GCSE November 2018 Edexcel Higher Paper 2 – Interactive Practice

Universal Set \( \mathscr{E} \): Even numbers between 1 and 25.

These are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

(Total = 12 numbers)

Strategy: Start from the center (intersection of all three) and work outwards.

• \( A \cap B \cap C \): Numbers in A, B, and C.
  Common to all: 8.
• \( A \cap B \) (excluding C): Numbers in A and B but not C.
  A has 2, 8, 10, 14. B has 6, 8, 20.
  Intersection is {8, …} wait, 20 is in B and C but not A.
  Only 8 is common to A and B. Since 8 is in C too, this region is empty.
• \( B \cap C \) (excluding A): Numbers in B and C but not A.
  B={6, 8, 20}, C={8, 18, 20, 22}. Common: 8, 20.
  8 is in A. 20 is NOT in A. So 20 goes here.
• \( A \cap C \) (excluding B): Numbers in A and C but not B.
  A={2, 8, 10, 14}, C={8, 18, 20, 22}. Common: 8.
  No other number. Empty.
• Remaining A: {2, 10, 14}.
• Remaining B: {6}.
• Remaining C: {18, 22}.
• Outside circles (in \( \mathscr{E} \)): Numbers not used yet.
  Used: 2, 6, 8, 10, 14, 18, 20, 22.
  Left: 4, 12, 16, 24.

Formula: \( P(\text{Event}) = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} \)

Total numbers in \( \mathscr{E} = 12 \).

Members of \( A \cap B \): This is the intersection of circle A and circle B. It includes the center (8) and the region \( A \cap B \) excluding C (empty). So just {8}.

Count = 1.

Observation: Look at the line drawn. It starts exactly at the corner (140, 85) and goes to the top right.

Rule: A line of best fit must follow the trend of the data points and have roughly equal numbers of points on either side. It does not need to start at the origin or corner.

Error 1: The line of best fit is incorrect (it does not suit the points / it is too high).

✓ (C1)

Observation: Look at the x-axis labels.

• The first label is 140.
• The next label is 160. The gap is 2 big squares.
• The next label is 170. The gap is 1 big square.

Analysis: 1 big square represents 10 units (160 to 170). Therefore, 2 big squares should represent 20 units. 140 + 20 = 160. This is actually mathematically linear.

However: The label “150” is missing in the sequence, which is poor practice, but technically linear spacing.

Look at the y-axis: 85, 90, 95… This is linear.

Mark Scheme Guidance: The mark scheme accepts “Height scale not linear” or “150 missing”. This suggests the examiners consider the missing label or the visual spacing of the first number relative to the axis start as an error.

Error 2: Height scale is not linear (or 150 is missing).

✓ (C1)

Strategy: Use parallel line rules. We have co-interior (allied) angles between parallel lines \( ABC \) and \( DEF \) connected by transversal \( BE \).

Angle \( ABE \) and Angle \( BED \) (which is \( 110^\circ \)) are alternate (Z-angles)? No, they are co-interior (C-angles) if on same side? No, A and D are on left. So they are Alternate Interior Angles? No, Alternate angles are equal. \( ABE \) and \( DEB \) are Alternate.

Wait, \( A \) is left, \( D \) is left. \( ABE \) and \( DEB \) are co-interior (add to 180)?

Actually, let’s look at the shape. \( A \to B \to E \to D \) forms a ‘Z’ shape? No, \( A \to B \to E \to F \) would be a Z. \( D \to E \to B \to A \) is a ‘C’ shape (allied).

Reason: Co-interior angles (or allied angles) add up to \( 180^\circ \).

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

We have \( \text{Angle } ABE = 70^\circ \) and \( \text{Angle } CBG = 35^\circ \).

Strategy: Angles on a straight line at \( E \) (line \( DEF \)).

Strategy: Angles in a triangle sum to \( 180^\circ \).

