GCSE Edexcel Higher Paper 3 – June 2024

💡 What are we asked to do?

We need to find the largest number that divides exactly into both 63 and 105.

Strategy: We can use prime factor trees to find the prime factors of each number, then multiply the common prime factors.

Break down each number into products of prime numbers.

Identify the numbers that appear in both lists.

💡 Why we do this: A positive power of 4 means we move the decimal point 4 places to the right to make the number bigger.

💡 Why we do this: A negative power of -5 means we move the decimal point 5 places to the left to make the number smaller.

💡 Strategy: To add numbers in standard form without a calculator, it’s safer to convert them to ordinary numbers or the same power of 10 first. Since this is a calculator paper, you can use your calculator, but let’s show the manual check.

💡 Analysis: Rana drew a rectangle that is 7cm wide and 4cm high.

The width of the side view is indeed the length of the prism (7cm). However, the height of the rectangle must be the perpendicular height of the triangle, NOT the slanted side length (4cm).

Because the triangle is slanted, the vertical height must be less than the slanted side of 4cm.

💡 What is a Plan View? It is the view looking straight down from the top.

Looking down, we see two rectangular faces sloping away from the top ridge.

• The total width is the base of the triangle: 6 cm.
• The total length is the length of the prism: 7 cm.
• There is a ridge running down the middle (since it’s an isosceles triangle), so we draw a line in the center.

💡 Understanding Percentage Increase: An increase of 6% means we start with 100% and add 6%.

💡 Formula: \( \text{New Amount} = \text{Original Amount} \times (\text{Multiplier})^{\text{Number of Years}} \)

Since we are counting workers (people), we usually round to the nearest whole number. Though in exams, the exact calculation is often accepted if it’s a clean decimal, usually people can’t be 0.4.

💡 Why? We know the mass and density of the flour, and it fills the tin completely. This lets us find the total volume of the tin.

Formula: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \)

💡 Insight: The salt does not fill the tin. There is 700 cm³ of empty space.

\( \text{Volume of Salt} = \text{Total Volume} – \text{Empty Space} \)

We know the mass of the salt (600g) and now its volume (300 cm³).

Formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)

💡 Rule: The probabilities on branches coming from a single point must add up to 1.

💡 Rule: To find the probability of two independent events both happening, we multiply their probabilities (follow the path along the branches).

Path: Heads on A \(\rightarrow\) Heads on B

💡 Formula: Volume of a cylinder = \( \pi r^2 h \)

We need to fill this volume at a rate of \( 250 \text{ cm}^3 \) per second.

Formula: \( \text{Time} = \frac{\text{Volume}}{\text{Rate}} \)

Divide by 60 to convert seconds to minutes.

💡 Method: Multiply vector \(\mathbf{a}\) by 2, then add vector \(\mathbf{b}\).

The vector \(\begin{pmatrix} 5 \\ 8 \end{pmatrix}\) means: 5 units Right and 8 units Up.

Start anywhere on the grid where there is enough space (e.g., bottom left) and draw a line.

💡 Formula: Volume of cube = \( \text{side}^3 \)

💡 Formula: Volume of pyramid = \( \frac{1}{3} \times \text{Base Area} \times \text{Height} \)

The base is a square with side 8 cm.

We know the volumes are equal.

Rearrange to find \( h \).

💡 Information: The ratio of Red : Yellow is 3 : 5.

We are told there are \( x \) red counters.

Since the red part of the ratio is 3 parts, then 3 parts = \( x \).

💡 Information: Counters in Bag B = Half of Bag A.

Add the totals from Bag A and Bag B.

💡 Identify the 5 Key Values:

• Lowest Speed (Min): 25
• Lower Quartile (LQ): 35
• Median: 40
• Upper Quartile (UQ): LQ + IQR = \( 35 + 12 = 47 \)
• Highest Speed (Max): Lowest + Range = \( 25 + 37 = 62 \)

Check: We plot these 5 points on the grid.

