Pearson Edexcel GCSE Maths (Higher) – Paper 3 (Calculator) 2020

💡 Why we do this:

When multiplying terms with the same base (here, \(n\)), we add the powers.

💡 Strategy:

We handle each variable separately. When dividing terms with the same base, we subtract the powers.

Remember that \(d\) is the same as \(d^1\).

💡 Strategy:

We treat the inequality sign \( > \) just like an equals sign \( = \). We need to isolate \(x\) by performing the inverse operations.

First, remove the fraction by multiplying by 2.

💡 Recall:

Speed = Distance ÷ Time

Rearranging for Time: Time = Distance ÷ Speed

💡 Strategy:

Add the decimal hours, then convert the decimal part to minutes.

Remember: 0.25 hours is NOT 25 minutes. We multiply by 60.

💡 Strategy:

When rounding to 1 decimal place (0.1), the “gap” between numbers is 0.1.

The bounds are half this gap (0.05) above and below the rounded value.

• Lower Bound = Value – 0.05
• Upper Bound = Value + 0.05

💡 Strategy:

1. Calculate the area of the original lawn and the rate of seed usage (kg per m²).
2. Calculate the area of the new lawn.
3. Calculate how much seed is needed for the new lawn.
4. Compare this with how much seed she has in the 5 boxes.

💡 What changed?

One box only has 1 kg. Let’s calculate her new total and compare it to the requirement (9.33 kg).

💡 Strategy:

The spinners are “3-sided” with numbers 1, 2, 3.

• Probability of getting a 2 = \( \frac{1}{3} \) (only one ‘2’ out of three numbers).
• Probability of NOT getting a 2 = \( \frac{2}{3} \) (numbers 1 and 3).

We fill these fractions onto the branches.

💡 Strategy:

We need “Spinner A lands on 2” AND “Spinner B does not land on 2”.

In probability, “AND” means we MULTIPLY along the branches.

Path: Top Branch (A=2) → Down Branch (B≠2).

💡 Strategy:

The solution to a pair of simultaneous equations is the point where their graphs intersect (cross).

Look at the graph in part (a). Find the point where the two lines cross.

💡 Strategy:

We need to solve \( x^2 – 4x + 2 = 0 \).

The graph shows \( y = x^2 – 4x + 2 \).

So we are looking for where \( y = 0 \). This corresponds to the x-intercepts (where the curve crosses the x-axis).

💡 Strategy:

Mean = Total ÷ Count. Therefore, Total = Mean × Count.

We need to find the total age of the boys and the total age of the girls separately.

💡 Strategy:

Add the totals together and divide by the total number of people (45).

💡 Strategy:

Probabilities must add up to 1.

First, find what remains for Red and Yellow.

💡 Strategy:

We are told \( P(\text{Red}) = 5 \times P(\text{Yellow}) \).

Let \( x \) be the probability of Yellow.

Then the probability of Red is \( 5x \).

💡 Strategy:

Multiply the total number of counters by the probability of getting a yellow one.

💡 Strategy:

The volume of any prism is: Area of Cross Section × Length.

The cross section (front face) is a compound shape made of:

1. A rectangle (10 cm wide, 12 cm high).
2. A triangle on top.

We know the triangle is isosceles because it has “exactly one line of symmetry”.

💡 Strategy:

We split the isosceles triangle in half to create a right-angled triangle.

• The base of the whole triangle is 10 cm, so the base of the right-angled triangle is \( 5 \text{ cm} \).
• The angle at the base is \( 40^\circ \).
• We need the height \( h \).

Using Trigonometry (SOH CAH TOA):

\( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \)

\( \tan(40) = \frac{h}{5} \)

💡 Strategy:

Multiply: Beats per day × Days per year × Number of years.

💡 Strategy:

Divide Total Mass by Number of Cells.

\( \text{Mass of one} = \frac{\text{Total Mass}}{\text{Count}} \)

💡 Strategy:

Let’s track the coordinates of the vertices.

Step 1: Rotate P 180° about (0,0) to get Q

Rule for 180° rotation: \( (x, y) \rightarrow (-x, -y) \)

• (1, 2) → (-1, -2)
• (1, 5) → (-1, -5)
• (2, 5) → (-2, -5)

Step 2: Translate Q by \( \begin{pmatrix} 5 \\ -2 \end{pmatrix} \) to get R

Rule: Add 5 to x, subtract 2 from y.

• (-1, -2) → (4, -4)
• (-1, -5) → (4, -7)
• (-2, -5) → (3, -7)

Step 3: Compare P to R

P: (1, 2), (1, 5), (2, 5)

R: (4, -4), (4, -7), (3, -7)

Notice the orientation is rotated 180°. A rotation of 180° maps P onto R.

To find the centre of rotation, find the midpoint of the line segment joining corresponding points (e.g., (1,2) and (4,-4)).

💡 What is an invariant point?

An invariant point is a point that does not move under the transformation.

For a rotation, the only point that doesn’t move is the centre of rotation.

💡 Strategy:

To add fractions, we need a common denominator.

The common denominator will be the product of the two denominators: \( (x+2)(x-4) \).

💡 Strategy:

Expand the first two brackets first, then multiply the result by the third bracket.

💡 Strategy:

Draw the boundary lines first. Use solid lines because the inequalities are inclusive (\(\leq, \geq\)).

1. \( x = 0 \): The y-axis. Region is to the right.
2. \( x = 2 \): Vertical line at x=2. Region is to the left.
3. \( y = x + 3 \): Diagonal line crossing y-axis at 3 and x-axis at -3. Test (0,0): \( 0 \leq 3 \) (True), so shade below.
4. \( 2x + 3y = 6 \): Find intercepts. If x=0, y=2. If y=0, x=3. Connect (0,2) and (3,0). Test (0,0): \( 0 \geq 6 \) (False), so shade above/away from origin.

💡 Strategy:

Check if the point (2, 4) satisfies the inequalities. Alternatively, check if it lies on a solid boundary line.

Boundary lines for inclusive inequalities (\(\leq, \geq\)) are PART of the solution set.

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

Quadrilateral \( BDEF \) is cyclic (all points on the circle).

Angle \( DEF \) is opposite Angle \( DBF \). Wait, looking at the diagram, \( BDEF \) is the cyclic quad.

So Angle \( DEF \) (100°) is opposite Angle \( DBF \)? No, vertices are B, D, E, F in order.

Opposite angles are \( \angle DEF \) and \( \angle DBF \).

Wait, looking at the diagram again. The diagonal \( DF \) is drawn.

The cyclic quad is \( BDEF \). The opposite angles are \( \angle DEF \) and \( \angle DBF \).

Calculation: \( \angle DBF = 180 – 100 = 80^\circ \).

Check diagram: The angle marked \( 40^\circ \) is \( \angle DBF \)?? No, let’s re-read carefully.

Looking at the screenshot (Page 16):

• The angle \( 100^\circ \) is at \( E \) (\( \angle DEF \)).
• The angle \( 40^\circ \) is \( \angle BDF \) or \( \angle DBF \)? It looks like \( \angle FBD \). Wait, looking closer at crop 16…
• The \( 40^\circ \) is \( \angle DBF \). That conflicts with cyclic quad rule if opposite angle is 100.
• Ah, the vertices might be B, F, E, D? Let’s trace the perimeter. B -> F -> E -> D -> B.
• Opposite angles are B and E. So \( \angle DBF + \angle DEF = 180 \).
• Wait, the angle marked 40 is adjacent to D? It’s inside the triangle BDF.
• Let’s assume the question asks for ABD.

Correction from Mark Scheme:

Mark scheme says: “for angle DBF, eg 180 – 100 (= 80)”.

So \( \angle DBF = 80^\circ \). The angle marked 40 must be \( \angle BDF \)? Or \( \angle BFD \)?

Mark scheme says: “for angle BFD, eg 180 – 80 – 40 (= 60)”.

This implies the 40 degree angle is indeed given in the question.

Let’s look at the diagram crop again. The 40 is at D. It’s angle \( BDF \).

So: Given \( \angle BDF = 40^\circ \) and \( \angle DEF = 100^\circ \).

💡 Strategy:

Let \( x \) equal the recurring decimal.

Multiply by powers of 10 to align the recurring parts, then subtract to eliminate them.

💡 Strategy:

The distance travelled is the area under the speed-time graph.

For the first 30 seconds, the shape is roughly a triangle.

Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \).

💡 Strategy:

Compare the triangle we used to the actual curve.

The curve bends outwards (concave down), so it is above the straight line of our triangle hypotenuse.

💡 Concepts:

• Average Acceleration = Change in Speed ÷ Time taken.
• Instantaneous Acceleration (at a specific time) = Gradient of the tangent at that point.

Julian calculated average acceleration from t=0 to t=60.

💡 Strategy:

In a histogram, Frequency = Frequency Density × Class Width.

Calculate the frequency (area) for each bar.

💡 Strategy:

To estimate the mean from grouped data, we use the midpoint of each class interval as the representative value.

💡 Formula:

\( \text{Mean} = \frac{\sum (f \times x)}{\sum f} \)

💡 Strategy:

The measurement is to the nearest mm (0.1 cm). The error interval is half of this unit (0.05 cm).

We want the Lower Bound of the edge, so we start with the Lower Bound of the diagonal AH.

💡 Formula:

For a cube of side length \( a \):

• Face diagonal \( d^2 = a^2 + a^2 = 2a^2 \)
• Space diagonal (AH) \( D^2 = a^2 + a^2 + a^2 = 3a^2 \)

So \( AH = \sqrt{3a^2} = a\sqrt{3} \).

💡 Strategy:

Substitute the Lower Bound of AH into the equation to find the Lower Bound of \( a \).

💡 Strategy:

A regular hexagon consists of 6 equilateral triangles of side length \( x \).

Area of one equilateral triangle with side \( x \) = \( \frac{1}{2} x^2 \sin(60) \).

💡 Strategy:

The large hexagon is an enlargement of scale factor \( p \).

The side length is \( px \).

Alternatively, use the Area Scale Factor principle: If linear scale factor is \( p \), area scale factor is \( p^2 \).

💡 Principle:

The Lowest Common Multiple (LCM) of a set of numbers with the same base is simply the number with the highest power.

Why? Because the highest power contains all the factors of the lower powers.

For example, LCM of \( 2^3 \) and \( 2^5 \) is \( 2^5 \), because \( 2^5 \) is a multiple of \( 2^3 \).

💡 Strategy:

Isolate terms so we have \( c \) on one side and \( d \) on the other.

Then express as a fraction \( \frac{c}{d} \) to find the ratio.

💡 Strategy:

This is a quadratic equation in two variables.

Rearrange to equal zero: \( 6x^2 – 7xy – 20y^2 = 0 \).

We can factorise this into two brackets: \( (ax + by)(cx + dy) = 0 \).

Alternatively, divide everything by \( y^2 \) to get a quadratic in \( (\frac{x}{y}) \).