Edexcel GCSE Maths Foundation Paper 2 (Calculator) – Nov 2022

💡 What are we being asked to do?

We need to arrange a set of integers (whole numbers) from the smallest value to the largest value.

Important: With negative numbers, the larger the number looks, the smaller its value (further left on the number line).

💡 What does “percent” mean?

“Percent” comes from “per centum”, which means “out of 100”. So, any percentage can be written as that number over 100.

Check: Can we simplify this fraction? 37 is a prime number, so it cannot be divided by anything other than 1 and itself. Since 100 is not divisible by 37, the fraction is already in simplest form.

💡 What is an odd number?

Odd numbers are integers that cannot be divided exactly by 2. They end in 1, 3, 5, 7, or 9. The sequence starts at 1.

Alternative Method: The formula for the \(n\)th odd number is \(2n – 1\).

\( 2(7) – 1 = 14 – 1 = 13 \)

💡 How many millimetres are in a centimetre?

There are 10 millimetres in 1 centimetre. This is the key conversion fact.

To go from cm to mm (a larger unit to a smaller unit), we multiply by 10.

💡 We have two conditions to satisfy:

1. The number must be smallest.
2. The number must be even.
3. We must use all 4 digits: 4, 7, 6, 3.

To make a number small, we want the smallest digits at the start (Thousands place).

To make a number even, it must end in an even digit (0, 2, 4, 6, 8). Our available even digits are 4 and 6.

Let’s try to put the smallest digits first to satisfy the “smallest” condition.

Order of digits from smallest to largest: 3, 4, 6, 7.

We need to swap the last digit to make it even. The number must end in 4 or 6.

Let’s keep the start as small as possible: 3.

Next smallest available: 4.

Now we have digits 6 and 7 left. One must go at the end to make it even.

If we put 7 then 6: 3476 (Ends in 6 -> Even).

If we put 6 then 7: 3467 (Not Even).

💡 How to do this:

Place a ruler along the line connecting point A and point B. Start the 0 mark at A and read the value at B.

💡 How to do this:

1. Place the centre of the protractor on the vertex D.

2. Align the baseline of the protractor with the line AD.

3. Read the scale that starts at 0° on the line AD.

4. Follow the scale around to the line DC.

Visually, angle \( x \) is obtuse (greater than 90°), so we expect a value over 90.

💡 Strategy: To find the distance travelled, we need to find the difference between the end reading and the start reading.

\( \text{Distance} = \text{End Reading} – \text{Start Reading} \)

💡 Strategy: Multiply the distance by the cost per mile (13p).

💡 Strategy: There are 100 pence in 1 pound. We need to divide by 100.

💡 Formula: Cost = £45 × number of days

We are given that the number of days is 7.

💡 Vertical Axis (y-axis): Needs to go up to at least 10 (the maximum value). A scale of 1 large square = 1 or 2 students works well. Let’s use 1 large square = 1 student.

💡 Horizontal Axis (x-axis): Label with the days of the week: Mon, Tue, Wed, Thu, Fri.

Rules for Bar Charts:

• Bars must be equal width.
• Gaps between bars must be equal width.
• Height represents the frequency.

💡 Step 1: Find the times

Look at the first column (the 07 20 bus from Wigan).

Depart Wigan: 07 20

Arrive Lostock: 08 09

💡 Step 2: Calculate difference

From 07 20 to 08 00 is 40 minutes.

From 08 00 to 08 09 is 9 minutes.

Total = 40 + 9 = 49 minutes.

Step 5: Conclusion

She arrives at 09 13.

She needs to be there by 09 20.

Is 09 13 before 09 20? Yes.

💡 Strategy: Total people minus adults equals children.

💡 Method A: Find 35% (wearing hats) and subtract from total children.

💡 Method B: If 35% represent those wearing hats, then \( 100\% – 35\% = 65\% \) represents those NOT wearing hats.

Let’s use Method B as it’s more direct.

We need to find \( 65\% \) of 200.

Calculator Steps: \( 0.65 \times 200 \)

💡 Strategy: Divide by the denominator (bottom), then multiply by the numerator (top).

Alternatively, use the calculator directly.

💡 Strategy: To compare fractions, give them all a common denominator.

Look at the denominators: 8, 32, 4, 64.

64 is a multiple of all of them. Let’s convert everything to be over 64.

Compare the numerators: 24, 18, 16, 21.

Smallest to largest: 16, 18, 21, 24.

Offer details: Pay for 2, get 1 free. Total 3 bottles.

Total pints: \( 3 \text{ bottles} \times 2 \text{ pints} = 6 \text{ pints} \)

Total cost: Pay for 2 bottles = \( 2 \times 75\text{p} = 150\text{p} \)

Offer details: Pay for 1, get 1 half price. Total 2 bottles.

Total pints: \( 2 \text{ bottles} \times 4 \text{ pints} = 8 \text{ pints} \)

Total cost:

First bottle: £1.28

Second bottle: \( £1.28 \div 2 = £0.64 \)

Total = \( 1.28 + 0.64 = £1.92 \)

Convert to pence: 192p.

Offer 1: 25p per pint

Offer 2: 24p per pint

💡 Strategy: Collect like terms. Group the \( c \)s and group the \( d \)s.

💡 Method 1: Expand first

💡 Method 2: Divide first

💡 Strategy: Multiply the quantity by the number of items.

3 boxes with \( x \) sweets = \( 3 \times x = 3x \)

2 packets with \( y \) sweets = \( 2 \times y = 2y \)

“Total” means add them together.

💡 Rules:

1. The “Stem” is the first digit(s) (Tens/Hundreds).

2. The “Leaf” is the last digit (Units).

3. Leaves must be ordered from smallest to largest.

4. Include a Key.

💡 Definition: The median is the middle value when data is ordered.

Total items (n) = 17.

Position of median = \( \frac{n+1}{2} = \frac{17+1}{2} = \frac{18}{2} = 9 \text{th} \) value.

💡 Understanding: If you have fewer pipes, it takes more time. If you have more pipes, it takes less time. This is inverse proportion.

First, let’s find out how long it would take 1 pipe to fill the lake.

Now we share that work between 6 pipes. It will go 6 times faster.

We only need to fill a quarter of the lake, so it will take a quarter of the time.

💡 Why? We don’t know the exact height of each student, just the range. We use the midpoint of the range as an estimate.

Sum of \( f \times x \) (Total estimated height) divided by Total Frequency (Total students).

💡 Definition: An outlier is a point that does not fit the pattern of the rest of the data.

Most points show that when rainfall is low, sunshine is high (top left), and when rainfall is high, sunshine is low (bottom right).

There is one point at (2, 1) (Low rainfall, Low sunshine) that is far away from the main group.

💡 Describe the trend: As one variable increases, what does the other do?

As the amount of rainfall increases, the number of hours of sunshine decreases.

This is called negative correlation.

💡 Method: Line of Best Fit

1. Draw a straight line through the middle of the main group of points (ignoring the outlier).

2. Find “7 hours of sunshine” on the vertical y-axis.

3. Go across to your line of best fit.

4. Go down to the horizontal x-axis to read the rainfall.

Front Elevation: Shows the height and width from the front.

Plan: Shows the width and depth from above.

Side Elevation: Shows the height and depth from the side.

The arrow is pointing from the Right. This means we are drawing what we see if we stand on the right and look left.

Width of Side Elevation: This corresponds to the depth of the Plan. The plan is 3 squares deep.

Height of Side Elevation: This corresponds to the height of the Front Elevation. The highest point on the Front is 5 squares high.

Shape: From the right side, we see the rectangular profile of the object. Since the object has a consistent width from the side view (based on the Plan), it will look like a rectangle.

💡 Step 1: Find the common difference

\( 13 – 7 = 6 \)

\( 19 – 13 = 6 \)

The sequence goes up by 6 each time. So the expression starts with \( 6n \).

💡 Strategy: Set the formula equal to \( -58 \) and solve for \( n \). If \( n \) is a positive whole number (integer), then yes, it is in the sequence.

Since \( n = 11 \) is a whole number, -58 is the 11th term of the sequence.

1. Rectangle ABCO:

Width \( AB = 11 \text{ m} \), Height \( BC = 7 \text{ m} \).

Area = \( 11 \times 7 = 77 \text{ m}^2 \).

(Note: This implies OA = 7 m)

2. Rectangle DEFO:

Height \( EF = 9 \text{ m} \), Width \( ED = 7 \text{ m} \).

Area = \( 9 \times 7 = 63 \text{ m}^2 \).

(Note: This implies OF = 7 m, which matches OA = 7 m, consistent with the sector radius)

3. Triangle CDO:

The legs of the right-angled triangle are \( OC \) and \( OD \).

\( OC = AB = 11 \text{ m} \).

\( OD = EF = 9 \text{ m} \).

Area = \( \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 11 \times 9 = 49.5 \text{ m}^2 \).

4. Sector AFO:

This is a quarter circle with radius \( r = 7 \text{ m} \).

Area = \( \frac{1}{4} \times \pi \times r^2 = 0.25 \times \pi \times 7^2 \)

Each bag covers \( 14 \text{ m}^2 \). Divide total area by 14.

Hypotenuse (H): Opposite the right angle. Length = 14.5 cm.

Adjacent (A): Next to the angle 53°. Length = \( x \).

Opposite (O): Opposite the angle. Not needed.

We have A and H.

We use CAH: \( \cos(\theta) = \frac{\text{Adj}}{\text{Hyp}} \)

Multiply both sides by 14.5.

Round to 3 significant figures.

8.726… becomes 8.73.

Multiplier Method: An increase of 3% means multiplying by \( 1 + 0.03 = 1.03 \).

The interest rate is now 1.5%. The multiplier is \( 1 + 0.015 = 1.015 \).

We apply this to the new amount (£7210).

Definition: Where the graph crosses the vertical y-axis (where \( x=0 \)).

Looking at the equation \( y = x^2 – 6x + 4 \), if \( x=0 \), \( y=4 \).

Looking at the graph, it crosses the y-axis at 4.

Definition: The turning point (vertex) is the lowest point of this U-shaped curve.

Look at the bottom of the curve.

x-coordinate: Halfway between 2 and 4, which is 3.

y-coordinate: -5.

Definition: The roots are where \( y = 0 \). This is where the graph crosses the horizontal x-axis.

Look for the two points where the curve hits the line \( y=0 \).

Point 1: Between 0 and 1. Approx 0.7 or 0.8.

Point 2: Between 5 and 6. Approx 5.2 or 5.3.

Definition: The reciprocal of a number \( n \) is \( \frac{1}{n} \).

Method:

1. Identify the degree of accuracy: “2 significant figures”.

The number 4700 to 2 s.f. represents the hundreds column (47|00).

The unit of accuracy is 100.

2. Halve the unit of accuracy: \( 100 \div 2 = 50 \).

3. Subtract 50 for the lower bound, Add 50 for the upper bound.

💡 Warning: We are given the final amount (2019) and asked for the original amount (2018). We cannot just subtract 9%.

This is a Reverse Percentage problem.

Increase of 9% means:

\( 100\% + 9\% = 109\% \)

Multiplier = 1.09