ESAT & TMUA practice

Challenging Questions for UK Admissions Tests

The hardest practice questions for ESAT and TMUA, with worked solutions.
Official Mocks every two weeks and live, competitive leaderboards.
The support of a course at a fraction of the cost.

London School of Economics
University of Warwick
University of Cambridge
University of Oxford
Imperial College London
University College London
Durham University
London School of Economics
University of Warwick
University of Cambridge
University of Oxford
Imperial College London
University College London
Durham University
London School of Economics
University of Warwick
University of Cambridge
University of Oxford
Imperial College London
University College London
Durham University
London School of Economics
University of Warwick
University of Cambridge
University of Oxford
Imperial College London
University College London
Durham University

Try a sample question

MATHSESAT

Define Q=∑k=12024k(k+1)!Q = \displaystyle\sum_{k=1}^{2024} \dfrac{k}{(k+1)!}. What is the value of QQ?

A1−12024!1 - \dfrac{1}{2024!}
B1−12025!1 - \dfrac{1}{2025!}
C1−22025!1 - \dfrac{2}{2025!}
D20242025!\dfrac{2024}{2025!}
E1−12023!⋅20251 - \dfrac{1}{2023! \cdot 2025}
F2025!−12025!\dfrac{2025! - 1}{2025!}
Answer B. Write k(k+1)!=1k!−1(k+1)!\dfrac{k}{(k+1)!} = \dfrac{1}{k!} - \dfrac{1}{(k+1)!}, so the sum telescopes to 11!−12025!=1−12025!\dfrac{1}{1!} - \dfrac{1}{2025!} = 1 - \dfrac{1}{2025!}. The answer is B.
PAPER 1TMUA

The positive integers are grouped into blocks as follows: {1}\{1\}, {2,3,4}\{2,3,4\}, {5,6,7,8,9}\{5,6,7,8,9\}, {10,11,…,16}\{10,11,\ldots,16\}, and so on, where the kkth block contains 2k−12k-1 integers. Let BkB_k denote the sum of all integers in the kkth block. Find B10B_{10}.

A15401540
B16401640
C17291729
D18201820
E19001900
F20102010
Answer C. The first block starts at 11 and block kk has 2k−12k-1 terms, so there are (k−1)2(k-1)^2 integers before block kk; hence block kk runs from (k−1)2+1(k-1)^2+1 to k2k^2. The sum is 2k−12((k−1)2+1+k2)=(2k−1)(k2−k+1)\frac{2k-1}{2}\bigl((k-1)^2+1+k^2\bigr) = (2k-1)(k^2-k+1). For k=10k=10: 19×91=172919 \times 91 = 1729. The answer is C.
PAPER 2TMUA

A formal system operates on strings of the letters PP and QQ. The system has exactly three rules: R1 PPPP is a valid string. R2 If XX is a valid string, then XQXQ is a valid string. R3 If XX is a valid string, then QXPQXP is a valid string.

A string is called "reachable" if it can be produced by applying the rules finitely many times, starting from R1. Consider the following three statements:

I.\textbf{I.} The number of PPs in any reachable string is always even.

II.\textbf{II.} The string QQPPQQQPPQ is reachable.

III.\textbf{III.} There exists a reachable string containing QPQQPQ as a substring.

Which of these statements are true?

AI only
BII only
CIII only
DI and II only
EI and III only
FII and III only
GAll three
HNone of them
Answer C. Statement I is false: R3 applied to PPPP gives QPPPQPPP, which has 3 PPs. Statement II is false: working backwards, QQPPQQQPPQ requires predecessor QQPPQQPP via R2, which requires predecessor QPQP via R3, but QPQP is unreachable. Statement III is true: applying R2 then R3 then R2 to PPPP yields QPPQPQQPPQPQ, which contains QPQQPQ. The answer is C.
Get started

No commitment. Cancel at any time.