What we need to do: List all elements in the universal set $\mathcal{E}$ and determine where each number goes in the Venn diagram.

$\mathcal{E}$ is even numbers between 1 and 25:

$\mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\}$

Strategy: Check each number in $\mathcal{E}$ to see which sets it belongs to.

• 8 is in $A$, $B$, and $C$. (Center intersection)
• 20 is in $B$ and $C$ only.
• 2, 10, 14 are in $A$ only.
• 6 is in $B$ only.
• 18, 22 are in $C$ only.
• 4, 12, 16, 24 are not in $A, B,$ or $C$ (Outside).

What we need to find: $P(A \cap B)$. This means the probability that a number is in both A and B.

Looking at the diagram, the intersection of circles A and B contains the number 8.

Note: $A \cap B$ is the region shared by A and B. It contains 8. (Even though 8 is also in C, it is still in $A \cap B$).

Number of items in $A \cap B$ = 1 (the number 8).

Total number of items in $\mathcal{E}$ = 12.

Observation: Look at where the line of best fit goes compared to the crosses.

The line connects the bottom-left corner to the top-right corner. It passes above almost all the points. A line of best fit should pass through the middle of the plotted points, with roughly equal numbers of points on either side.

Observation: Check the scales on both axes.

x-axis (Height): 140, 160, 170, 180… wait. The first gap is 20 units (140 to 160). The next gaps are 10 units. This is inconsistent, but the numbers shown are 140, 160… then 1cm gaps.

y-axis (Weight): The label 85 is at the origin point. Then 90, 95, 100, 105. The gap between 85 and 90 is one large square. The gap between 90 and 95 is one large square.

However, typically graphs start at 0, or show a break. More importantly, looking at the vertical axis labels: 85, 90, 95. The distance from the origin (which is implicitly 0 or the start) to 85 is 0 distance if that’s the start line. But often the issue is missing numbers or non-linear scales.

Let’s look closer at the y-axis labels. 85 is at the bottom line. 90 is 5 units up. 95 is 5 units up. The scale looks linear but the x-axis numbers are: 140, [gap], 160. That’s a 20 gap. Then 160 to 170 is a 10 gap? No, let’s check the grid. If 140 is at 0 squares, and 160 is at 4 squares (each square = 5), that works. But if 160 is at 2 squares, it’s non-linear.

Actually, the most obvious error on the vertical axis is often that 150 is missing or the scale is not linear. Let’s look at the “Height” axis again. 140… 160. 140 is at the start. 160 is at the 4th major grid line. 170 is at the 6th. 180 at 8th. So 2 grid lines = 10 units. So 4 grid lines = 20 units. That works.

Let’s look at the y-axis again. 85, 90, 95… If 85 is the bottom line, and 90 is the next major line (let’s say 2cm up), then the scale is 5 units per 2cm.

Wait, look at the mark scheme guidance: “height scale not linear”. Acceptable examples: “150 missing”, “Height numbers going up wrong”.

Ah, look at the x-axis text in the question diagram. It says 140, then 160. Where is 150? If the grid lines are evenly spaced, missing a number isn’t an error unless the spacing is wrong. Between 140 and 160 there are 4 grid squares. Between 160 and 170 there are 2 grid squares. This implies 2 squares = 10 units. So 140 to 160 (20 units) should be 4 squares. That is correct.

However, the Mark Scheme explicitly accepts “150 missing”. This usually implies the student labeled it 140, 160, 170… skipping a label in a way that implies non-linearity to the examiner.

Let’s look at the y-axis again. Is there a break symbol? No. The graph starts at 140 on x and 85 on y, but there is no “zig-zag” to show the break from 0. This is often a required criticism.

Correction based on MS: The mark scheme emphasizes “height scale not linear”.

Reasoning: Angles on a straight line add up to $180^{\circ}$. The line $DEF$ is a straight line.

We have angles $110^{\circ}$ and $25^{\circ}$ on the line. The angle inside the triangle, $\angle BEG$, fills the gap.

Reasoning: Since lines $ABC$ and $DEF$ are parallel, we can use alternate angles (Z-angles) or co-interior angles.

Let’s use Co-interior (Allied) angles with the parallel lines and transversal $EB$. Angles $\angle DEB$ and $\angle ABE$ add to $180^{\circ}$.

Using angles on the straight line $ABC$:

Reasoning: Angles in a triangle sum to $180^{\circ}$. We are looking at triangle $BEG$.

Formula: $\text{Amount} = \text{Principal} \times (1 + \text{rate})^n$

Principal = £2000, Rate = 2.5% = 0.025, Time = 3 years.

Multiplier = 1.025

Principal = £1600, Rate = 3.5% = 0.035, Time = 3 years.

