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TMUA practice questions, with worked answers

By The STEM-Tutor team · Updated October 2026

These are original questions written in the style of the TMUA, seven for Paper 1 and six for Paper 2. Give yourself about four minutes each, with no calculator, then open the worked answer.

On this page
  1. How to use these questions
  2. Paper 1 style: applications of mathematical knowledge
  3. Paper 2 style: mathematical reasoning
  4. What these questions teach
  5. Common mistakes
  6. Frequently asked questions
  7. Key takeaways
  8. Related guides

How to use these questions

The TMUA is two papers of 20 multiple-choice questions, 75 minutes each, with no calculator. Paper 1 tests whether you can use GCSE and first-year A level pure maths in unfamiliar ways. Paper 2 tests logic and proof: reading statements precisely, spotting counterexamples and finding the flaw in an argument. The questions below cover the topics that come up most, and each worked answer shows the quick route the test rewards. When you have finished, move on to the official past papers, which are the best practice there is.

Paper 1 style: applications of mathematical knowledge

1. A hidden quadratic

How many real solutions does x4 − 5x2 + 4 = 0 have?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show the worked solution
  • Let y = x2: y2 − 5y + 4 = 0, so (y − 1)(y − 4) = 0 and y = 1 or 4.
  • x2 = 1 gives x = ±1 and x2 = 4 gives x = ±2.
  • Answer: E, four solutions. Both values of y are positive, so each gives two values of x.

2. Stationary points

The curve y = x3 − 3x2 − 9x + 5 has two stationary points. What is the sum of their x-coordinates?

  1. −2
  2. −1
  3. 2
  4. 3
  5. 6
Show the worked solution
  • dy/dx = 3x2 − 6x − 9 = 3(x2 − 2x − 3).
  • Shortcut: the sum of the roots of x2 − 2x − 3 = 0 is 2, with no need to solve. (The roots are 3 and −1.)
  • Answer: C, 2.

3. A geometric series

A geometric series has sum to infinity 12, and the sum of its first two terms is 9. What is the sum of all the possible values of the first term?

  1. 6
  2. 12
  3. 18
  4. 24
  5. 30
Show the worked solution
  • a/(1 − r) = 12 gives a = 12(1 − r), and a(1 + r) = 9.
  • So 12(1 − r)(1 + r) = 9, giving 1 − r2 = 3/4 and r = ±1/2. Both satisfy |r| < 1.
  • r = 1/2 gives a = 6, and r = −1/2 gives a = 18. Answer: D, 24.

4. Logarithms

Solve log2x + log2(x − 2) = 3.

  1. x = −2 only
  2. x = 4 only
  3. x = −2 or x = 4
  4. x = 6 only
  5. no solution
Show the worked solution
  • Combine: log2[x(x − 2)] = 3, so x2 − 2x = 8 and (x − 4)(x + 2) = 0.
  • A logarithm needs a positive argument, so x > 2 and x = −2 is rejected.
  • Answer: B, x = 4 only. Option C is the trap.

5. Area between curves

What is the area of the region enclosed by y = x2 and y = 2x?

  1. 2/3
  2. 1
  3. 4/3
  4. 8/3
  5. 4
Show the worked solution
  • They meet where x2 = 2x, at x = 0 and x = 2. Between them, 2x is above x2.
  • Area = ∫02 (2x − x2) dx = [x2 − x3/3] from 0 to 2 = 4 − 8/3.
  • Answer: C, 4/3.

6. Counting trig solutions

How many solutions does sin 2x = sin x have for 0 ≤ x ≤ 2π?

  1. 2
  2. 3
  3. 4
  4. 5
  5. 6
Show the worked solution
  • sin 2x = 2 sin x cos x, so 2 sin x cos x − sin x = sin x (2 cos x − 1) = 0. Never divide by sin x: you would lose solutions.
  • sin x = 0 gives x = 0, π, 2π. cos x = 1/2 gives x = π/3, 5π/3.
  • Answer: D, 5.

