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GCSE May 2024 Edexcel Higher Paper 1 (Non-Calculator)
📝 Guidance
- Paper 1: Non-Calculator. You must show all your working out.
- Diagrams: Are NOT accurately drawn unless otherwise indicated.
- Marks: The marks for each question are shown in brackets.
📚 Table of Contents
- Question 1 (Arithmetic Sequence)
- Question 2 (Fractions)
- Question 3 (Area & Estimation)
- Question 4 (Venn Diagrams)
- Question 5 (Estimation)
- Question 6 (Linear Graphs)
- Question 7 (Ratios & Percentages)
- Question 8 (Mean)
- Question 9 (Percentages)
- Question 10 (Simultaneous Equations)
- Question 11 (Transformations)
- Question 12 (Graph Recognition)
- Question 13 (Histograms)
- Question 14 (Algebra Expansion)
- Question 15 (Sector Area)
- Question 16 (Probability)
- Question 17 (Surds)
- Question 18 (Recurring Decimals)
- Question 19 (Similar Triangles)
- Question 20 (Indices)
- Question 21 (Problem Solving – Cuboid)
- Question 22 (Trig Graphs)
- Question 23 (Coordinate Geometry)
Question 1 (2 marks)
Here are the first four terms of an arithmetic sequence.
1 5 9 13
Find an expression, in terms of \( n \), for the \( n \)th term of this sequence.
Worked Solution
Step 1: Find the Common Difference
What are we looking for?
An arithmetic sequence increases by the same amount each time. This amount is called the “common difference”.
Difference between terms:
\( 5 – 1 = 4 \)
\( 9 – 5 = 4 \)
\( 13 – 9 = 4 \)
The sequence goes up by 4 each time.
What this tells us:
The formula must start with \( 4n \), similar to the 4 times table.
Step 2: Adjust the Formula
Strategy:
Compare our sequence to the 4 times table (\( 4n \)) to find the adjustment.
\( n \) (Position): 1, 2, 3, 4
\( 4n \) (4 × table): 4, 8, 12, 16
Our Sequence: 1, 5, 9, 13
How do we get from 4 to 1? We subtract 3.
How do we get from 8 to 5? We subtract 3.
Conclusion:
We need to subtract 3 from \( 4n \).
Final Answer:
\( 4n – 3 \)
✓ (Total: 2 marks)
Question 2 (3 marks)
(a) Work out \( 3\frac{4}{5} – 1\frac{2}{3} \)
(2 marks)
Kevin was asked to work out \( 2\frac{1}{3} \times \frac{5}{8} \)
Here is his working and his answer.
\( 2\frac{1}{3} \times \frac{5}{8} = \frac{7}{3} \times \frac{5}{8} \)
\( = \frac{35}{24} \)
\( = 1\frac{9}{24} \)
Kevin’s answer is wrong.
(b) What mistake has Kevin made?
(1 mark)
Worked Solution
Part (a): Subtracting Mixed Numbers
Strategy:
It is often safer to convert mixed numbers to improper fractions first, then find a common denominator.
Convert to improper fractions:
\( 3\frac{4}{5} = \frac{(3 \times 5) + 4}{5} = \frac{19}{5} \)
\( 1\frac{2}{3} = \frac{(1 \times 3) + 2}{3} = \frac{5}{3} \)
Now calculate \( \frac{19}{5} – \frac{5}{3} \)
Part (a): Common Denominator
Why do we do this?
We cannot subtract fractions with different denominators (5 and 3). The lowest common multiple of 5 and 3 is 15.
Multiply top and bottom to get denominator 15:
\( \frac{19 \times 3}{5 \times 3} = \frac{57}{15} \)
\( \frac{5 \times 5}{3 \times 5} = \frac{25}{15} \)
Now subtract:
\( \frac{57}{15} – \frac{25}{15} = \frac{32}{15} \)
Convert back to mixed number (optional but good practice):
How many 15s in 32? 2, remainder 2.
\( 2\frac{2}{15} \)
✓ (M1) for correct method to subtract using common denominator
✓ (A1) for correct answer
Part (b): Identifying the Error
Analysis of Kevin’s work:
Line 1: \( \frac{7}{3} \times \frac{5}{8} \) is correct.
Line 2: \( \frac{35}{24} \) is correct (\( 7 \times 5 = 35, 3 \times 8 = 24 \)).
Line 3: He converts \( \frac{35}{24} \) to \( 1\frac{9}{24} \).
Let’s check this conversion:
\( 35 – 24 = 11 \), so the remainder is 11.
It should be \( 1\frac{11}{24} \), but he wrote \( 1\frac{9}{24} \).
Final Answer (a): \( 2\frac{2}{15} \) (or \( \frac{32}{15} \))
Final Answer (b): Kevin calculated the remainder incorrectly when converting to a mixed number. \( 35 – 24 = 11 \), so the numerator should be 11, not 9.
✓ (Total: 3 marks)
Question 3 (5 marks)
The diagram shows a plan of a floor.
Petra is going to cover the floor with paint.
Petra has 3 tins of paint.
There are 2.5 litres of paint in each tin.
Petra thinks 1 litre of paint will cover \( 10 \text{ m}^2 \) of floor.
(a) Assuming Petra is correct, does she have enough paint to cover the floor?
You must show all your working.
(4 marks)
Actually, 1 litre of paint will cover \( 11 \text{ m}^2 \) of floor.
(b) Does this affect your answer to part (a)?
You must give a reason for your answer.
(1 mark)
Worked Solution
Step 1: Calculate the Total Area
Strategy:
The shape is a composite polygon (an L-shape). We can split it into two rectangles to find the area.
Let’s split it horizontally into a top rectangle and a bottom rectangle.
Method 1: Split Vertically
Left Rectangle: Height is 8m. What is the width? Wait, the bottom is 6m. The full top is 10m. The diagram shows the left side is 8m and bottom left part is 6m.
Let’s re-read the diagram carefully. Left wall = 8m. Top wall = 10m. Bottom-left wall = 6m. Top-right wall = 5m.
Let’s split vertically into two rectangles:
- Left Rectangle: Width 6m, Height 8m? No, looking at the diagram, the bottom-left segment is 6m wide. The left vertical is 8m. So we have a 6m x 8m rectangle on the left? No, because the top is 10m.
Let’s calculate the missing lengths first:
Total Width = 10m. Bottom part is 6m. So the “missing” width on the right is \( 10 – 6 = 4 \text{ m} \).
Total Height (Left) = 8m. Right side is 5m. So the “missing” height at the bottom right is \( 8 – 5 = 3 \text{ m} \).
Area Calculation (Two Rectangles):
- Top Rectangle: \( 10 \text{ m} \times 5 \text{ m} = 50 \text{ m}^2 \)
- Bottom Rectangle: \( 6 \text{ m} \times (8 – 5) \text{ m} = 6 \text{ m} \times 3 \text{ m} = 18 \text{ m}^2 \)
Total Area: \( 50 + 18 = 68 \text{ m}^2 \)
✓ (P1) Process to find area of one shape
✓ (P1) Process to find total area (68)
Step 2: Calculate Paint Needed vs. Paint Available
Given:
1 litre covers \( 10 \text{ m}^2 \).
Petra has 3 tins, each 2.5 litres.
Paint Required:
Area \( = 68 \text{ m}^2 \)
Litres needed \( = 68 \div 10 = 6.8 \text{ litres} \)
Paint Available:
\( 3 \text{ tins} \times 2.5 \text{ litres/tin} \)
\( 3 \times 2 = 6 \)
\( 3 \times 0.5 = 1.5 \)
\( 6 + 1.5 = 7.5 \text{ litres} \)
Comparison:
She has 7.5 litres. She needs 6.8 litres.
\( 7.5 > 6.8 \)
✓ (P1) Process to find litres needed OR total paint available
Final Conclusion for Part (a)
Yes, she has enough paint.
She has 7.5 litres but only needs 6.8 litres.
✓ (A1) Correct conclusion with supporting figures
Part (b): Effect of better coverage
Analysis:
If 1 litre covers \( 11 \text{ m}^2 \) instead of \( 10 \text{ m}^2 \), the paint goes further.
She will need less paint than calculated in part (a).
No, it does not affect the answer.
Reason: She already had enough paint. Since the paint covers more area, she will still have enough (she will just have more leftover).
✓ (C1) Correct reasoning (must reference answer to (a))
Question 4 (3 marks)
Here is a Venn diagram.
(a) Write down the numbers that are in set \( P’ \)
(1 mark)
A number is chosen at random from the universal set, \( \mathcal{E} \)
(b) Find the probability that this number is in the set \( P \cup Q \)
(2 marks)
Worked Solution
Part (a): Set Notation \( P’ \)
Understanding \( P’ \):
\( P’ \) (P prime or P complement) means everything NOT inside the circle P.
Look at the diagram and list all numbers outside the left circle.
Numbers in P: 12, 15, 18
Numbers NOT in P:
- Inside Q only: 10, 14
- Outside circles: 11, 13, 16, 17
Final Answer (a): 10, 11, 13, 14, 16, 17
✓ (B1)
Part (b): Probability of \( P \cup Q \)
What is \( P \cup Q \)?
The Union (\( \cup \)) means P OR Q. It includes all numbers inside either circle (including the overlap).
Step 1: Count total numbers in Universal Set (\( \mathcal{E} \))
Inside P only: 2 numbers (12, 15)
Intersection: 1 number (18)
Inside Q only: 2 numbers (10, 14)
Outside: 4 numbers (11, 13, 16, 17)
Total = \( 2 + 1 + 2 + 4 = 9 \) numbers.
Step 2: Count numbers in \( P \cup Q \)
These are the numbers in the circles: 12, 15, 18, 10, 14.
Count = 5.
Step 3: Write as probability
Probability = \( \frac{\text{Successful Outcomes}}{\text{Total Outcomes}} = \frac{5}{9} \)
Final Answer (b): \( \frac{5}{9} \)
✓ (M1) for identifying correct numbers
✓ (A1) for \( \frac{5}{9} \)
Question 5 (4 marks)
Sophie drives a distance of 513 kilometres on a motorway in France.
She pays 0.81 euros for every 10 kilometres she drives.
(a) Work out an estimate for the total amount that Sophie pays.
(3 marks)
(b) Is your answer to part (a) an underestimate or an overestimate?
Give a reason for your answer.
(1 mark)
Worked Solution
Step 1: Round the numbers (Estimation)
Rule for Estimation:
Round numbers to 1 significant figure to make the calculation easy to do in your head.
513 km \( \approx 500 \) km
0.81 euros \( \approx 0.80 \) (or 0.8) euros
Step 2: Calculate the cost
Method:
Find how many “10 km chunks” are in the total distance, then multiply by the cost per chunk.
Number of 10 km chunks = \( \frac{500}{10} = 50 \)
Total Cost = \( 50 \times 0.8 \)
Calculation Trick: \( 50 \times 8 = 400 \), so \( 50 \times 0.8 = 40.0 \)
Or: \( 5 \times 8 = 40 \)
Estimated Cost = 40 euros.
✓ (P1) Using rounded values
✓ (P1) Full process
✓ (A1) 40
Part (b): Evaluate the Estimate
Comparison:
Actual Distance: 513 (We rounded DOWN to 500)
Actual Cost: 0.81 (We rounded DOWN to 0.8)
Since we rounded both numbers down, our result will be lower than the true answer.
Final Answer (b): Underestimate
Reason: We rounded both the distance (513 to 500) and the cost (0.81 to 0.8) down.
✓ (C1)
Question 6 (4 marks)
Here is a straight line L drawn on a grid.
(a) Find an equation for L.
(3 marks)
M is a different straight line with equation \( y = 5x \)
(b) Write down the equation of a straight line parallel to M.
(1 mark)
Worked Solution
Part (a): Equation of a Line
What do we need?
The equation of a straight line is usually written as \( y = mx + c \), where:
- \( m \) is the gradient (slope)
- \( c \) is the y-intercept (where it crosses the vertical axis)
Step 1: Find the y-intercept (c)
Look at the graph. Where does line L cross the y-axis?
It crosses at \( y = 3 \).
So, \( c = 3 \).
Step 2: Find the gradient (m)
Method: Pick two clear points on the line and calculate \( \frac{\text{Change in } y}{\text{Change in } x} \) (Rise over Run).
Point 1: \( (0, 3) \)
Point 2: \( (-2, 0) \)
Rise (Change in y): \( 3 – 0 = 3 \)
Run (Change in x): \( 0 – (-2) = 2 \)
Gradient \( m = \frac{3}{2} \) (or 1.5)
Step 3: Write the Equation
Substitute \( m = \frac{3}{2} \) and \( c = 3 \) into \( y = mx + c \).
\( y = \frac{3}{2}x + 3 \)
(Also acceptable: \( y = 1.5x + 3 \))
✓ (3 marks)
Part (b): Parallel Lines
Rule: Parallel lines have the same gradient.
Line M is \( y = 5x \). This is in the form \( y = mx + c \) where \( m = 5 \) and \( c = 0 \).
The gradient of M is 5.
Any parallel line must also have a gradient of 5.
The equation will look like \( y = 5x + c \).
We can choose any value for \( c \) (except 0, though usually any different line counts).
Example: \( y = 5x + 1 \)
Final Answer (b): \( y = 5x + 1 \) (or any form \( y = 5x + c \))
✓ (1 mark)
Question 7 (5 marks)
Kasim has some small jars, some medium jars and some large jars.
He has a total of 400 jars.
\( \frac{3}{8} \) of the 400 jars are empty.
For the empty jars,
number of small jars : number of medium jars = 3 : 4
number of medium jars : number of large jars = 1 : 2
Work out the percentage of Kasim’s jars that are empty small jars.
Worked Solution
Step 1: Calculate Total Empty Jars
Why? We are told \( \frac{3}{8} \) of the total jars are empty. We need the actual number.
\( \frac{3}{8} \text{ of } 400 \)
First, find \( \frac{1}{8} \):
\( 400 \div 8 = 50 \)
Then find \( \frac{3}{8} \):
\( 50 \times 3 = 150 \)
There are 150 empty jars.
Step 2: Combine the Ratios
Strategy: We have two separate ratios. We need to link them.
Small : Medium = 3 : 4
Medium : Large = 1 : 2
The “Medium” part is 4 in the first ratio and 1 in the second. We must make them the same number.
Multiply the second ratio by 4 so the Medium part becomes 4:
Medium : Large = \( 1 \times 4 : 2 \times 4 = 4 : 8 \)
Now combine them:
Small : Medium : Large = 3 : 4 : 8
Step 3: Find the Number of Small Empty Jars
Method: Share the total empty jars (150) in the ratio 3 : 4 : 8.
Total parts = \( 3 + 4 + 8 = 15 \)
Value of one part = \( 150 \div 15 = 10 \)
Number of Small Jars = \( 3 \text{ parts} \times 10 = 30 \)
Step 4: Calculate the Percentage
Question asks for: The percentage of Kasim’s jars (Total 400) that are empty small jars (30).
Percentage = \( \frac{\text{Small Empty}}{\text{Total Jars}} \times 100 \)
\( \frac{30}{400} \times 100 \)
Simplify fraction:
\( \frac{30}{400} = \frac{3}{40} \)
Convert to percentage:
\( \frac{3}{40} \times 100 = \frac{300}{40} = \frac{30}{4} = 7.5 \% \)
Final Answer: 7.5 %
✓ (Total: 5 marks)
Question 8 (3 marks)
Len has 8 parcels.
The mean weight of the 8 parcels is 2.5 kg.
The mean weight of 3 of the parcels is 2 kg.
Work out the mean weight of the other 5 parcels.
Worked Solution
Step 1: Calculate Total Weights
Formula: \( \text{Total} = \text{Mean} \times \text{Count} \)
We need to find the total weight of all parcels first.
Total weight of 8 parcels:
\( 8 \times 2.5 \)
\( 8 \times 2 = 16 \)
\( 8 \times 0.5 = 4 \)
\( 16 + 4 = 20 \text{ kg} \)
Step 2: Calculate Weight of Known Parcels
Total weight of the 3 parcels:
\( 3 \times 2 = 6 \text{ kg} \)
Step 3: Find Remaining Weight and Mean
Logic: If we take the weight of the 3 parcels away from the total, we get the weight of the other 5.
Weight of other 5 parcels:
\( 20 – 6 = 14 \text{ kg} \)
Mean of other 5 parcels:
\( \frac{\text{Total Weight}}{\text{Count}} = \frac{14}{5} \)
To divide by 5, divide by 10 and multiply by 2:
\( 14 \div 10 = 1.4 \)
\( 1.4 \times 2 = 2.8 \)
Final Answer: 2.8 kg
✓ (3 marks)
Question 9 (1 mark)
In a sale, the normal price of a coat is reduced by \( R \% \)
Given that
sale price = 0.7 × normal price
find the value of \( R \).
Worked Solution
Understanding the Multiplier
Concept: A multiplier of 0.7 means the sale price is 70% of the original price.
If you pay 70%, what percentage was taken off?
\( 100\% – 70\% = 30\% \)
So the reduction \( R \) is 30.
Final Answer: \( R = 30 \)
✓ (1 mark)
Question 10 (4 marks)
Solve the simultaneous equations
\[ \begin{aligned} 5x – 2y &= 23 \\ 2x – 3y &= 18 \end{aligned} \]
Worked Solution
Step 1: Match Coefficients
Strategy: We want to eliminate either \( x \) or \( y \). Let’s eliminate \( y \) by making the number in front of \( y \) the same (ignoring the sign for now).
Multiples of 2 and 3: The LCM is 6.
Label the equations:
(1) \( 5x – 2y = 23 \)
(2) \( 2x – 3y = 18 \)
Multiply (1) by 3:
\( 15x – 6y = 69 \) (3)
Multiply (2) by 2:
\( 4x – 6y = 36 \) (4)
Step 2: Eliminate
Same Sign Subtract (SSS): Both \( y \) terms are \( -6y \). To cancel them out, we subtract the equations.
Do (3) – (4):
15x - 6y = 69
- (4x - 6y = 36)
----------------
11x = 33
\( x = 33 \div 11 \)
\( x = 3 \)
Step 3: Substitute to find y
Put \( x = 3 \) back into one of the original equations. Equation (2) looks slightly easier numbers.
Using \( 2x – 3y = 18 \):
\( 2(3) – 3y = 18 \)
\( 6 – 3y = 18 \)
Subtract 6 from both sides:
\( -3y = 12 \)
Divide by -3:
\( y = -4 \)
Step 4: Check
Check in Equation (1): \( 5x – 2y = 23 \)
\( 5(3) – 2(-4) \)
\( 15 – (-8) \)
\( 15 + 8 = 23 \)
It works!
Final Answer:
\( x = 3 \)
\( y = -4 \)
✓ (Total: 4 marks)
Question 11 (3 marks)
Triangle A is translated by the vector \( \begin{pmatrix} 6 \\ -4 \end{pmatrix} \) to give triangle B.
Triangle B is rotated 90° clockwise about the point (1, 2) to give triangle C.
Describe fully the single transformation that maps triangle A onto triangle C.
Worked Solution
Step 1: Find Triangle B (Translation)
Transformation: Translation by \( \begin{pmatrix} 6 \\ -4 \end{pmatrix} \) means move 6 units right and 4 units down.
Original Points of A: \( (-3, 2), (-2, 2), (-2, 4) \)
Add 6 to x, subtract 4 from y:
\( (-3, 2) \to (3, -2) \)
\( (-2, 2) \to (4, -2) \)
\( (-2, 4) \to (4, 0) \)
Triangle B is at vertices \( (3, -2), (4, -2), (4, 0) \).
Step 2: Find Triangle C (Rotation)
Transformation: Rotate B 90° clockwise about \( (1, 2) \).
Let’s track one point, e.g., \( (4, 0) \).
From center \( (1, 2) \) to point \( (4, 0) \):
Vector is \( \begin{pmatrix} 3 \\ -2 \end{pmatrix} \) (Right 3, Down 2)
Rotate 90° clockwise: \( (x, y) \to (y, -x) \) is a common rule for rotation about origin, but here we visualize “Right 3, Down 2” becoming “Down 3, Left 2”.
Vector becomes \( \begin{pmatrix} -2 \\ -3 \end{pmatrix} \) (Left 2, Down 3).
Add to center \( (1, 2) \):
\( (1 – 2, 2 – 3) = (-1, -1) \)
Let’s do another point \( (4, -2) \):
Vector from \( (1, 2) \) is \( \begin{pmatrix} 3 \\ -4 \end{pmatrix} \). Rotated 90° CW becomes \( \begin{pmatrix} -4 \\ -3 \end{pmatrix} \).
New point: \( (1-4, 2-3) = (-3, -1) \)
Third point \( (3, -2) \):
Vector \( \begin{pmatrix} 2 \\ -4 \end{pmatrix} \). Rotated 90° CW becomes \( \begin{pmatrix} -4 \\ -2 \end{pmatrix} \).
New point: \( (1-4, 2-2) = (-3, 0) \)
Triangle C is at vertices \( (-1, -1), (-3, -1), (-3, 0) \).
Step 3: Map A to C
Compare Triangle A and Triangle C.
Triangle A: \( (-3, 2), (-2, 2), (-2, 4) \)
Triangle C: \( (-3, 0), (-3, -1), (-1, -1) \)
Look at the orientation. Triangle A points up. Triangle C points right? No, look carefully.
Let’s plot them mentally or sketch.
Triangle C has been rotated relative to A.
Center of rotation can be found by intersecting perpendicular bisectors of connecting lines (e.g. A’s top vertex to C’s corresponding vertex).
Vertices map: \( (-2, 4) \to (-1, -1) \), \( (-2, 2) \to (-3, -1) \), \( (-3, 2) \to (-3, 0) \).
Let’s try finding the center.
It’s a rotation of 90° clockwise.
Center is usually integer coordinates.
Let’s guess \( (-4, 1) \)?
Test \( (-2, 4) \) about \( (-4, 1) \): Vector \( (2, 3) \). Rotated 90° CW \( (3, -2) \). Add to \( (-4, 1) \to (-1, -1) \). Correct!
Final Answer:
Rotation 90° clockwise about (-4, 1)
✓ (3 marks)
Question 12 (3 marks)
Here are some graphs.
Write down the letter of the graph that could have the equation
(i) \( y = x^2 – 4 \)
(1 mark)
(ii) \( y = -x^3 \)
(1 mark)
(iii) \( y = -\frac{5}{x} \)
(1 mark)
Worked Solution
Part (i): \( y = x^2 – 4 \)
Recognition:
\( y = x^2 \) is a U-shaped parabola.
The \( -4 \) moves it down by 4 units. So the minimum point (turning point) is below the x-axis.
Look for a U-shape that crosses the x-axis twice and has a minimum below the axis.
Answer: Graph H
✓ (1 mark)
Part (ii): \( y = -x^3 \)
Recognition:
\( y = x^3 \) goes from bottom-left to top-right (increasing).
\( y = -x^3 \) is a reflection, so it goes from top-left to bottom-right (decreasing).
It passes through the origin (0,0).
Answer: Graph F
✓ (1 mark)
Part (iii): \( y = -\frac{5}{x} \)
Recognition:
\( y = \frac{k}{x} \) is a reciprocal graph (hyperbola).
If positive (\( \frac{1}{x} \)), it is in the top-right and bottom-left quadrants.
If negative (\( -\frac{5}{x} \)), it flips to the top-left and bottom-right quadrants.
Answer: Graph J
✓ (1 mark)
Question 13 (5 marks)
The table gives information about the amount of time that each of 150 people were in a shop.
| Time (\( t \) minutes) | Frequency |
|---|---|
| \( 0 < t \leq 10 \) | 20 |
| \( 10 < t \leq 30 \) | 70 |
| \( 30 < t \leq 35 \) | 22 |
| \( 35 < t \leq 50 \) | 30 |
| \( 50 < t \leq 60 \) | 8 |
(a) On the grid, draw a histogram for this information.
(3 marks)
(b) Work out an estimate for the fraction of these 150 people who were in the shop for between 20 minutes and 40 minutes.
(2 marks)
Worked Solution
Part (a): Frequency Density Calculation
Formula: \( \text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} \)
| Interval | Width | Freq | FD Calculation | FD |
|---|---|---|---|---|
| 0-10 | 10 | 20 | 20 ÷ 10 | 2 |
| 10-30 | 20 | 70 | 70 ÷ 20 | 3.5 |
| 30-35 | 5 | 22 | 22 ÷ 5 | 4.4 |
| 35-50 | 15 | 30 | 30 ÷ 15 | 2 |
| 50-60 | 10 | 8 | 8 ÷ 10 | 0.8 |
✓ (3 marks) for fully correct histogram
Part (b): Estimate Fraction (20 to 40 mins)
Strategy: We need to find the number of people in the range 20-40.
This range cuts across two bars:
- The 10-30 bar (we need the part from 20 to 30)
- The 30-35 bar (we need all of it)
- The 35-50 bar (we need the part from 35 to 40)
1. Part of the 10-30 interval (from 20-30):
Width = \( 30 – 20 = 10 \)
FD = 3.5
Frequency = \( 10 \times 3.5 = 35 \)
2. Full 30-35 interval:
Frequency = 22
3. Part of the 35-50 interval (from 35-40):
Width = \( 40 – 35 = 5 \)
FD = 2
Frequency = \( 5 \times 2 = 10 \)
Total people:
\( 35 + 22 + 10 = 67 \)
Fraction:
\( \frac{67}{150} \)
Final Answer (b): \( \frac{67}{150} \)
✓ (Total: 5 marks)
Question 14 (3 marks)
Expand and simplify \( (3x – 1)(2x + 3)(x – 5) \)
Worked Solution
Step 1: Expand the first two brackets
Method: FOIL (First, Outer, Inner, Last) for \( (3x – 1)(2x + 3) \).
\( (3x)(2x) = 6x^2 \)
\( (3x)(3) = 9x \)
\( (-1)(2x) = -2x \)
\( (-1)(3) = -3 \)
Combine: \( 6x^2 + 9x – 2x – 3 = 6x^2 + 7x – 3 \)
Step 2: Multiply by the third bracket
Now multiply \( (6x^2 + 7x – 3) \) by \( (x – 5) \).
Multiply every term in the first bracket by \( x \), then by \( -5 \).
Multiply by \( x \):
\( x(6x^2 + 7x – 3) = 6x^3 + 7x^2 – 3x \)
Multiply by \( -5 \):
\( -5(6x^2 + 7x – 3) = -30x^2 – 35x + 15 \)
Add them together:
\( 6x^3 + 7x^2 – 30x^2 – 3x – 35x + 15 \)
Step 3: Simplify
\( x^2 \) terms: \( 7x^2 – 30x^2 = -23x^2 \)
\( x \) terms: \( -3x – 35x = -38x \)
Result: \( 6x^3 – 23x^2 – 38x + 15 \)
Final Answer: \( 6x^3 – 23x^2 – 38x + 15 \)
✓ (3 marks)
Question 15 (4 marks)
OAB is a sector of a circle with centre O and radius 6 cm.
The length of the arc AB is \( 5\pi \) cm.
Work out, in terms of \( \pi \), the area of the sector.
Give your answer in its simplest form.
Worked Solution
Step 1: Find the Fraction of the Circle
Formula: Circumference \( C = 2\pi r \)
Radius \( r = 6 \)
Total Circumference = \( 2 \times \pi \times 6 = 12\pi \) cm
We are told the Arc Length is \( 5\pi \) cm.
Fraction of circle = \( \frac{\text{Arc Length}}{\text{Circumference}} = \frac{5\pi}{12\pi} = \frac{5}{12} \)
Step 2: Calculate Sector Area
Formula: Area of Circle \( A = \pi r^2 \)
Sector Area = Fraction × Total Area
Total Area = \( \pi \times 6^2 = 36\pi \) cm²
Sector Area = \( \frac{5}{12} \times 36\pi \)
Calculation: \( 36 \div 12 = 3 \)
\( 5 \times 3\pi = 15\pi \)
Final Answer: \( 15\pi \) cm²
✓ (Total: 4 marks)
Question 16 (2 marks)
There are only \( n \) orange sweets and 1 white sweet in a bag.
Saira takes at random a sweet from the bag and eats the sweet.
She then takes at random another sweet from the bag and eats this sweet.
Show that the probability that Saira eats two orange sweets is \( \frac{n}{n+1} \times \frac{n-1}{n} \)
(Wait, the question asks to show the result is \( \frac{n-1}{n+1} \)? Or derivation?)
(Looking at mark scheme/question flow: Show that probability is… let’s check the PDF text… “Show that the probability that Saira eats two orange sweets is \( \frac{n-1}{n+1} \)” seems incorrect. Let’s re-read carefully.)
Text says: “Show that the probability that Saira eats two orange sweets is \( \frac{n}{n+1} \times \frac{n-1}{n} \)” No, looking at image 17… “Show that the probability that Saira eats two orange sweets is \( \frac{n-1}{n+1} \)”… Wait, the OCR says “Show that the probability that Saira eats two orange sweets is \( \frac{n}{n+1} \)”. No, let’s derive it.
Let’s look at the equation in the image 17: \( \frac{n-1}{n+1} \)
Worked Solution
Step 1: Define Totals
Initial State:
Orange sweets = \( n \)
White sweets = 1
Total sweets = \( n + 1 \)
Step 2: Probability of First Orange
P(First is Orange) = \( \frac{\text{Number of Orange}}{\text{Total Sweets}} = \frac{n}{n+1} \)
Step 3: Probability of Second Orange
She eats the sweet, so it is not replaced.
New number of Orange = \( n – 1 \)
New Total = \( (n + 1) – 1 = n \)
P(Second is Orange) = \( \frac{n-1}{n} \)
Step 4: Combined Probability
P(Orange AND Orange) = P(1st Orange) \( \times \) P(2nd Orange)
\[ = \frac{n}{n+1} \times \frac{n-1}{n} \]
We can cancel the \( n \) from the numerator and denominator:
\[ = \frac{\cancel{n}}{n+1} \times \frac{n-1}{\cancel{n}} \]
\[ = \frac{n-1}{n+1} \]
This matches the required expression.
✓ (2 marks)
Question 17 (3 marks)
(a) Rationalise the denominator of \( \frac{1}{\sqrt{7}} \)
(1 mark)
(b) Simplify fully \( \sqrt{80} – \sqrt{5} \)
(2 marks)
Worked Solution
Part (a): Rationalising
Method: Multiply top and bottom by the square root in the denominator.
\[ \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7} \]
✓ (1 mark)
Part (b): Simplifying Surds
Method: Find a square number factor in 80.
Square numbers: 4, 9, 16, 25, …
16 goes into 80 (\( 16 \times 5 = 80 \)).
\( \sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5} \)
Now calculate:
\( 4\sqrt{5} – \sqrt{5} \)
(Think of it like \( 4x – x = 3x \))
\( = 3\sqrt{5} \)
Final Answer: \( 3\sqrt{5} \)
✓ (2 marks)
Question 18 (3 marks)
Show that \( 0.1\dot{5} + 0.2\dot{2}\dot{7} \) can be written in the form \( \frac{m}{66} \) where \( m \) is an integer.
Worked Solution
Step 1: Convert \( 0.1\dot{5} \) to a fraction
Let \( x = 0.15555… \)
\( 10x = 1.5555… \)
\( 100x = 15.5555… \)
Subtract:
\( 100x – 10x = 15.55… – 1.55… \)
\( 90x = 14 \)
\( x = \frac{14}{90} = \frac{7}{45} \)
Step 2: Convert \( 0.2\dot{2}\dot{7} \) to a fraction
Let \( y = 0.2272727… \)
\( 10y = 2.2727… \)
\( 1000y = 227.2727… \)
Subtract:
\( 1000y – 10y = 227.27… – 2.27… \)
\( 990y = 225 \)
\( y = \frac{225}{990} \)
Simplify (divide by 5): \( \frac{45}{198} \)
Divide by 9: \( \frac{5}{22} \)
Step 3: Add the fractions
We need to add \( \frac{7}{45} + \frac{5}{22} \)
Let’s find a common denominator. Wait, the question asks for form \( \frac{m}{66} \).
Let’s check our fractions again.
\( \frac{7}{45} \) and \( \frac{5}{22} \). The denominator 66 suggests something involving 11, 2, 3.
Let’s re-read the first decimal conversion. \( 0.1\dot{5} \).
Is there a simpler way? Maybe convert both to something over 990 or similar.
Wait, \( \frac{7}{45} = \frac{?}{?}\). 45 doesn’t go into 66.
Let’s re-check the question numbers. \( 0.1\dot{5} \) and \( 0.2\dot{2}\dot{7} \).
\( 0.2\dot{2}\dot{7} = 0.22727… \)
Let \( y = 0.22727… \)
\( 100y = 22.72727… \)
\( y = 0.22727… \)
\( 99y = 22.5 \) -> \( y = \frac{22.5}{99} = \frac{225}{990} = \frac{5}{22} \). Correct.
Now \( 0.1\dot{5} = 0.151515… \)?
Looking at the dot notation: \( 0.1\dot{5} \) usually means \( 0.15555… \). If it meant \( 0.\dot{1}\dot{5} \) it would be \( 0.1515… \).
Let’s assume \( 0.1\dot{5} \) means \( 0.1555… \).
\( \frac{7}{45} + \frac{5}{22} \)
Common denominator of 45 and 22 is \( 45 \times 22 = 990 \).
\( \frac{7 \times 22}{990} + \frac{5 \times 45}{990} = \frac{154 + 225}{990} = \frac{379}{990} \).
Does \( 379/990 \) simplify to \( m/66 \)? \( 990 \div 15 = 66 \). Is 379 divisible by 15? No.
Alternative Interpretation: Maybe the question meant \( 0.\dot{1}\dot{5} \)?
\( 0.\dot{1}\dot{5} = \frac{15}{99} = \frac{5}{33} \).
\( \frac{5}{33} + \frac{5}{22} \).
Common denom 66.
\( \frac{5 \times 2}{66} + \frac{5 \times 3}{66} = \frac{10}{66} + \frac{15}{66} = \frac{25}{66} \).
This works perfectly! The dot must be over the 1 and 5 (or ambiguous in text).
Let’s look at the image again. Image 18 shows \( 0.1\dot{5} + 0.2\dot{2}\dot{7} \).
Actually, looking closely at Image 18, the dot is clearly over the 5 in 0.15. Wait, looking really closely at the crop… it looks like \( 0.\dot{1}\dot{5} \). There is a dot over 1 and 5. Ah, OCR said \( 0.1\dot{5} \). But the image has two dots.
Correction: The first number is \( 0.\dot{1}\dot{5} \).
Step 1: Convert \( 0.\dot{1}\dot{5} \)
Let \( x = 0.151515… \)
\( 100x = 15.151515… \)
\( 100x – x = 15 \)
\( 99x = 15 \)
\( x = \frac{15}{99} \)
Divide by 3: \( x = \frac{5}{33} \)
Step 2: Convert \( 0.2\dot{2}\dot{7} \)
Wait, looking at image 18 again. \( 0.2\dot{2}\dot{7} \)?
The dots are over the 2 and 7. So \( 0.22727… \). No, if dots are over 2 and 7 it means the 27 repeats? Or is the 0.2 separate? Usually \( 0.\dot{2}\dot{7} \) means 0.272727.
Let’s look at the gap. It is \( 0.2\dot{2}\dot{7} \). This usually means the repeating part starts at the first dot and ends at the second.
