 90 minutes | ✅ Calculator

90:00

Section A — Number, Algebra & Functions (24 marks)

Q1 4 marks

The force F between two magnets is inversely proportional to the square of the distance d between them. When d = 3 cm, F = 40 N.

(a) Find F when d = 6 cm.

(b) Find d when F = 160 N. Give the positive value.

Q1a: F =N

Q1b: d =cm

Q1c lower bound:N

Q1c upper bound:N

Q2 4 marks

Simplify fully: (2x² + 5x − 3) / (4x² − 1)
Substitute x = 2 into your simplified expression. Enter the result as a fraction — enter as a decimal.

Also: Solve the equation 3/(x+2) − 2/(x−1) = 1. Enter the positive root (if two roots exist, enter the larger positive one).

Q2 simplify at x=2:

Q2 solve — positive root ≈

Q3 4 marks

f(x) = 2x + 3 and g(x) = x² − 1.

(a) Find fg(x). State the coefficient of x² in fg(x).

(b) Solve fg(x) = 17. Enter the positive solution.

(d) Find the value of x for which f(x) = g(x). Enter the larger root.

Q3a: coeff of x² =

Q3b: positive x ≈

Q3c: f⁻¹(−5) =

Q3d: larger root ≈

Q4 4 marks

Write 3x² − 12x + 7 in the form a(x − p)² + q.

(a) State the value of q (the minimum value).

(b) Hence solve 3x² − 12x + 7 = 0. Enter the larger root to 2 d.p.

(c) The graph of y = 3x² − 12x + 7 is translated by vector (2, −3). Write the equation of the new graph in the form y = a(x − h)² + k. Enter k.

Q4a: q =

Q4b: larger root ≈

Q4c: k =

Q5 4 marks

A geometric series has first term 8 and common ratio r where |r| < 1.

(a) The sum to infinity is 20. Find r.

(b) Find the sum of the first 4 terms. Give your answer as a fraction — enter as exact decimal if possible, else 3 s.f.

Q5a: r =

Q5b: S₄ =

Q5c: n =

Q6 4 marks

Prove that n³ − n is always divisible by 6 for all positive integers n.

💡 Hint: factorise n³−n, then consider the three consecutive integers.

To self-check: the factored form is n(n−1)(n+1) = (n−1)n(n+1). This is a product of 3 consecutive integers. One must be divisible by 2 and one by 3, so the product is divisible by 6.

Enter the value of 6³ − 6 divided by 6 (to verify divisibility):

(a) Value of (6³ − 6) ÷ 6.

(b) Prove: "n² + 3n + 2 is always even." The simplified factored form is (n+1)(n+2). These are consecutive integers. One is always even, so product is even. Enter the sum of the roots of n² + 3n + 2 = 0 (i.e. the values of n that make it zero).

Q6a: (6³−6)÷6 =

Q6b: sum of roots =

Section B — Graphs & Calculus (18 marks)

Q7 4 marks

The curve y = x³ − 3x² − 9x + 2 has derivative dy/dx = 3x² − 6x − 9.

(a) Find the x-coordinates of both stationary points.

(b) Find the y-coordinate of the local maximum.

Q7a: smaller x =

larger x =

Q7b: local max y =

Q7c: y-intercept of tangent =

Q8 3 marks

Estimate the area under y = 1/x between x = 1 and x = 5 using 4 strips of equal width. Use the trapezium rule.

(a) State the width h of each strip.

(b) Calculate the trapezium rule estimate.

Q8a: h =

Q8b: estimate ≈

Q8c: 1=over / 2=under =

Q9 4 marks

The graph of y = f(x) has vertex at (3, −2) and passes through (5, 6).

(a) The graph is reflected in the x-axis. State the new vertex.

(b) The graph y = f(x) is transformed to y = f(2x). State the new x-coordinate of the vertex.

(c) The transformation y = f(x−1) + 4 is applied. State the new vertex coordinates. Enter the y-coordinate of the new vertex.

(d) Given f(x) = a(x−3)² − 2, use the point (5, 6) to find a.

Q9a: new vertex y =

Q9b: new vertex x =

Q9c: new vertex y =

Q9d: a =

Q10 4 marks

A particle moves so that its velocity v m/s at time t seconds is given by v = 6t − t².

(a) Find the acceleration at t = 2. (Recall: a = dv/dt = 6 − 2t)

(b) Find the time when the particle is momentarily at rest (v = 0, t > 0).

Q10a: acceleration =m/s²

Q10b: t =s

Q10c: distance ≈m

Q11 3 marks

Find the value of k such that the line y = kx + 3 is tangent to the curve y = x² + 1.

💡 Substitute y=kx+3 into y=x²+1, form a quadratic in x. Set discriminant = 0 for tangency.

(a) Form the quadratic in x after substitution.

(b) Find both values of k for which the line is tangent.

