Pearson Edexcel GCSE (9-1) Mathematics
Paper 1 (Non-Calculator) Higher Tier
November 2022

💡 Strategy: We need to repeatedly divide by prime numbers (2, 3, 5, 7, etc.) until we reach 1. A factor tree is a good way to visualize this.

Since the number ends in 0, we know it’s divisible by 2 and 5 (or 10).

💡 Concept: Group the same prime numbers together and write them using index notation (powers).

🔍 Check: \( 2^2 = 4 \) and \( 5^3 = 125 \). \( 4 \times 125 = 500 \). The answer is correct.

💡 Strategy: It’s often safer to convert mixed numbers to improper fractions before adding.

💡 Concept: The common multiple of 5 and 4 is 20.

💡 Strategy: For division, we MUST use improper fractions.

💡 Concept: Dividing by a number is the same as multiplying by its reciprocal. Dividing by 6 is multiplying by \( \frac{1}{6} \).

💡 Strategy: Divide top and bottom by their greatest common divisor (2) to show it equals the required value.

💡 Concept: When multiplying terms with the same base, add the powers: \( a^m \times a^n = a^{m+n} \).

💡 Concept: When raising a power to another power, multiply the indices: \( (a^m)^n = a^{m \times n} \).

💡 Strategy: Ignore the decimal points initially. Calculate \( 4 \times 32 \).

💡 Concept: Count the total decimal places in the question and apply them to the answer.

• \( 0.004 \) has 3 decimal places.
• \( 0.32 \) has 2 decimal places.
• Total: \( 3 + 2 = 5 \) decimal places needed.

💡 Strategy: We know that 15 out of 80 people chose Model B. We can write this as a fraction.

💡 Concept: To find the expected number for the whole factory, multiply the fraction by the total number of cars (40,000).

Alternative check: \( 15 \times 5 = 75 \), add two zeros \( \rightarrow 7500 \).

💡 Strategy: To combine two ratios, the common variable (in this case \( b \)) must have the same value in both.

• Ratio 1: \( a:b = 1:3 \)
• Ratio 2: \( b:c = 6:5 \)

In the first ratio \( b=3 \), and in the second \( b=6 \). We need to make them the same.

💡 Concept: To find a fraction of the total, we first need the sum of all parts in the ratio \( a:b:c \).

💡 Strategy: We want the ratio \( m:p \). We know relationships between \( m, n \) and \( p \). Let’s substitute to remove \( n \).

💡 Concept: The formula requires Area. The base is a rectangle, so \( \text{Area} = \text{length} \times \text{width} \).

💡 Concept: Substitute the Force (10000) and Area (8) into the given formula.

💡 Strategy: Since the HCF is 7, both numbers must be multiples of 7.

• For \( m \): It must be a multiple of 5 AND a multiple of 7.
• For \( n \): It must be even (multiple of 2) AND a multiple of 7.

🔍 Check: We need to make sure the HCF is exactly 7, not something higher (like 14 or 35).

💡 Strategy: Substitute the \( x \) values into \( y = 6x – x^3 \). Watch out for negatives!

💡 Strategy: Plot the points from the table and join them with a smooth curve.

Points: \( (-3, 9), (-2, -4), (-1, -5), (0, 0), (1, 5), (2, 4), (3, -9) \)

💡 Concept: The spinner is biased, so we use the experimental results (relative frequency) as an estimate for probability.

Score 5 appeared 10 times out of 40 spins.

💡 Strategy: For independent events occurring one after another, we multiply their probabilities.

💡 Strategy: Score 1 appeared 6 times out of 40.

💡 Strategy: Multiply the fraction by 100.

💡 Concept: The shape has changed size (Q is smaller than P) but kept the same orientation. This is an Enlargement.

💡 Strategy: Compare the lengths of corresponding sides.

P (Object): Bottom width is from \( x=9 \) to \( x=12 \), so width is \( 3 \).

Q (Image): Bottom width is from \( x=3 \) to \( x=4 \), so width is \( 1 \).

Calculation: \( \text{Scale Factor} = \frac{\text{Image Length}}{\text{Object Length}} = \frac{1}{3} \).

💡 Strategy: Draw lines through corresponding vertices (e.g., top-left of P to top-left of Q). Where they cross is the centre.

• Top-left P: \( (6, 14) \) \(\to\) Top-left Q: \( (2, 6) \)
• Bottom-left P: \( (9, 8) \) \(\to\) Bottom-left Q: \( (3, 4) \)

Let’s check the vector from Q to P to find the centre. P is 3 times further away than Q.

Or, look at the coordinates. If we go back from Q (3,4) towards (0,2):

Vector P to Q is huge reduction. Let’s assume centre is \( (0, 2) \) and check:

Vector from \( (0,2) \) to \( P(9,8) \) is \( \binom{9}{6} \).

