GCSE Nov 2022 Edexcel Higher Paper 3 (Calculator)

What are we being asked to do?

We need to rearrange the formula so that it starts with \( a = \dots \). Currently, \( a \) is mixed in with other numbers.

Why do we do this?

To get \( a \) on its own, we first move everything else to the other side of the equation. We start with the \( -9 \).

What is the final step?

Currently \( a \) is multiplied by 3. To isolate \( a \), we do the opposite operation: divide by 3.

How do we share in a ratio?

To share an amount in a ratio like \( 3:5 \), we first need to find the total number of parts. Rob should have added \( 3 + 5 = 8 \) parts in total.

What did Rob do?

Rob divided the total amount (120) by each part of the ratio individually (3 and 5). This is incorrect because he isn’t sharing the total into equal shares of the sum of the parts.

Why do this?

This is a “frequency tree” or “two-way table” problem. It’s best to set up a table to fill in the missing gaps logically.

What do we know?

• Total students = 200
• Total Boys = 90. (So Total Girls = \( 200 – 90 = 110 \))
• Total Spanish = 70.
• Total French = 104.
• Boys who chose French = 60.
• Girls who chose German = 18.

How do we find Boys who chose Spanish?

We can find the number of German students first. Or we can work row by row.

1. Find Total German: \( 200 – 104 – 70 = 26 \).

2. Find Boys choosing German: Total German (26) – Girls choosing German (18) = \( 26 – 18 = 8 \).

3. Find Boys choosing Spanish: Total Boys (90) – Boys French (60) – Boys German (8).

What is the formula?

The volume of a cylinder is \( V = \pi r^2 h \).

We are given the diameter \( d = 80 \text{ cm} \), so the radius \( r = 40 \text{ cm} \).

Height \( h = 160 \text{ cm} \).

Why?

Karina has 4 identical tanks.

Why convert?

The fertiliser amount is given in litres. We know \( 1 \text{ litre} = 1000 \text{ cm}^3 \).

How much is fertiliser?

The ratio of fertiliser to water is \( 1 : 100 \). This means for every \( 1+100 = 101 \) parts of mixture, 1 part is fertiliser.

Fraction of fertiliser = \( \frac{1}{101} \).

Does she have enough?

She needs approximately 31.85 litres. She has 32 litres.

How do we find the scale factor?

We look for corresponding sides where we have values for both triangles.

Side \( AC \) corresponds to side \( DF \).

\( AC = 5 \text{ cm} \), \( DF = 20 \text{ cm} \).

How to find EF?

Side \( EF \) corresponds to side \( BC \) (the base).

\( BC = 4 \text{ cm} \).

Since \( DEF \) is larger, we multiply by the scale factor.

How to find AB?

Side \( AB \) corresponds to side \( DE \).

\( DE = 22 \text{ cm} \).

Since \( ABC \) is smaller, we divide by the scale factor.

How do we fill the gaps?

Probabilities on branches originating from the same point must sum to 1.

Which path do we take?

We need “Win Sports” AND “Win Music”. In probability, “AND” means we multiply along the branches.

What is the conversion factor?

\( 1 \text{ m} = 100 \text{ cm} \).

For volume (cubed), we cube the conversion factor: \( 1 \text{ m}^3 = 100^3 \text{ cm}^3 = 1,000,000 \text{ cm}^3 \).

Convert km to m:

\( 1 \text{ km} = 1000 \text{ m} \).

Convert hours to seconds:

\( 1 \text{ hour} = 60 \text{ minutes} \times 60 \text{ seconds} = 3600 \text{ seconds} \).

Why find the total?

\( \text{Total} = \text{Mean} \times \text{Count} \).

We know the mean for all 50 people.

Find the men’s contribution:

We know the mean for the 20 men.

Subtract men from total to get women’s total:

How do we move the decimal?

The power is \(-4\), so we move the decimal point 4 places to the left.

Use calculator or group terms:

Group numbers and powers of 10.

\( \frac{2.56 \times 4.12}{1.6} \times \frac{10^6 \times 10^{-3}}{10^{-2}} \)

Look at the expansion of the negative sign:

When subtracting a bracket, the negative sign applies to every term inside the bracket.

\( – (x^2 – 2x – 4) \) should become \( -x^2 + 2x + 4 \).

Peter wrote \( -x^2 – 2x – 4 \). He failed to change the signs of \( -2x \) and \( -4 \).

What integers satisfy the inequalities?

Remember that \( < \) means strictly less than (do not include the boundary numbers).

Which pairs add up to 14?

We check each possible \( x \) to see if the required \( y \) (where \( y = 14 – x \)) is in our allowed list.

What does this mean?

If the calculator screen starts with “1.2…”, the number implies truncation, not rounding.

The number could be exactly 1.2, or 1.20001, or 1.29999…

It cannot be 1.3 or higher, and it cannot be less than 1.2.

Lower Bound: The smallest number starting with 1.2 is 1.2.

Upper Bound: The number must be less than 1.3.

Where should cumulative frequency be plotted?

Cumulative frequency points should be plotted at the upper bound of the class interval (e.g., at \( t=10 \), not \( t=5 \)).

Looking at the graph, the points are plotted at the midpoints (5, 15, 25, etc.). This is incorrect.

Are the axes labelled correctly?

The y-axis is labelled “Cumulative frequency”.

The x-axis has numbers but is missing the label “Time (seconds)”.