💡 Rule for Comparisons: Always compare two things:

1. Average (Median): Who was faster?
2. Spread (IQR or Range): Who was more consistent?

Data extraction:

• Friday Median: 40 | Sunday Median: 42
• Friday IQR: 12 | Sunday IQR: 15

Scale Factor -2 means:

• The size will be 2 times larger.
• The position will be on the opposite side of the centre of enlargement.
• The orientation will be rotated 180°.

Centre of enlargement is \((0,0)\).

Multiply each coordinate of T by \(-2\).

💡 “Choose 2”: This is a combinations problem. The order doesn’t matter (choosing Alice then Bob is the same pair as Bob then Alice).

Choice 1: There are 30 students.

Choice 2: There are 29 students left.

Since order doesn’t matter, we divide by 2.

Final state: £2937.14

This was after taking £1000.

So before taking £1000, the amount was: \( 2937.14 + 1000 = 3937.14 \)

This amount (£3937.14) includes 3.5% interest added during 2023.

To find the value at the start of 2023, we reverse the interest multiplier.

Multiplier for 3.5% increase = 1.035.

The amount £3804 was after taking £750 at the end of 2022.

So before taking £750, the amount was: \( 3804 + 750 = 4554 \)

This amount (£4554) includes 3.5% interest added during 2022.

Divide by the multiplier again to find the original investment.

Cancel the common factor \( (a-3) \).

Factorising a Quadratic: We need two brackets that expand to \( 3k^2 + 11k – 4 \).

Start with \( (3k \dots)(k \dots) \).

We need numbers that multiply to -4 and add to make the middle term 11k when combined.

Step 1: Factorise everything.

• \( 4-x^2 \) is a difference of two squares: \( (2-x)(2+x) \)
• \( x^2+3x \) factorises to \( x(x+3) \)

Step 2: Flip the second fraction (divide becomes multiply).

Replace \( x \) with \( -3 \) in the function \( f(x) \).

💡 Order Matters: \( fg(1) \) means find \( g(1) \) first, then put that result into \( f \).

💡 Method: \( g^{-1}(4) \) asks “What input \( x \) gives an output of 4?”

Set \( g(x) = 4 \) and solve for \( x \).

We are given \( h_1 = 8 \).

The multiplier is \( 0.55 \). This is a geometric sequence.

We need to find \( h_4 \).

💡 Why? We know the area of \( \Delta ABC \) is 54, and we have side \( AB = 17 \) and \( \angle BAC = 35^\circ \).

We can use the area formula: \( \text{Area} = \frac{1}{2}ab \sin C \).

In \( \Delta ACD \), we know \( \angle ACD = 48^\circ \) and \( \angle ADC = 57^\circ \).

We can find the third angle \( \angle CAD \).

We need side \( AD \) (or \( CD \)) to calculate the area.

Using Sine Rule in \( \Delta ACD \): \( \frac{a}{\sin A} = \frac{d}{\sin D} \)

Using Area formula again with the angle between \( AC \) and \( AD \).

P (3 s.f.): \( 5.88 \). Lower Bound is \( 5.875 \).

Q (2 s.f.): \( 3.6 \). Upper Bound is \( 3.65 \).

💡 Strategy: To get the Lower Bound of a division \( \frac{P}{Q} \), we need the Smallest Numerator divided by the Largest Denominator.

💡 Property: The difference between terms is constant. \( u_2 – u_1 = u_3 – u_2 \).

💡 Property: The ratio between terms is constant. \( \frac{u_2}{u_1} = \frac{u_3}{u_2} \).

Cross multiply: \( (u_2)^2 = u_1 \times u_3 \).

The inner hexagon consists of 6 equilateral triangles with side length \( r \).

The outer hexagon has sides tangent to the circle. The distance from the center to the side (apothem) is \( r \).

Consider the small right-angled triangle formed by the radius, half the side, and the line to the vertex.

The angle at the center is \( 60^\circ / 2 = 30^\circ \).

\( \tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{half-side}}{r} \)

The circumference of the circle lies between the perimeter of the inner hexagon and the perimeter of the outer hexagon.

\( \text{Perimeter}_{\text{inner}} < \text{Circumference} < \text{Perimeter}_{\text{outer}} \)