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GCSE Nov 2019 Edexcel Mathematics Higher Paper 3 (Calculator)
🎸 How to use this page
- 📝 Try it first: Attempt the question before looking at the solution.
- 📲 Calculator allowed: You can use a scientific calculator for this paper.
- 👁️ Visual learning: Diagrams are redrawn for clarity.
- 💡 Three-layer solutions: We explain Why, show How, and Check What it means.
📚 Table of Contents
- Question 1 (Algebra: Expand/Factorise)
- Question 2 (Calculator Use)
- Question 3 (Scatter Graph)
- Question 4 (Mean from Frequency Table)
- Question 5 (Unit Conversion)
- Question 6 (Speed, Distance, Time)
- Question 7 (Standard Form)
- Question 8 (Geometry: Polygons)
- Question 9 (Transformations: Enlargement)
- Question 10 (Solving Equations)
- Question 11 (Probability Tree)
- Question 12 (Box Plot)
- Question 13 (Density & Mass)
- Question 14 (Ratio & Fraction)
- Question 15 (Algebraic Proof)
- Question 16 (Proportion)
- Question 17 (Quadratic Graph)
- Question 18 (Trigonometry)
- Question 19 (Distance-Time Gradient)
- Question 20 (Sequences)
- Question 21 (Probability)
- Question 22 (Algebraic Fractions)
- Question 23 (Cone Geometry)
- Question 24 (Vector Geometry)
Question 1 (4 marks)
(a) Expand and simplify \( (x + 5)(x – 9) \)
(b) Factorise fully \( 9x^2 + 6x \)
Worked Solution
Part (a): Expand and Simplify
💡 What we are doing: We need to multiply every term in the first bracket by every term in the second bracket using the FOIL method (First, Outside, Inside, Last).
✏ Working:
\[ \begin{aligned} (x + 5)(x – 9) &= x \times x + x \times (-9) + 5 \times x + 5 \times (-9) \\ &= x^2 – 9x + 5x – 45 \end{aligned} \]💡 Simplify: Now combine the like terms (the x terms).
✓ (M1) for 3 of 4 terms correct
✓ (A1) for correct final answer
Part (b): Factorise Fully
💡 What we are doing: “Factorise fully” means we need to find the highest common factor (HCF) of both numbers and algebra terms.
Look at \( 9x^2 + 6x \):
- Numbers: The highest factor of 9 and 6 is 3.
- Algebra: The highest factor of \( x^2 \) and \( x \) is \( x \).
So the common factor is \( 3x \).
Divide each term by \( 3x \):
\[ \frac{9x^2}{3x} = 3x \quad \text{and} \quad \frac{6x}{3x} = 2 \]Put the factor outside the bracket:
\[ 3x(3x + 2) \]✓ (B2) for 3x(3x + 2)
Final Answer:
(a) \( x^2 – 4x – 45 \)
(b) \( 3x(3x + 2) \)
Question 2 (3 marks)
(a) Use your calculator to work out
\[ \frac{29^2 – 4.6}{\sqrt{35 – 1.9^3}} \]Write down all the figures on your calculator display.
(b) Write your answer to part (a) correct to 4 significant figures.
Worked Solution
Part (a): Calculator Work
💡 Strategy: Calculate the numerator (top) and denominator (bottom) separately or use the fraction button on your calculator carefully.
Numerator:
\[ 29^2 – 4.6 = 841 – 4.6 = 836.4 \]Denominator:
\[ \sqrt{35 – 1.9^3} = \sqrt{35 – 6.859} = \sqrt{28.141} \approx 5.3048… \]Division:
\[ \frac{836.4}{5.3048…} \approx 157.668255… \]✓ (M1) for 836.4 or 5.304… seen
✓ (A1) for 157.668255…
Part (b): Rounding
💡 Strategy: We need 4 significant figures. Count 4 digits from the start (ignoring leading zeros, which we don’t have here).
Value: \( 157.668255… \)
1st sf: 1
2nd sf: 5
3rd sf: 7
4th sf: 6
Check the next digit (5th digit) which is 6. Since 6 \(\geq\) 5, we round up the 4th digit.
\( 157.6 \rightarrow 157.7 \)
✓ (B1) for 157.7
Final Answer:
(a) \( 157.668255 \)
(b) \( 157.7 \)
Question 3 (2 marks)
The scatter graph shows information about the marks a group of students got in a Science test and in a Maths test.
Jamie got a mark of 34 in the Science test.
Using the scatter graph, find an estimate for Jamie’s mark in the Maths test.