Formula: \( \text{Amount} = \text{Principal} \times (1 + \text{rate})^n \)

Principal = 2000, Rate = 2.5% (0.025), Time = 3 years.

Principal = 1600, Rate = 3.5% (0.035), Time = 3 years.

Logic: Ben already received more interest when the rate was 3.5%. If the rate increases to 4% in the 3rd year, his interest will be even higher.

Formula: Area of trapezium = \( \frac{a + b}{2} \times h \)

\( a = 10, b = 16, h = 7 \)

1 litre covers 2 m². We need to find total litres.

Tins are sold in 5 litres. We must buy whole tins.

Formula: \( \text{Gradient} = \frac{y_2 – y_1}{x_2 – x_1} \)

We know Gradient = 3, \( (x_1, y_1) = (5, 9) \), and \( (x_2, y_2) = (d, 15) \).

Convert to ordinary numbers or use calculator.

Let’s take one vertex of \( P \) to track the transformation.
Let’s use point \( A(1, 3) \).

We need to map \( R \) back to \( P \).

Point on \( R \): \( (1, -1) \)

Point on \( P \): \( (1, 3) \)

Let’s check another point. \( P(3, 4) \).

• Reflect in \( y = -x \rightarrow (-4, -3) \) (on Q)
• Reflect in \( x = -1 \): Distance from -1 is 3 (left). Image is 3 right of -1 \(\rightarrow x = 2\).
• Point on \( R \): \( (2, -3) \).

Mapping \( R(1, -1) \rightarrow P(1, 3) \) and \( R(2, -3) \rightarrow P(3, 4) \).

Let’s visualize. This looks like a rotation.

Center of rotation? Midpoint logic or trial.

Try rotating \( R(1, -1) \) 90° anticlockwise about \( (-1, 1) \).

• Vector from \( (-1, 1) \) to \( (1, -1) \) is \( (2, -2) \).
• Rotate vector 90° anti \(\rightarrow (2, 2)\).
• Add to center: \( (-1+2, 1+2) = (1, 3) \). Matches P!

Truncation means cutting off the digits after a certain point without rounding up.

If \( N \) is truncated to 1 digit and gives 7, it means the first digit is 7.

• 7.0 works
• 7.9 works (truncates to 7)
• 7.999… works
• 8.0 does NOT work (truncates to 8)

Total green paint = 50 litres.

Ratio Yellow : Blue = \( 2 : 3 \).

Total parts = \( 2 + 3 = 5 \).

Yellow: 20 litres needed. Sold in 5L tins.

\( 20 \div 5 = 4 \) tins.

Cost = \( 4 \times 26 = £104 \).

Blue: 30 litres needed. Sold in 10L tins.

\( 30 \div 10 = 3 \) tins.

Cost = \( 3 \times 48 = £144 \).

He sells 50 litres in 10 litre tins.

\( 50 \div 10 = 5 \) tins of green paint.

Selling price per tin = £66.96.

Formula: \( \frac{\text{Profit}}{\text{Cost}} \times 100 \)

💡 What are we calculating?

We need to find the total number of ways Janet can choose her meal. There are three distinct “scenarios” allowed:

1. Starter AND Main
2. Main AND Dessert
3. Starter AND Main AND Dessert

Mathematical Rule:

• When we say “AND” (making multiple choices together), we multiply the number of options.
• When we say “OR” (choosing between different scenarios), we add the results together.

Let’s break it down:

• Starters (S) = 9
• Mains (M) = 15
• Desserts (D) = 8

Summing them up:

Since she chooses option 1 OR option 2 OR option 3, we add the totals.

Strategy: To simplify algebraic fractions being multiplied, we must factorise numerators and denominators first to cancel common terms.

• \( 4x^2 – 9 \) is a difference of two squares: \( (2x)^2 – 3^2 \)
• \( 6x + 9 \) has a common factor of 3
• \( x^2 – 3x \) has a common factor of \( x \)

Strategy: To add/subtract fractions, we need a common denominator. Here, it is the product of all three denominators: \( x(x+1)(x-2) \).

