ESAT Paper 3 sample ยท Advanced Mathematics
ESAT Paper 3 Advanced Mathematics Sample Questions
Five questions from ESAT Paper 3, written to the depth of a full paper, with a worked solution for every one. This is a sample: a full module runs to 27 questions, and these five are drawn from the same question bank that writes the unseen papers schools commission here. Part of the ESAT preparation guide.
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Take ESAT Paper 3 Advanced Mathematics under the clock
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Question 1
Back to top โFor some $x > 0$ it is given that $\dfrac{x^{2}\cdot x^{-1/3}}{x^{4/3}} = 2$. Correct to 3 significant figures, what does $x^{-2}$ equal?
Key Idea (๐ก): Powers of the same base combine by adding indices when they are multiplied and by subtracting when they are divided, and those rules hold for negative and fractional indices exactly as they do for whole ones. Once a whole expression has collapsed to a single power of the unknown, the equation is solved by undoing that power, after which any other power of the same unknown follows by direct substitution. A negative index means a reciprocal, never a change of sign.
ESAT specification: MM1.1
Reveal the answer & worked solution: commit to an option first
Correct Answer: A. 0.0156
Step-by-Step Breakdown:
1. Collapse the left side to a single power of $x$
Powers of one base multiply by ADDING indices, so the numerator combines first:
$$x^{2}\cdot x^{-1/3} = x^{5/3}$$
Dividing by the denominator subtracts its index:
$$\frac{x^{5/3}}{x^{4/3}} = x^{1/3}$$
Run in a single pass, the indices total $2 - \tfrac{1}{3} - \tfrac{4}{3} = \tfrac{1}{3}$, so the condition in the stem is simply $x^{1/3} = 2$.
2. Solve for $x$
$x^{1/3}$ is the cube root of $x$, and the stem fixes $x > 0$, so raising both sides to the power $3$ is safe and leaves no second solution to check:
$$x = 2^{3} = 8$$
3. Evaluate the requested power
A negative index means a reciprocal, not a negative value:
$$x^{-2} = \frac{1}{x^{2}} = \frac{1}{8^{2}} = \frac{1}{64} = 0.0156$$
to 3 significant figures, which is the form the stem asks for.
Sanity check
Put $x = 8$ back into the left side. Its indices are the same chain as before, $2 - \tfrac{1}{3} - \tfrac{4}{3} = \tfrac{1}{3}$, so the expression is $8^{1/3}$, and the cube root of $8$ is $2$, as the stem requires. Since $x = 8 > 1$, every negative power of $x$ has to be a small positive fraction, so the two whole number options were never candidates.
The key is $0.0156$.
Why the Other Options Are Wrong (โ):
Common Mistake (โ ๏ธ):
Treating the division as though it removed the denominator's index by addition rather than by subtraction. The chain $2 - \tfrac{1}{3} - \tfrac{4}{3} = \tfrac{1}{3}$ only lands on $1/3$ because $4/3$ is taken away; add it instead and the left side becomes a power of $x$ that no whole number solves, which is the clearest warning available that the sign went the wrong way.
Takeaway (๐):
One base means one index. Total the indices first and the whole stem shrinks to $x^{1/3} = 2$; from there $x = 8$, and the power requested is arithmetic rather than algebra. The minus sign is the last thing to apply, and it inverts.
Question 2
Back to top โSimplify $\dfrac{\sqrt{21}-\sqrt{11}}{\sqrt{21}+\sqrt{11}}$ by clearing every surd out of its denominator. Which of the values below is the result, correct to 3 significant figures?
Key Idea (๐ก): A denominator built from a sum of two surds is cleared by multiplying above and below by its conjugate, because a conjugate pair multiplies out as a difference of two squares and leaves a rational number behind. When the numerator is that same conjugate, the multiplication squares it, so the cross term of the expansion survives and the simplified quotient still carries a surd on top. Dividing by the rational denominator then means dividing every term of the numerator, not the surd term alone.