Apply the power to everything inside the bracket:

\( (3x^5 y^6)^4 = 3^4 \times (x^5)^4 \times (y^6)^4 \)

Recall: \( (a^m)^n = a^{m \times n} \)

Start with \( (x + 2)(x – 3) \):

Multiply \( (x^2 – x – 6) \) by \( (x + 4) \):

Multiply each term in the first expression by each term in the second.

Multiplication Rule: To find the number of ways to choose one from set A AND one from set B, multiply the counts.

Addition Rule: Since he chooses option 1 OR option 2 OR option 3, we add the results.

Use the given ratio \( 2:1 \):

Let angle \( BCD = x \).

Then angle \( EAB = 2x \).

Use Triangle BCD and Straight Lines:

In triangle \( BCD \), the angles sum to \( 180^\circ \).

We need angle \( BDC \) first. But wait, we can find angle \( BDE \) directly using angles on a straight line?

Let’s look at the exterior angle of the cyclic quad, or just calculate step-by-step.

In \( \Delta BCD \): \( \angle BDC = 180 – 60 – x = 120 – x \).

Angles on straight line \( EDC \): \( \angle BDE + \angle BDC = 180 \).

\( \angle BDE = 180 – (120 – x) = 60 + x \).

Opposite angles sum to \( 180^\circ \):

In cyclic quadrilateral \( ABDE \), angles \( A \) and \( D \) (specifically \( \angle EAB \) and \( \angle BDE \)) are opposite.

Ratio A : B

\( A = \frac{2}{3} B \implies \frac{A}{B} = \frac{2}{3} \implies A:B = 2:3 \).

Total parts for A+B = \( 2+3=5 \) parts.

Ratio C : D

\( C = 75\% D = \frac{3}{4} D \implies C:D = 3:4 \).

Total parts for C+D = \( 3+4=7 \) parts.

✓ (P1)

We are told: \( (A+B) = 3 \times (C+D) \).

Let the total weight of \( (C+D) \) be \( W \). Then \( (A+B) \) is \( 3W \).

We need a number for \( W \) that is divisible by 7 (for C+D parts) and where \( 3W \) is divisible by 5 (for A+B parts).

Using the totals we just found:

What happens when you reflect in two parallel lines?

Reflecting in \( x = a \) then \( x = b \) results in a translation.

The translation vector is \( \begin{pmatrix} 2(b-a) \\ 0 \end{pmatrix} \).

First line: \( a = 2 \). Second line: \( b = 6 \).

Distance between lines is \( 6 – 2 = 4 \).

The translation is double this distance in the positive x direction.

Share the remaining probability:

The ratio Red : Green is 7 : 8.

Total parts = \( 7 + 8 = 15 \).

Fraction that is Green = \( \frac{8}{15} \).

Multiply probability by number of trials:

Goal: Prove \( \Delta ACE \cong \Delta DBE \).

We look for equal sides and angles.

Use the Ratio \( 1:2:1 \):

Let the length of one part be \( x \).

\( AB = x \), \( BC = 2x \), \( CD = x \).

Length \( AC = AB + BC = x + 2x = 3x \).

Length \( DB = DC + CB = x + 2x = 3x \).

Therefore, \( AC = DB \).

✓ (C1 – Reasoning for sides)

Triangle ADE is Isosceles:

Given \( AE = DE \), the base angles of isosceles triangle \( ADE \) are equal.

Therefore, \( \angle CAE = \angle BDE \) (or \( \angle A = \angle D \)).

✓ (C1 – Reasoning for angles)

Why? To complete the square, it is easier if the coefficient of \( x^2 \) is 1.

How? Take half of the coefficient of \( x \) (which is -14), square it, and subtract it.

Half of -14 is -7. Square of -7 is 49.

Expand the outer bracket: Multiply everything by 4.

The common denominator is \( (x – 5)(x + 5) \), which expands to \( x^2 – 25 \).

We need to write “3” as a fraction with this denominator too.

What does \( f(x + 2) \) mean?

A change inside the bracket affects the x-axis.

\( x + a \) is a shift to the LEFT by \( a \).

So, we shift the entire graph 2 units to the left.

How is Graph B related to Graph A?

Graph A is a U-shape. Graph B is an n-shape.

Graph B is a mirror image of Graph A across the x-axis.

A reflection in the x-axis changes \( y \) to \( -y \).

So if \( y = g(x) \), the new equation is \( y = -g(x) \).

To go from \( C \) to \( F \), we can go \( C \to D \to E \to F \).

First find \( \vec{CE} \) to locate point \( P \).

\( \vec{CE} = \vec{CD} + \vec{DE} = 2\mathbf{a} – \mathbf{b} \).

Since \( CE = EP \) and it’s a straight line, \( \vec{EP} = \vec{CE} = 2\mathbf{a} – \mathbf{b} \).

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

We know the average density of the whole pyramid.

Formula: Sum of interior angles = \( (n-2) \times 180 \).

For a heptagon (\( n=7 \)): \( 5 \times 180 = 900^\circ \).

One interior angle (e.g., angle \( GAB \)) = \( \frac{900}{7} \approx 128.57^\circ \).

Use Area of Triangle Formula:

\( \text{Area} = \frac{1}{2} ab \sin(C) \).

In \( \Delta ABG \), sides \( AG \) and \( AB \) are equal (regular polygon). Let length \(= x\).

\( 30 = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{900}{7}) \).

Use Cosine Rule on \( \Delta ABG \):

\( a^2 = b^2 + c^2 – 2bc \cos(A) \).

\( GB^2 = x^2 + x^2 – 2(x)(x) \cos(128.57…) \).