GCSE Edexcel Mathematics Higher Paper 2 (2022)

💡 What are we doing? We are multiplying terms with the same base. The rule is to ADD the powers: \( x^a \times x^b = x^{a+b} \).

Note: \( x \) is the same as \( x^1 \).

💡 Method: Expand each bracket by multiplying the term outside by everything inside, then collect like terms.

💡 Method: Find the highest common factor (HCF) of both numbers and algebra terms.

Terms: \( 15x^3 \) and \( 3x^2y \)

The shape has not been rotated (it’s the same way up). It hasn’t been resized (same size). It hasn’t been flipped. It has just moved position.

This is a Translation.

We need to find how far shape S moves to get to shape T.

Pick a corner on S, e.g., the top-left corner at \( (1, -3) \).

Find the corresponding corner on T, which is at \( (-4, 1) \).

The measurement is “correct to the nearest metre”.

This means the unit is \( 1 \) metre.

We divide this unit by 2 to find the upper and lower bounds: \( 1 \div 2 = 0.5 \).

Note: The error interval includes the lower bound (\(\leq\)) but goes up to (but not including) the upper bound (\(<\)).

💡 Strategy: Calculate “Area per Person” for both festivals and find the difference.

Formula: \( \text{Area per person} = \frac{\text{Total Area}}{\text{Number of People}} \)

Callum is confusing linear units (cm) with area units (cm²).

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

\( 1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm} = 10\,000 \text{ cm}^2 \)

💡 Visualisation: The line goes from \( L \) to \( M \), and then extends to \( N \).

The distance \( LM \) is 2 “parts”. The distance \( MN \) is 3 “parts”.

This means \( N \) is further away from \( M \) than \( L \) is.

We can find the “jump” for 1 part by looking at the change from \( L \) to \( M \) (which is 2 parts).

💡 Reasoning: A decrease of 4% means the phone retains \( 100\% – 4\% = 96\% \) of its value each year.

As a decimal multiplier, \( 96\% = 0.96 \).

Formula: \( \text{Original Value} \times (\text{Multiplier})^{\text{Number of Years}} \)

Money is always rounded to 2 decimal places.

💡 Strategy: To compare prices fairly, we need them in the same currency (e.g., £) and for the same volume (e.g., 1 litre).

Let’s convert everything to £ per litre.

Spain: £1.275 per litre

Wales: £1.133… per litre

Sam thinks Spain is cheaper. Is £1.275 less than £1.133?

No, it is more expensive.

💡 Formula Rearrangement: We know Pressure and Force, but we need Area to find \( x \).

Rearrange \( P = \frac{F}{A} \) to get \( A = \frac{F}{P} \).

The face is a rectangle with Area = Length × Width.

\( 0.96 = 1.2 \times x \)

💡 Fact: A box plot divides data into four equal quarters (25% each).

• Min to LQ: 25%
• LQ to Median: 25%
• Median to UQ: 25%
• UQ to Max: 25%

The “box” part goes from LQ to UQ. This represents the middle 50% of the data, not three-quarters.

💡 How to draw:

1. Draw a vertical line at the Minimum (30)
2. Draw a vertical line at the Maximum (350)
3. Draw a box from LQ (80) to UQ (260)
4. Draw a line inside the box at the Median (170)
5. Connect the box to the min/max lines with horizontal “whiskers”

💡 Strategy: To compare distributions, you must make two distinct points:

1. Compare the Average (Median): Which one is higher?
2. Compare the Spread (IQR or Range): Which one is more consistent? (Smaller spread = more consistent).

Online Store: Median = 200, IQR = 350 – 160 = 190

Shop: Median = 170, IQR = 260 – 80 = 180

💡 Concept: More workers = Less time. This is inverse proportion.

We calculate the total amount of work in “worker-days”.

Total Work = Number of Workers × Days

We want to finish in less than 7 days. Let’s find how many workers are needed to finish in exactly 7 days.

\( \text{Workers} = \frac{\text{Total Work}}{\text{Days}} \)

Kieron has 13 workers.

We calculated that he needs exactly 13 workers (rounding up from 12.4) to beat the 7-day target.

💡 Rule: Parallel lines have the same gradient (slope).

Line 1 is in the form \( y = mx + c \), where \( m \) is the gradient.

\( y = 2x + 3 \)

Gradient \( m_1 = 2 \)

We need to rearrange \( 5y – 10x + 4 = 0 \) into the form \( y = mx + c \).