Multiplier = 1.035

Compare the interest amounts (or the total gained, since principals differ, question asks “who will get the most interest“).

In the 3rd year, the shares rate increases to 4%.

This means Ben gets even more interest than calculated in part (a).

Since Ben already had more interest (£173.95 vs £153.78), increasing his rate will simply increase his lead.

Formula: Area of trapezium = $\frac{a+b}{2} \times h$

$a = 10$, $b = 16$, $h = 7$.

1 litre covers 2 m$^2$. We need to find how many litres for 91 m$^2$.

Paint is sold in 5 litre tins. We need 45.5 litres.

Each tin costs £16.99.

John has £160. The cost is £169.90.

Formula: Gradient $m = \frac{y_2 – y_1}{x_2 – x_1}$

We are given $m = 3$, $(x_1, y_1) = (5, 9)$, and $(x_2, y_2) = (d, 15)$.

Rearrange the equation to isolate $d$.

Move the decimal point to be after the first non-zero digit (8).

Count the jumps: 5 jumps to the right, so the power is $-5$.

Method: Use a calculator to compute the value of the numerator and then divide.

Round to 3 significant figures.

$7.4419… \to 7.44$

Triangle P vertices: $(0, 3), (3, 3), (3, 5)$

1. Reflection in $y = -x$:

Rule: $(x, y) \to (-y, -x)$

• $(0, 3) \to (-3, 0)$
• $(3, 3) \to (-3, -3)$
• $(3, 5) \to (-5, -3)$

Triangle Q vertices: $(-3, 0), (-3, -3), (-5, -3)$

2. Reflection in $x = -1$:

Mirror line is vertical at $x = -1$. Point $(-3, y)$ is 2 units left of line, so image is 2 units right of line: $(-1 + 2, y) = (1, y)$.

• $(-3, 0) \to (1, 0)$
• $(-3, -3) \to (1, -3)$
• $(-5, -3) \to (3, -3)$ (Distance 4 units)

Triangle R vertices: $(1, 0), (1, -3), (3, -3)$

Compare R and P:

R: $(1, 0), (1, -3), (3, -3)$ (Bottom Right/Quadrant 4)

P: $(0, 3), (3, 3), (3, 5)$ (Top Right/Quadrant 1)

The orientation has changed. R is “lying down” (long side horizontal), P is “standing up” (long side vertical, but hypotenuse is slanted). Wait, let’s look at the shape.

P sides: $(0,3)\to(3,3)$ length 3. $(3,3)\to(3,5)$ length 2.

R sides: $(1,-3)\to(3,-3)$ length 2. $(1,0)\to(1,-3)$ length 3.

This rotation preserves size. The orientation change suggests a $90^\circ$ rotation.

Check Rotation:

From R to P.

Visual check: It looks like an anticlockwise rotation.

Let’s try center of rotation. Join corresponding points $(1,0) \to (0,3)$ and $(1,-3) \to (3,3)$.

Perpendicular bisector of $(1,0)$ and $(0,3)$: Midpoint $(0.5, 1.5)$. Gradient $-3$, perp gradient $1/3$. Equation $y – 1.5 = 1/3(x – 0.5)$.

Perpendicular bisector of $(1,-3)$ and $(3,3)$: Midpoint $(2, 0)$. Gradient $6/2 = 3$, perp gradient $-1/3$. Equation $y – 0 = -1/3(x – 2)$.

Solve intersection:

\[ \frac{1}{3}x – \frac{1}{6} + \frac{3}{2} = -\frac{1}{3}x + \frac{2}{3} \] \[ \frac{2}{3}x = \frac{2}{3} – \frac{9}{6} + \frac{1}{6} = \frac{4}{6} – \frac{8}{6} = -\frac{4}{6} \] \[ x = -1 \] \[ y = -1/3(-1 – 2) = -1/3(-3) = 1 \]

Center is $(-1, 1)$.

Check: Rotate $(1,0)$ about $(-1,1)$ $90^\circ$ ACW?

Vector: $(2, -1) \to (1, 2)$. New point $(-1+1, 1+2) = (0, 3)$. Correct.

Truncation means cutting off the number after a certain digit, without rounding.

If $N$ truncated to 1 digit is 7, the first digit is 7.

The number could be $7.0, 7.1, 7.9, 7.999…$

It cannot be 8, because truncating 8 gives 8.

It cannot be less than 7 (e.g., 6.9), because truncating 6.9 gives 6.

Ratio Yellow : Blue = $2 : 3$. Total parts = $2 + 3 = 5$.

Total volume = 50 litres.

Yellow: 20 litres needed. Sold in 5L tins.

Tins needed = $20 \div 5 = 4$ tins.

Cost = $4 \times 26 =$ £104.

Blue: 30 litres needed. Sold in 10L tins.

Tins needed = $30 \div 10 = 3$ tins.

Cost = $3 \times 48 =$ £144.

Robert sells the paint in 10 litre tins.

Total volume 50L implies $50 \div 10 = 5$ tins of green paint.

Selling price per tin = £66.96.

$\text{Percentage Profit} = \frac{\text{Profit}}{\text{Original Cost}} \times 100$

Rule: Multiply the number of choices for each course in a combination.

Add the number of ways for each separate option.

Factorise each numerator and denominator where possible.

• $4x^2 – 9$: Difference of two squares $\to (2x – 3)(2x + 3)$
• $6x + 9$: Common factor 3 $\to 3(2x + 3)$
• $x^2 – 3x$: Common factor $x \to x(x – 3)$

Cancel common factors.

• $(2x + 3)$ cancels from top and bottom.
• $x$ cancels from the second fraction.

The common denominator is the product of the individual denominators: $x(x+1)(x-2)$.

Expand the numerators and combine them.

Formula: Circumference $C = \pi d$

We know $C = 44$.

The side length of the triangle is equal to the diameter $d$.

Formula: Area = $\frac{1}{2} a b \sin(C)$

For an equilateral triangle with side $d$: Area = $\frac{1}{2} d^2 \sin(60^{\circ})$.

Features of $y = 2^x$:

• It is an exponential growth curve.
• When $x = 0$, $y = 2^0 = 1$. The y-intercept is $(0, 1)$.
• As $x \to \infty$, $y$ increases rapidly.
• As $x \to -\infty$, $y$ approaches 0 but never touches it (Asymptote $y=0$).

Equation: $x^2 + y^2 = r^2$

We are given $r^2 = 42.25$.

Different colours means:

• Joe picks Red AND Mary picks Blue
• OR Joe picks Blue AND Mary picks Red

Given $P(\text{Red}) = 0.65$.

Therefore $P(\text{Blue}) = 1 – 0.65 = 0.35$.

We know that Red counters represent 0.65 of the total.

Number of Red counters = 78.

Blue counters = Total – Red

Statement 1: $(p – 5) : (q – 5) = 5 : 1$

This means $\frac{p – 5}{q – 5} = \frac{5}{1}$.

Cross-multiplying: $1(p – 5) = 5(q – 5) \implies p – 5 = 5q – 25$.

Rearranging: $p = 5q – 20$. (Equation 1)

(P1) Associating expressions with ratio

Statement 2: $(p + 20) : (q + 20) = 5 : 2$

This means $\frac{p + 20}{q + 20} = \frac{5}{2}$.

Cross-multiplying: $2(p + 20) = 5(q + 20) \implies 2p + 40 = 5q + 100$.

Rearranging: $2p – 5q = 60$. (Equation 2)

(P1) Correct equations formed

Substitute Equation 1 ($p = 5q – 20$) into Equation 2.

Substitute $q = 20$ back into Equation 1.

Ratio $p : q = 80 : 20$.

Simplify by dividing by 20.

Gradient: $m = \frac{y_2 – y_1}{x_2 – x_1}$

Points $(4, 6)$ and $(12, 2)$.

Passes through origin $(0, 0)$ $\implies c = 0$.

Gradient = $-3$.

Equation of $\mathbf{L}_2$: $y = -3x$.

Equate the two expressions for $y$.

Multiply the entire inequality by 4.

Subtract 7 from all parts.

We need to square root. Remember that if $m^2 > 81$, then $m > 9$ OR $m < -9$.

Critical values are $\pm 9$ and $\pm 11$.

Method: Volume of Large Cone – Volume of Small Cone.

Large Cone: Height $H = 6.4$, Diameter = 7.2 $\implies$ Radius $R = 3.6$.

Small Cone: Height $h = 3.2$. Radius $r$? Since height is half (3.2 is half of 6.4), radius is half of 3.6 $\implies r = 1.8$.

Radius $R = 3.6$. Volume = $\frac{1}{2} \times \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3$.

Formula: Mass = Density $\times$ Volume.

Formula: Average Density = Total Mass $\div$ Total Volume.

Consider the cyclic quadrilateral $AORC$ in the right-hand circle.

Let $\angle CAB = x$. This is the same as $\angle CAO$.

Opposite angles in a cyclic quadrilateral sum to $180^{\circ}$.

$CRB$ is a straight line. Angles on a straight line sum to $180^{\circ}$.

In the left circle with centre $O$, $OB$ and $OR$ are radii.

Therefore, triangle $ORB$ is an isosceles triangle.

Base angles of an isosceles triangle are equal.

We started with $\angle CAB = x$.

We proved $\angle ABC = x$.