GCSE Nov 2023 Edexcel Higher Paper 1 (Non-Calculator)

Why: It’s easier to multiply whole numbers. We will put the decimal point back at the end.

How: Count the decimal places.

• \( 6.3 \) has 1 decimal place.
• \( 2.4 \) has 1 decimal place.
• Total decimal places = \( 1 + 1 = 2 \).

Now multiply \( 63 \times 24 \).

Why: The original question was decimals, so the answer must be adjusted.

How: We need 2 decimal places in the answer.

\( 1512 \rightarrow 15.12 \)

Rule: Any non-zero number to the power of 0 is 1.

\( a^0 = 1 \)

Rule: A negative power creates a reciprocal (fraction).

\( a^{-n} = \frac{1}{a^n} \)

Rule: When multiplying terms with the same base, ADD the powers.

\( a^x \times a^y = a^{x+y} \)

Rule: When dividing terms with the same base, SUBTRACT the powers.

\( \frac{a^x}{a^y} = a^{x-y} \)

We need to break 156 down into prime numbers (numbers divisible only by 1 and themselves: 2, 3, 5, 7, 11…).

First, find prime factors of 130.

Identify the common factors in both lists.

• Both have a 2.
• Both have a 13.

Multiply the common factors to find the HCF.

Why: The mean is calculated by \( \text{Total} \div \text{Count} \). To find the Total, we multiply.

We know one stick is 7 cm. We need the mean of the other 4.

Concept: “The mean length of 5 sticks is 4.2 cm.” This was the True Mean of the set of sticks.

If the mean of 4.2 was correct for the 5 sticks, then the Total Length was definitively 21 cm.

One stick was actually 17 cm, not 7 cm.

Let’s check the true sum of the other 4 sticks:

Comparison: Our answer in (a) was 3.5. The true answer is 1.

Therefore, the correct answer is less than our answer in (a).

Goal: Construct a perpendicular line passing through P.

1. Place compass point on \( P \).
2. Draw two arcs on the line \( AB \) (one left, one right) at equal distance from \( P \).
3. Open the compass wider.
4. Place compass point on the left intersection. Draw an arc above \( P \).
5. Place compass point on the right intersection. Draw an arc crossing the previous one.
6. Draw a line from \( P \) through the crossing point.

Why: The question states \( BA = BD \). This means triangle \( ABD \) is isosceles.

Property: In an isosceles triangle, the angles opposite the equal sides are equal.

• Side \( BD \) is opposite angle \( \text{BAD} \) (which is \( x^\circ \)).
• Side \( BA \) is opposite angle \( \text{BDA} \).
• Therefore, angle \( \text{BDA} = x^\circ \).

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

Given: \( x : y = 2 : 1 \)

This means \( x \) is twice as big as \( y \).

\( x = 2y \)

Substitute this into our equation from Step 2:

Why: \( ABC \) is a straight line. Angles on a straight line add to \( 180^\circ \).

Why: We know the total number of books is 44. We can form an equation.

How: Substitute \( x = 8 \) into the expression for Shelf B.

Arithmetic Tip: \( 75 \div 25 = 3 \), so \( 7500 \div 25 = 300 \).

How: Shelf A has \( x \) books. We know \( x = 8 \).

Why: This is a “Reverse Percentage” question because we have the final price and want the original price.

If the price is reduced by 40%, the sale price represents:

So, £660 is 60% of the normal price.

How: Use the ratio 4 : 5.

• When \( x = 0 \), \( y = 0 \) (0 rice needs 0 water). Plot (0, 0).
• When \( x = 4 \), \( y = 5 \). Plot (4, 5).
• When \( x = 8 \), \( y = 10 \). Plot (8, 10).

Connect these points with a straight line.

(i) Gradient: \( \text{Gradient} = \frac{\text{change in } y}{\text{change in } x} \)

Using point (4, 5):

(ii) Meaning: The gradient represents the relationship between units of \( y \) and units of \( x \).

Since \( y \) is water and \( x \) is rice, it means “1.25 cups of water are needed for every 1 cup of rice”.

Formula: Circumference \( C = 2\pi r \).

We know \( C = 10 \).

Formula: Area \( A = \pi r^2 \).

Substitute \( r = \frac{5}{\pi} \).

1. Analyze the Table:

• Least number (Min): 300. This is the start of the left whisker.
• Median: 900. This is the line inside the box.
• Range: 1000. Range = Max – Min.
• \( \text{Max} = \text{Range} + \text{Min} = 1000 + 300 = 1300 \). This is the end of the right whisker.

2. Analyze the Box Plot:

• The drawn box starts at 780. This is the Lower Quartile (LQ).
• The drawn box ends at 1200. This is the Upper Quartile (UQ).

What it tells us: The Upper Quartile (UQ) represents the value below which 75% of the data lies.

From part (a), the UQ is 1200.

This means 75% of the 80 days had fewer than 1200 cars.

Why: To work with gradients, we need the equation in the form \( y = mx + c \).

Rule: Perpendicular gradients are negative reciprocals. \( m_{\perp} = -\frac{1}{m} \).

“Flip the fraction and change the sign.”

How: Use \( y – y_1 = m(x – x_1) \) with point \( (6, -5) \).