ESAT specification: MM1.2
Reveal the answer & worked solution: commit to an option first
Correct Answer: E. 0.160
Step-by-Step Breakdown:
1. Multiply above and below by the conjugate of the denominator
The denominator is $\sqrt{21}+\sqrt{11}$, whose conjugate is $\sqrt{21}-\sqrt{11}$. Multiplying numerator and denominator by the same quantity leaves the value unchanged:
$$\dfrac{\sqrt{21}-\sqrt{11}}{\sqrt{21}+\sqrt{11}} = \frac{(\sqrt{21}-\sqrt{11})^{2}}{(\sqrt{21})^{2}-(\sqrt{11})^{2}} = \frac{(\sqrt{21}-\sqrt{11})^{2}}{21-11}$$
The denominator is now the rational number $10$.
2. Expand the numerator, then divide every term of it
$$(\sqrt{21}-\sqrt{11})^{2} = 21 - 2\sqrt{21}\sqrt{11} + 11 = 32 - 2\sqrt{231}$$
so the quotient is $\dfrac{32-2\sqrt{231}}{10}$. Both $21$ and $11$ are odd, so $32$ and $10$ are both even and the factor $2$ comes out of the whole numerator and out of the denominator together:
$$\frac{32-2\sqrt{231}}{10} = \frac{2\left(16-\sqrt{231}\right)}{2 \times 5} = \frac{16-\sqrt{231}}{5}$$
The factor cancels out of the whole bracket, not out of one term of it.
3. Put a number to the exact form
$\sqrt{231}$ is close to $15.2$, and it is smaller than $16$, since $16$ is the arithmetic mean of $21$ and $11$ while $\sqrt{231}$ is their geometric mean. The numerator is therefore positive, and to 3 significant figures the fraction $\dfrac{16-\sqrt{231}}{5}$ is $0.160$.
Sanity check
Take $\sqrt{21} \approx 4.58$ and $\sqrt{11} \approx 3.32$. The original quotient is then roughly $4.58 - 3.32$ over $4.58 + 3.32$, which is close to $0.160$. The value also has to be positive and below $1$, because $\sqrt{21}-\sqrt{11}$ is the smaller of two positive quantities and $\sqrt{21}+\sqrt{11}$ is the larger, and it is.
The key is $0.160$.
Why the Other Options Are Wrong (โ):
Common Mistake (โ ๏ธ):
Multiplying numerator and denominator by the denominator itself instead of by its conjugate. That produces $\dfrac{(\sqrt{21}-\sqrt{11})(\sqrt{21}+\sqrt{11})}{(\sqrt{21}+\sqrt{11})^{2}} = \dfrac{10}{32+2\sqrt{231}}$, which still has a surd underneath, so nothing has been rationalised and the work has to start again.
Takeaway (๐):
When the top and the bottom of a surd fraction are conjugates of one another, one multiplication does everything: the denominator collapses to the integer $10$ and the numerator becomes the perfect square $32 - 2\sqrt{231}$. Divide every term of that numerator by the integer, then estimate the surd once to choose between the options.
Question 3
Back to top โFind the least value of $k$ for which the horizontal line $y=k$ meets the curve $y=x^2 + 4x + 3$.
Key Idea (๐ก): A horizontal line $y=k$ crosses a curve exactly where the curve's expression equals $k$, so the points of contact are the real roots of $x^2 + 4x + 3=k$. Exactly one point of contact means a repeated root, and a repeated root means a discriminant of zero. Completing the square says the same thing in a different language: an upward parabola written as a square plus a constant reaches that constant and never goes below it, so the one height a horizontal line can touch without crossing is that constant. Both routes must give the same $k$, and using one to check the other is the cheapest guard against a sign slip.
ESAT specification: MM1.3
Reveal the answer & worked solution: commit to an option first
Correct Answer: F. -1
Step-by-Step Breakdown:
1. Complete the square
Halve the coefficient of $x$: half of $4$ is $2$, so the bracket is $(x+2)^2$. Expanding that bracket introduces an extra $4$ that the original expression does not have, so the same amount has to come back off:
The two constants collect to $-4 + 3 = -1$. A square is never negative and $(x+2)^2$ is $0$ at $x=-2$, so the lowest point of the curve is $(-2, -1)$ and its least $y$-value is $-1$.
2. Set the discriminant to zero
The line $y=k$ meets the curve wherever $x^2 + 4x + 3=k$. In standard form that is
so, with $a=1$, $b=4$ and $c=3-k$, the discriminant is
One point of contact is a repeated root, which needs $D=0$:
3. Check the value
The two routes agree, which is the check worth doing: the completed square puts the least value of the curve at $-1$, and the discriminant puts the repeated root at the same height. Substituting $x=-2$ into $x^2 + 4x + 3$ gives $-1$ directly, so the line $y=-1$ touches the curve at $(-2, -1)$. Any $k$ above $-1$ gives $D>0$ and two crossings; any $k$ below $-1$ gives $D<0$ and no crossing at all.
The key is $-1$.
Why the Other Options Are Wrong (โ):
Common Mistake (โ ๏ธ):
Stopping at the constant term. The completed square form is $(x+2)^2 - 4 + 3$, and the correction $-4$ has to be carried through: quoting $3$ gives the height of the curve at $x=0$, not its least value $-1$. Substituting $x=-2$ back into $x^2 + 4x + 3$ catches it in one line.
Takeaway (๐):
For a curve $y=x^2+bx+c$ the horizontal line $y=k$ touches it at exactly one point when $k$ is the height of the vertex, and that height is the constant left after completing the square. Halve the coefficient of $x$, square it, subtract: $3 - 4 = -1$. The discriminant gives the same number and takes longer.
Question 4
Back to top โThe graphs of $y = 5x^2 + 8x - 1$ and $y = -2x - 3$ meet at two points. Correct to 3 significant figures, by how much do the two $x$-coordinates differ?
Key Idea (๐ก): At an intersection the line and the curve share both coordinates, so setting their two expressions for $y$ equal removes $y$ and leaves one quadratic in $x$. Every term of the line has to cross over, the constant as well as the $x$ term, or the quadratic is not the right one. Once it is collected, the two roots sit symmetrically about the axis of symmetry, so their separation is fixed by the discriminant and the leading coefficient alone.
ESAT specification: MM1.4
Reveal the answer & worked solution: commit to an option first
Correct Answer: A. 1.55
Step-by-Step Breakdown:
1. Substitute the line into the curve
Both equations give $y$, so at an intersection
$-2x - 3 = 5x^2 + 8x - 1$
Taking every term of the line across, the $x$ term and the constant together, leaves one quadratic:
$5x^2 + 10x + 2 = 0$
Its $x$ coefficient is $10$ and its constant is $2$.
2. Evaluate the discriminant
With leading coefficient $5$,
$(10)^2 - 4(5)(2) = 60$
It is positive, which is consistent with the two crossings the question describes.
3. Take the difference of the roots
$x = \dfrac{-(10) \pm \sqrt{60}}{2(5)}$, so the two roots differ by
$\dfrac{2\sqrt{60}}{2(5)} = \dfrac{\sqrt{60}}{5} = 1.55$ to 3 significant figures.
The centre the two roots sit either side of cancels in that subtraction and never has to be worked out.
Sanity check
The roots are near $-1.77$ and $-0.225$, which are roughly $1.55$ apart. Between them $5x^2 + 10x + 2$ is negative and outside them it is positive, as a quadratic with a positive $x^2$ coefficient and two real roots must be.
The key is $1.55$.
Why the Other Options Are Wrong (โ):
Common Mistake (โ ๏ธ):
Subtracting only part of the line. Collecting $-2x - 3 = 5x^2 + 8x - 1$ means moving the $x$ term and the constant together; stopping after the $x$ term leaves $5x^2 + 10x - 1 = 0$, whose constant is still $-1$ instead of $2$ and whose discriminant is $120$ instead of $60$. The substitution is one move, not two independent ones.