Reasoning: We are given the circumference of the circle. We know that one side of the triangle is the diameter. Let’s find the diameter first.

Formula: \( C = \pi d \)

Strategy: The triangle is equilateral, and its side length \( s \) is equal to the diameter \( d \).

We can use the formula for the area of a triangle: \( \text{Area} = \frac{1}{2}ab\sin(C) \).

For an equilateral triangle, sides are \( d \) and angle is \( 60^\circ \).

What does \( y = 2^x \) look like?

• It is an exponential growth curve.
• As \( x \) increases, \( y \) increases rapidly.
• As \( x \) decreases (negative), \( y \) gets closer and closer to 0 but never touches it (asymptote at \( y=0 \)).

Y-intercept (when \( x = 0 \)):

So it crosses the y-axis at (0, 1).

X-intercept:

\( 2^x \) is never 0, so it never crosses the x-axis.

Equation of a circle:

The standard equation for a circle centered at the origin is \( x^2 + y^2 = r^2 \), where \( r \) is the radius.

\( P(\text{Red}) = 0.65 \)

\( P(\text{Blue}) = 1 – 0.65 = 0.35 \)

Since the counter is replaced, the probabilities remain the same for Mary.

There are two ways they can pick different colours:

1. Joe picks Red AND Mary picks Blue.
2. Joe picks Blue AND Mary picks Red.

We know \( P(\text{Red}) = \frac{\text{Number of Red}}{\text{Total Counters}} \)

Red counters = 78.

Ratio 1: \( (p – 5) : (q – 5) = 5 : 1 \)

This can be written as a fraction: \( \frac{p – 5}{q – 5} = \frac{5}{1} \)

Cross multiply: \( 1(p – 5) = 5(q – 5) \)

\( p – 5 = 5q – 25 \)

\( p = 5q – 20 \) — (Equation 1)

Ratio 2: \( (p + 20) : (q + 20) = 5 : 2 \)

\( \frac{p + 20}{q + 20} = \frac{5}{2} \)

\( 2(p + 20) = 5(q + 20) \)

\( 2p + 40 = 5q + 100 \)

\( 2p – 5q = 60 \) — (Equation 2)

✓ (P1) Setting up algebraic equations

Find Gradient: \( m = \frac{y_2 – y_1}{x_2 – x_1} \)

Passes through origin (0,0), so \( c = 0 \).

Gradient is -3.

Set \( y = y \) to find \( x \).

Multiply the entire inequality by 4.

Subtract 7 from all parts.

Take the square root. Remember that \( m^2 \) can come from a positive or a negative number.

Values: \( \sqrt{81} = 9 \) and \( \sqrt{121} = 11 \).

Strategy: Volume of Frustum = Volume of Large Cone – Volume of Small Cone.

Large Cone: Height \( H = 6.4 \), Radius \( R = 3.6 \) (half of 7.2).

Small Cone: Height \( h = 3.2 \). Since heights are in ratio \( 1:2 \), radius \( r = 1.8 \).

Formula: \( V = \frac{1}{3} \pi r^2 h \)

Formula: \( \frac{2}{3} \pi r^3 \)

Radius \( R = 3.6 \).

Mass = Volume × Density

Average Density = Total Mass ÷ Total Volume

Let \( \angle CAB = x \).

Goal: We want to show that \( \angle ABC = x \).

Consider the circle passing through \( A, C, R, O \).

The points \( A, C, R, O \) form a Cyclic Quadrilateral.

Property: Opposite angles in a cyclic quadrilateral sum to \( 180^\circ \).

We are given that \( CRB \) is a straight line.

Angles on a straight line add to \( 180^\circ \).

Consider the triangle \( ORB \).

\( O \) is the centre of the circle containing \( R \) and \( B \).

Therefore, \( OR \) and \( OB \) are both radii.

This means triangle \( ORB \) is an isosceles triangle.