Worked Solution
Step 1: Draw a Line of Best Fit
💡 Strategy: To estimate a value, we first need to model the trend of the data. We draw a straight line that goes through the middle of the points, keeping roughly the same number of points above and below the line.
Step 2: Read from the graph
- Locate 34 on the Science axis (x-axis).
- Go up to your line of best fit.
- Go across to the Maths axis (y-axis).
- Read the value.
Based on a standard line of best fit, the value is typically between 35 and 42.
✓ (M1) for drawing a suitable line of best fit
✓ (A1) for answer in range 35 – 42
Final Answer:
Any value between 35 and 42 (e.g., 38)
Question 4 (3 marks)
The table gives information about the times taken, in seconds, by 18 students to run a race.
| Time (t seconds) | Frequency |
|---|---|
| \( 5 < t \leq 10 \) | 1 |
| \( 10 < t \leq 15 \) | 2 |
| \( 15 < t \leq 20 \) | 7 |
| \( 20 < t \leq 25 \) | 8 |
Work out an estimate for the mean time.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Find Midpoints
💡 Why? We don’t know the exact times of the students, only the group they are in. We estimate by assuming everyone in the group took the midpoint time.
Midpoint = \( \frac{\text{Start} + \text{End}}{2} \)
- 5 to 10: \( \frac{5+10}{2} = 7.5 \)
- 10 to 15: \( \frac{10+15}{2} = 12.5 \)
- 15 to 20: \( \frac{15+20}{2} = 17.5 \)
- 20 to 25: \( \frac{20+25}{2} = 22.5 \)
Step 2: Calculate \( \text{Frequency} \times \text{Midpoint} \)
💡 Why? This gives us the estimated total time for all students in that group.
| MP | Freq | \( f \times x \) |
|---|---|---|
| 7.5 | 1 | \( 7.5 \times 1 = 7.5 \) |
| 12.5 | 2 | \( 12.5 \times 2 = 25.0 \) |
| 17.5 | 7 | \( 17.5 \times 7 = 122.5 \) |
| 22.5 | 8 | \( 22.5 \times 8 = 180.0 \) |
✓ (M1) for finding 4 products fx
Step 3: Calculate the Mean
💡 Formula: \( \text{Mean} = \frac{\text{Total Time}}{\text{Total Students}} \)
Total Time: \( 7.5 + 25.0 + 122.5 + 180.0 = 335 \)
Total Frequency: \( 1 + 2 + 7 + 8 = 18 \)
Mean:
\[ \frac{335}{18} = 18.6111… \]✓ (M1) for total fx ÷ total frequency
Step 4: Rounding
Round \( 18.6111… \) to 3 significant figures.
1st: 1, 2nd: 8, 3rd: 6. Next is 1 (round down).
\( 18.6 \)
✓ (A1)
Final Answer:
18.6 seconds
Question 5 (1 mark)
Write \( 37 \text{ cm}^3 \) in \( \text{mm}^3 \)
Worked Solution
Step 1: Understand Volume Conversions
💡 Concept: Length conversions are simple (\( 1 \text{ cm} = 10 \text{ mm} \)), but volume is length cubed.
Length: \( 1 \text{ cm} = 10 \text{ mm} \)
Volume: \( 1 \text{ cm}^3 = 1 \text{ cm} \times 1 \text{ cm} \times 1 \text{ cm} \)
Convert each side to mm:
\[ 10 \text{ mm} \times 10 \text{ mm} \times 10 \text{ mm} = 1000 \text{ mm}^3 \]So, the conversion factor is 1000.
Step 2: Calculate
✓ (B1)
Final Answer:
\( 37,000 \text{ mm}^3 \)
Question 6 (4 marks)
Nimer was driving to a hotel.
He looked at his Sat Nav at 13:30.
| Time | 13 30 |
| Distance to destination | 65 miles |
Nimer arrived at the hotel at 14:48.
Work out the average speed of the car from 13:30 to 14:48.
You must show all your working.
Worked Solution
Step 1: Calculate Time Taken
💡 Strategy: Find the difference between the arrival time and the start time.
Start: 13:30
Finish: 14:48
Hours: 13:30 to 14:30 is 1 hour.
Minutes: 14:30 to 14:48 is 18 minutes.
Total time = 1 hour 18 minutes.
Step 2: Convert Time to Decimals
💡 Why? To calculate speed in miles per hour (mph), the time needs to be a decimal number of hours, not minutes.
18 minutes as a fraction of an hour is \( \frac{18}{60} \).
\[ \frac{18}{60} = 0.3 \text{ hours} \]Total time in hours = \( 1 + 0.3 = 1.3 \text{ hours} \).