Q11: discriminant = 0 gives (4+k)² =

Q11: larger value of k =

Q11: smaller value of k =

Section C — Geometry, Vectors & Matrices (18 marks)

Q12 4 marks

Triangle ABC: AB = 15 cm, BC = 11 cm, angle ABC = 42°.

(a) Find AC using the cosine rule. Give to 3 s.f.

(b) Find angle BAC using the sine rule. Give to 1 d.p.

(d) A point D lies on BC such that AD bisects angle BAC. Find BD:DC using the angle bisector theorem: BD/DC = AB/AC. Enter BD:DC as a single decimal (BD÷DC to 2 d.p.).

Q12a: AC ≈cm

Q12b: angle BAC ≈°

Q12c: area ≈cm²

Q12d: BD÷DC ≈

Q13 4 marks

A rectangular room has length 8 m, width 5 m, height 3 m.

(a) Find the length of the longest diagonal (space diagonal).

(b) Find the angle this diagonal makes with the floor (base). Give to 1 d.p.

(c) A ladder leans against the 8 m wall. The base of the ladder is 2 m from the wall and the top reaches 5 m up the wall. Find the angle of inclination to the floor. Give to 1 d.p.

(d) A fly rests at one corner of the ceiling. Find its straight-line distance to the diagonally opposite corner on the floor (the space diagonal). Enter to 3 s.f.

Q13a: diagonal ≈m

Q13b: angle ≈°

Q13c: angle ≈°

Q13d: distance ≈m

Q14 4 marks

OABC is a quadrilateral where OA = a, OC = c. D is the midpoint of AC and E is the point on OB such that OE:EB = 2:1.

(a) Express OD in terms of a and c.

(b) Given OB = a + c, express OE in terms of a and c.

(c) Show that D, E, and B are collinear by finding DE and DB and showing they are parallel. Enter the ratio DE:EB as a decimal.

(d) If |a|=3 and |c|=4 and a·c=6, find |OD|².

Q14a: coeff of a in OD =

Q14b: coeff of (a+c) in OE =

Q14c: DE÷EB =

Q14d: |OD|² =

Q15 3 marks

Matrix P = [3, 1; −1, 2].

(a) Find det(P).

(b) Find P⁻¹.

Q15a: det(P) =

Q15c: x-component of X =

Q16 3 marks

O is the centre of a circle. A, B, C, D are points on the circumference. TP and TQ are tangents from external point T. Angle PTQ = 48°.

(a) Find angle POQ.

(b) Find the angle in the major arc PQ (i.e., angle PAQ where A is on the major arc). This is the angle subtended by the minor arc PQ at the circumference.

Q16a: angle POQ =°

Q16b: angle PAQ =°

Q16c: radius OP =cm

Section D — Statistics & Probability (10 marks)

Q17 4 marks

A test for a virus has sensitivity 0.92 (P(positive | infected)) and specificity 0.88 (P(negative | not infected)). The prevalence of infection is 5%.

(a) Find P(positive | not infected) (false positive rate).

(b) Find P(positive) using the law of total probability.

(d) Comment: is a positive test a reliable indicator? (Enter 1 = yes reliable, 0 = no not reliable)

Q17a: false positive =

Q17b: P(+) =

Q17c: P(inf|+) ≈

Q17d: reliable? (1/0) =

Q18 3 marks

The histogram below shows the distribution of waiting times (minutes) for 120 patients at a clinic:

Class	Freq. Density
0 – 5	3.6
5 – 10	8.4
10 – 20	4.2
20 – 40	1.5

(a) Verify the total frequency is 120.

(b) Estimate the mean waiting time. Use midpoints 2.5, 7.5, 15, 30.

Q18a: total =

Q18b: mean ≈min

Q18c: median class lower bound =

Q19 3 marks

Two groups of students sat a maths test. Their results are summarised:

Group	Median	IQR	Range
A	62	18	55
B	71	12	70

(a) Which group performed better on average? Enter 1 for Group A, 2 for Group B.

(b) Which group was more consistent? Enter 1 for A, 2 for B. (Use IQR as measure of spread.)

(c) An outlier is defined as a value more than 1.5×IQR above the UQ or below the LQ. For Group B: LQ = 65, UQ = 77. Find the lower outlier fence.

Q19a: better average (1=A, 2=B) =

Q19b: more consistent (1=A, 2=B) =

Q19c: lower outlier fence =

⭐ Bonus Questions (4 marks — push yourself!)

B1 2 marks

A sequence satisfies uₙ₊₁ = (uₙ + k/uₙ)/2 for some constant k, with u₁ = 1. If this converges to √5, find k.

B1: k =

B2 2 marks

The transformation matrix M = [cosθ, −sinθ; sinθ, cosθ] represents a rotation by angle θ. If M maps (1, 0) to (0.6, 0.8), find θ in degrees (to 1 d.p.).

B2: θ ≈°

Grade 10 — Practice Paper 3

Final Score