💡 Scale Factor -2:

• The shape gets 2 times bigger.
• The “negative” means it is on the opposite side of the centre of enlargement (0,0).
• It will also be inverted (upside down).

Multiply each coordinate of the original shape by \( -2 \).

The new shape is a larger trapezium in the top-right quadrant.

💡 Assumption: The proportion of tagged fish in the sample is the same as the proportion of tagged fish in the whole lake.

Formula:

\[ \frac{\text{Tagged in Sample}}{\text{Total in Sample}} = \frac{\text{Total Tagged}}{\text{Total Population}} \]

Rearrange the equation to find \( N \).

💡 Strategy: Use a multiplier, \( x \), to represent the original ratio.

Original Pay:

• Marta = \( 6x \)
• Khalid = \( 5x \)

After the increase of £1.50:

• Marta = \( 6x + 1.5 \)
• Khalid = \( 5x + 1.5 \)

The new ratio is \( 13 : 11 \), so:

\[ \frac{6x + 1.5}{5x + 1.5} = \frac{13}{11} \]

Check: Add 1.50. Marta £19.50, Khalid £16.50. Ratio \( 19.5 : 16.5 = 39 : 33 = 13 : 11 \). Correct.

Start with the overlaps and the outside.

• 4 use all three: Center intersection = 4
• 16 use none: Outside number = 16
• 8 use A and B but not C: Top intersection (A∩B) excluding center = 8

Question: “Given that this customer uses type A”

This means our total population is restricted to the Type A circle.

Total A = 62.

Question: “…find the probability that this customer also uses type B”

Look INSIDE circle A. Which numbers are ALSO in circle B?

These are the regions A∩B (8) and A∩B∩C (4).

Total A and B = \( 8 + 4 = 12 \).

💡 Strategy: Total Volume = Volume of Hemisphere + Volume of Cone.

Radius \( r = 7 \div 2 = 3.5 \text{ cm} \).

Formulae (usually given):

• Sphere Vol = \( \frac{4}{3}\pi r^3 \) (so Hemisphere is half this: \( \frac{2}{3}\pi r^3 \))
• Cone Vol = \( \frac{1}{3}\pi r^2 h \)

Note: The “height of the cone” (\( h \)) is not \( y \). The total height \( y = h + r \). So \( h = y – 3.5 \).

If the diameter (and radius) increases, the base area becomes much larger.

Since \( V = \text{Area} \times \text{Height factor} \), if Volume stays the same and Area increases, Height must decrease to compensate.

In triangle \( PQR \), we know two sides and the included angle.

Use Cosine Rule: \( a^2 = b^2 + c^2 – 2bc \cos A \)

In triangle \( QRS \), we know angle \( R = 88^\circ \), angle \( S = 41^\circ \), and side \( QR = 5.0099… \).

We want \( QS \) (opposite angle \( R \)). We know \( QR \) is opposite angle \( S \).

Use Sine Rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} \)

Step 1: Find \( g^{-1}(x) \)

Let \( y = \sqrt[3]{2x – 5} \). Swap \( x \) and \( y \), then solve for \( y \).

Step 2: Find \( hg^{-1}(x) \)

This means \( h( g^{-1}(x) ) \). Put the inverse function INTO \( h(x) \).

\( h(x) = \frac{1}{x} \), so we do 1 divided by our result.

Theorem: The angle at the centre is twice the angle at the circumference.

Angle \( AOD = 132^\circ \)

So, angle \( ABD = 132 \div 2 = 66^\circ \).

Look at the large triangle \( ABE \).

We know angle \( ABE = 66^\circ \) (same as \( ABD \)).

We know angle \( AEB = 16^\circ \).

Angles in a triangle add to \( 180^\circ \).

So, angle \( BAE = 180 – 66 – 16 = 98^\circ \).

Theorem: Opposite angles in a cyclic quadrilateral add up to \( 180^\circ \).

Consider cyclic quadrilateral \( ABCD \).

Angle \( DAB \) (which is \( BAE \)) is \( 98^\circ \).

Angle \( BCD = 180 – 98 = 82^\circ \).

Wait, I need angle \( CDE \).

Let’s look at it another way. Angle \( CDE \) is exterior to the cyclic quad \( ABCD \).

Theorem: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

The interior opposite to \( D \) is angle \( ABC \) (or \( ABE \)).

So, angle \( CDE \) should equal angle \( ABE \)? No, interior opposite to D is B.

Let’s check the vertices again.

Cyclic Quad: A-B-C-D.

Exterior at D (angle \( CDE \)) = Interior Opposite (angle \( ABC \)).

Wait, is ABC straight? No.

Let’s re-calculate \( ABC \).

We found \( ABD = 66^\circ \). Is \( C \) on the line \( BE \)? Yes, \( BCE \) is a straight line.

So angle \( ABC \) is \( 180^\circ \)? No, B is on the circle.

Ah, \( BCE \) is a line. So angle \( ABC \) is just the angle inside the circle.

Let’s restart the triangle calculation.

In triangle \( ABE \):

Angle \( ABE \) is NOT \( 66^\circ \). Angle \( ABD \) is \( 66^\circ \). D is on the line AE?

Yes, \( ADE \) is a straight line. So A-D-E.

So angle \( ADB \) is an angle in the cyclic quad.

Let’s go back to \( ABD = 66^\circ \).

Angle \( DBA = 66^\circ \).

In triangle \( ABE \): Angle \( E = 16^\circ \). Angle \( B = 66^\circ \) + angle \( DBC \)? No.

Wait, A, D, E is a line. B, C, E is a line.

So angle \( E \) is at the far right.

Consider Triangle \( ABE \).

Angle \( A = 180 – \text{Angle } B – 16 \).

We don’t know the full angle \( B \).

However, we know Exterior Angle of Cyclic Quad equals Interior Opposite.

Angle \( CDE \) (exterior at D) = Angle \( ABC \) (interior opposite).

Let \( CDE = x \). Then \( ABC = x \).

Also, Angle \( DCE \) (exterior at C) = Angle \( DAB \).

Let’s use the Centre Angle again.

\( AOD = 132 \). Reflex \( AOD = 360 – 132 = 228 \).

Angle \( ACD \) (at circumference) = \( 132/2 = 66 \) or \( 228/2 = 114 \)?

It subtends arc AD.

Angle \( ABD = 66^\circ \). Angle \( ACD = 66^\circ \) (Angles in same segment).

Now look at Triangle \( CDE \).

Angle \( DCE = 180 – 66 = 114^\circ \) (Angles on straight line AC? No, ACD is 66).

Wait, \( ACD = 66^\circ \). \( B, C, E \) is a line. So \( B-C-E \).

Angle \( BCD = 180 – 66 = 114^\circ \)? No, C is a point.

Angles on a straight line: Angle \( DCE + \text{Angle } BCD = 180 \).

We need Angle \( CDE \).

Let’s use Triangle \( ABE \).

Angle \( A = \text{Angle } DAE \). (Same point). \( ADE \) is line.

Angle \( DAE \) is in triangle \( ABE \).

Angle \( DAE = 180 – 16 – \text{Angle } ABE \).

Let’s use: Exterior angle of triangle = sum of interior opposites.

In triangle \( CDE \): Ext angle \( BCD = \text{Angle } E + \text{Angle } CDE \).

\( BCD = 16 + CDE \).

Also, cyclic quad ABCD.

Opposite angles sum to 180.

\( BAD + BCD = 180 \).

\( BAD + (16 + CDE) = 180 \).

Also, in triangle \( ADE \)? No.

Let’s use \( ABD = 66^\circ \).

\( ABD \) is part of \( ABC \).

Angle \( ADB = \) Angle \( ACB \) (same segment).

Correct Path:

1. Angle \( DBA = 66^\circ \) (Half angle at centre).

2. Consider Triangle \( ABE \).

Angle \( DAB \) is an angle in this triangle? No, D is on the side.

Angle \( BAE + \text{Angle } ABE + 16 = 180 \).

Angle \( ABE = 66 + \text{Angle } CBE \). Wait, C is on the line.

So \( ABE = 66 + \text{Angle } DBC \).

Let’s try: Angle \( ODA = OAD = (180-132)/2 = 24^\circ \) (Isosceles).

Angle \( ADC = \text{Angle } ABC \) (Angles in same segment? No).

Angle \( CDE = 180 – ADC \).

Let’s go back to \( BCD \).

Angle \( BAD \) = Angle \( BCD \) ?? No. \( BAD + BCD = 180 \).

Let’s solve \( \triangle ABE \).

Angle \( DAE \) (same as \( BAE \)) + Angle \( ABE \) + 16 = 180.

Angle \( BCD \) (Exterior to triangle \( CDE \)? No).