✓ (P1) for converting minutes to hours (1.3)
Step 3: Calculate Speed
💡 Formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
Calculation:
\[ 65 \div 1.3 = 50 \]✓ (P1) for using Distance ÷ Time
✓ (A1) for 50
Final Answer:
50 mph
Question 7 (3 marks)
(a) Write 32 460 000 in standard form.
(b) Write \( 4.96 \times 10^{-3} \) as an ordinary number.
Asma was asked to compare the following two numbers.
\[ A = 6.212 \times 10^8 \quad \text{and} \quad B = 4.73 \times 10^9 \]She says,
“6.212 is bigger than 4.73 so A is bigger than B.”
(c) Is Asma correct? You must give a reason for your answer.
Worked Solution
Part (a)
Place the decimal point after the first non-zero digit: \( 3.246 \)
Count how many places the decimal moved (from the end): 7 places.
\[ 3.246 \times 10^7 \]✓ (B1)
Part (b)
Move the decimal point 3 places to the left (because the power is negative).
\[ 4.96 \rightarrow 0.496 \rightarrow 0.0496 \rightarrow 0.00496 \]✓ (B1)
Part (c)
💡 Concept: In standard form, the power of 10 is the most important factor for size. A higher power of 10 means a much larger number.
Compare the powers:
- A has power \( 10^8 \)
- B has power \( 10^9 \)
\( 10^9 \) is 10 times bigger than \( 10^8 \).
Conclusion: No, Asma is incorrect. B is bigger because the power of 10 is higher (\( 10^9 > 10^8 \)).
✓ (C1) for “No” with correct explanation
Final Answer:
(a) \( 3.246 \times 10^7 \)
(b) 0.00496
(c) No, because B has a higher power of 10 ($10^9$) than A ($10^8$).
Question 8 (4 marks)
The diagram shows a regular pentagon and a parallelogram.
Work out the size of the angle marked \( x \).
You must show all your working.
Worked Solution
Step 1: Find Interior Angle of Pentagon
💡 Formula: Sum of interior angles = \( (n-2) \times 180 \). Then divide by \( n \) for a regular polygon.
For a pentagon, \( n = 5 \).
\[ \text{Sum} = (5 – 2) \times 180 = 3 \times 180 = 540^\circ \] \[ \text{One angle} = \frac{540}{5} = 108^\circ \]✓ (P1) for process to find interior angle (108)
Step 2: Find Angle of Parallelogram
💡 Property: Consecutive angles in a parallelogram add up to \( 180^\circ \) (co-interior angles).
The angle marked \( 117^\circ \) is obtuse.
The angle at the vertex (inside the pentagon corner) is adjacent to \( 117^\circ \).
\[ \text{Parallelogram angle} = 180^\circ – 117^\circ = 63^\circ \]✓ (P1) for 180 – 117 = 63
Step 3: Calculate x
💡 Reasoning: The corner of the pentagon is made up of the parallelogram angle and angle \( x \).
✓ (P1) for complete process (108 – 63)
✓ (A1) for 45
Final Answer:
\( 45^\circ \)
Question 9 (2 marks)
Enlarge triangle A by scale factor 2.5 with centre (0, 1).
Worked Solution
Step 1: Identify Key Information
Object: Triangle A with points (1, 1), (1, 3), and (2, 3).
Centre: (0, 1)
Scale Factor: 2.5
Step 2: Calculate Vectors from Centre
💡 Method: Find the distance (vector) from the centre (0, 1) to each point, multiply by 2.5, then add back to the centre.
Point 1: (1, 1)
- Vector from (0, 1): \( (1-0, 1-1) = (1, 0) \)
- Multiply by 2.5: \( (2.5, 0) \)
- Add to centre: \( (0+2.5, 1+0) = \mathbf{(2.5, 1)} \)
Point 2: (1, 3)
- Vector from (0, 1): \( (1-0, 3-1) = (1, 2) \)
- Multiply by 2.5: \( (2.5, 5) \)
- Add to centre: \( (0+2.5, 1+5) = \mathbf{(2.5, 6)} \)
Point 3: (2, 3)
- Vector from (0, 1): \( (2-0, 3-1) = (2, 2) \)
- Multiply by 2.5: \( (5, 5) \)
- Add to centre: \( (0+5, 1+5) = \mathbf{(5, 6)} \)
Step 3: Draw the New Triangle
Plot the points (2.5, 1), (2.5, 6), and (5, 6).
Connect them to form a larger triangle.
✓ (B2) for vertices at (2.5, 1), (2.5, 6), (5, 6)
Final Answer:
Triangle drawn at coordinates (2.5, 1), (2.5, 6), (5, 6).
Question 10 (4 marks)
(a) Solve \( \frac{9 + x}{7} = 11 – x \)
(b) Simplify \( \frac{4(y + 3)^3}{(y + 3)^2} \)
Worked Solution
Part (a): Solve Equation
💡 Strategy: First, remove the fraction by multiplying both sides by 7.
Multiply by 7:
\[ 9 + x = 7(11 – x) \]Expand the bracket:
\[ 9 + x = 77 – 7x \]✓ (M1) for multiplying by 7 correctly
💡 Isolate x: Get all x terms on one side and numbers on the other.
Add \( 7x \) to both sides:
\[ 9 + 8x = 77 \]Subtract 9 from both sides:
\[ 8x = 68 \]Divide by 8:
\[ x = \frac{68}{8} \] \[ x = 8.5 \]✓ (A1)
Part (b): Simplify
💡 Strategy: Use index laws. \( \frac{a^3}{a^2} = a^{3-2} = a^1 \).
The term \( (y+3) \) is common.
\[ 4 \times (y+3)^{3-2} \] \[ 4(y+3)^1 \] \[ 4(y+3) \]✓ (B1)
Final Answer:
(a) \( x = 8.5 \)
(b) \( 4(y+3) \) or \( 4y + 12 \)
Question 11 (3 marks)
The probability tree diagram shows the probabilities that Bismah will be late for work on two days next week.
Calculate the probability that Bismah will be late on exactly one of the two days.
Worked Solution
Step 1: Identify the Valid Paths
💡 Strategy: “Exactly one day” means either:
- Late on Monday AND Not Late on Tuesday (Late, Not Late)
- OR
- Not Late on Monday AND Late on Tuesday (Not Late, Late)
We need to calculate both probabilities and add them together.
Step 2: Calculate Probabilities
Path 1: Late, Not Late
\[ P(\text{L, NL}) = 0.07 \times 0.98 = 0.0686 \]Path 2: Not Late, Late
\[ P(\text{NL, L}) = 0.93 \times 0.11 = 0.1023 \]✓ (M1) for one correct product
Step 3: Add the Probabilities
✓ (M1) for adding correct products
✓ (A1) for 0.1709
Final Answer:
0.1709
Question 12 (3 marks)
The stem and leaf diagram shows information about the heights, in cm, of 23 sunflowers.
| 17 | 3 4 9 |
| 18 | 6 8 8 |
| 19 | 0 0 1 4 6 7 8 |
| 20 | 1 4 7 7 9 9 |
| 21 | 4 8 8 9 |
Key: 17|3 represents 173 cm
On the grid, draw a box plot for this information.
Worked Solution
Step 1: Find the 5 Key Values
💡 Needed: Lowest Value, Lower Quartile (LQ), Median, Upper Quartile (UQ), Highest Value.
Total data points \( n = 23 \).
1. Lowest Value: First number = 173
2. Highest Value: Last number = 219
3. Median: Position = \( \frac{n+1}{2} = \frac{24}{2} = 12^{\text{th}} \) value.
- Count 12 numbers in the stem and leaf.
- 170s: 3 values
- 180s: 3 values (Total 6)
- 190s: We need the 6th value in this row (since 6+6=12).
- Row 19: 0, 0, 1, 4, 6, 7
Median = 197
4. Lower Quartile: Median of the lower half (first 11 numbers).
Position = \( \frac{11+1}{2} = 6^{\text{th}} \) value.
Count 6: 173, 174, 179, 186, 188, 188.
LQ = 188
5. Upper Quartile: Median of the upper half (last 11 numbers).
Position = 6th value from the end.
Count back 6: 219, 218, 218, 214, 209, 209.
UQ = 209
✓ (M1) for correctly identifying LQ (188), Median (197), or UQ (209)
Step 2: Draw the Box Plot
💡 Checklist:
- Whiskers extend to Lowest (173) and Highest (219).
- Box starts at LQ (188) and ends at UQ (209).
- Line inside box at Median (197).
✓ (A1) for fully correct box plot
Final Answer:
Box plot drawn with Min=173, LQ=188, Med=197, UQ=209, Max=219.
Question 13 (4 marks)
Liquid A and liquid B are mixed together in the ratio 2 : 13 by volume to make liquid C.
Liquid A has density \( 1.21 \text{ g/cm}^3 \)
Liquid B has density \( 1.02 \text{ g/cm}^3 \)
A cylindrical container is filled completely with liquid C.
The cylinder has radius 3 cm and height 25 cm.
Work out the mass of the liquid in the container.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Calculate Volume of Cylinder (Liquid C)
💡 Formula: Volume of cylinder = \( \pi r^2 h \)
✓ (P1) for process to find volume of C
Step 2: Split Volume into A and B
💡 Ratio: 2 : 13. Total parts = \( 2 + 13 = 15 \).
Volume of A = \( \frac{2}{15} \times 225\pi = 30\pi \approx 94.248 \text{ cm}^3 \)
Volume of B = \( \frac{13}{15} \times 225\pi = 195\pi \approx 612.61 \text{ cm}^3 \)
✓ (P1) for process to find volume of A or B
Step 3: Calculate Mass of A and B
💡 Formula: \( \text{Mass} = \text{Density} \times \text{Volume} \)
Mass of A:
\[ 1.21 \times 30\pi = 36.3\pi \approx 114.04 \text{ g} \]Mass of B:
\[ 1.02 \times 195\pi = 198.9\pi \approx 624.86 \text{ g} \]✓ (P1) for process to find mass of C (sum of parts)
Step 4: Total Mass and Rounding
Round to 3 significant figures: 739 g
✓ (A1)
Final Answer:
739 g
Question 14 (4 marks)
A group of people went to a restaurant. Each person chose one starter and one main course.
| starter | main course |
|---|---|
| soup | lasagne |
| prawns | curry |
The number of people who chose soup : the number of people who chose prawns = 2 : 3
Of those who chose soup, the number of people who chose lasagne : the number of people who chose curry = 5 : 3
Of those who chose prawns, the number of people who chose lasagne : the number of people who chose curry = 1 : 5
What fraction of the people chose curry?
Worked Solution
Step 1: Use Fractions or a Total Number
💡 Strategy: Working with fractions is direct, but assuming a total number of people can make the numbers easier to visualize.
Fraction Method:
- Probability of choosing Soup: \( \frac{2}{5} \)
- Probability of choosing Prawns: \( \frac{3}{5} \)
✓ (P1) for finding fraction for soup/prawns or sharing in ratio 2:3
Step 2: Calculate Fractions for Curry
From Soup Group:
Ratio Lasagne : Curry = 5 : 3 (Total 8 parts)
Fraction choosing Curry = \( \frac{3}{8} \)
Total Fraction (Soup AND Curry) = \( \frac{2}{5} \times \frac{3}{8} = \frac{6}{40} \)
From Prawn Group:
Ratio Lasagne : Curry = 1 : 5 (Total 6 parts)
Fraction choosing Curry = \( \frac{5}{6} \)
Total Fraction (Prawns AND Curry) = \( \frac{3}{5} \times \frac{5}{6} = \frac{15}{30} = \frac{1}{2} \)
✓ (P1) for process to find proportion for curry in either group
Step 3: Add Fractions Together
Total Curry = (Soup & Curry) + (Prawns & Curry)
\[ \frac{6}{40} + \frac{1}{2} \]Common denominator is 40:
\[ \frac{6}{40} + \frac{20}{40} = \frac{26}{40} \]Simplify:
\[ \frac{13}{20} \]✓ (P1) for complete process
✓ (A1)
Final Answer:
\( \frac{13}{20} \)
Question 15 (3 marks)
Prove algebraically that the sum of the squares of any two consecutive even numbers is always a multiple of 4.
Worked Solution
Step 1: Define the Terms
💡 Reasoning: An even number can be written as \( 2n \). The next consecutive even number is \( 2n + 2 \).
Let the first number be \( 2n \).
Let the second number be \( 2n + 2 \).
✓ (M1) for correct expressions
Step 2: Square and Sum
Sum of squares = \( (2n)^2 + (2n + 2)^2 \)
Expand:
\[ (2n)^2 = 4n^2 \] \[ (2n + 2)^2 = (2n + 2)(2n + 2) = 4n^2 + 4n + 4n + 4 = 4n^2 + 8n + 4 \]Add them together:
\[ 4n^2 + (4n^2 + 8n + 4) = 8n^2 + 8n + 4 \]✓ (M1) for expanding both correctly
Step 3: Factorise to Prove
💡 Goal: To show it’s a multiple of 4, we need to show we can pull out a factor of 4.
Since \( 4(2n^2 + 2n + 1) \) is 4 times an integer, the result is always a multiple of 4.
✓ (A1) for complete proof
Final Answer:
Proof shown: \( 4(2n^2 + 2n + 1) \)
Question 16 (3 marks)
\( y \) is inversely proportional to the square of \( x \).
\( y = 8 \) when \( x = 2.5 \)
Find the negative value of \( x \) when \( y = \frac{8}{9} \)
Worked Solution
Step 1: Set up the Formula
💡 Principle: “Inversely proportional” means \( y = \frac{k}{x^n} \). Here, “square of x” means \( x^2 \).
Substitute the given values to find \( k \):
\[ 8 = \frac{k}{2.5^2} \] \[ 8 = \frac{k}{6.25} \] \[ k = 8 \times 6.25 \] \[ k = 50 \]So the formula is \( y = \frac{50}{x^2} \).
✓ (M1) for correct relationship \( y = \frac{k}{x^2} \)
✓ (A1) for \( k = 50 \)
Step 2: Calculate x for new y
Substitute \( y = \frac{8}{9} \) into the formula:
\[ \frac{8}{9} = \frac{50}{x^2} \]Cross multiply:
\[ 8x^2 = 9 \times 50 \] \[ 8x^2 = 450 \] \[ x^2 = \frac{450}{8} \] \[ x^2 = 56.25 \]Find the square root:
\[ x = \sqrt{56.25} \] \[ x = \pm 7.5 \]The question asks for the negative value.
\[ x = -7.5 \]✓ (A1) for -7.5
Final Answer:
\( x = -7.5 \)
Question 17 (4 marks)
Here is the graph of \( y = x^2 – 3 \)
Use the graph to find estimates for the solutions to the equation \( x^2 – 2x – 2 = 0 \)
You must show how you get your solutions.
Worked Solution
Step 1: Rearrange the Equation
💡 Strategy: We are given the graph of \( y = x^2 – 3 \). We need to solve \( x^2 – 2x – 2 = 0 \).
We need to make the equation look like \( x^2 – 3 = \text{something} \).
Equation to solve: \( x^2 – 2x – 2 = 0 \)
Add \( 2x \) to both sides:
\[ x^2 – 2 = 2x \]Subtract 1 from both sides to get \( x^2 – 3 \):
\[ x^2 – 3 = 2x – 1 \]So, we need to find where the curve \( y = x^2 – 3 \) intersects the line \( y = 2x – 1 \).
✓ (M1) for relating to \( 2x – 1 \)
Step 2: Draw the Line \( y = 2x – 1 \)
💡 Plotting: Choose a few x-values to find points for the line.
- If \( x = 0 \), \( y = -1 \). Point (0, -1).
- If \( x = 2 \), \( y = 3 \). Point (2, 3).
- If \( x = 3 \), \( y = 5 \). Point (3, 5).
Draw a straight line through these points on the graph.
✓ (M1) for drawing the line \( y = 2x – 1 \)
Step 3: Read the Intersection Points
Look where the red line crosses the curve.
Read the x-values down to the axis.
- First point is between -0.6 and -0.8. Estimate: -0.7
- Second point is between 2.6 and 2.8. Estimate: 2.7
✓ (M1) for identifying intersection points
✓ (A1) for answers in range -0.6 to -0.8 and 2.6 to 2.8
Final Answer:
\( x = -0.7 \) and \( x = 2.7 \)
Question 18 (5 marks)
The diagram shows triangle ABC.
\( AB = 3.4 \text{ cm} \quad AC = 6.2 \text{ cm} \quad BC = 6.1 \text{ cm} \)
\( D \) is the point on \( BC \) such that
size of angle \( DAC = \frac{2}{5} \times \) size of angle \( BCA \)
Calculate the length \( DC \).
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Find Angle C (BCA)
💡 Method: Use the Cosine Rule because we know all three sides of the big triangle ABC.
\( a^2 = b^2 + c^2 – 2bc \cos A \) (Standard form)
Rearranged for angle C: \( \cos C = \frac{a^2 + b^2 – c^2}{2ab} \)
Here, relative to angle C:
- Adjacent sides: \( AC = 6.2 \), \( BC = 6.1 \)
- Opposite side: \( AB = 3.4 \)
✓ (P1) for cosine rule substitution
✓ (P1) for angle C = 32.08…
Step 2: Find Angle DAC and ADC
💡 Given: Angle \( DAC = \frac{2}{5} \times \) Angle \( BCA \)
Angle \( DAC = \frac{2}{5} \times 32.08… = 12.83^\circ \)
Now consider triangle ADC. We know:
- Angle \( C = 32.08^\circ \)
- Angle \( DAC = 12.83^\circ \)
Angles in a triangle add to 180, so Angle \( ADC = 180 – (32.08 + 12.83) = 135.09^\circ \)
Step 3: Calculate Length DC
💡 Method: Use the Sine Rule in triangle ADC.
\( \frac{a}{\sin A} = \frac{b}{\sin B} \)
Substitute \( AC = 6.2 \):
\[ \frac{DC}{\sin(12.83)} = \frac{6.2}{\sin(135.09)} \] \[ DC = \frac{6.2 \times \sin(12.83)}{\sin(135.09)} \] \[ DC = \frac{6.2 \times 0.222…}{0.706…} \] \[ DC = 1.951… \]✓ (P1) for sine rule setup
✓ (A1) for 1.95 cm (3 s.f.)
Final Answer:
1.95 cm
Question 19 (3 marks)
The graph shows information about part of a cyclist’s journey.
Work out an estimate of the speed, in m/s, of the cyclist at time 6 seconds.
Worked Solution
Step 1: Understand Gradient = Speed
💡 Concept: On a Distance-Time graph, the gradient (slope) represents the speed.
Since the line is curved, the speed is changing. To find the speed at a specific time (t=6), we must draw a tangent.
Step 2: Draw a Tangent at t = 6
💡 Action: Draw a straight line that touches the curve exactly at the point where \( t = 6 \). The line should follow the direction of the curve at that point.
✓ (M1) for drawing a suitable tangent at t = 6
Step 3: Calculate Gradient
Pick two points on your tangent line to calculate \( \frac{\text{Change in y}}{\text{Change in x}} \).
Example points from tangent:
- Point 1: \( (4, 4) \)
- Point 2: \( (8, 18) \)
Change in distance (y) = \( 18 – 4 = 14 \text{ m} \)
Change in time (x) = \( 8 – 4 = 4 \text{ s} \)
\[ \text{Speed} = \frac{14}{4} = 3.5 \text{ m/s} \](Acceptable range is typically 3.05 to 3.7)
✓ (M1) for method to find gradient (change in y / change in x)
✓ (A1) for answer in range 3.05 – 3.7
Final Answer:
3.5 m/s (estimate)
Question 20 (2 marks)
Here are the first five terms of a sequence.
-1 0 3 8 15
Find an expression, in terms of \( n \), for the nth term of this sequence.
Worked Solution
Step 1: Check Differences
💡 Strategy: Find the first and second differences.
Sequence: -1, 0, 3, 8, 15
1st Diff: +1, +3, +5, +7
2nd Diff: +2, +2, +2
Since the second difference is constant (2), it is a quadratic sequence (\( an^2 \)).
Coefficient of \( n^2 \) is half the second difference: \( \frac{2}{2} = 1 \).
So the sequence starts with \( n^2 \).
✓ (M1) for deducing \( n^2 \)
Step 2: Compare with \( n^2 \)
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sequence | -1 | 0 | 3 | 8 | 15 |
| \( n^2 \) | 1 | 4 | 9 | 16 | 25 |
| Difference | -2 | -4 | -6 | -8 | -10 |
Step 3: Find the Linear Part
The difference sequence is: -2, -4, -6, -8, -10
This is the \( -2n \) times table.
So the full expression is \( n^2 – 2n \).
✓ (A1)
Final Answer:
\( n^2 – 2n \)
Question 21 (2 marks)
When a biased coin is thrown 4 times, the probability of getting 4 heads is \( \frac{16}{81} \).
Work out the probability of getting 4 tails when the coin is thrown 4 times.
Worked Solution
Step 1: Find Probability of Heads
💡 Reasoning: Let \( P(H) = p \). Since the coin is thrown 4 times independently, the probability of 4 heads is \( p^4 \).
To find \( p \), take the 4th root:
\[ p = \sqrt[4]{\frac{16}{81}} \] \[ p = \frac{\sqrt[4]{16}}{\sqrt[4]{81}} \] \[ p = \frac{2}{3} \]So \( P(H) = \frac{2}{3} \).
✓ (M1) for finding probability of heads (2/3)
Step 2: Find Probability of Tails
💡 Complement Rule: \( P(T) = 1 – P(H) \).
Step 3: Calculate Probability of 4 Tails
We need \( P(T, T, T, T) = (\frac{1}{3})^4 \).
\[ (\frac{1}{3})^4 = \frac{1^4}{3^4} = \frac{1}{81} \]✓ (A1)
Final Answer:
\( \frac{1}{81} \)
Question 22 (4 marks)
Show that
\[ \frac{7x – 14}{x^2 + 4x – 12} \div \frac{x – 6}{x^3 – 36x} \]simplifies to \( ax \) where \( a \) is an integer.
Worked Solution
Step 1: Factorise Everything Possible
💡 Strategy: Factorise numerators and denominators first. Look for common factors, difference of two squares, and quadratics.