So \( 0.2272727… \)? No, that would be \( 0.\dot{2}\dot{7} \).
If it is \( 0.2\dot{2}\dot{7} \), does it mean \( 0.22727… \)?
Let’s re-read standard notation. \( 0.1\dot{2}\dot{3} \) means \( 0.1232323… \).
If the number is \( 0.2\dot{2}\dot{7} \), the repeating part is 27? But there is a 2 before it?
Maybe it is \( 0.\dot{2}\dot{2}\dot{7} \)? No.
Let’s try the value we found earlier: \( \frac{5}{22} \). That came from \( 0.22727… \). This matches \( 0.2\dot{2}\dot{7} \) if the repeating part is ’27’ starting after the 0.2? No, \( 0.2\dot{2}\dot{7} \) usually means digits between dots repeat. So 27 repeats.
Is it \( 0.\dot{2}\dot{2}\dot{7} \) (227 repeats)? \( 227/999 \).
Let’s trust the “Simple Answer” heuristic: The result is \( \frac{25}{66} \).
\( \frac{25}{66} = \frac{5}{33} + \frac{5}{22} \).
\( \frac{5}{33} \) corresponds to \( 0.\dot{1}\dot{5} \).
\( \frac{5}{22} \) corresponds to \( 0.2272727… \). This is written as \( 0.2\dot{2}\dot{7} \) in some notations, or \( 0.2\dot{2}7\dot{?} \). Actually \( 0.2\dot{2}\dot{7} \) is standard for \( 0.22727… \).
So we proceed with \( 0.22727… \).
Let \( y = 0.22727… \)
\( 10y = 2.2727… \)
\( 1000y = 227.2727… \)
\( 990y = 225 \)
\( y = \frac{225}{990} = \frac{5}{22} \)
Step 3: Combine
\[ \frac{5}{33} + \frac{5}{22} \]
Common denominator 66:
\[ \frac{5 \times 2}{66} + \frac{5 \times 3}{66} \]
\[ \frac{10}{66} + \frac{15}{66} = \frac{25}{66} \]
This is in the form \( \frac{m}{66} \) with \( m = 25 \).
✓ (Total: 3 marks)
Question 19 (3 marks)
ABC and DAB are similar isosceles triangles.
\( AB = AC \)
\( AD = BD \)
\( BC : CD = 4 : 21 \)
Find the ratio \( AB : AD \)
Worked Solution
Step 1: Set up lengths from the ratio
We are given \( BC : CD = 4 : 21 \).
Let \( BC = 4x \) and \( CD = 21x \).
Total length \( BD = BC + CD = 4x + 21x = 25x \).
Step 2: Use Similarity
Triangles \( ABC \) and \( DAB \) are similar.
Match the sides:
- Base of \( ABC \) is \( BC \) (length \( 4x \)).
- Base of \( DAB \)? Wait, let’s look at the equal sides.
- In \( ABC \), equal sides are \( AB \) and \( AC \). Base is \( BC \).
- In \( DAB \), equal sides are \( AD \) and \( BD \). Wait, \( AD=BD \)? The text says \( AD=BD \). So \( D \) is the vertex? No, \( D \) is on the line. The triangle is \( DAB \). Vertices D, A, B.
- If \( AD = BD \), then the equal sides are \( AD \) and \( BD \). The base is \( AB \).
So we compare:
Triangle 1 (ABC): Base \( BC \), Side \( AB \).
Triangle 2 (DAB): Base \( AB \), Side \( BD \).
Ratio of sides must be equal:
\[ \frac{\text{Side of DAB}}{\text{Side of ABC}} = \frac{\text{Base of DAB}}{\text{Base of ABC}} \]
\[ \frac{BD}{AB} = \frac{AB}{BC} \]
We know \( BD = 25x \) and \( BC = 4x \).
Let \( AB = y \).
\[ \frac{25x}{y} = \frac{y}{4x} \]
Step 3: Solve for AB
Cross multiply:
\( y^2 = 25x \times 4x \)
\( y^2 = 100x^2 \)
\( y = \sqrt{100x^2} = 10x \)
So \( AB = 10x \).
Step 4: Find Ratio AB : AD
We need \( AB : AD \).
We found \( AB = 10x \).
We know \( AD = BD \) (given). And \( BD = 25x \). So \( AD = 25x \).
Ratio \( AB : AD = 10x : 25x \)
Divide by \( 5x \):
\( 2 : 5 \)
Final Answer: 2 : 5
✓ (Total: 3 marks)
Question 20 (3 marks)
\( 2^x = \frac{2^n}{\sqrt[3]{2}} \)
\( 2^y = (\sqrt{2})^5 \)
Given that \( x + y = 8 \)
work out the value of \( n \).
Worked Solution
Step 1: Simplify \( x \)
Express everything as powers of 2.
\( \sqrt[3]{2} = 2^{1/3} \)
\[ 2^x = \frac{2^n}{2^{1/3}} \]
Using division law (\( \frac{a^m}{a^n} = a^{m-n} \)):
\[ 2^x = 2^{n – 1/3} \]
So, \( x = n – \frac{1}{3} \)
Step 2: Simplify \( y \)
\( \sqrt{2} = 2^{1/2} \)
\[ 2^y = (2^{1/2})^5 \]
Power law (\( (a^m)^n = a^{mn} \)):
\[ 2^y = 2^{5/2} \]
So, \( y = \frac{5}{2} \) (or 2.5)
Step 3: Solve for \( n \)
We are given \( x + y = 8 \).
Substitute our expressions:
\[ (n – \frac{1}{3}) + \frac{5}{2} = 8 \]
Let’s clear fractions (multiply by 6):
\[ 6n – 2 + 15 = 48 \]
\[ 6n + 13 = 48 \]
\[ 6n = 35 \]
\[ n = \frac{35}{6} \]
(Or \( 5\frac{5}{6} \))
Final Answer: \( n = \frac{35}{6} \)
✓ (Total: 3 marks)
Question 21 (5 marks)
A solid cuboid has a volume of 300 cm³.
The cuboid has a total surface area of 370 cm².
The length of the cuboid is 20 cm.
The width of the cuboid is greater than the height of the cuboid.
Work out the height of the cuboid.
You must show all your working.
Worked Solution
Step 1: Set up the equations
Let length = \( l = 20 \).
Let width = \( w \).
Let height = \( h \).
We know Volume \( V = l \times w \times h \).
We know Surface Area \( SA = 2lw + 2lh + 2wh \).
Volume Equation:
\( 20 \times w \times h = 300 \)
\( 20wh = 300 \)
Divide by 20:
\( wh = 15 \)
So \( w = \frac{15}{h} \)
Surface Area Equation:
\( 2(20w) + 2(20h) + 2(wh) = 370 \)
\( 40w + 40h + 2wh = 370 \)
Step 2: Substitute and Solve
We know \( wh = 15 \). Let’s plug that into the surface area equation.
\( 40w + 40h + 2(15) = 370 \)
\( 40w + 40h + 30 = 370 \)
\( 40w + 40h = 340 \)
Divide by 40 (or first by 10 then 4):
\( 4w + 4h = 34 \)
\( 2w + 2h = 17 \)
\( w + h = 8.5 \)
Step 3: Solve the System
We have two simple equations:
1) \( w \times h = 15 \)
2) \( w + h = 8.5 \)
We are looking for two numbers that multiply to 15 and add to 8.5.
Let’s form a quadratic or guess.
\( w = 8.5 – h \)
\( (8.5 – h)h = 15 \)
\( 8.5h – h^2 = 15 \)
\( h^2 – 8.5h + 15 = 0 \)
Multiply by 2 to remove decimal:
\( 2h^2 – 17h + 30 = 0 \)
Step 4: Factorise
We need factors of \( 2 \times 30 = 60 \) that add to -17.
Factors of 60: 1,60; 2,30; 3,20; 4,15; 5,12…
\( -5 \) and \( -12 \) add to -17.
\( 2h^2 – 12h – 5h + 30 = 0 \)
\( 2h(h – 6) – 5(h – 6) = 0 \)
\( (2h – 5)(h – 6) = 0 \)
So \( h = 6 \) or \( h = 2.5 \)
Step 5: Choose the correct value
If \( h = 6 \), then \( w = 15 \div 6 = 2.5 \). (Here width < height)
If \( h = 2.5 \), then \( w = 15 \div 2.5 = 6 \). (Here width > height)
The question says: “The width of the cuboid is greater than the height”.
So we must choose the pair where \( w > h \).
\( w = 6, h = 2.5 \)
Final Answer: Height = 2.5 cm
✓ (Total: 5 marks)
Question 22 (4 marks)
(a) Sketch the graph of \( y = \sin x^\circ \) for \( 0 \leq x \leq 360 \)
(2 marks)
(b) Solve the equation \( 2 \sin x^\circ = -1 \) for \( 0 \leq x \leq 360 \)
(2 marks)
Worked Solution
Part (a): Sketching Sin(x)
Key Features:
- Starts at (0, 0).
- Peak at (90, 1).
- Crosses x-axis at 180.
- Trough (minimum) at (270, -1).
- Ends at (360, 0).
- Correct “wave” shape.
✓ (2 marks)
Part (b): Solve Equation
\( 2 \sin x = -1 \)
\( \sin x = -\frac{1}{2} \)
Step 2: Find Base Angle
Ignore the negative sign first. When is \( \sin x = \frac{1}{2} \)?
From exact values, \( x = 30^\circ \).
Step 3: Apply CAST Rule
Since \( \sin x \) is negative (-0.5), the solutions are in the 3rd and 4th quadrants (T and C).
Quadrant 3: \( 180 + \text{base} \)
Quadrant 4: \( 360 – \text{base} \)
\( x_1 = 180 + 30 = 210^\circ \)
\( x_2 = 360 – 30 = 330^\circ \)
Final Answer: \( 210^\circ, 330^\circ \)
✓ (2 marks)
Question 23 (5 marks)
C is a circle with centre (0, 0)
L is a straight line.
The circle C and the line L intersect at the points \( P \) and \( Q \).
The coordinates of \( P \) are (5, 10)
The x coordinate of \( Q \) is -2
L has a positive gradient and crosses the y-axis at the point (0, k)
Find the value of \( k \).
Worked Solution
Step 1: Find Equation of Circle
Formula: \( x^2 + y^2 = r^2 \)
Point \( P(5, 10) \) is on the circle.
\( r^2 = 5^2 + 10^2 \)
\( r^2 = 25 + 100 = 125 \)
Equation of circle: \( x^2 + y^2 = 125 \)
Step 2: Find Coordinate of Q
We know \( Q \) is on the circle and its x-coordinate is -2.
Substitute \( x = -2 \) into \( x^2 + y^2 = 125 \):
\( (-2)^2 + y^2 = 125 \)
\( 4 + y^2 = 125 \)
\( y^2 = 121 \)
\( y = \sqrt{121} = \pm 11 \)
So \( Q \) is either \( (-2, 11) \) or \( (-2, -11) \).
How do we decide? The line L connects \( P(5, 10) \) and \( Q \). It has a positive gradient.
Let’s check the gradient for both options.
Option 1: \( Q(-2, 11) \). Gradient \( = \frac{11 – 10}{-2 – 5} = \frac{1}{-7} \) (Negative). Reject.
Option 2: \( Q(-2, -11) \). Gradient \( = \frac{-11 – 10}{-2 – 5} = \frac{-21}{-7} = 3 \) (Positive). Accept.
So \( Q \) is \( (-2, -11) \) and gradient \( m = 3 \).
Step 3: Find Equation of Line to get k
We have gradient \( m = 3 \) and point \( P(5, 10) \).
Equation \( y = mx + c \)
\( 10 = 3(5) + c \)
\( 10 = 15 + c \)
\( c = -5 \)
Equation is \( y = 3x – 5 \).
The line crosses y-axis at \( (0, k) \), which is the y-intercept \( c \).
So \( k = -5 \).
Final Answer: \( k = -5 \)
✓ (Total: 5 marks)