Takeaway (๐):
Collect the line and the curve into a single quadratic, then the horizontal gap between the crossings is the square root of the discriminant divided by the size of the leading coefficient, here $\dfrac{\sqrt{60}}{5}$. Neither root is ever needed, and the axis of symmetry never has to be found.
Question 5
Back to top โThe terms of a sequence obey the recurrence $u_{n+1}=\frac{u_{n} - 14}{2}$. Given that $u_{4}=-13$, what is the first term?
Key Idea (๐ก): Everything about $u_{n+1}=\frac{u_{n} - 14}{2}$ is easier in terms of the distance from its fixed point. Solving $L=\frac{L - 14}{2}$ gives the one value that reproduces itself, and writing each term as $L$ plus a difference turns the recurrence into a plain multiplication: forwards the difference is divided by $2$, backwards it is multiplied by $2$. Walking from $u_{4}$ to $u_{1}$ is then a single power of $2$, and the count of steps is the difference of the indices rather than the number of terms mentioned.
ESAT specification: MM2.1
Reveal the answer & worked solution: commit to an option first
Correct Answer: C. -6
Step-by-Step Breakdown:
1. Invert the rule
Make the earlier term the subject:
$$u_{n+1}=\frac{u_{n} - 14}{2} \implies 2u_{n+1} = u_{n} - 14 \implies u_{n} = 2u_{n+1} + 14$$
This rule takes any term and returns the one before it, so it can be applied as many times as needed.
2. Find the fixed point
A term equal to $L$ would reproduce itself, so $L=\frac{L - 14}{2}$, giving $2L = L - 14$, then $(2-1)L = -14$ and $L = -14$. Writing every term as $L$ plus a difference is what turns the chain into one multiplication.
3. Walk back from $u_{4}$ to $u_{1}$
The difference between the given term and the fixed point is $u_{4} - L = -13 - (-14) = 1$. Each application of the inverse rule multiplies that difference by $2$, and the number of applications is the difference of the indices, $4 - 1 = 3$:
$$u_{1} = L + 2^{3} \times (1) = -14 + (8) = -6$$
Check it forwards through the original rule: $\frac{-6 - 14}{2} = -10$, which is $u_{2}$, and a further $2$ forward steps return $u_{4} = -13$ as stated.
The key is $-6$.
Why the Other Options Are Wrong (โ):
Common Mistake (โ ๏ธ):
Counting the steps by the number of terms rather than the gap between the indices. $4$ terms are involved, so $2$ or $4$ applications of the inverse rule feel natural, but moving from $u_{4}$ to $u_{1}$ is exactly $4 - 1 = 3$ of them.
Takeaway (๐):
For $u_{n+1}=\frac{u_{n}+c}{k}$, solve $L=\frac{L+c}{k}$ for the fixed point first. The difference from $L$ is divided by $k$ at every forward step and multiplied by $k$ at every backward one, so $u_{1} = L + k^{m}(u_{1+m} - L)$ replaces the whole chain of substitutions. Here that reads $u_{1} = -14 + 2^{3}(-13 - (-14)) = -6$.
Where to go next
- Next: ESAT Paper 4 Advanced Maths, five more questions at the same standard.
- Five questions at test pace in Advanced Maths: practice set 1A and set 1B, each worked in full.
- Twenty sample questions in Advanced Maths across the four papers, five on each module page, each page with a timed test at the top.
- Every paper and practice set across all five ESAT subjects is indexed on the ESAT preparation guide.
- Teaching a cohort rather than sitting the test? A free ESAT sample pack holds 10 of the 27 questions in every module, with the worked solutions and mark schemes in full, and unseen packs are written for individual schools.
- If the method is the problem rather than the answer, Lucas runs 1-on-1 ESAT tutoring for Cambridge, Oxford and Imperial applicants: apply for admissions tutoring.
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