Concept:

• Length Scale Factor (LSF) = \( k \)
• Area Scale Factor (ASF) = \( k^2 \)
• Volume Scale Factor (VSF) = \( k^3 \)

We are given the length (height) ratio A:B is \( 2:5 \).

How: Multiply Volume of A by the VSF.

Rule: \( a^{\frac{m}{n}} = (\sqrt[n]{a})^m \).

The bottom number (3) is the root. The top number (2) is the power.

Rule: A negative power flips the fraction.

\( (\frac{a}{b})^{-n} = (\frac{b}{a})^n \).

Why: The gradient of a curve changes constantly. To find the gradient at a specific point (\( t=3 \)), we must draw a straight line that touches the curve only at that point (a tangent).

Pick two points on the red tangent line.

• Point 1: \( (2, 13.5) \)
• Point 2: \( (4, 67.5) \)

Why: We have the formula linking year 1 and year 2. We can substitute the known values to find \( k \).

\( P_1 = 8 \), \( P_2 = 6 \).

How: Use the formula again with \( P_2 \) and \( k \) to find \( P_3 \).

Why: Population growth depends on the multiplier \( k \).

• If \( k > 1 \), population increases.
• If \( k < 1 \), population decreases.

Calculate new k:

Since \( 1.05 > 1 \), the population will be multiplied by a number larger than 1 each year.

Method: “AC” Method (Splitting the middle term).

1. Multiply \( a \times c \): \( 6 \times -4 = -24 \).

2. Find two numbers that multiply to -24 and add to -5 (the \( b \) term).

• Factors of 24: (1,24), (2,12), (3,8), (4,6)
• To get -5: Use \( -8 \) and \( +3 \).

3. Rewrite the middle term and factorise by grouping.

Step 1: Find Critical Values

Set the factors to zero.

• \( 2x + 1 = 0 \Rightarrow 2x = -1 \Rightarrow x = -0.5 \)
• \( 3x – 4 = 0 \Rightarrow 3x = 4 \Rightarrow x = \frac{4}{3} \)

Step 2: Determine the Region

The quadratic \( 6x^2 \) opens upwards (U-shape).

We want where it is less than 0 (below the x-axis).

This occurs between the two roots.

Why: The events are independent. \( P(A \cap B) = P(A) \times P(B) \).

There are two ways this can happen:

1. Case 1: A is Red AND B is Not Red.
2. Case 2: A is Not Red AND B is Red.

We need the “Not Red” probabilities:

• \( P(A_{\text{Not}}) = 1 – \frac{1}{4} = \frac{3}{4} \)
• \( P(B_{\text{Not}}) = 1 – \frac{1}{6} = \frac{5}{6} \)

How: Draw a horizontal line at \( y = 0.3 \).

Find where this line crosses the curve.

The first intersection is a small angle after 0 (approx 17°).

The second intersection is symmetric: \( 180 – 17 = 163^\circ \).

Logic: Where does the sine graph equal 0?

At \( 0^\circ, 180^\circ, -180^\circ \).

So, the angle inside the bracket must equal one of these.

• \( x + 20 = 0 \Rightarrow x = -20 \)
• \( x + 20 = 180 \Rightarrow x = 160 \)
• \( x + 20 = -180 \Rightarrow x = -200 \) (Too small, range is -180 to 180)

Why: We have 3 sides and need 1 angle. This is the Cosine Rule.

Formula: \( a^2 = b^2 + c^2 – 2bc \cos A \)

Rearranged for the angle at B: \( b^2 = a^2 + c^2 – 2ac \cos B \)

Where:

• \( b \) is the side opposite B (\( 5\sqrt{7} \))
• \( a \) and \( c \) are the other sides (5 and 10)

Memory Recall: \( \cos(60^\circ) = 0.5 \).

Since the value is negative (\(-0.5\)), the angle is obtuse (in the second quadrant).

\( 180^\circ – 60^\circ = 120^\circ \).

Equation: \( x^2 + y^2 = r^2 \).

This is a circle with centre \( (0,0) \).

Radius: \( r^2 = 169 \), so \( r = \sqrt{169} = 13 \).

How to draw: Place compass at origin. Draw circle passing through 13 on all axes.

Equation: \( 2y = 3x \).

Rearrange to \( y = \frac{3}{2}x \) or \( y = 1.5x \).

Plot Points:

• If \( x = 0, y = 0 \).
• If \( x = 4, y = 1.5 \times 4 = 6 \). Point \( (4, 6) \).
• If \( x = 8, y = 1.5 \times 8 = 12 \). Point \( (8, 12) \).
• Also plot negative values: \( (-8, -12) \).

Draw the straight line and find where it crosses the circle.

Read the coordinates where the line meets the circle.

Intersection 1: \( x \approx 7.2, y \approx 10.8 \)

Intersection 2: \( x \approx -7.2, y \approx -10.8 \)

Why: In a geometric sequence, you multiply by the common ratio \( r \) to get the next term.

Therefore, \( r = \frac{\text{3rd Term}}{\text{2nd Term}} \).

How: Multiply top and bottom by the conjugate of the denominator (\( 3 – 2\sqrt{2} \)).