7. A tangent to a circle

The line y = x + k is a tangent to the circle x2 + y2 = 8. What are the possible values of k?

  1. k = ±2
  2. k = ±2√2
  3. k = ±4
  4. k = ±8
  5. k = 4 only
Show the worked solution
  • A tangent's distance from the centre (0, 0) equals the radius, √8 = 2√2.
  • The distance from (0, 0) to x − y + k = 0 is |k|/√2, so |k|/√2 = 2√2 and |k| = 4.
  • Answer: C, k = ±4. Substituting and setting the discriminant to zero gives the same result, more slowly.

Paper 2 style: mathematical reasoning

8. Negation

Which is the negation of “every prime number greater than 2 is odd”?

  1. Every prime number greater than 2 is even.
  2. There is a prime number greater than 2 that is even.
  3. No prime number greater than 2 is odd.
  4. There is a prime number greater than 2 that is odd.
  5. Every even number greater than 2 is not prime.
Show the worked solution
  • The negation of “for all x, P(x)” is “there exists an x for which P(x) is false”.
  • Answer: B. Option A is a common wrong answer: it is stronger than the negation needs to be.

9. Necessary or sufficient?

For a real number x, the condition x2 > 4 is … for x > 2.

  1. necessary and sufficient
  2. necessary but not sufficient
  3. sufficient but not necessary
  4. neither necessary nor sufficient
Show the worked solution
  • If x > 2 then x2 > 4, so x2 > 4 must hold whenever x > 2: it is necessary.
  • But x = −3 has x2 > 4 without x > 2, so it is not sufficient.
  • Answer: B.

10. A counterexample

Which value of n is a counterexample to “if n2 is a multiple of 8, then n is a multiple of 8”?

  1. n = 2
  2. n = 4
  3. n = 8
  4. n = 16
Show the worked solution
  • A counterexample must make the “if” part true and the “then” part false.
  • n = 4: n2 = 16 is a multiple of 8, but 4 is not. n = 2 fails the “if” part, and 8 and 16 satisfy both parts.
  • Answer: B.

11. Find the flaw

A ‘proof’ that 2 = 1: (I) let a = b; (II) then a2 = ab; (III) so a2 − b2 = ab − b2; (IV) so (a + b)(a − b) = b(a − b); (V) so a + b = b; (VI) so 2b = b, and 2 = 1. Which step is the first that is not valid?

  1. II
  2. III
  3. IV
  4. V
  5. VI
Show the worked solution
  • Step V divides both sides by a − b, which is zero because a = b. Division by zero is not allowed.
  • Answer: D. (Step VI also divides by b, which fails if b = 0, but V is the first invalid step.)

12. Order of quantifiers

Which of these statements is true? (I) For every real x, there is a real y with y > x2. (II) There is a real y such that, for every real x, y > x2.

  1. neither
  2. I only
  3. II only
  4. both
Show the worked solution
  • I is true: given any x, take y = x2 + 1.
  • II is false: no single y is bigger than every x2, because x2 can be as large as you like.
  • Answer: B. Swapping “for every” and “there is” changes the meaning.

13. ‘Only if’

“A only if B” means the same as which of these?

  1. If B then A
  2. If A then B
  3. A if and only if B
  4. If not A then not B
Show the worked solution
  • “A only if B” says A cannot happen without B, so whenever A is true, B is true: if A then B.
  • Answer: B. Option D is the inverse, which is equivalent to option A, not to the statement.

What these questions teach

Common mistakes

Frequently asked questions

Are these official TMUA questions?

No. They are original questions written in the TMUA style. For official questions, use the past papers and worked answers from UAT-UK, which we link on our TMUA past papers page.

How long should each question take?

About 3 minutes 45 seconds on average, since each paper is 20 questions in 75 minutes. Some take one minute, others five.

Is a calculator allowed in the TMUA?

No. Both papers are taken without a calculator.

Key takeaways

More TMUA and ESAT guides

All TMUA and ESAT guides