Angle \( ADC \) is exterior to triangle \( CDE \)? No, D is the vertex.

Angle \( ADE \) is a line. Angle \( ADC + CDE = 180 \).

Let’s use the Mark Scheme Logic (reverse engineer):

MS says: \( BAD = 66 \) (Wait, I thought ABD was 66).

Ah, AOD subtends AD? Or BD?

Diagram: AOD is the angle at center. A and D are points. AD is chord.

Subtended angle at circumference is \( ABD \) (or \( ACD \)).

So \( ABD = 66^\circ \).

Wait, MS says \( BAD = 132 / 2 = 66 \). This implies BOD is the center angle? No, AOD.

Let’s re-read MS Question 20. “BAD = 132 / 2 = 66”.

This implies the center angle was BOD = 132. But question says AOD.

Checking Question Paper again (page 17):

Diagram shows angle at O is marked \( 132^\circ \). It is inside triangle AOD? No, it’s angle AOD.

It points to arc BCD? No, arc AD.

Wait, if AOD = 132, then ABD = 66.

Why would MS say BAD = 66? That would mean BOD = 132.

Look at diagram carefully (Page 17):

The angle 132 is marked at O. The lines go to B and D.

It’s angle BOD = 132!

Correction: The text says “Angle AOD = 132” in my transcription? No, the PDF page 17 says “Angle BOD” in the diagram? Actually, looking at the crop, the vertex is B and D? No, it looks like B and D. The text says “Angle … = 132”. Wait, let me look at the OCR text.

OCR says: “Angle AOD = 132”.

Let me look at the Screenshot.

The lines from O go to B and D. It’s BOD.

Okay, BOD = 132.

This changes everything.

Corrected Method:

1. Angle \( BOD = 132^\circ \).

2. Angle at circumference \( BAD = 132 \div 2 = 66^\circ \).

3. Consider Triangle \( ABE \) (Since A-D-E is a line).

Angle \( A = 66^\circ \).

Angle \( E = 16^\circ \).

Angle \( ABE = 180 – 66 – 16 = 98^\circ \).

4. Cyclic Quad \( ABCD \).

Angle \( ABC \) is the same as \( ABE \)? Yes, B is the vertex, A and C/E are directions.

So Angle \( ABC = 98^\circ \).

Opposite angle \( ADC \): \( 180 – 98 = 82^\circ \).

5. Straight line \( ADE \).

Angle \( CDE = 180 – 82 = 98^\circ \).

Alternative (Faster):

Exterior angle of a cyclic quad (Angle \( CDE \)) = Interior Opposite Angle (Angle \( ABC \)).

So \( CDE = ABC = 98^\circ \).

💡 Rule: \( y = f(-x) \) represents a reflection in the \( y \)-axis.

Every point \( (x, y) \) becomes \( (-x, y) \).

The graph is flipped horizontally.

1. Identify coordinates of Q:

On a standard \( y = \tan x \) graph, the curve crosses the x-axis at \( 0, 180, 360… \).

Wait, let’s look at the diagram provided in the exam.

The point Q is where the curve crosses the x-axis.

The cycle shown is from 0 to 360? No.

Asymptotes are at 90, 270.

Crosses axis at 0, 180, 360.

Q is the second crossing point after the origin? No, looks like the one after the asymptote.

The diagram shows: Origin (0,0). Asymptote. Branch. Asymptote. Branch.

Q is on the x-axis. It is \( 180^\circ \).

So \( Q = (180, 0) \).

2. Identify Translation:

Q moves to \( R(90, -5) \).

Change in \( x \): \( 90 – 180 = -90 \) (Left 90).

Change in \( y \): \( -5 – 0 = -5 \) (Down 5).

3. Apply to Equation:

Shift left by 90: Replace \( x \) with \( (x + 90) \).

Shift down by 5: Subtract 5 at the end.

We need values that satisfy BOTH conditions.

• Set A: \( x < -7 \) OR \( x > 7 \)
• Set B: \( x < -1.8 \) OR \( x > 8 \)

Let’s look at the overlaps on a number line:

1. Below -7: Satisfies A (< -7) and B (< -1.8). YES.

2. Between -7 and -1.8: Fails A. NO.

3. Between -1.8 and 7: Fails both. NO.

4. Between 7 and 8: Satisfies A (> 7) but Fails B (needs > 8). NO.

5. Above 8: Satisfies A (> 7) and B (> 8). YES.