Pearson Edexcel GCSE (9-1) Mathematics
Paper 3 (Calculator) Higher Tier
November 2023

What is standard form?

Standard form is written as \( A \times 10^n \), where \( 1 \leq A < 10 \) and \( n \) is an integer.

We need to move the decimal point so that the number falls between 1 and 10.

Understanding the negative power:

A power of \( -4 \) means the number is small. We need to move the decimal point 4 places to the left.

What is the probability of landing on A?

From the table, the probability for A is given as \( 0.4 \).

How do we estimate the count?

To find the estimated number of times an event happens, we multiply the total number of trials by the probability of that event.

Why is it an “estimate”?

We only have grouped data, not the exact temperatures. We must use the midpoint of each group to represent that group.

Method:

1. Find the midpoint for each interval.
2. Multiply the midpoint by the frequency (\( f \times x \)).
3. Sum these values (\( \sum fx \)).
4. Divide by the total frequency (\( \sum f \)).

Analysis:

Seija claims the median is 22.5 because it’s the middle of the “middle group” (20-25).

To find the median position: \( \frac{n+1}{2} = \frac{51}{2} = 25.5\text{th} \) value.

Cumulative frequencies:

• 10-15: 2
• 15-20: 2+8 = 10
• 20-25: 10+13 = 23
• 25-30: 23+21 = 44 (The 25.5th value is in this group!)

The median lies in the 25-30 group, NOT the 20-25 group. Also, Seija’s reasoning about “middle number in the middle group” is statistically incorrect; the median depends on frequency distribution, not just group position.

We need to compare Jenna’s diagram to the inequality \( -3 < x \leq 4 \).

• The inequality symbol for 4 is \( \leq \) (less than or equal to). This requires a filled (solid) circle. Jenna used an empty circle.
• The inequality symbol for -3 is \( < \) (greater than -3). This requires an empty circle, which she did correctly. Wait, let's look closer at the question logic.
• Usually, mistakes involve the circle type or the direction/position.
• Let’s check the start and end points. She drew from -3 to 4. This is correct.
• Mistake 1: The circle at 4 should be filled/colored in because of the \( \leq \) sign.
• Mistake 2: Is there another? Jenna’s line is correct.
• Let’s re-read the diagram. Ah, strictly speaking, \( -3 < x \) means an empty circle at -3. She has drawn that.
• Is it possible she started at the wrong place? No, starts at -3.
• Wait, is the inequality \( -3 < x \leq 4 \)? Yes.
• Mistake 1: Circle at 4 should be solid.
• Mistake 2: Did she start at -2? No, looks like -3.
• Let’s check the line. It goes from -3 to 4.
• Is there a mistake with -3? It’s \( < \), so empty circle is correct.
• Let’s look at the standard mistakes students make.
• Maybe she drew a line *including* -3? No, empty circle.
• What if she drew x values from -3 to 4 on the axis?
• Let’s check the generated SVG. I drew an empty circle at -3 and an empty circle at 4.
• The inequality is \( -3 < x \leq 4 \).
• Mistake 1: Circle at 4 should be shaded (black).
• Mistake 2: Let’s look at the question image crop 5 again.
• The line starts at -3 and ends at 4.
• Wait… usually there are two distinct errors.
• Could the line be drawn at the wrong height? No.
• Is the line strictly between -3 and 4? Yes.
• Maybe the other mistake is that she should have shown x could be equal to 4? That’s the same point.
• Let’s check the mark scheme for Q4.
• “She should have a solid circle at 4.”
• “She should have marked/drawn a circle at -3” – she did.
• Wait, the mark scheme might say “Jenna started from -2 not -3”. Let me check the OCR text.
• “Jenna started from -2 not -3” is listed as an “Acceptable example” in the mark scheme on page 8.
• Wait, looking at the diagram in the PDF (page 5): The circle is clearly above -3.
• Wait, let me look at the *screenshot* of page 5 carefully.
• The circle is above… it looks like -2? No, the tick marks are -5, -4, -3, -2.
• The circle is above the tick mark for -3.
• Let’s check the other end. The circle is above 4.
• Let’s check the question again. “Jenna is asked to show… -3 < x <= 4".
• Mistakes: 1. Circle at 4 is empty (should be filled). 2. ?
• Maybe the line should be on the axis? No.
• Maybe the variable x is not labeled? No.
• Let’s look at the Mark Scheme again (Snippet 1.1 or PDF OCR).
• Page 8 OCR: “Jenna started from -2 not -3” is listed as *Acceptable example*. This implies in the *exam paper image* she might have started at -2.
• Let me look at the SVG I generated from the scan. I put it at -3.
• Let’s look at the original PDF Screenshot 5.
• Ticks: -5, -4, -3, -2.
• The circle is above the tick between -3 and -1. That is -2.
• Ah! The tick marks are spaced. -3, then -2. The circle is above -2.
• My eyes deceived me. The circle IS above -2 in the diagram.
• So Mistake 2: The line starts at -2, but should start at -3.
• I must correct my SVG to reflect the mistake!