Multiply by scale factor \( \frac{1}{3} \): \( \binom{9}{6} \times \frac{1}{3} = \binom{3}{2} \).

Add to centre: \( (0,2) + (3,2) = (3,4) \). This matches Q’s bottom-left corner.

💡 Strategy: We need to make the number of \( y \)’s (or \( x \)’s) the same in both equations so we can eliminate one variable.

Let’s match \( y \). The multiples of 2 and 3 meet at 6.

💡 Concept: Subtract equation (B) from equation (A) because the signs of \( 6y \) are the same.

💡 Strategy: Put \( x = 3 \) back into one of the original equations.

🔍 Check: Use the second equation \( 4x + 3y = 6 \).

It works!

💡 Concept: Inverse proportion means \( p = \frac{k}{t} \) or \( p \times t = k \) (where \( k \) is a constant).

We use the completed column to find \( k \).

💡 Concept: In a histogram, the area of the bar represents frequency. The height represents Frequency Density.

Formula: \( \text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} \)

💡 Strategy: Plot the bars using the Frequency Densities as heights. The widths must match the class intervals.

Max FD is 6. We can use a scale where 50px = 1 unit on the Y axis.

💡 Concept: The arc length is a fraction of the full circle’s circumference.

💡 Strategy: Expand both squared brackets individually. Recall that \( (a+b)^2 = a^2 + 2ab + b^2 \).

💡 Warning: Be careful with the negative sign when subtracting the second expression!

💡 Concept: An odd number is generally written as \( 2k+1 \) or \( 2k-1 \). The proof in (a) used \( 2m+1 \) and \( 2n-1 \), which represent any two odd numbers.

The result is \( 4 \times (\dots) \). Any integer multiplied by 4 is a multiple of 4.

💡 Concept: A power of \( \frac{4}{3} \) works in two parts:

• The denominator (3) is the cube root.
• The numerator (4) is the power.

Rule: \( x^{\frac{a}{b}} = (\sqrt[b]{x})^a \)

💡 Strategy: It’s easier to find the root of the numbers before raising them to the power of 4 (keeps numbers smaller).

💡 Reason: \( OA = OB \) because they are both radii of the same circle. This means triangle \( OAB \) is an isosceles triangle.

💡 Reason: The tangent is perpendicular to the radius at the point of contact.

💡 Insight: While the method above is standard, you might notice the Alternate Segment Theorem implies the angle between tangent and chord (\( \angle ABC \)) equals the angle in the alternate segment.

The angle at the center (\( x \)) is twice the angle at the circumference. So the angle at circumference is \( \frac{1}{2}x \). Therefore \( \angle ABC = \frac{1}{2}x \).

💡 Strategy: Find a common denominator to combine the LHS fractions. The common denominator is \( x(x+1) \).

💡 Concept: For \( ax^2 + bx + c = 0 \), use \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \).

Here \( a=4, b=4, c=-1 \).

💡 Strategy: Simplify \( \sqrt{32} \) by finding square factors.

💡 Strategy: It is usually easier to calculate \( 1 – P(\text{Same Colour}) \) than to calculate all the different colour combinations (Blue/Green, Blue/White, Green/Blue, etc.).

Since there is only 1 white card, he cannot pick two white cards. He can only pick two Blue or two Green.

💡 Concept: The graph of \( y = \cos x \) repeats every 360°.

• Starts at maximum: \( (0, 1) \)
• Crosses x-axis: \( (90, 0) \)
• Reaches minimum: \( (180, -1) \)
• Crosses x-axis: \( (270, 0) \)
• Returns to maximum: \( (360, 1) \)

💡 Observation: Point P is the first minimum point on the graph after the origin.

Based on the standard cosine wave properties, this occurs at \( x = 180 \).

The y-value at the minimum is \( -1 \).

💡 Strategy: We have a “matching pair” of an angle and an opposite side: Angle \( 30^\circ \) and side \( 10.7 \) cm.

We want to find the angle at \( B \), and we know the side opposite it is \( 6.5 \) cm.

This means we use the Sine Rule.

💡 Strategy: Eliminate decimals by multiplying top and bottom by 100.

💡 Concept: In a geometric sequence, divide any term by the previous one to find the common ratio \( r \).

💡 Concept: The \( n \)-th term is \( ar^{n-1} \).

💡 Strategy: Divide the 6th term equation by the 4th term equation to eliminate \( a \).

💡 Strategy: We need \( r:h \). Since \( h = 4r \), for every 1 unit of \( r \), there are 4 units of \( h \).

💡 Warning: The question specifies the total surface area of the cone. This includes the curved part (\( \pi rl \)) AND the circular base (\( \pi r^2 \)).

💡 Concept: In a cone, the slant height \( l \), radius \( r \), and height \( h \) form a right-angled triangle. By Pythagoras: