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GCSE Summer 2022 Pearson Edexcel Higher Paper 1 (Non-Calculator)
Mark Scheme Legend
- M1 – Method mark
- A1 – Accuracy mark
- B1 – Independent accuracy mark
- C1 – Communication mark
- oe – Or Equivalent
- ft – Follow Through
Table of Contents
- Question 1 (Algebra: Inequalities)
- Question 2 (Number: Prime Factors)
- Question 3 (Number: Ratio)
- Question 4 (Number: Standard Form)
- Question 5 (Geometry: Polygons)
- Question 6 (Algebra: Graphs)
- Question 7 (Geometry: Density)
- Question 8 (Statistics: Mean)
- Question 9 (Geometry: Surface Area)
- Question 10 (Statistics: Cumulative Frequency)
- Question 11 (Probability: Ratio)
- Question 12 (Number: Recurring Decimals)
- Question 13 (Geometry: Proof)
- Question 14 (Graphs: Gradients)
- Question 15 (Vectors: Proof)
- Question 16 (Probability: Tree Diagrams)
- Question 17 (Algebra: Direct/Inverse Proportion)
- Question 18 (Algebra: Indices)
- Question 19 (Algebra: Algebraic Fractions)
- Question 20 (Geometry: Circle Equations)
- Question 21 (Geometry: Areas)
Question 1 (2 marks)
Solve \( 7x – 27 < 8 \)
Worked Solution
Step 1: Isolate the term with \( x \)
Why we do this: To solve the inequality, we want \( x \) on its own. First, we need to remove the constant term by doing the inverse operation. The inverse of subtracting 27 is adding 27.
Add 27 to both sides:
\[ 7x – 27 + 27 < 8 + 27 \] \[ 7x < 35 \]✓ (M1) Adding 27 to both sides
Step 2: Solve for \( x \)
What this tells us: Now we have \( 7x \). To get \( x \), we divide by 7.
Final Answer:
\[ x < 5 \]✓ (A1)
Question 2 (2 marks)
Write 124 as a product of its prime factors.
Worked Solution
Step 1: Break down the number using a factor tree
Method: We divide 124 by small prime numbers (2, 3, 5, etc.) until we are left with only prime numbers.
124 is even, so divide by 2:
\[ 124 \div 2 = 62 \]62 is even, so divide by 2:
\[ 62 \div 2 = 31 \]31 is a prime number (it has no factors other than 1 and itself).
So the prime factors are 2, 2, and 31.
✓ (M1) Complete method to find prime factors
Step 2: Write as a product
Or in index form:
\[ 2^2 \times 31 \]Final Answer:
\[ 2 \times 2 \times 31 \quad \text{or} \quad 2^2 \times 31 \]✓ (A1)
Question 3 (5 marks)
A delivery company has a total of 160 cars and vans.
The number of cars : the number of vans = \( 3 : 7 \)
Each car and each van uses electricity or diesel or petrol.
- \( \frac{1}{8} \) of the cars use electricity.
- 25% of the cars use diesel.
- The rest of the cars use petrol.
Work out the number of cars that use petrol.
You must show all your working.
Worked Solution
Step 1: Find the total number of cars
Why we do this: We know the total vehicles (160) and the ratio of cars to vans (\( 3:7 \)). We can find the number of cars by sharing the total in the given ratio.
Total parts in ratio \( = 3 + 7 = 10 \)
Value of one part:
\[ 160 \div 10 = 16 \]Number of cars (3 parts):
\[ 3 \times 16 = 48 \text{ cars} \]✓ (P1) Finding number of cars (48)
Step 2: Calculate cars using electricity and diesel
Method: Apply the given fractions and percentages to the number of cars (48).
Electricity: \( \frac{1}{8} \) of 48
\[ 48 \div 8 = 6 \text{ cars} \]Diesel: 25% of 48
25% is \( \frac{1}{4} \), so divide by 4:
\[ 48 \div 4 = 12 \text{ cars} \]✓ (P1) Correct step using 48 (finding 6 or 12)
Step 3: Calculate cars using petrol
What this tells us: The “rest” use petrol. We subtract the electric and diesel cars from the total cars.
Total non-petrol cars:
\[ 6 + 12 = 18 \]Petrol cars:
\[ 48 – 18 = 30 \]✓ (P1) Process to find petrol cars
Final Answer:
\[ 30 \]✓ (A1)
Question 4 (4 marks)
(a) Write \( 1.63 \times 10^{-3} \) as an ordinary number.
(b) Write 438 000 in standard form.
(c) Work out \( (4 \times 10^3) \times (6 \times 10^{-5}) \)
Give your answer in standard form.
Worked Solution
Part (a): Ordinary Number
Method: The power is \( -3 \), so we move the decimal point 3 places to the left.
Answer: 0.00163
✓ (B1)
Part (b): Standard Form
Method: Standard form is \( A \times 10^n \) where \( 1 \leq A < 10 \). We place the decimal after the first non-zero digit (4).
(Moved decimal 5 places to the right)
Answer: \( 4.38 \times 10^5 \)
✓ (B1)
Part (c): Calculation
Method: Multiply the numbers and add the powers of 10.
Multiply coefficients:
\[ 4 \times 6 = 24 \]Multiply powers of 10:
\[ 10^3 \times 10^{-5} = 10^{3 + (-5)} = 10^{-2} \]Combine:
\[ 24 \times 10^{-2} \]Convert to standard form (must be between 1 and 10):
\[ 24 = 2.4 \times 10^1 \] \[ 2.4 \times 10^1 \times 10^{-2} = 2.4 \times 10^{-1} \]✓ (M1) for \( 24 \times 10^{-2} \) or \( 0.24 \)
Final Answer (c):
\[ 2.4 \times 10^{-1} \]✓ (A1)
Question 5 (3 marks)
Here is a regular hexagon and a regular pentagon.
Work out the size of the angle marked \( x \).
You must show all your working.
Worked Solution
Step 1: Calculate interior angle of the Hexagon
Formula: Sum of interior angles \( = (n-2) \times 180 \). One angle \( = \frac{(n-2) \times 180}{n} \).
For a Hexagon (\( n=6 \)):
\[ \text{Sum} = (6-2) \times 180 = 4 \times 180 = 720^\circ \] \[ \text{Interior Angle} = 720 \div 6 = 120^\circ \]✓ (M1) Finding interior angle 120 (or exterior 60)
Step 2: Calculate interior angle of the Pentagon
For a Pentagon (\( n=5 \)):
\[ \text{Sum} = (5-2) \times 180 = 3 \times 180 = 540^\circ \] \[ \text{Interior Angle} = 540 \div 5 = 108^\circ \]✓ (M1) Finding interior angle 108 (or exterior 72)
Step 3: Calculate angle \( x \)
Reasoning: Angles around a point sum to \( 360^\circ \). The angle \( x \) meets the interior angles of the hexagon and pentagon at the vertex.
Final Answer:
\[ 132^\circ \]✓ (A1)
Question 6 (6 marks)
(a) Complete the table of values for \( y = x^2 – 3x + 1 \)
| \( x \) | -1 | 0 | 1 | 2 | 3 | 4 |
| \( y \) | 1 | -1 |
(b) On the grid, draw the graph of \( y = x^2 – 3x + 1 \) for values of \( x \) from -1 to 4
(c) Using your graph, find estimates for the solutions of the equation \( x^2 – 3x + 1 = 0 \)
Worked Solution
Part (a): Complete the Table
Substitute values of \( x \) into \( y = x^2 – 3x + 1 \):
- If \( x = -1 \): \( (-1)^2 – 3(-1) + 1 = 1 + 3 + 1 = 5 \)
- If \( x = 2 \): \( (2)^2 – 3(2) + 1 = 4 – 6 + 1 = -1 \)
- If \( x = 3 \): \( (3)^2 – 3(3) + 1 = 9 – 9 + 1 = 1 \)
- If \( x = 4 \): \( (4)^2 – 3(4) + 1 = 16 – 12 + 1 = 5 \)
Completed Table:
| \( x \) | -1 | 0 | 1 | 2 | 3 | 4 |
| \( y \) | 5 | 1 | -1 | -1 | 1 | 5 |
✓ (B2) All 4 values correct
Part (b): Draw the Graph
Method: Plot the points from the table: \( (-1, 5), (0, 1), (1, -1), (2, -1), (3, 1), (4, 5) \). Join them with a smooth curve.
✓ (B2) Fully correct graph
Part (c): Estimate Solutions
What this tells us: The solutions to \( x^2 – 3x + 1 = 0 \) are the \( x \)-values where the graph crosses the \( x \)-axis (where \( y=0 \)).
Looking at the graph, the curve crosses the \( x \)-axis at two points.
1. Between 0 and 1: Approx \( x = 0.4 \)
2. Between 2 and 3: Approx \( x = 2.6 \)
Final Answer:
Estimates: \( 0.4 \) and \( 2.6 \)
(Accept range \( 0.3-0.5 \) and \( 2.5-2.7 \))
✓ (M1) Marking intercepts or stating one solution
✓ (A1) Both values in range
Question 7 (3 marks)
Here are two cubes, A and B.
Cube A has a mass of 81 g.
Cube B has a mass of 128 g.
Work out
the density of cube A : the density of cube B
Give your answer in the form \( a : b \), where \( a \) and \( b \) are integers.
Worked Solution
Step 1: Calculate the Volume of each cube
Formula: Volume of a cube = \( \text{side} \times \text{side} \times \text{side} = s^3 \).
Cube A: Side = 3 cm
\[ V_A = 3^3 = 3 \times 3 \times 3 = 27 \, \text{cm}^3 \]Cube B: Side = 4 cm
\[ V_B = 4^3 = 4 \times 4 \times 4 = 64 \, \text{cm}^3 \]✓ (P1) Process to find either volume (27 or 64)
Step 2: Calculate the Density of each cube
Formula: Density = \( \frac{\text{Mass}}{\text{Volume}} \).
Density of A:
\[ D_A = \frac{81}{27} \]Using short division or recognizing factors:
\[ 27 \times 3 = 81 \implies D_A = 3 \, \text{g/cm}^3 \]Density of B:
\[ D_B = \frac{128}{64} \] \[ 64 \times 2 = 128 \implies D_B = 2 \, \text{g/cm}^3 \]✓ (P1) Finding densities (3 and 2)
Step 3: Write the Ratio
Density A : Density B
\[ 3 : 2 \]Final Answer:
\[ 3 : 2 \]✓ (A1)
Question 8 (3 marks)
The table shows the amount of snow, in cm, that fell each day for 30 days.
| Amount of snow (\( s \) cm) | Frequency |
|---|---|
| \( 0 \leq s < 10 \) | 8 |
| \( 10 \leq s < 20 \) | 10 |
| \( 20 \leq s < 30 \) | 7 |
| \( 30 \leq s < 40 \) | 2 |
| \( 40 \leq s < 50 \) | 3 |
Work out an estimate for the mean amount of snow per day.
Worked Solution
Step 1: Find the Midpoints
Why we do this: For grouped data, we use the midpoint of each group to represent the values in that group.
- \( 0 \leq s < 10 \): Midpoint = \( \frac{0+10}{2} = 5 \)
- \( 10 \leq s < 20 \): Midpoint = \( 15 \)
- \( 20 \leq s < 30 \): Midpoint = \( 25 \)
- \( 30 \leq s < 40 \): Midpoint = \( 35 \)
- \( 40 \leq s < 50 \): Midpoint = \( 45 \)
Step 2: Calculate \( \text{Frequency} \times \text{Midpoint} \) (\( fx \))
Group | Mid(x) | Freq(f) | fx ------------------------------------ 0-10 | 5 | 8 | 5 × 8 = 40 10-20 | 15 | 10 | 15 × 10 = 150 20-30 | 25 | 7 | 25 × 7 = 175 30-40 | 35 | 2 | 35 × 2 = 70 40-50 | 45 | 3 | 45 × 3 = 135
✓ (M1) Finding products \( fx \)
Step 3: Calculate the Totals and Mean
Formula: Mean = \( \frac{\text{Sum of } fx}{\text{Sum of } f} \)
Sum of frequencies (Total days):
\[ 8 + 10 + 7 + 2 + 3 = 30 \]Sum of \( fx \):
\[ 40 + 150 + 175 + 70 + 135 = 570 \]Calculate Mean:
\[ \frac{570}{30} = \frac{57}{3} = 19 \]✓ (M1) Division by 30
Final Answer:
\[ 19 \text{ cm} \]✓ (A1)
Question 9 (3 marks)
A cube is placed on top of a cuboid, as shown in the diagram, to form a solid.
The cube has edges of length 4 cm.
The cuboid has dimensions 7 cm by 6 cm by 5 cm.
Work out the total surface area of the solid.
Worked Solution
Step 1: Understand Surface Area of Composite Solids
Strategy: The total surface area is the sum of all exposed faces.
Usually, when you place one shape on another, the contact area is hidden. However, notice something important:
- The base of the cube covers a part of the top of the cuboid.
- The top of the cube adds an area exactly equal to the base that was covered.
Therefore, the total surface area is simply:
(Total Surface Area of the Cuboid) + (Area of the 4 vertical faces of the Cube)
Step 2: Surface Area of the Cuboid
Dimensions: \( l=7, w=6, h=5 \)
Faces:
- Bottom/Top: \( 7 \times 6 = 42 \) (2 faces)
- Front/Back: \( 7 \times 5 = 35 \) (2 faces)
- Sides: \( 6 \times 5 = 30 \) (2 faces)
✓ (P1) Correct process for cuboid area
Step 3: Add Exposed Faces of the Cube
The cube has side length 4 cm.
We need to add the 4 vertical faces (Front, Back, Left, Right). The top face effectively replaces the hidden patch on the cuboid.
Area of one face: \( 4 \times 4 = 16 \, \text{cm}^2 \)
Area of 4 faces:
\[ 4 \times 16 = 64 \, \text{cm}^2 \]Step 4: Total Surface Area
Alternative Method:
SA Cuboid = 214. SA Cube = \( 6 \times 4^2 = 96 \).
Overlap area (base of cube) = \( 4 \times 4 = 16 \).
This overlap is lost from BOTH the cuboid and the cube.
\[ \text{Total} = 214 + 96 – 2(16) \] \[ = 310 – 32 = 278 \]Final Answer:
\[ 278 \, \text{cm}^2 \]✓ (A1)
Question 10 (6 marks)
The table shows some information about the profit made each day at a cricket club on 100 days.
| Profit (£\( x \)) | Frequency |
|---|---|
| \( 0 \leq x < 50 \) | 10 |
| \( 50 \leq x < 100 \) | 15 |
| \( 100 \leq x < 150 \) | 25 |
| \( 150 \leq x < 200 \) | 30 |
| \( 200 \leq x < 250 \) | 5 |
| \( 250 \leq x < 300 \) | 15 |
(a) Complete the cumulative frequency table.
| Profit (£\( x \)) | Cumulative frequency |
|---|---|
| \( 0 \leq x < 50 \) | |
| \( 0 \leq x < 100 \) | |
| \( 0 \leq x < 150 \) | |
| \( 0 \leq x < 200 \) | |
| \( 0 \leq x < 250 \) | |
| \( 0 \leq x < 300 \) |
(b) On the grid, draw a cumulative frequency graph for this information.
(c) Use your graph to find an estimate for the number of days on which the profit was less than £125.
(d) Use your graph to find an estimate for the interquartile range.
Worked Solution
Part (a): Cumulative Frequency Table
Method: Add the frequencies as you go down the table (Running total).
- 0 to 50: \( 10 \)
- 0 to 100: \( 10 + 15 = 25 \)
- 0 to 150: \( 25 + 25 = 50 \)
- 0 to 200: \( 50 + 30 = 80 \)
- 0 to 250: \( 80 + 5 = 85 \)
- 0 to 300: \( 85 + 15 = 100 \)
Values: 10, 25, 50, 80, 85, 100
✓ (B1) Correct table
Part (b): Draw Graph
Method: Plot points using the upper bound of each interval and the cumulative frequency. Join with a smooth curve starting from (0,0).
Points: (50, 10), (100, 25), (150, 50), (200, 80), (250, 85), (300, 100).
✓ (M1) Plotting points correctly
✓ (A1) Fully correct graph
Part (c): Estimate days < £125
Locate 125 on the x-axis (Profit).
Go up to the curve, then across to the Cumulative Frequency axis.
At £125 (midway between 100 and 150), the graph is typically around 37-38.
Estimate: 37 (Accept 35-39)
✓ (B1)
Part (d): Interquartile Range
Formula: \( \text{IQR} = \text{Upper Quartile} (Q_3) – \text{Lower Quartile} (Q_1) \)
Lower Quartile (\( 25\% \) of 100) = 25th value.
Read from y=25 across to the curve, then down. Matches exactly £100 (from table).
Upper Quartile (\( 75\% \) of 100) = 75th value.
Read from y=75 across to the curve, then down. Approx £190-195.
\[ \text{IQR} \approx 192 – 100 = 92 \]Answer: 85 to 93
✓ (M1) Finding difference between reading at CF 75 and CF 25
✓ (A1)
Question 11 (3 marks)
Cormac has some sweets in a bag.
The sweets are lime flavoured or strawberry flavoured or orange flavoured.
In the bag
number of lime flavoured sweets : number of strawberry flavoured sweets : number of orange flavoured sweets = \( 9 : 4 : x \)
Cormac is going to take at random a sweet from the bag.
The probability that he takes a lime flavoured sweet is \( \frac{3}{7} \)
Work out the value of \( x \).
Worked Solution
Step 1: Relate Probability to Ratio
Why we do this: The probability of picking a lime sweet is the number of lime parts divided by the total number of parts.
Ratio = \( 9 : 4 : x \)
Number of lime parts = 9
Total parts = \( 9 + 4 + x = 13 + x \)
Probability of lime = \( \frac{9}{13 + x} \)
We are given that this probability is \( \frac{3}{7} \).
\[ \frac{9}{13 + x} = \frac{3}{7} \]✓ (P1) Setting up equation
Step 2: Solve for \( x \)
Cross-multiply or equivalent fractions:
\[ 9 \times 7 = 3(13 + x) \] \[ 63 = 39 + 3x \]Subtract 39 from both sides:
\[ 63 – 39 = 3x \] \[ 24 = 3x \]Divide by 3:
\[ x = 8 \]Alternative Method:
\( \frac{3}{7} = \frac{9}{21} \) (Multiplying top and bottom by 3)
So Total Parts = 21.
\[ 13 + x = 21 \] \[ x = 21 – 13 = 8 \]✓ (P1) Forming correct equation without fractions or finding total is 21
Final Answer:
\[ x = 8 \]✓ (A1)
Question 12 (3 marks)
Express \( 0.1\dot{1}\dot{7} \) as a fraction.
You must show all your working.
Worked Solution
Step 1: Define the Variable
Notation: \( 0.1\dot{1}\dot{7} \) means the digits 17 repeat. Note the first 1 is not under a dot, but the recurring part usually starts after non-recurring decimals. However, strictly \( 0.1\dot{1}\dot{7} \) often implies \( 0.1171717… \).
Let’s verify the recurring pattern. If the dots are on the second 1 and the 7, it is \( 0.1171717… \).
Let \( x = 0.1171717… \)
We need to shift the decimal point so the recurring parts align.
Step 2: Eliminate the Recurring Part
Multiply by 10 to get the repeating part right after the decimal:
\[ 10x = 1.171717… \]Multiply by 1000 to move one full cycle (17) past the decimal:
\[ 1000x = 117.171717… \]Now subtract the two equations:
\[ \begin{aligned} 1000x &= 117.1717… \\ – \quad 10x &= \phantom{00}1.1717… \\ \hline 990x &= 116 \end{aligned} \]✓ (M1) Setting up correct equations (e.g., \( 1000x \) and \( 10x \))
✓ (M1) Subtracting to eliminate decimals
Step 3: Simplify the Fraction
Both numbers are even, divide by 2:
\[ x = \frac{58}{495} \]Final Answer:
\[ \frac{116}{990} \text{ or } \frac{58}{495} \]✓ (A1)
Question 13 (3 marks)
A right-angled triangle is formed by the diameters of three semicircular regions, A, B and C as shown in the diagram.
Show that area of region A = area of region B + area of region C
Worked Solution
Step 1: Define Variables
Let the diameter of semicircle A be \( a \) (hypotenuse).
Let the diameter of semicircle B be \( b \).
Let the diameter of semicircle C be \( c \).
Since the triangle is right-angled, by Pythagoras’ theorem:
\[ a^2 = b^2 + c^2 \]✓ (M1) Use of Pythagoras’ theorem
Step 2: Express the Area of each Semicircle
Area of a circle is \( \pi r^2 \). Area of a semicircle is \( \frac{1}{2}\pi r^2 \).
Radius is \( \frac{\text{diameter}}{2} \).
Area A: \( \frac{1}{2}\pi \left(\frac{a}{2}\right)^2 = \frac{1}{2}\pi \frac{a^2}{4} = \frac{\pi a^2}{8} \)
Area B: \( \frac{1}{2}\pi \left(\frac{b}{2}\right)^2 = \frac{\pi b^2}{8} \)
Area C: \( \frac{1}{2}\pi \left(\frac{c}{2}\right)^2 = \frac{\pi c^2}{8} \)
✓ (M1) Correct expressions for areas
Step 3: Combine and Prove
We need to show Area B + Area C = Area A.
\[ \text{Area B} + \text{Area C} = \frac{\pi b^2}{8} + \frac{\pi c^2}{8} \]Factor out \( \frac{\pi}{8} \):
\[ = \frac{\pi}{8} (b^2 + c^2) \]From Step 1, we know \( b^2 + c^2 = a^2 \). Substitute this in:
\[ = \frac{\pi}{8} (a^2) = \frac{\pi a^2}{8} \]This is exactly the expression for Area A.
Therefore, Area A = Area B + Area C.
✓ (C1) Fully correct chain of reasoning
Question 14 (4 marks)
Here is a speed-time graph.
(a) Work out an estimate of the gradient of the graph at \( t = 2 \)
(b) What does the area under the graph represent?
Worked Solution
Part (a): Estimate Gradient
Method: The gradient of a curve at a specific point is the gradient of the tangent at that point. Draw a straight line touching the curve at \( t = 2 \).
Draw a tangent at \( t = 2 \).
Read two points on your tangent line, for example:
- Point 1: \( (2, 2.7) \) (The point of contact)
- Point 2: \( (3, 3.6) \)
Calculate Gradient = \( \frac{\text{Change in } y}{\text{Change in } x} \)
\[ m = \frac{3.6 – 2.7}{3 – 2} = \frac{0.9}{1} = 0.9 \]Acceptable Range: \( 0.7 \) to \( 1.1 \)
✓ (M1) Drawing tangent
✓ (M1) Calculation of gradient
✓ (A1) Answer in range [0.7, 1.1]
Part (b): Area Interpretation
Concept: In a speed-time graph:
- Gradient = Acceleration
- Area under graph = Distance travelled
Answer: Distance (travelled)
✓ (C1)
Question 15 (5 marks)
\( A, B \) and \( C \) are three points such that
\[ \overrightarrow{AB} = 3\mathbf{a} + 4\mathbf{b} \] \[ \overrightarrow{AC} = 15\mathbf{a} + 20\mathbf{b} \](a) Prove that \( A, B \) and \( C \) lie on a straight line.
\( D, E \) and \( F \) are three points on a straight line such that
\[ \overrightarrow{DE} = 3\mathbf{e} + 6\mathbf{f} \] \[ \overrightarrow{EF} = -10.5\mathbf{e} – 21\mathbf{f} \](b) Find the ratio
length of \( DF \) : length of \( DE \)
Worked Solution
Part (a): Proving Collinearity
Method: To prove points are on a straight line, we must show that vectors connecting them are parallel (scalar multiples of each other) and they share a common point.
Factorise \( \overrightarrow{AC} \):
\[ \overrightarrow{AC} = 15\mathbf{a} + 20\mathbf{b} = 5(3\mathbf{a} + 4\mathbf{b}) \]Compare with \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = 3\mathbf{a} + 4\mathbf{b} \]So:
\[ \overrightarrow{AC} = 5 \overrightarrow{AB} \]Conclusion:
- \( \overrightarrow{AC} \) is a multiple of \( \overrightarrow{AB} \), so they are parallel.
- They both share the point \( A \).
- Therefore, \( A, B, \) and \( C \) lie on a straight line.
✓ (C1) Correct relationship and reason
Part (b): Finding the Ratio
We need the ratio of lengths \( DF : DE \).
First, find vector \( \overrightarrow{DF} \) using vector addition: \( \overrightarrow{DF} = \overrightarrow{DE} + \overrightarrow{EF} \).
Let’s compare magnitudes. Factor out to find the scale factor relative to \( \overrightarrow{DE} = 3(\mathbf{e} + 2\mathbf{f}) \).
\[ \overrightarrow{DF} = -7.5(\mathbf{e} + 2\mathbf{f}) \]The “length” is proportional to the scalar multiplier.
Length \( DE \) corresponds to multiplier 3.
Length \( DF \) corresponds to multiplier 7.5 (ignoring sign, which indicates direction).
Ratio \( DF : DE \)
\[ 7.5 : 3 \]Multiply by 2 to clear decimals:
\[ 15 : 6 \]Divide by 3:
\[ 5 : 2 \]✓ (P1) Finding vector \( DF \)
✓ (A1) \( 5 : 2 \) or \( 2.5 : 1 \)
Question 16 (4 marks)
A first aid test has two parts, a theory test and a practical test.
The probability of passing the theory test is 0.75.
The probability of passing only one of the two parts is 0.36.
The two events are independent.
Work out the probability of passing the practical test.
Worked Solution
Step 1: Define Variables
Method: Let \( P(T) \) be the probability of passing the Theory test and \( P(P) \) be the probability of passing the Practical test.
Given: \( P(T) = 0.75 \)
Therefore, probability of failing theory: \( P(T’) = 1 – 0.75 = 0.25 \)
Let \( x \) be the probability of passing the practical test, i.e., \( P(P) = x \).
Probability of failing practical: \( P(P’) = 1 – x \)
✓ (P1) Setting up variable for unknown probability
Step 2: Form an Equation
Reasoning: “Passing only one” means either:
- Passing Theory AND Failing Practical
- Failing Theory AND Passing Practical
Since events are independent, we multiply probabilities.
Substitute the values:
\[ 0.36 = 0.75(1 – x) + 0.25(x) \]✓ (P1) Forming correct equation
Step 3: Solve for \( x \)
Expand the brackets:
\[ 0.36 = 0.75 – 0.75x + 0.25x \]Simplify:
\[ 0.36 = 0.75 – 0.5x \]Rearrange to find \( x \):
\[ 0.5x = 0.75 – 0.36 \] \[ 0.5x = 0.39 \] \[ x = \frac{0.39}{0.5} = 0.39 \times 2 = 0.78 \]✓ (P1) Process to solve equation
Final Answer:
\[ 0.78 \]✓ (A1)
Question 17 (4 marks)
\( y \) is directly proportional to the square root of \( t \).
\( y = 15 \) when \( t = 9 \).
\( t \) is inversely proportional to the cube of \( x \).
\( t = 8 \) when \( x = 2 \).
Find a formula for \( y \) in terms of \( x \).
Give your answer in its simplest form.
Worked Solution
Step 1: Find formula for \( y \) in terms of \( t \)
Direct Proportion: \( y \propto \sqrt{t} \) means \( y = k\sqrt{t} \).
Substitute \( y=15, t=9 \):
\[ 15 = k\sqrt{9} \] \[ 15 = 3k \] \[ k = 5 \]So, \( y = 5\sqrt{t} \)
✓ (P1) Setting up \( y=k\sqrt{t} \) and finding \( k=5 \)
Step 2: Find formula for \( t \) in terms of \( x \)
Inverse Proportion: \( t \propto \frac{1}{x^3} \) means \( t = \frac{c}{x^3} \).
Substitute \( t=8, x=2 \):
\[ 8 = \frac{c}{2^3} \] \[ 8 = \frac{c}{8} \] \[ c = 64 \]So, \( t = \frac{64}{x^3} \)
✓ (P1) Setting up \( t=c/x^3 \) and finding \( c=64 \)
Step 3: Combine formulas
Substitute \( t = \frac{64}{x^3} \) into \( y = 5\sqrt{t} \):
\[ y = 5\sqrt{\frac{64}{x^3}} \]Simplify:
\[ y = 5 \times \frac{\sqrt{64}}{\sqrt{x^3}} \] \[ y = 5 \times \frac{8}{x^{1.5}} \] \[ y = \frac{40}{\sqrt{x^3}} \quad \text{or} \quad y = \frac{40}{x^{1.5}} \]✓ (P1) Substitution process
Final Answer:
\[ y = \frac{40}{\sqrt{x^3}} \]✓ (A1)
Question 18 (4 marks)
Work out the value of
\[ \frac{ \left(\frac{49}{9}\right)^{-\frac{1}{2}} \times 4\frac{2}{3} }{ 2^{-3} } \]You must show all your working.
Worked Solution
Step 1: Simplify the first term in numerator
Method: Negative power means reciprocal (flip the fraction). Fractional power \( \frac{1}{2} \) means square root.
✓ (M1) Dealing with negative index or square root
Step 2: Convert mixed number and multiply
Convert \( 4\frac{2}{3} \) to an improper fraction:
\[ 4\frac{2}{3} = \frac{12+2}{3} = \frac{14}{3} \]Multiply numerator terms:
\[ \frac{3}{7} \times \frac{14}{3} \]Cancel 3s and simplify \( 14 \div 7 \):
\[ = 2 \]✓ (M1) Simplifying numerator to 2
Step 3: Simplify denominator and divide
Denominator is \( 2^{-3} \):
\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \]Now divide numerator by denominator:
\[ \frac{2}{ \frac{1}{8} } = 2 \times 8 = 16 \]✓ (M1) Processing the division
Final Answer:
\[ 16 \]✓ (A1)
Question 19 (4 marks)
Solve \( \frac{1}{2x – 1} + \frac{3}{x – 1} = 1 \)
Give your answer in the form \( \frac{p \pm \sqrt{q}}{2} \) where \( p \) and \( q \) are integers.
Worked Solution
Step 1: Combine algebraic fractions
Method: Find a common denominator, which is \( (2x-1)(x-1) \).
Expand the top:
\[ (x – 1) + (6x – 3) = 7x – 4 \]So:
\[ \frac{7x – 4}{(2x-1)(x-1)} = 1 \]✓ (M1) Using common denominator
Step 2: Form Quadratic Equation
Multiply up the denominator:
\[ 7x – 4 = (2x – 1)(x – 1) \]Expand the right side:
\[ 7x – 4 = 2x^2 – 2x – x + 1 \] \[ 7x – 4 = 2x^2 – 3x + 1 \]Rearrange to form \( ax^2 + bx + c = 0 \):
\[ 0 = 2x^2 – 3x – 7x + 1 + 4 \] \[ 2x^2 – 10x + 5 = 0 \]✓ (M1) Expanding and rearranging to quadratic
Step 3: Solve using Quadratic Formula
Formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Here \( a = 2, b = -10, c = 5 \).
Simplify the surd \( \sqrt{60} \):
\[ \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \]Substitute back:
\[ x = \frac{10 \pm 2\sqrt{15}}{4} \]Divide top and bottom by 2:
\[ x = \frac{5 \pm \sqrt{15}}{2} \]✓ (M1) Method to solve quadratic
Final Answer:
\[ \frac{5 \pm \sqrt{15}}{2} \]✓ (A1)
Question 20 (4 marks)
The centre of a circle is the point with coordinates \( (-1, 3) \).
The point \( A \) with coordinates \( (6, 8) \) lies on the circle.
Find an equation of the tangent to the circle at \( A \).
Give your answer in the form \( ax + by + c = 0 \) where \( a, b \) and \( c \) are integers.
Worked Solution
Step 1: Find the gradient of the radius
Method: Use \( m = \frac{y_2 – y_1}{x_2 – x_1} \) for points \( (-1, 3) \) and \( (6, 8) \).
✓ (P1) Process to find gradient of radius
Step 2: Find the gradient of the tangent
Theory: The tangent is perpendicular to the radius. Perpendicular gradients multiply to -1.
✓ (P1) Using \( mn = -1 \)
Step 3: Find equation of the tangent
Method: Use \( y – y_1 = m(x – x_1) \) with point \( A(6, 8) \).
Multiply by 5 to remove fraction:
\[ 5(y – 8) = -7(x – 6) \] \[ 5y – 40 = -7x + 42 \]Rearrange to \( ax + by + c = 0 \):
\[ 7x + 5y – 40 – 42 = 0 \] \[ 7x + 5y – 82 = 0 \]✓ (P1) Substitution into line equation
Final Answer:
\[ 7x + 5y – 82 = 0 \]✓ (A1)
Question 21 (5 marks)
The diagram shows three circles, each of radius 4 cm.
The centres of the circles are \( A, B \) and \( C \) such that \( ABC \) is a straight line and \( AB = BC = 4 \) cm.
Work out the total area of the two shaded regions.
Give your answer in terms of \( \pi \).
Worked Solution
Step 1: Analyze the geometry of one intersection
Strategy: The two shaded regions are identical due to symmetry. We calculate the area of one region (between A and B) and double it.
The centers A and B are 4 cm apart. The radius is 4 cm.
This means triangle ABX (where X is an intersection point) is an equilateral triangle because all sides are 4 cm.
Let the intersection points of Circle A and Circle B be \( P \) (top) and \( Q \) (bottom).
Triangle \( ABP \) is equilateral (sides 4, 4, 4).
Therefore, angle \( PAB = 60^\circ \).
By symmetry, angle \( QAB = 60^\circ \).
Total angle of the sector \( PAQ \) in circle A is \( 120^\circ \).
✓ (P1) Identifying 60° or 120° angle
Step 2: Calculate Area of Sector and Triangle
The shaded area is composed of two segments. The area of one “lens” is \( 2 \times (\text{Area of Sector} – \text{Area of Triangle}) \). Wait, actually simpler:
One shaded region (lens) is made of two segments back-to-back.
Area of one segment = Area of Sector (120°) – Area of Triangle (Isosceles with 120°).
Wait, let’s use the 60° sector approach. The lens is the overlap.
Alternative: The lens area = 2 * (Area of Sector (60°) – Area of Equilateral Triangle). No, that’s for half the lens.
Let’s find the area of the segment cut by chord PQ in circle A.
Angle at center A is \( 120^\circ \).
Area of Sector (120°):
\[ \text{Area} = \frac{120}{360} \times \pi r^2 = \frac{1}{3} \times \pi (4^2) = \frac{16\pi}{3} \]Area of Triangle APQ:
Using \( \frac{1}{2}ab \sin C \):
\[ \text{Area} = \frac{1}{2}(4)(4) \sin 120^\circ \] \[ = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} \]Area of Segment (One half of the lens):
\[ \text{Segment} = \frac{16\pi}{3} – 4\sqrt{3} \]✓ (P1) Area of sector
✓ (P1) Area of triangle
Step 3: Total Area
One lens consists of TWO such segments (one from circle A, one from circle B).
\[ \text{Area of one lens} = 2 \times \left( \frac{16\pi}{3} – 4\sqrt{3} \right) \]There are TWO shaded regions (two lenses).
\[ \text{Total Area} = 2 \times \text{Area of one lens} \] \[ = 4 \times \left( \frac{16\pi}{3} – 4\sqrt{3} \right) \] \[ = \frac{64\pi}{3} – 16\sqrt{3} \]Wait, let’s check the Mark Scheme method.
Mark scheme says: \( 16\pi – 16\sqrt{3} \) or equivalent.
Let’s re-evaluate. Is the lens formed by 120 sectors?
Yes. Overlap of A and B.
Area = Area(Sector A) + Area(Sector B) – Area(Rhombus APBQ)? No.
Area = 2 * Segment.
Segment Area = \( \frac{16\pi}{3} – 4\sqrt{3} \).
One lens = \( 2(\frac{16\pi}{3} – 4\sqrt{3}) = \frac{32\pi}{3} – 8\sqrt{3} \).
Two lenses = \( \frac{64\pi}{3} – 16\sqrt{3} \).
Let’s check the Mark Scheme “Answer” in derivation.
Mark scheme: \( \frac{16\pi}{3} – 4\sqrt{3} \) for segment.
Result: \( \frac{16\pi}{3} \times 4 – 16\sqrt{3} \) … Wait.
Ah, the “Area of sector” in MS uses 60 degrees? No, 120 degrees is \( 16\pi/3 \).
Let’s re-read the MS carefully. “16pi – 16sqrt(3)” is not listed as the final answer in MS, it says “16sqrt(3) – … ” No.
MS Answer A1: \( \frac{16\pi}{3} \times 2 – … \)? No.
MS A1: \( 16\pi – \dots \) or \( \frac{16\pi}{3} – \dots \).
Let’s look at the MS image provided earlier.
A1: \( 16\pi – \frac{16\pi}{6} – 4\sqrt{3} + \dots \) ?
Actually, let’s stick to the math derived. It is robust.
Total Area = \( 2 \times [ 2 \times (\text{Sector}_{120} – \text{Triangle}_{120}) ] \)
= \( 4 \times (\frac{16\pi}{3} – 4\sqrt{3}) \)
= \( \frac{64\pi}{3} – 16\sqrt{3} \)
Let’s check if there is a simpler way.
Area of Lens = Area of 2 Sectors (120) – Area of 2 Triangles? No.
Area of intersection of two circles = \( 2r^2 \cos^{-1}(d/2r) – \frac{1}{2}d\sqrt{4r^2-d^2} \).
Here \( d=r=4 \). \( \cos^{-1}(0.5) = \pi/3 \) (60 deg? No, half angle). Total angle 120.
Area = \( r^2 (\theta – \sin \theta) \) where \( \theta \) in radians.
\( \theta = 2\pi/3 \).
Area Lens = \( 16 ( \frac{2\pi}{3} – \frac{\sqrt{3}}{2} ) = \frac{32\pi}{3} – 8\sqrt{3} \).
Total for 2 lenses = \( \frac{64\pi}{3} – 16\sqrt{3} \).
This matches.
Final Answer:
\[ \frac{64\pi}{3} – 16\sqrt{3} \]✓ (A1)