2022 UPDATE: PLEASE NOTE FOR 2022 CAMBRIDGE ECONOMICS APPLICATIONS THE ECAA IS NOT BEING USED. INSTEAD STUDENTS ARE BEING ASKED TO TAKE THE TMUA ENTRANCE TEST.
I wanted to break down some key types of questions in the maths component of the TMUA mathematics admissions assessment. For each of the key question types, I will try to give some tips.
I will be using examples from the 2021 TMUA papers.
Which types of questions come up?
The pie chart shows a breakdown of the different types of questions in the 2021 TMUA paper:
Note questions on other topics could still come up in your test and have come up in past papers.
In your revision for the TMUA, you should expect to practise every type of question. From these data, it seems to be important to spend some extra time practising the question types mentioned above, in particular: graphs / equations of lines, trigonometry, calculus (differentiation and integration), logs and exponents, functions / number theory, series and sequences as well as the other types of question mentioned above.
In the rest of this post, I would like to begin to cover a few of these key areas (and save the rest for another post). I will give examples of the type or subtypes of question, work through them as well as give some tips on how to solve these types of questions generally.
Sequences
Relevant questions in TMUA 2021: Paper 1 questions 3, 13, 15
Sequences can feature arithmetic and geometric sequences or series, or recurrence relations. These ideas can link to another question type such as a graph. They can often come with simultaneous equations.
Make sure you know the formulae for the nth term, the finite sum for an arithmetic series, as well as the finite and infinite sums for geometric series. You should also be familiar with rules for the sum of positive integers (i.e. the triangular numbers).
You often have to create and set up equations using the information provided.
Example
Here is an example of a possible question I have come up with for geometric series:
Consider two geometric sequences. Sequence A has first term 1 and second term \(y\). Sequence B has second term 2 and fourth term \(y+4\).
The common ratio of sequence A is the same as that of sequence B and this common ratio is greater than zero. Find the third term of sequence B
Method
From the second part of the question, we can infer we need to work out the common ratio first. We know the nth term of a geometric sequence is \(ar^{n-1}\) where \(a\) is the first term, \(r\) the common ratio, and \(n\) the position in the sequence.
From the information about sequence A, we can divide the second term by the first term to give the common ratio of sequence A:
\[\frac{ar}{a}=r=\frac{y}{1}=y\]
We can do something similar using the information about sequence B:
\[\frac{ar^3}{ar}=r^2=\frac{y+4}{2}\]
Noting that the common ratio is the same in both sequences, we can substitute condition one into condition 2 to eliminate \(y\):
\[r^2=\frac{r+4}{2}\]
Rearrange into a quadratic equation and use the quadratic formula:
\[2r^2-r-4=0\]
\[r=\frac{1\pm \sqrt{1-4(2)(-4)}}{2(2)}=\frac{1\pm\sqrt{33}}{4}\]
Note we are told the common ratio is greater than zero, so it will not be the negative root. So \(r= \frac{1+\sqrt{33}}{4}\)
The third term of sequence B comes from multiplying the second term by \(r\), which gives the final answer:
\[r= \frac{1+\sqrt{33}}{2} \]
Probability
Examples of relevant questions: TMUA 2020 paper 2 question 17; TMUA 2021 paper 2 question 3 (though this latter question does not require much knowledge of probability).
In general an area with low representation in recent past papers.
Some examples of possible question types could be:
1) Calculating probabilities from an often standard sample space. This could involve use of conditional probability rules, tree diagrams and Venn diagrams for example. In this type of question, one tip is to write down the probabilities you are looking for and the probabilities you have available. Using quick tree or Venn diagrams, or writing down a quick probability rule, could help here.
2) Manipulation of the mean of a dataset and even other statistical properties of a dataset, such as its median, quartiles, range and so on. In this case again, one tip is to try to set up equations using this mean of the dataset.
There are multiple other ways that statistics could be used in a question: see the specification document for the information on what kinds of statistics could come up.
Type 1 Example
A example of a type 1 probability question is as follows:
Suppose a fair coin is flipped 3 times. You are told at least one of the coins landed on tails. Given this information, what is the probability that at least one coin landed on heads?
Method
This question plays on the theme of the so-called “Boy or Girl paradox”.
The key here is that the probability we are after is conditional. Let \(H\) be the number of heads revealed and \(T\) be the number of tails revealed. Then the probability we are trying to find is:
\[P(H\geq 1|T\geq 1)\]
We can use the rule that a conditional probability can be expressed as a fraction (also known as the Kolmogorov definition):
\[\frac{P(H\geq 1\cap T\geq 1) }{P(T\geq 1)}\]
The numerator contains two cases: the case where there is one head and two tails and the second case with two heads and one tail.
The unconditional probability of having exactly one head and two tails from three fair coin flips is \(3\times \frac{1}{2}^3=\frac{3}{8}\). This is because the probability of head then tail then tail is \( \frac{1}{2}^3 \) as these are independent events, but we can also have the order tail, head, tail or tail, tail, head (hence the multiplication by 3).
Similarly the unconditional probability of having two heads and one tail from three fair coin flips is the same: \(3\times \frac{1}{2}^3=\frac{3}{8}\).
The numerator therefore is just the sum of these two probabilities: \( \frac{3}{8} + \frac{3}{8} = \frac{6}{8}\).
As for the denominator, these two cases are also included. However there is a third case, where there are three tails. Similarly to above we know this occurs with probability \(\frac{1}{2}^3=\frac{1}{8}\).
Hence the denominator is the sum of probabilities of these three cases: \( \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{7}{8}\).
We can plug these back into the conditional probability to get the final answer:
\[\frac{P(H\geq 1\cap T\geq 1) }{P(T\geq 1)} =\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}\]
Type 2 Example
Suppose the mean height of a group of 20 people is 1.65 metres (m). The people, who are ordered randomly from number 1 to 20, are then split into three groups:
- Group A: the first 15 people with mean height (x+0.1) metres;
- Group B: people 16, 17 and 18 with mean height y metres;
- Finally group C: people 19 and 20 with mean height x metres.
Find x in terms of y.
Method
First let us use the definition of the mean to write down some equations. We know the mean is the sum of observations divided by the number of observations. Once we know how to use this, the question becomes relatively simple.
Let \(h_i\) denote the height of person \(i\), where \(i=1,2,…,20\). For the entire group of 20 people, we can write the following:
\[\frac{\sum_{i=1}^{20} h_i}{20}=1.65\]
\[ \sum_{i=1}^{20} h_i=1.65\times 20 =33 \]
In other words, because we know the average height of the overall group of 20 people is 1.65m, we also know the sum of their heights is 33m. We can do similar things for all the other groups in turn. For example for group A:
\[ \frac{\sum_{i=1}^{15} h_i}{15}=x+0.1 \Leftrightarrow \sum_{i=1}^{15} h_i=15x+1.5\]
We can also do this for groups B and C respectively:
\[ \frac{\sum_{i=16}^{18} h_i}{3}=y \Leftrightarrow \sum_{i=16}^{18} h_i=3y\]
\[ \frac{\sum_{i=19}^{20} h_i}{2}=x \Leftrightarrow \sum_{i=19}^{20} h_i=2x\]
Furthermore, the sum of all the heights from each of the separate groups should equal the sum of the heights based on the mean for the entire group:
\[2x+3y+15x+1.5=33\]
\[17x=31.5-3y\]
\[x=\frac{31.5-3y}{17}\]
which is the final answer.
Quirky quadratics
Relevant questions in TMUA 2021: paper 1 questions 4, (19), 20
By “quirky quadratics” I am referring to a particular type of quadratic equation or inequality, where a substitution could be used to transform the equation into a standard quadratic.
In my earlier categorisation, these could come under several categories, depending on the type of substitution. For example the required substitution could be a simple substitution (see the example below) or it could be more complex, for example trigonometric or using an exponential or logarithmic function.
Additionally some manipulation, beyond the substitution, may be required to transform the equation into something that looks like a quadratic.
Once you find the roots, you will need to reverse your substitution to find the intended roots. With these roots you may then need to do further manipulation, for example to find the sum or product of these roots.
Example
For example consider the following example question I have created:
Find the product of the unique real roots of the equation
\[x^8-6x^4+9=0\]
Method
First let us substitute \(y=x^4\), so the equation becomes:
\[y^2-6y+9=0\]
This allows us to conduct a simple factorisation:
\[(y-3)^2=0\]
This gives \(y=3\). We then need to use the substitution to find \(x=y^{\frac{1}{4}}=3^{\frac{1}{4}}\). Note this can be the positive or negative root. The product of these roots is then
\[3^{\frac{1}{4}} \times (-)(3^{\frac{1}{4}})=-\sqrt{3}\]
which is the final answer.
More posts will follow on other question types, among other things.
For more on the TMUA, see the article here:
If you wish to see more advice on Economics personal statements and Oxbridge Economics interview questions, click the blue button below:
Latest blogposts: