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GCSE Maths Edexcel Foundation Paper 2 (Calculator) – Summer 2022
Guidance
- Calculator Allowed: This is a calculator paper. Show your button sequences.
- Diagrams: All diagrams are redrawn for clarity.
- Navigation: Use the Table of Contents below to jump to questions.
Table of Contents
- Question 1 (Rounding)
- Question 2 (Fractions)
- Question 3 (Mode)
- Question 4 (Multiples)
- Question 5 (Percentages)
- Question 6 (Ordering Numbers)
- Question 7 (Polygons)
- Question 8 (Coordinates)
- Question 9 (Line Graphs)
- Question 10 (Finance/Arithmetic)
- Question 11 (Angles)
- Question 12 (Number Machines)
- Question 13 (Two-way Tables)
- Question 14 (Inequalities)
- Question 15 (Frequency Tables)
- Question 16 (Scale Drawings)
- Question 17 (Linear Graphs)
- Question 18 (Mean)
- Question 19 (Profit/Percentages)
- Question 20 (Probability Trees)
- Question 21 (Algebra)
- Question 22 (Transformations)
- Question 23 (Error Intervals)
- Question 24 (Area/Ratio)
- Question 25 (Coordinates)
- Question 26 (Depreciation)
- Question 27 (Currency Conversion)
- Question 28 (Simultaneous Equations)
Question 1 (1 mark)
Write \( 1476 \) to the nearest \( 10 \).
Worked Solution
Step 1: Understanding Rounding
Why we do this: We need to round to the nearest “10”. This means we look at the digit in the “tens” column and the digit to its right (the “units” column) to decide whether to round up or down.
The number is \( 1476 \).
- The tens digit is \( 7 \).
- The units digit is \( 6 \).
Since the units digit (\( 6 \)) is \( 5 \) or more, we round up.
The \( 7 \) in the tens column becomes \( 8 \), and the unit becomes \( 0 \).
\[ 1476 \rightarrow 1480 \]Final Answer:
1480
✓ (B1)
Question 2 (1 mark)
Write a fraction in the box to make the calculation correct.
\[ 1 – \frac{3}{10} = \boxed{\phantom{000}} \]Worked Solution
Step 1: Subtracting Fractions
Why we do this: The number \( 1 \) represents a whole. To subtract \( \frac{3}{10} \), it helps to write \( 1 \) as a fraction with the same denominator.
Write \( 1 \) as \( \frac{10}{10} \).
\[ \frac{10}{10} – \frac{3}{10} = \frac{10-3}{10} = \frac{7}{10} \]Final Answer:
\(\frac{7}{10}\)
✓ (B1)
Question 3 (1 mark)
Here is a list of numbers.
3 3 3 3 4 4 5 7 8
Write down the mode of the numbers.
Worked Solution
Step 1: Definition of Mode
What is the mode? The mode is the number that appears most often.
Let’s count how many times each number appears:
- 3 appears 4 times
- 4 appears 2 times
- 5 appears 1 time
- 7 appears 1 time
- 8 appears 1 time
The number 3 appears the most.
Final Answer:
3
✓ (B1)
Question 4 (1 mark)
Write down a 3 digit number that is a multiple of \( 5 \).
Worked Solution
Step 1: Identifying Multiples of 5
Rule: Any number that ends in \( 0 \) or \( 5 \) is a multiple of \( 5 \).
The question asks for a 3 digit number (any number between 100 and 999).
We can pick any 3 digit number ending in 0 or 5. Examples include:
- 100
- 105
- 250
- 995
Any of these are correct.
Final Answer:
e.g., 725
(Any 3-digit number ending in 0 or 5 is accepted)
✓ (B1)
Question 5 (1 mark)
Write \( 0.4 \) as a percentage.
Worked Solution
Step 1: Converting Decimal to Percentage
Method: To convert a decimal to a percentage, multiply by \( 100 \).
Check: \( 0.4 \) is the same as \( \frac{4}{10} \) or \( \frac{40}{100} \), which is \( 40\% \).
