GCSE November 2021 Edexcel Higher Paper 1 (Non-Calculator)

Step 1: Ignore decimals temporarily

To make the multiplication easier, we remove the decimal points. We can put them back at the end.

We need to calculate \( 367 \times 42 \).

Step 2: Replace the decimal points

Count the total decimal places in the original question:

• \( 3.67 \) has 2 decimal places.
• \( 4.2 \) has 1 decimal place.
• Total = 3 decimal places.

We place the decimal point in our answer so it has 3 decimal places.

Step 1: Adjust to divide by a whole number

It is difficult to divide by a decimal like \( 1.6 \). We can multiply both numbers by 10 to make the divisor a whole number (16).

Step 2: Perform division (Bus Stop Method)

Now we divide \( 598.4 \) by \( 16 \).

What are we including?

The universal set \( \mathscr{E} \) contains all even numbers strictly less than 19.

Find numbers in both lists

Look at sets A and B to see which numbers appear in both.

\( A = \{\mathbf{6}, 12, \mathbf{18}\} \)

\( B = \{2, \mathbf{6}, 14, \mathbf{18}\} \)

Remaining parts of A: Numbers in A that are not in the intersection.

\( \{12\} \) goes in the left circle only.

Remaining parts of B: Numbers in B that are not in the intersection.

\( \{2, 14\} \) go in the right circle only.

Outside the circles: Numbers in \( \mathscr{E} \) that are not in A or B.

Used: \( \{2, 6, 12, 14, 18\} \)

Leftover: \( \{4, 8, 10, 16\} \)

Why do this?

Subtracting mixed numbers can be messy if the fractional part of the second number is larger than the first. Improper fractions avoid this.

We need a number divisible by both 5 and 3. The lowest common multiple is 15.

Decrease by 20%

We can find 10% first, then 20%, and subtract it.

Original Value: £220,000

Increase by 30%

We can find 10% first, then 30%, and add it.

Original Value: £160,000

What do we know?

Ibrahim has 15 parts and Matilda has 7 parts.

The difference in stickers is 24.

We need to find how many more stickers Ibrahim has than Rosie.

Ibrahim: 15 parts

Rosie: 4 parts

What is the formula for a prism?

\[ \text{Volume} = \text{Area of Cross Section} \times \text{Length} \]

We know the Volume (\(750\)) and the Length (\(25\)). We need to find the Area of the Cross Section first.

Divide 750 by 25

We know that \( 25 \times 3 = 75 \), so \( 25 \times 30 = 750 \).

Area of a Triangle

The cross section is a right-angled triangle.

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

We know Area = 30 and base = 5.

Use the given formula

Radius \( r = 3 \).

Formula: \( A = 4\pi r^2 \)

Cube Properties

A cube has 6 identical square faces.

Each face has side length \( x \), so area of one face is \( x^2 \).

We are told the surface areas are equal.

Get equation equal to zero

Subtract \( 5x \) and \( 24 \) from both sides to get \( ax^2 + bx + c = 0 \).

Find two numbers

We need two numbers that:

• Multiply to make \( -24 \)
• Add to make \( -5 \)

Pairs for 24: (1, 24), (2, 12), (3, 8), (4, 6).

To get \( -5 \), we use \( -8 \) and \( +3 \).

\( -8 \times 3 = -24 \) and \( -8 + 3 = -5 \).

Either bracket could be zero.

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

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

Remember that \( 3 \) is the same as \( 3^1 \).

Rule: A negative power means the reciprocal (1 over).

\[ x^{-n} = \frac{1}{x^n} \]

Rule: A fractional power \( \frac{1}{n} \) means the \( n \)-th root.

\[ x^{\frac{1}{3}} = \sqrt[3]{x} \]

Step 1: Estimate the Total Area

The area is \( 5406 \). We need a number that is easy to divide by 6.

\( 54 \) is a multiple of \( 6 \), so \( 5400 \) is a good estimate.

Step 2: Find Area of One Square

There are 6 identical squares.

Step 3: Find Side Length

Area of square = \( \text{side}^2 \).

Reasoning

We rounded the area down from 5406 to 5400.

This means our calculation used a smaller number than the real one.

Area = length × width

We are told the areas are equal.

We need \( y \) as the subject. Move all terms containing \( y \) to one side and others to the other side.

Factor out \( y \) on the left side.

Rules for Cumulative Frequency Graphs:

• Plot points at the upper bound of each interval.
• Points: \((5, 4), (10, 11), (15, 24), (20, 34), (25, 38), (30, 40)\).
• Join with a smooth curve starting from \((0, 0)\).

Step 1: Locate Median Position

Median position = \( \frac{1}{2} \times \text{Total Frequency} \).

\( \frac{1}{2} \times 40 = 20 \).

Step 2: Read from Graph

Go to 20 on the vertical axis (Cumulative Frequency). Go across to the curve, then down to the horizontal axis.

What went wrong?

Ted’s goal is to eliminate the recurring decimal part.

• \( x = 0.43333… \)
• \( 10x = 4.3333… \)

If he subtracts \( 10x – x \), he gets:

\( 4.333… – 0.433… \)

The decimal parts do not match perfectly at the start (\( .333 \) vs \( .433 \)).

He needs to multiply by a power of 10 that moves the non-recurring part past the decimal point, and another to align the recurring part.

Split into two rectangles.

Option 1: Vertical split.

• Rectangle 1 (Left): Width \( x + 1 \), Height \( x + 11 \).
• Rectangle 2 (Right): Need dimensions.

Total Width = \( 2x + 6 \). Width of Left = \( x + 1 \).
Width of Right = \( (2x + 6) – (x + 1) = x + 5 \).

Height of Right: The top indent is 4 down from top of left? No, look at the diagram. The side labelled 4 is the vertical drop.

Height of Right = Total Height (\( x + 11 \)) – Top Step (4) = \( x + 7 \).
Wait, looking at the diagram, the side length 4 is vertical. So the right rectangle has height \( x + 11 – 4 = x + 7 \) if splitting horizontally? Or is 4 horizontal?

Let’s look at the labels again. The “4” is next to the vertical drop. So the height of the right section is \( (x+11) – 4 = x + 7 \)?? No, let’s assume vertical split.

Actually, simpler split: Horizontal Split.

• Top Rectangle: Width \( x + 1 \), Height \( 4 \). (Wait, is 4 the height? The label is vertical). Let’s assume the vertical segment is 4.
• If the vertical segment is 4, then the top-left part is higher.

Let’s try standard decomposition:

Large Rectangle – Missing Corner

Total Width: \( 2x + 6 \)

Total Height: \( x + 11 \)

Area of Large = \( (2x + 6)(x + 11) \)

Missing Rectangle (Top Right):

Width = \( (2x + 6) – (x + 1) = x + 5 \)

Height = 4.

Subtract Area: \( 4(x + 5) \).

The denominators are \( 2x \) and \( 5 \).

Common denominator = \( 2x \times 5 = 10x \).

Totals

Red = 3, Yellow = 5.

Total = 8 counters.

Selection

He picks 3 counters without replacement (implied by “takes 3 counters”).

We want exactly one red.

This means 1 Red and 2 Yellows.

Possible orders:

• Red, Yellow, Yellow (RYY)
• Yellow, Red, Yellow (YRY)
• Yellow, Yellow, Red (YYR)

Line 1: \( 2y + 4 = x \)

Rearrange for y: \( 2y = x – 4 \implies y = \frac{1}{2}x – 2 \).

Intercept: -2. Gradient: 0.5.

Points: (0, -2), (4, 0).

Line 2: \( x = 3 \)

Vertical line passing through x = 3.

Line 3: \( y = 6 – 3x \)

Intercept: 6. Gradient: -3.

Points: (0, 6), (2, 0).

Since the inequalities are strictly \( < \), we draw dashed lines.

Testing Region:

• \( 2y + 4 < x \): Region "below/right" line 1. (Test (0,0): \( 4 < 0 \) False. So shade away from origin, or region is where (0,0) isn't).
• \( x < 3 \): Region left of line 2.
• \( y < 6 - 3x \): Region below line 3. (Test (0,0): \( 0 < 6 \) True. So (0,0) is in this region).

We need the region that satisfies all three.

Note: The region R is the area below both diagonal lines and to the left of the vertical line.

Use Trigonometry

Drop a perpendicular from B to line CD. Let’s call the point on CD “M”.

Triangle BMC is a right-angled triangle.

• Hypotenuse \( BC = 6 \).
• Angle \( C = 30^\circ \).
• Height \( h = BM \) is the opposite side.

\( \sin(30^\circ) = \frac{h}{6} \)

Use the Ratio

\( AB : CD = 2 : 3 \)

Let \( AB = 2x \) and \( CD = 3x \).

Area of Trapezium

\[ \text{Area} = \frac{1}{2}(a + b)h \]

Here \( a = 2x \), \( b = 3x \), \( h = 3 \), Area = 66.

Simplify \( \sqrt{12} \).

\( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \).

Multiply top and bottom by \( 5 – \sqrt{3} \).

Draw the line \( y – 2x = 1 \)

Rearrange: \( y = 2x + 1 \).

Find points:

• If \( x = 0 \), \( y = 1 \). Point (0, 1).
• If \( x = 2 \), \( y = 5 \). Point (2, 5).
• If \( x = -2 \), \( y = -3 \). Point (-2, -3).

Draw a straight line through these points on the graph.

Identify where the line crosses the circle.

Point 1 (Top Right):

\( x \approx 2.1 \), \( y \approx 5.1 \)

Point 2 (Bottom Left):

\( x \approx -2.9 \), \( y \approx -4.7 \)

Order of Operations

\( gf(1) \) means \( g(f(1)) \). We calculate \( f(1) \) first, then put the result into \( g \).

Step 1: Find fg(x)

\( fg(x) \) means \( f(g(x)) \). We substitute \( g(x) \) into \( f(x) \).

\( f(x) = 3x^2 + 1 \), so replace \( x \) with \( \frac{4}{x^2} \).

Let \( y = \frac{48}{x^4} + 1 \). Rearrange to make \( x \) the subject.

Completing the Square

First, write the terms in descending order of powers of \( x \).

\( y = -3x^2 + 18x + 9 \)

Factor out the coefficient of \( x^2 \) (which is -3) from the first two terms.

For \( x^2 – 6x \):

Halve the coefficient of \( x \) (-6 becomes -3).

\( (x – 3)^2 = x^2 – 6x + 9 \).

So, \( x^2 – 6x = (x – 3)^2 – 9 \).

The vertex form is \( y = a(x – h)^2 + k \).

The turning point is at \( (h, k) \).

Here, we have \( (x – 3)^2 \), so \( h = 3 \).

The constant is \( +36 \), so \( k = 36 \).