40% off

Ends in

--d--h--mLock in £90
← All past papers

TMUA Mock Tmua-maths-tmm4-1

3 questions3 marks75Updated August 2026

The TMUA Mock Tmua-maths-tmm4-1 paper in full: all 3 questions, each with its answer. TMUA is the Test of Mathematics for University Admission. Sit it cold under exam timing, mark it, then work back through anything you missed using the solutions below.

Questions and answers are free. Full step-by-step worked solutions unlock with a free account. Start practising.

Question 1

1 mark
A sector of a circle has radius rr and angle θ\theta radians, where 0<θ<π0 < \theta < \pi. The perimeter of the sector (the two radii plus the arc) equals half the circumference of the full circle of radius rr. What is θ\theta?
  • A.22
  • B.π\pi
  • C.π−2\pi - 2
  • D.2π−42\pi - 4
  • E.π−1\pi-1
  • F.π2\frac{\pi}{2}

Answer: C

Full step-by-step worked solution

Locked. Available with a free account.

Unlock worked solutions →

Question 2

1 mark
A circular sector has radius rr and angle θ\theta radians, where 0<θ<π0<\theta<\pi. The chord joining the two ends of the arc has length equal to the radius rr. Let SS be the area of the minor segment cut off by this chord, and let TT be the area of the sector. What is ST\frac{S}{T}?
  • A.1−3π1-\frac{\sqrt{3}}{\pi}
  • B.332π\frac{3\sqrt{3}}{2\pi}
  • C.2π−332π\frac{2\pi-3\sqrt{3}}{2\pi}
  • D.1−332π1-\frac{3\sqrt3}{2\pi}
  • E.1−33π1-\frac{3\sqrt{3}}{\pi}
  • F.π−332π\frac{\pi-3\sqrt3}{2\pi}

Answer: C

Full step-by-step worked solution

Locked. Available with a free account.

Unlock worked solutions →

Question 3

1 mark
A circle has centre OO and radius rr. Two radii OAOA and OBOB are drawn so that the minor arc ABAB has length equal to the length of the chord ABAB. Let θ\theta be the angle ∠AOB\angle AOB in radians (with 0<θ<π0 < \theta < \pi). Which of the following statements about θ\theta is correct?
  • A.θ\theta satisfies θ=2sin⁡(θ/2)\theta = 2\sin(\theta/2) and has a unique nonzero solution in (0,π)(0,\pi), but it cannot be expressed in closed form using only elementary radicals
  • B.θ=π/2\theta = \pi/2 exactly
  • C.θ=π/3\theta = \pi/3 exactly
  • D.θ\theta satisfies sin⁡θ=θ\sin\theta = \theta and has no solution other than θ=0\theta=0 in (0,π)(0,\pi)
  • E.θ=3\theta = \sqrt{3} exactly
  • F.θ\theta satisfies θ=2sin⁡(θ/2)\theta = 2\sin(\theta/2), and the only solution in (0,π)(0,\pi) is θ=0\theta = 0, so no such sector exists

Answer: A

Full step-by-step worked solution

Locked. Available with a free account.

Unlock worked solutions →