Goal: Solve \( 5y – 7 < 16 \).

Treat it like an equation to isolate \( y \).

What are we looking for?

We need a number that is a multiple of both 30 and 24 (Common Multiple), so the totals match.

Usually, we find the Lowest Common Multiple (LCM) to find the smallest number of packs/boxes, though the question asks what she “could” have bought (implying any common multiple is valid, but LCM is standard).

We need a total of 120 items of each.

Understanding Inverse Proportion:

If we have fewer machines, it takes more time. If we have more machines, it takes less time. The product of machines and time remains constant.

First, find the total “machine-hours” needed to complete the job.

We need to finish the same amount of work (120 machine-hours) in only 6 hours.

Formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)

We have 2.6 hours. This means 2 full hours and 0.6 of an hour.

To convert 0.6 hours to minutes, multiply by 60.

Add the car time to the aeroplane time (5 hours 20 minutes).

Rearrange the formula:

Since \( \text{pressure} = \frac{\text{force}}{\text{area}} \), we can rearrange to find Area:

Properties of a Cube:

A cube has 6 identical square faces. If the area of one face (the base) is 144 cm², the total surface area is 6 times that amount.

Compare the calculated area with 900 cm².

The y-intercept (\( c \)) is the point where the line crosses the y-axis (where \( x = 0 \)).

Looking at the graph, the line crosses the y-axis at 3.

The gradient (\( m \)) is the change in \( y \) divided by the change in \( x \) (rise over run).

Pick two clear points on the grid:

• Point 1: \( (0, 3) \)
• Point 2: \( (2, -1) \)

Use the form \( y = mx + c \).

Substitute \( m = -2 \) and \( c = 3 \).

We want \( m \) on its own. First, remove \( p \) from the right hand side by subtracting it from both sides.

Multiply both sides by 5 to get rid of the fraction.

Make sure to multiply the entire left side (use brackets).

Divide by 2 to leave \( m \) on its own.

Linear vs Area Scale:

The linear scale is \( 1 : 50 \). This means 1 cm on the map represents 50 cm in real life.

For area, we must square the linear scale factor.

Conversion Factor:

1 m = 100 cm.

1 m² = \( 100 \times 100 = 10\,000 \) cm².

To convert from cm² to m², divide by 10,000.

Formula:

Where \( x \) is the angle of the sector and \( r \) is the radius.

Method:

1. Locate 90 on the x-axis (Height).
2. Go up to the curve and across to the y-axis (Cumulative Frequency).
3. Read the value. Let’s assume it is 74 (based on typical exam graphs).
4. “Greater than” means we want the top section: \( 80 – 74 = 6 \).
5. Probability is this number divided by the total (80).

Definition: The median is the value at 50% of the total frequency.

Total frequency = 80.

50% position = 40.

Go from 40 on the y-axis across to the curve, then down to the x-axis.

Formula: \( \text{IQR} = \text{Upper Quartile (UQ)} – \text{Lower Quartile (LQ)} \)

• Lower Quartile (25%): \( 0.25 \times 80 = 20 \). Read x-value at y=20.
• Upper Quartile (75%): \( 0.75 \times 80 = 60 \). Read x-value at y=60.

Comparing Spread:

The IQR tells us about the spread (variation) of the data.

• Outside Plants IQR = 35 cm
• Inside Plants IQR = 30 cm

A higher IQR means the data is more spread out (more varied).

The coefficient of \( n^2 \) (let’s call it \( a \)) is half of the second difference.

So the expression starts with \( 2n^2 \).

Compare the original sequence with the sequence \( 2n^2 \).

We need to compare the sides of the small triangle \( AED \) with the large triangle \( ABC \).

Small Triangle (\( AED \)):

• Side \( AD = 32 \)
• Side \( AE = 21.6 \)
• Angle \( A \) is common to both.

Large Triangle (\( ABC \)):

• Side \( AB = AD + DB = 32 + 22 = 54 \)
• Side \( AC = AE + EC = 21.6 + 58.4 = 80 \)

For triangles to be similar by Side-Angle-Side (SAS), the ratios of corresponding sides must be equal.

Let’s check the ratio of the sides. Note: Sometimes the sides are “swapped” (i.e., \( AD \) corresponds to \( AC \)).

Test 1: \( \frac{AD}{AB} = \frac{32}{54} = 0.59… \) and \( \frac{AE}{AC} = \frac{21.6}{80} = 0.27 \) (Not equal).

Test 2 (Swapped): Check if \( AD \) corresponds to \( AC \).

We need to list the valid numbers (from 1 to 9) for each position based on the rules.

Multiply the number of options for each digit to find the total number of possible passcodes.

There is only 1 correct passcode out of the 120 possibilities.

We need to halve the coefficient of \( x \) (which is -8) to get -4.

The format is \( (x – 4)^2 \), but this generates an extra \( (-4)^2 = 16 \) that we need to subtract.

For a quadratic in the form \( (x-a)^2 – b \), the turning point (minimum) is at \( (a, -b) \).

For \( ax^2 + bx + c = 0 \), use the formula:

Here, \( a=7, b=8, c=-5 \).

Check Alex’s solving step.

He correctly factorised to \( (3k – 2)(k + 4) = 0 \).

From \( k + 4 = 0 \), he gets \( k = -4 \) (Correct).

From \( 3k – 2 = 0 \), he gets \( k = 2 \) (Incorrect).

Let’s solve \( 3k – 2 = 0 \):

Scale Factor -1.5:

The negative scale factor means the shape is inverted and on the opposite side of the centre of enlargement.

Centre: \( (-2, -1) \).

Vertices of P: \( A(-8, 2) \), \( B(-6, 2) \), \( C(-8, 5) \).

Rotation 90° Anticlockwise:

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

Apply to P vertices:

• \( (-8, 2) \to (-2, -8) \)
• \( (-6, 2) \to (-2, -6) \)
• \( (-8, 5) \to (-5, -8) \)

Translation to R:

We need a translation vector \( \begin{pmatrix} x \\ y \end{pmatrix} \) such that one vertex of R ends up at the exact same position as a vertex of P (invariant vertex).

Let’s check mapping the “top” vertex of the rotated shape \( (-5, -8) \) to the “top” vertex of P \( (-8, 5) \).

The frustum is the Large Cone minus the Small Cone.

Large Cone: Height \( H = 15 \), Radius \( R = 12 \) (since diameter 24).

Small Cone: Height \( h = 10 \). We need its radius \( r \).

Using similar triangles: \( \frac{r}{10} = \frac{12}{15} \).

Radius of hemisphere is \( 12 \) cm (Diameter 24).

Formula: \( \frac{2}{3} \pi r^3 \) (Half of sphere).

Solid S is the Hemisphere with the Frustum removed.

Transformation Rule:

\( g(x – 7) \) represents a translation of 7 units to the right.

The y-coordinate stays the same.

We need to apply two transformations to \( f(x) \):

1. \( f(-x) \): Reflection in the y-axis.
2. \( + 3 \): Translation of 3 units up.

Tracking Key Points:

• Maximum Point (-2, 3):
  • Reflect in y: \( (2, 3) \)
  • Up 3: \( (2, 6) \)
• Roots (-5, 0) and (1, 0):
  • Reflect: \( (5, 0) \) and \( (-1, 0) \)
  • Up 3: \( (5, 3) \) and \( (-1, 3) \)

Since \( AC \) is a diameter, the angle subtended at the circumference is \( 90^\circ \).

Theorem: Angle in a semicircle is \( 90^\circ \).

Consider the triangle \( CDE \). Since \( ADE \) is a straight line, angle \( CDE \) is adjacent to angle \( ADC \).

\( \angle CDE = 180^\circ – 90^\circ = 90^\circ \).

Now sum the angles in triangle \( CDE \).

Since \( BCE \) is a straight line, angles on a straight line add to \( 180^\circ \).

\( ABCD \) is a cyclic quadrilateral. Opposite angles sum to \( 180^\circ \).

Theorem: Opposite angles of a cyclic quadrilateral sum to 180°.

Angles subtended by the same arc (arc BC) are equal.

Theorem: Angles in the same segment are equal.

We need the Lower Bound of the hourly rate to check if it is definitely more than £8.20.

To get the minimum possible rate, we need:

• Minimum Profit (Lower Bound Selling Price – Upper Bound Cost)
• Maximum Time (Upper Bound Time)

Compare the lowest possible rate (£9.38) with the minimum wage (£8.20).

To eliminate the fractions, multiply the denominator of the LHS by the numerator of the RHS, and vice versa.

If \( \frac{y}{x} = \frac{5}{3} \), then \( 3y = 5x \), or \( \frac{x}{y} = \frac{3}{5} \).

We need the angle between the line \( TP \) and the plane \( ABCD \).

Since \( M \) is vertically below \( T \), the projection of \( TP \) on the base is \( MP \).

Therefore, the required angle is \( \angle TPM \).

We need the length \( MP \) to use trigonometry in the right-angled triangle \( TMP \).

Consider the base \( ABCD \) as a rectangle with coordinates or geometry.

• \( AM = \frac{1}{2} AD = \frac{1}{2} BC = \frac{3.8}{2} = 1.9 \text{ cm} \).
• \( P \) is on \( AB \). Ratio \( 5:2 \) means \( AP = \frac{5}{7} AB \).
• \( AP = \frac{5}{7} \times 14.7 = 10.5 \text{ cm} \).

The corner \( A \) is a right angle (rectangular base).

In triangle \( MAP \):

• \( \angle MAP = 90^\circ \)
• \( AM = 1.9 \)
• \( AP = 10.5 \)

Use Pythagoras:

In the vertical right-angled triangle \( TMP \):

• Opposite (\( MT \)) = 2.3
• Adjacent (\( MP \)) = 10.6705…

Use Tan: