Edexcel GCSE Foundation Paper 2 (Calculator) – June 2023

💡 What are we looking for?

We need to round to the nearest hundred. This means we look at the hundreds column and the digit to its right (the tens column) to decide whether to round up or down.

💡 Why we do this:

Decimals are just another way of writing fractions with denominators of 10, 100, 1000, etc.

The first digit after the decimal point is the “tenths” column.

💡 Key Fact:

There are \( 100 \) centimetres in \( 1 \) metre.

\[ 1 \text{ m} = 100 \text{ cm} \]

💡 Strategy:

In algebra, we can multiply the numbers together, but the letter (variable) stays attached.

💡 What is a multiple?

A multiple of \( 25 \) is a number found in the \( 25 \) times table.

The multiples of \( 25 \) are: \( 25, 50, 75, 100, 125, \dots \)

💡 Reasoning:

An ordinary dice has numbers \( 1, 2, 3, 4, 5, 6 \).

It is impossible to roll a 7.

The probability of an impossible event is \( 0 \).

💡 Reasoning:

The numbers on a dice are \( 1, 2, 3, 4, 5, 6 \).

The even numbers are \( 2, 4, 6 \).

There are 3 even numbers out of 6 total outcomes.

💡 Method:

Using a ruler, measure the straight line distance from point A to point C.

Note: On a digital screen, physical measurement is not possible. The accepted value from the exam paper is provided below.

💡 Method:

Using a protractor, measure the angle at vertex B formed by lines AB and BC.

💡 Identification:

The question states \( QP = QR \). This means two sides are equal in length.

A triangle with two equal sides (and two equal base angles) is called an isosceles triangle.

💡 Method:

We need the total length on the map from \( P \) to \( R \).

Distance \( P \) to \( R \) = Distance \( P \) to \( Q \) + Distance \( Q \) to \( R \).

💡 Scale Rule:

The scale is \( 1 \text{ cm} = 4 \text{ km} \).

To find the real distance, we multiply the map distance by 4.

💡 Pattern Spotting:

Let’s look at the difference between the terms:

• \( 8 – 3 = 5 \)
• \( 13 – 8 = 5 \)

The rule is “add 5”.

💡 Identify the Terms:

• The second term is \( 8 \).
• The fourth term is \( 18 \).

We need the ratio \( 8 : 18 \).

💡 Method:

1. Find \( 4 \) on the horizontal (x) axis (“Number of hours”).

2. Go straight up to the line.

3. Go straight across to the vertical (y) axis (“Cost”).

💡 Step 1: Find the duration.

Find £9 on the vertical (Cost) axis.

Go across to the line, then down to the hours axis.

Justin arrived at 08:00.

He stayed for 6 hours.

💡 Strategy:

Multiply each weight by the number of people with that weight to find the total for each group. Then sum them all up.

The question asks to show that the weight is less than \( 1200 \text{ kg} \).

💡 Method:

The mirror line is the perpendicular bisector of the line joining corresponding points on shape A and shape B.

Looking at the positions, the shapes are reflected diagonally.

💡 Reasoning:

Alex was asked to reflect in the \( x \)-axis.

The \( x \)-axis is the horizontal line at \( y=0 \).

Reflecting in the \( x \)-axis would move shape P from the top-right (Quadrant 1) to the bottom-right (Quadrant 4).

Alex has moved the shape to the left (Quadrant 2), which is a reflection in the \( y \)-axis.

💡 Visualise the Fraction:

If \( \frac{1}{16} \) of the school is \( 50 \) people, imagine the school divided into 16 equal parts.

One part is equal to \( 50 \).

💡 Problem: The units are different (mm and cm). We must convert them to be the same.

It is usually easier to work in cm. Note that \( 10 \text{ mm} = 1 \text{ cm} \).

💡 Strategy:

We need to see how many packets fit along each side of the box. We divide the box dimension by the packet dimension.

Event A: Number less than 3 on a 6-sided dice.

Numbers less than 3 are: \( 1, 2 \).

There are 2 numbers.

Total outcomes on dice = 6.

Event B: Number greater than 5 on an 8-sided spinner.

Numbers greater than 5 are: \( 6, 7, 8 \).

There are 3 numbers.

Total outcomes on spinner = 8.

💡 Problem: The speed is in km/h, but the time is in hours and minutes.

We must convert \( 45 \text{ minutes} \) into hours.

💡 Formula:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

💡 Mean Formula:

\[ \text{Total} = \text{Mean} \times \text{Number of items} \]

We know the mean is \( 380 \) for \( 3 \) cinemas.

We know the total seats for A, B, and C combined is \( 1140 \).

Subtract the known seats in A and B to find C.

Step 1: Find number of packs.

Asha buys \( 180 \) cans, and there are \( 12 \) cans per pack.

\[ 180 \div 12 = 15 \text{ packs} \]

Step 2: Calculate cost.

Each pack costs £\( 3 \).

\[ 15 \times 3 = 45 \]

💡 Strategy:

Find the total volume of lemonade and the total cost in pence, then find the cost per \( 100 \text{ ml} \).

Round to the nearest penny (whole number).

\( 8.838… \) rounds to \( 9 \).

Step 1: Car Branches

Total people = \( 240 \).

“\( 150 \) have a car”. So the top branch is \( 150 \).

People without a car = \( 240 – 150 = 90 \). This is the bottom branch.

Step 2: Bicycle Branches (No Car)

“\( 65 \) of the people who have a bicycle do not have a car”.

This goes in the “No Car” -> “Have Bicycle” circle (Bottom-Top).

The “No Car” -> “No Bicycle” circle is \( 90 – 65 = 25 \).

Step 3: Bicycle Branches (Have Car)

Total people with a bicycle = \( 110 \).

We know \( 65 \) of these don’t have a car.

So, people with a car AND a bicycle = \( 110 – 65 = 45 \).

This goes in “Have Car” -> “Have Bicycle” (Top-Top).

The “Have Car” -> “No Bicycle” circle is \( 150 – 45 = 105 \).

💡 Problem: “Percentage of the 150 people who have a car also have a bicycle”.

Base number = \( 150 \) (people with cars).

Target number = \( 45 \) (people with cars AND bicycles).

💡 Calculator Tip: Use brackets for the numerator and denominator if entering in one line.

\[ (25 – \sqrt{43.87}) \div (6 + 2.1^2) \]

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

💡 Strategy: Break \( 60 \) down into pairs of factors until you reach prime numbers (numbers that only have factors of 1 and themselves).

Circle the prime numbers as you go.

💡 Understanding Ratio parts:

A ratio of \( 1:2 \) means for every \( 1 \) red counter, there are \( 2 \) blue counters.

Total parts = \( 1 + 2 = 3 \) parts.

Helen assumed that because the ratio numbers are 1 and 2, red is half of the total. This is incorrect. Red is half of blue, but it is one third of the total.

💡 Interpreting the Symbols:

\( n < 5 \) means \( n \) must be strictly less than 5.

The integers below 5 are \( 4, 3, 2, \dots \)

💡 Rule:

• \( \leq \) means “less than or equal to” → Filled Circle (Solid dot).
• \( < \) means "strictly less than" → Empty Circle (Hollow ring).

We need a line connecting -4 and 1.

At -4: Filled circle.

At 1: Empty circle.

Triangle Area: \( \frac{1}{2} \times \text{base} \times \text{height} \)

Rectangle Area: \( \text{length} \times \text{width} \)

The problem states: “Area of triangle is 10 greater than Area of rectangle”.

So: \( \text{Triangle} = \text{Rectangle} + 10 \)

The ratio is \( \text{Paper} : \text{Glass} = 12 : 7 \).

Total parts = \( 12 + 7 = 19 \).

We need to split \( 456 \text{ kg} \) into 19 parts, then find the value of 7 parts (Glass).

💡 Strategy:

The number is rounded to 1 decimal place (0.1).

To find the bounds, we halve the degree of accuracy.

\[ 0.1 \div 2 = 0.05 \]

The error interval includes the lower bound but excludes the upper bound.

💡 Multiplier Method:

Increase by \( 4\% \) means \( 100\% + 4\% = 104\% \).

Multiplier = \( 1.04 \).

For 2 years, we multiply by \( 1.04^2 \).

💡 Multiplier Method:

Increase by \( 1.5\% \) means \( 100\% + 1.5\% = 101.5\% \).

Multiplier = \( 1.015 \).

For 2 years, we multiply by \( 1.015^2 \).

Graph A: Straight line sloping downwards. Negative gradient. Linear equation.

Graph B: U-shaped curve. Parabola. Quadratic equation (\( x^2 \)).

Graph C: Two curves in opposite corners. Asymptotes. Reciprocal equation (\( \frac{1}{x} \)).

Graph D: Straight line sloping upwards. Positive gradient. Linear equation.

Graph E: S-shaped curve. Cubic equation (\( x^3 \)).