Final Answer:
40%
✓ (B1)
Question 6 (1 mark)
Write the following numbers in order of size. Start with the smallest number.
\( -11 \quad -2 \quad 8 \quad -7 \quad 3 \quad 10 \)
Worked Solution
Step 1: Ordering Negative and Positive Numbers
Strategy: Imagine a number line. The further left a number is (more negative), the smaller it is.
Let’s separate the negative and positive numbers:
- Negatives: \( -11, -2, -7 \)
- Positives: \( 8, 3, 10 \)
Order the negatives (most negative is smallest):
\( -11 \) is the smallest, followed by \( -7 \), then \( -2 \).
Order the positives:
\( 3 \) is smaller than \( 8 \), which is smaller than \( 10 \).
Combining them:
\( -11, -7, -2, 3, 8, 10 \)
Final Answer:
-11, -7, -2, 3, 8, 10
✓ (B1)
Question 7 (3 marks)
Here is polygon \( ABCDEF \) on a square grid.
(a) Write down the mathematical name of the polygon.
(b) Which side of the polygon is parallel to the side \( CD \)?
(c) Write down a side of the polygon that is perpendicular to the side \( AF \).
Worked Solution
Part (a): Naming the Polygon
Method: Count the number of sides.
- \( AB \)
- \( BC \)
- \( CD \)
- \( DE \)
- \( EF \)
- \( FA \)
There are 6 sides. A 6-sided polygon is called a hexagon.
Answer: Hexagon ✓ (B1)
Part (b): Parallel Sides
Definition: Parallel lines run in the same direction and never meet. They are like train tracks.
Look at side \( CD \). It is a horizontal line at the top.
Look for another horizontal line in the shape.
The side \( AF \) (the bottom base) is also horizontal.
Therefore, \( AF \) is parallel to \( CD \).
Answer: AF ✓ (B1)
Part (c): Perpendicular Sides
Definition: Perpendicular lines meet at a right angle (\( 90^\circ \)).
Side \( AF \) is horizontal.
We need to find a vertical side attached to \( AF \).
Side \( AB \) is vertical and meets \( AF \) at \( A \).
Side \( EF \) is vertical and meets \( AF \) at \( F \).
Either answer is correct.
Answer: AB (or EF) ✓ (B1)
Question 8 (4 marks)
Here is a centimetre grid.
(a) Write down the coordinates of point \( A \).
(b) On the grid, mark with a cross (\( \times \)) the point with coordinates \( (-4, 3) \). Label this point \( B \).
(c) On the grid, draw the circle with centre \( (1, -1) \) and radius \( 4 \) cm.
Worked Solution
Part (a): Coordinates of A
Rule: Coordinates are written as \( (x, y) \).
Start at the origin \( (0,0) \).
- Go along the x-axis to align with \( A \): We move right to \( 3 \).
- Go up the y-axis to reach \( A \): We move up to \( 2 \).
Answer: (3, 2) ✓ (B1)
Part (b): Plotting Point B
Coordinates: \( (-4, 3) \)
- x-coordinate is \( -4 \): Move left 4 units.
- y-coordinate is \( 3 \): Move up 3 units.
Mark this point with a cross and label it \( B \).
(Visual approximation)
Result: Cross marked at \( (-4, 3) \). ✓ (B1)
Part (c): Drawing the Circle
Instructions:
- Locate the centre \( (1, -1) \).
- Set your compass radius to \( 4 \) cm (which is 4 grid squares).
- Place the compass point on \( (1, -1) \) and draw the circle.
The circle should pass through:
- \( (5, -1) \) (Right side)
- \( (1, 3) \) (Top)
- \( (-3, -1) \) (Left side)
- \( (1, -5) \) (Bottom)
Result: Correct circle drawn. ✓ (B2)
Question 9 (3 marks)
The graph shows information about the number of houses sold by an estate agent in each of six months last year.
(a) How many houses were sold by the estate agent in February?
(b) For this estate agent, write down the ratio of the number of houses sold in January to the number of houses sold in June.
Worked Solution
Part (a): Reading the Graph
Scale Check: On the y-axis, the gap between \( 0 \) and \( 10 \) is divided into 5 small squares.
\[ \text{Value of 1 small square} = \frac{10}{5} = 2 \]
Locate “February” on the x-axis and look up to the cross.
- The cross is above the line for \( 20 \).
- It is in the middle of the second small square, which usually implies halfway values, but counting strictly by lines: 20, 22, 24. It is exactly in the middle of 22 and 24.
- Wait, careful reading: The cross is at \( 23 \).
Answer: 23 ✓ (B1)
Part (b): Ratio Calculation
Step 1: Get values
- January: Locate cross at \( 10 \).
- June: Locate cross. It is 2 small squares below \( 60 \).
- Calculation: \( 60 – (2 \times 2) = 56 \).
Step 2: Write the Ratio
Ratio is “January to June” = \( 10 : 56 \)
The question asks to “write down the ratio”, so \( 10:56 \) is sufficient. You can simplify it to \( 5:28 \) but it’s not required unless specified.
Answer: 10 : 56 ✓ (A1)
Question 10 (3 marks)
Sonia wants to book a holiday. The holiday will cost \( £1428 \).
Sonia will pay a deposit of \( £150 \).
She will then pay the rest of the cost in \( 6 \) equal monthly payments.
How much is each monthly payment?
Worked Solution
Step 1: Calculate “The Rest of the Cost”
Sonia pays a deposit first. This amount is subtracted from the total cost.
\[ \text{Remaining Cost} = 1428 – 150 = 1278 \]
Step 2: Divide into Payments
The remaining \( £1278 \) is split equally into \( 6 \) payments.
\[ 1278 \div 6 \]
Calculator: 1278 ÷ 6 =
\[ 213 \]
Final Answer:
£213
✓ (A1)
Question 11 (3 marks)
The diagram shows a triangle \( ABC \).
\( ACD \) and \( BCE \) are straight lines.
Work out the size of the angle marked \( x \).
Give a reason for each stage of your working.
Worked Solution
Step 1: Find Angle ACB
Reasoning: Angles in a triangle add up to \( 180^\circ \).
In triangle \( ABC \), we know angle \( A = 116^\circ \) and angle \( B = 25^\circ \).
Step 2: Find Angle x
Reasoning: Vertically opposite angles are equal.
Angle \( x \) (which is \( \angle ECD \)) is vertically opposite to \( \angle ACB \).
Final Answer:
39°
Reasons:
- Angles in a triangle add up to 180°.
- Vertically opposite angles are equal.
✓ (C1)
Question 12 (3 marks)
Here is a number machine.
(a) Work out the output when the input is \( 28 \).
Here is a different number machine. The number machine is not complete.
When the input is \( 8 \), the output is \( 154 \).
(b) Complete the number machine.
Worked Solution
Part (a)
Method: Follow the operations from left to right with the input \( 28 \).
- First operation: Divide by 7. \[ 28 \div 7 = 4 \]
- Second operation: Add 5. \[ 4 + 5 = 9 \]
Answer: 9 ✓ (B1)
Part (b)
Method: Work backwards (inverse operations) from the output to find the missing operation.
Input: \( 8 \)
Output: \( 154 \)
1. Start at the end: The last step was \( \times 11 \) to get \( 154 \).
Inverse of \( \times 11 \) is \( \div 11 \).
\[ 154 \div 11 = 14 \]So, the number after the first box must be \( 14 \).
2. Find the missing operation:
The input is \( 8 \) and it becomes \( 14 \).
The box has a \( + \) sign.
\[ 8 + ? = 14 \] \[ ? = 14 – 8 = 6 \]Answer: 6 ✓ (B1 for methods, A1 for 6)
Question 13 (3 marks)
Sophie works in a bed shop. During the last three months she sold \( 198 \) beds.
\( 59 \) beds were sold without a mattress.
\( 45 \) beds were double beds.
\( 17 \) of the single beds were sold without a mattress.
\( 67 \) of the \( 83 \) king size beds were sold with a mattress.
Use this information to complete the two-way table.
| Single | Double | King size | Total | |
|---|---|---|---|---|
| With mattress | ||||
| Without mattress | ||||
| Total |
Worked Solution
Step 1: Enter Given Data
Fill in the numbers directly from the question:
- Total beds = \( 198 \) (Bottom-Right)
- Without mattress total = \( 59 \) (Row 2, Column “Total”)
- Double total = \( 45 \) (Bottom Row, Column “Double”)
- Single without mattress = \( 17 \) (Row 2, Column “Single”)
- King size total = \( 83 \) (Bottom Row, Column “King size”)
- King size with mattress = \( 67 \) (Row 1, Column “King size”)
Step 2: Calculate Missing Totals
Row 3 (Total Beds):
Total = Single + Double + King
\( 198 = \text{Single} + 45 + 83 \)
\( \text{Single} = 198 – 128 = 70 \)
Column (Total Mattress Types):
Total = With + Without
\( 198 = \text{With} + 59 \)
\( \text{With} = 198 – 59 = 139 \)
Step 3: Calculate Internal Cells
Column (Single):
Total Single = With + Without
\( 70 = \text{With} + 17 \)
\( \text{With} = 53 \)
Column (King size):
Total King = With + Without
\( 83 = 67 + \text{Without} \)
\( \text{Without} = 16 \)
Row 1 (With Mattress):
Total = Single + Double + King
\( 139 = 53 + \text{Double} + 67 \)
\( \text{Double} = 139 – 120 = 19 \)
Row 2 (Without Mattress):
Total = Single + Double + King
\( 59 = 17 + \text{Double} + 16 \)
\( \text{Double} = 59 – 33 = 26 \)
Check: \( 19 + 26 = 45 \) (Matches Double Total)
Completed Table:
| Single | Double | King size | Total | |
|---|---|---|---|---|
| With mattress | 53 | 19 | 67 | 139 |
| Without mattress | 17 | 26 | 16 | 59 |
| Total | 70 | 45 | 83 | 198 |
✓ (B3)
Question 14 (2 marks)
The box below contains three mathematical symbols.
From the box, choose a symbol to make each of the following statements correct.
(i) \( \frac{5}{8} \quad \dots \quad \frac{2}{8} \)
(ii) \( -2 \times -3 \quad \dots \quad -3 + 9 \)
Worked Solution
Part (i)
We are comparing \( \frac{5}{8} \) and \( \frac{2}{8} \).
Since the denominators are the same, we compare the numerators: \( 5 \) and \( 2 \).
\( 5 \) is greater than \( 2 \).
So, \( \frac{5}{8} > \frac{2}{8} \).
Answer: \( > \) ✓ (B1)
Part (ii)
Calculate the value of each side.
Left Side: \( -2 \times -3 \)
Negative \( \times \) Negative = Positive.
\( 2 \times 3 = 6 \). So, result is \( 6 \).
Right Side: \( -3 + 9 \)
Start at \( -3 \) and go up \( 9 \).
\( 9 – 3 = 6 \). So, result is \( 6 \).
Comparing sides: \( 6 \) is equal to \( 6 \).
Answer: \( = \) ✓ (B1)
Question 15 (4 marks)
The table shows information about the number of social media accounts used by each of \( 300 \) students.
| Number of social media accounts | Frequency |
|---|---|
| 0 | 3 |
| 1 | 57 |
| 2 | 84 |
| 3 | 75 |
| 4 | 81 |
(a) Work out the total number of social media accounts used by these students.
(b) Find the median number of social media accounts used by these students.
Worked Solution
Part (a): Total Number of Accounts
Method: Multiply the number of accounts by the frequency for each row, then sum them up.
- \( 0 \times 3 = 0 \)
- \( 1 \times 57 = 57 \)
- \( 2 \times 84 = 168 \)
- \( 3 \times 75 = 225 \)
- \( 4 \times 81 = 324 \)
Total = \( 0 + 57 + 168 + 225 + 324 \)
Total = \( 774 \)
Answer: 774 ✓ (M1, A1)
Part (b): Finding the Median
Definition: The median is the middle value.
There are \( 300 \) students. The middle is between the 150th and 151st student.
Let’s use cumulative frequency to find where the 150th student is:
- \( 0 \) accounts: \( 3 \) students (Positions 1-3)
- \( 1 \) account: \( 3 + 57 = 60 \) students (Positions 4-60)
- \( 2 \) accounts: \( 60 + 84 = 144 \) students (Positions 61-144)
- \( 3 \) accounts: \( 144 + 75 = 219 \) students (Positions 145-219)
The 150th student is in the group with \( 3 \) accounts.
The 151st student is in the group with \( 3 \) accounts.
Therefore, the median is \( 3 \).
Answer: 3 ✓ (M1, A1)
Question 16 (3 marks)
On a scale drawing, a building has length \( 12.4 \) cm and width \( 9.4 \) cm.
The real length of the building is \( 62 \) metres.
Work out, in metres, the real width of the building.
Worked Solution
Step 1: Determine the Scale Factor
Why we do this: We need to find how many real metres are represented by \( 1 \) cm on the drawing.
We know:
- Drawing Length = \( 12.4 \) cm
- Real Length = \( 62 \) m
Divide the real length by the drawing length:
\[ \text{Scale Factor} = \frac{62 \text{ m}}{12.4 \text{ cm}} \]Calculator: 62 ÷ 12.4 = \( 5 \)
This means \( 1 \) cm on the drawing represents \( 5 \) m in real life.
Step 2: Calculate Real Width
We know the drawing width is \( 9.4 \) cm.
Using the scale factor of \( 5 \) m per cm:
Calculator: 9.4 × 5 = \( 47 \)
Final Answer:
47 metres
✓ (P1, P1, A1)
Question 17 (3 marks)
On the grid below, draw the graph of \( y = 4 – x \) for values of \( x \) from \( -2 \) to \( 4 \).
Worked Solution
Step 1: Create a Table of Values
Equation: \( y = 4 – x \)
We calculate \( y \) for integer values of \( x \) from \( -2 \) to \( 4 \).
| \( x \) | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| \( 4 – x \) | \( 4 – (-2) = 6 \) | \( 4 – (-1) = 5 \) | 4 | 3 | 2 | 1 | 0 |
Step 2: Plot and Connect
Plot the points: \( (-2, 6) \), \( (-1, 5) \), \( (0, 4) \), \( (1, 3) \), \( (2, 2) \), \( (3, 1) \), \( (4, 0) \).
Join them with a straight line.
Answer: Correct straight line drawn. ✓ (B3)
Question 18 (3 marks)
This sign was in a doctor’s waiting room.
\( 115 \) appointments were missed last month.
These missed appointments were a total of \( 25.3 \) hours.
Work out the mean length of time for each missed appointment.
Give your answer in minutes.
Worked Solution
Step 1: Convert Total Time to Minutes
Why: The question asks for the answer in minutes, but the total time is in hours.
There are \( 60 \) minutes in \( 1 \) hour.
Calculator: 25.3 × 60 = \( 1518 \) minutes.
Step 2: Calculate Mean Time
Formula: \( \text{Mean} = \frac{\text{Total Time}}{\text{Number of Appointments}} \)
Calculator: 1518 ÷ 115 = \( 13.2 \)
Final Answer:
13.2 minutes
✓ (P1, P1, A1)
Question 19 (5 marks)
Nimra buys a \( 3 \) kg box of sweets for \( £17.60 \).
She puts the sweets into bags to sell.
Each bag contains \( 150 \) g of sweets.
Nimra fills as many bags as possible.
She will sell each bag for the same price.
Nimra wants to make a profit of at least \( 35\% \).
Assuming she sells all the bags, what is the lowest price Nimra should charge for each bag?
Worked Solution
Step 1: Calculate Total Number of Bags
Why: We need to know how many bags she sells. First, convert units so they match.
\( 3 \text{ kg} = 3000 \text{ g} \)
Calculator: 3000 ÷ 150 = \( 20 \)
She makes \( 20 \) bags.
Step 2: Calculate Target Total Income
Why: She wants a \( 35\% \) profit on her cost of \( £17.60 \).
We need to increase \( £17.60 \) by \( 35\% \).
Multiplier for \( 35\% \) increase = \( 1.35 \).
Calculator: 17.60 × 1.35 = \( 23.76 \)
She needs to make a total of \( £23.76 \).
Step 3: Calculate Price Per Bag
Divide the total target income by the number of bags.
Calculator: 23.76 ÷ 20 = \( 1.188 \)
Step 4: Rounding (Money Context)
In money, we usually round to 2 decimal places.
\( £1.188 \) rounds to \( £1.19 \).
Checking: \( 1.19 \times 20 = 23.80 \), which is more than \( 23.76 \), so profit condition is met.
Answer: £1.19 ✓ (A1)
Question 20 (4 marks)
Lorena gets a train at the same time each morning to go to work.
She gets a train at the same time each evening to come home.
The probability tree diagram shows the probabilities of each train arriving late.
(a) Complete the probability tree diagram.
(b) Work out the probability that the train to work and the train home will both arrive late.
Worked Solution
Part (a): Completing the Tree
Rule: Probabilities on branches from a single point must add up to \( 1 \).
First Set (Train to work):
- Late = \( 0.13 \)
- Not Late = \( 1 – 0.13 = 0.87 \)
Second Set (Train home):
- Late = \( 0.06 \) (Given)
- Not Late = \( 1 – 0.06 = 0.94 \)
These values are the same for both branches because the events are independent (probability stays the same).
Answer: 0.87, 0.94, 0.94 ✓ (B2)
Part (b): Both Late
Method: Follow the path “Late” AND “Late”.
When moving along branches (AND), we multiply the probabilities.
Calculator: 0.13 × 0.06 = \( 0.0078 \)
Answer: 0.0078 ✓ (M1, A1)
Question 21 (5 marks)
(a) Simplify \( (x^3)^5 \)
(b) Expand and simplify \( 4(x + 3) + 7(4 – 2x) \)
(c) Factorise fully \( 15x^3 + 3x^2y \)
Worked Solution
Part (a): Simplify \( (x^3)^5 \)
Rule: When raising a power to a power, multiply the indices: \( (a^m)^n = a^{m \times n} \).
Answer: \( x^{15} \) ✓ (B1)
Part (b): Expand and Simplify
Step 1: Expand the brackets
Multiply the number outside by each term inside.
\( 4(x + 3) = 4x + 12 \)
\( 7(4 – 2x) = 28 – 14x \)
Combined expression: \( 4x + 12 + 28 – 14x \)
Step 2: Collect like terms
Combine the \( x \) terms and the number terms.
\( x \) terms: \( 4x – 14x = -10x \)
Numbers: \( 12 + 28 = 40 \)
Result: \( 40 – 10x \)
Answer: \( 40 – 10x \) ✓ (M1, A1)
Part (c): Factorise Fully
Method: Find the highest common factor (HCF) of numbers and letters.
Terms: \( 15x^3 \) and \( 3x^2y \)
Numbers: HCF of \( 15 \) and \( 3 \) is \( 3 \).
Letters: Both terms have \( x \). The lowest power is \( x^2 \). So HCF is \( x^2 \).
Total Factor: \( 3x^2 \).
Divide each term by \( 3x^2 \):
- \( 15x^3 \div 3x^2 = 5x \)
- \( 3x^2y \div 3x^2 = y \)
Put it together: \( 3x^2(5x + y) \)
Answer: \( 3x^2(5x + y) \) ✓ (M1, A1)
Question 22 (2 marks)
Describe fully the single transformation that maps shape \( S \) onto shape \( T \).
Worked Solution
Step 1: Identify Transformation Type
Compare shape \( S \) and shape \( T \).
- Is it rotated? No, the orientation is the same.
- Is it reflected? No, it hasn’t flipped.
- Is it larger/smaller? No, same size.
- It has simply moved. This is a translation.
Step 2: Find the Vector
Pick a corner on shape \( S \) (e.g., top-left corner) and find how to get to the matching corner on shape \( T \).
Corner on \( S \): \( (1, -3) \)
Corner on \( T \): \( (-4, 3) \)
Movement in x (horizontal): From \( 1 \) to \( -4 \) is \( -5 \) (5 left).
Movement in y (vertical): From \( -3 \) to \( 3 \) is \( +6 \) (6 up).
Vector: \( \begin{pmatrix} -5 \\ 6 \end{pmatrix} \)
Final Answer:
Translation by vector \( \begin{pmatrix} -5 \\ 6 \end{pmatrix} \)
✓ (B1 for Translation, B1 for Vector)
Question 23 (2 marks)
The length of a football pitch is \( 90 \) metres, correct to the nearest metre.
Complete the error interval for the length of the football pitch.
\( \dots\dots\dots\dots \) m \( \leq \text{length} < \dots\dots\dots\dots \) m
Worked Solution
Step 1: Understanding Error Intervals
Why: When rounding to the nearest unit (1 metre), the actual value could be half a unit below or half a unit above.
Unit = \( 1 \). Half unit = \( 0.5 \).
Lower Bound: \( 90 – 0.5 = 89.5 \)
Upper Bound: \( 90 + 0.5 = 90.5 \)
Notation: The lower bound is inclusive (\( \leq \)), the upper bound is exclusive (\( < \)).
Answer: \( 89.5 \leq \text{length} < 90.5 \) ✓ (B1, B1)
Question 24 (5 marks)
Festival A will be in a rectangular field with an area of \( 80\,000 \text{ m}^2 \).
The greatest number of people allowed to attend Festival A is \( 425 \).
Festival B will be in a rectangular field \( 700 \text{ m} \) by \( 2000 \text{ m} \).
The greatest number of people allowed to attend Festival B is \( 6750 \).
The area per person allowed for Festival B is greater than the area per person allowed for Festival A.
(a) How much greater? Give your answer correct to the nearest whole number.
Callum says,
“\( 300 \text{ cm}^2 \) is the same as \( 3 \text{ m}^2 \) because there are \( 100 \text{ cm} \) in \( 1 \text{ m} \) so you divide by \( 100 \).”
Callum’s method is wrong.
(b) Explain why.
Worked Solution
Part (a): Comparing Areas Per Person
Step 1: Festival A
Area = \( 80\,000 \text{ m}^2 \). People = \( 425 \).
Area per person = \( \frac{80\,000}{425} \).
Calculator: \( 80000 \div 425 = 188.235… \)
Step 2: Festival B
First find the area: \( 700 \times 2000 \).
Area = \( 1\,400\,000 \text{ m}^2 \).
People = \( 6750 \).
Area per person = \( \frac{1\,400\,000}{6750} \).
Calculator: \( 1400000 \div 6750 = 207.407… \)
Step 3: Difference
Subtract A from B.
\( 207.407… – 188.235… = 19.17… \)
Round to nearest whole number: \( 19 \).
Answer: 19 m² ✓ (P1, P1, P1, A1)
Part (b): Area Conversion Explanation
Reasoning: Area is two-dimensional (length × width).
Conversion for length is \( \div 100 \).
Conversion for area is \( \div 100^2 \) (or \( \div 100 \div 100 \)).
Explanation: He should divide by \( 100^2 \) (or \( 10\,000 \)) because it is square units, not linear units.
Answer: He must square the conversion factor (divide by 10,000). ✓ (C1)
Question 25 (4 marks)
The points \( L \), \( M \) and \( N \) are such that \( LMN \) is a straight line.
The coordinates of \( L \) are \( (-3, 1) \).
The coordinates of \( M \) are \( (4, 9) \).
Given that \( LM : MN = 2 : 3 \), find the coordinates of \( N \).
Worked Solution
Step 1: Calculate the Vector LM
Find the change in x and y from \( L \) to \( M \).
\( L(-3, 1) \to M(4, 9) \)
Change in \( x = 4 – (-3) = 7 \)
Change in \( y = 9 – 1 = 8 \)
Vector \( LM = \begin{pmatrix} 7 \\ 8 \end{pmatrix} \)
Step 2: Use the Ratio to Find Vector MN
Ratio \( LM : MN = 2 : 3 \).
This means \( MN \) is \( \frac{3}{2} \) times (or \( 1.5 \) times) as long as \( LM \).
Multiply the vector \( LM \) by \( 1.5 \).
\( x \text{ change for } MN = 7 \times 1.5 = 10.5 \)
\( y \text{ change for } MN = 8 \times 1.5 = 12 \)
Step 3: Calculate Coordinates of N
Add the change for \( MN \) to the coordinates of \( M \).
\( M \) is at \( (4, 9) \).
\( x = 4 + 10.5 = 14.5 \)
\( y = 9 + 12 = 21 \)
Answer: (14.5, 21) ✓ (A1)
Question 26 (3 marks)
A new phone cost \( £679 \).
The value of the phone decreases at a rate of \( 4\% \) per year.
Work out the value of the phone at the end of \( 3 \) years.
Worked Solution
Step 1: Determine the Multiplier
Method: The value decreases by \( 4\% \), so \( 96\% \) of the value remains each year.
\( 100\% – 4\% = 96\% \)
Multiplier = \( 0.96 \).
Step 2: Calculate Compound Depreciation
Formula: \( \text{Original Amount} \times (\text{Multiplier})^{\text{Number of Years}} \)
Calculator: 679 × 0.96 ^ 3 = \( 600.73856 \)
Step 3: Round to Money
Money is rounded to \( 2 \) decimal places.
\( 600.738… \to 600.74 \)
Final Answer:
£600.74
✓ (M1, M1, A1)
Question 27 (4 marks)
In Spain, Sam pays \( 27 \) euros for \( 18 \) litres of petrol.
In Wales, Leo pays \( £40.80 \) for \( 8 \) gallons of the same type of petrol.
\( 1 \) euro = \( £0.85 \)
\( 4.5 \) litres = \( 1 \) gallon
Sam thinks that petrol is cheaper in Spain than in Wales.
Is Sam correct?
You must show how you get your answer.
Worked Solution
Step 1: Convert Spain Cost to £ per Litre
Convert Euros to Pounds:
Cost = \( 27 \text{ euros} \times 0.85 = £22.95 \).
Volume = \( 18 \text{ litres} \).
Unit Price (Spain):
\[ \frac{22.95}{18} = £1.275 \text{ per litre} \]
Step 2: Convert Wales Cost to £ per Litre
Convert Gallons to Litres:
Volume = \( 8 \text{ gallons} \times 4.5 = 36 \text{ litres} \).
Cost = \( £40.80 \).
Unit Price (Wales):
\[ \frac{40.80}{36} = £1.133… \text{ per litre} \]
Step 3: Compare and Conclude
Spain: \( £1.275 \)
Wales: \( £1.13 \)
\( 1.275 > 1.13 \)
Spain is more expensive.
Final Answer:
Sam is incorrect (No).
(Spain is more expensive)
✓ (P1, P1, P1, C1)
Question 28 (3 marks)
Solve the simultaneous equations
\[ 5x + 2y = 27 \]
\[ 6x + 4y = 28 \]
Worked Solution
Step 1: Equalise Coefficients
Multiply the first equation by \( 2 \) so the \( y \) coefficients become the same (\( 4y \)).
Eq 1: \( 5x + 2y = 27 \) \( (\times 2) \rightarrow 10x + 4y = 54 \)
Eq 2: \( 6x + 4y = 28 \)
Step 2: Subtract Equations
Subtract Eq 2 from the new Eq 1 to eliminate \( y \).
\( (10x – 6x) + (4y – 4y) = (54 – 28) \)
\( 4x = 26 \)
\( x = \frac{26}{4} = 6.5 \)
Step 3: Substitute and Solve for y
Substitute \( x = 6.5 \) into the original Eq 1.
\( 5(6.5) + 2y = 27 \)
\( 32.5 + 2y = 27 \)
\( 2y = 27 – 32.5 \)
\( 2y = -5.5 \)
\( y = -2.75 \)
Final Answer:
\( x = 6.5 \)
\( y = -2.75 \)
✓ (M1, M1, A1)