Top Left: \( 7x – 14 = 7(x – 2) \)
Bottom Left: \( x^2 + 4x – 12 = (x + 6)(x – 2) \)
Top Right: \( x – 6 \) (Cannot factorise)
Bottom Right: \( x^3 – 36x = x(x^2 – 36) = x(x + 6)(x – 6) \)
✓ (M1) for factorising numerator/denominator of 1st fraction
✓ (M1) for factorising denominator of 2nd fraction
Step 2: Rewrite the Division as Multiplication
💡 Rule: To divide by a fraction, multiply by its reciprocal (flip the second fraction).
✓ (M1) for multiplication by reciprocal
Step 3: Cancel Common Terms
Cancel terms present in both top and bottom:
- \( (x – 2) \) cancels with \( (x – 2) \)
- \( (x + 6) \) cancels with \( (x + 6) \)
- \( (x – 6) \) cancels with \( (x – 6) \)
Remaining terms:
\[ 7 \times x = 7x \]So \( a = 7 \).
✓ (A1) for 7x
Final Answer:
\( 7x \)
Question 23 (5 marks)
The diagram shows a sector \( OACB \) of a circle with centre \( O \).
The point \( C \) is the midpoint of the arc \( AB \).
The diagram also shows a hollow cone with vertex \( O \).
The cone is formed by joining \( OA \) and \( OB \).
Volume of cone \( = \frac{1}{3} \pi r^2 h \)
Curved surface area of cone \( = \pi r l \)
The cone has volume \( 56.8 \text{ cm}^3 \) and height \( 3.6 \text{ cm} \).
Calculate the size of angle \( AOB \) of sector \( OACB \).
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Find the Radius of the Cone Base
💡 Formula: \( V = \frac{1}{3} \pi r^2 h \)
We know \( V = 56.8 \) and \( h = 3.6 \).
Simplify:
\[ 56.8 = 1.2 \pi r^2 \] \[ r^2 = \frac{56.8}{1.2 \pi} \] \[ r^2 \approx 15.066… \] \[ r \approx 3.881… \text{ cm} \]✓ (P1) for finding base radius r
Step 2: Find the Slant Height (l)
💡 Reasoning: The slant height \( l \) of the cone is the radius of the original sector. Use Pythagoras theorem on the cone’s cross-section.
This \( l \) is the radius of the large sector.
✓ (P1) for finding slant height l
Step 3: Relate Arc Length to Circumference
💡 Concept: The circumference of the cone’s base is equal to the arc length of the sector.
Circumference of cone base = \( 2 \pi r \)
\[ C = 2 \times \pi \times 3.881… \]Arc Length of Sector = \( \frac{\theta}{360} \times 2 \pi L \)
(where L is the slant height/sector radius)
Set them equal:
\[ 2 \pi (3.881…) = \frac{\theta}{360} \times 2 \pi (5.294…) \]Divide both sides by \( 2 \pi \):
\[ 3.881… = \frac{\theta}{360} \times 5.294… \]✓ (P1) for setting up equation linking arc length and circumference
Step 4: Calculate Angle \( \theta \)
Round to 3 significant figures.
✓ (A1) for answer in range 263.9 – 264.1
Final Answer:
\( 264^\circ \)
Question 24 (5 marks)
\( OXYZ \) is a parallelogram.
\( \overrightarrow{OX} = \mathbf{a} \)
\( \overrightarrow{OY} = \mathbf{b} \)
\( P \) is the point on \( OX \) such that \( OP : PX = 1 : 2 \)
\( R \) is the point on \( OY \) such that \( OR : RY = 1 : 3 \)
Work out, in its simplest form, the ratio \( ZP : ZR \)
Worked Solution
Step 1: Define Basic Vectors
💡 Given:
- \( \overrightarrow{OX} = \mathbf{a} \)
- \( \overrightarrow{OY} = \mathbf{b} \)
Since \( OXYZ \) is a parallelogram, \( \overrightarrow{OZ} = \overrightarrow{XY} \).
However, we need \( \overrightarrow{OZ} \) first.
Vector path: \( \overrightarrow{OY} = \overrightarrow{OX} + \overrightarrow{XY} \)
\( \mathbf{b} = \mathbf{a} + \overrightarrow{XY} \)
\( \overrightarrow{XY} = \mathbf{b} – \mathbf{a} \)
Since \( \overrightarrow{OZ} \) is parallel and equal to \( \overrightarrow{XY} \), \( \overrightarrow{OZ} = \mathbf{b} – \mathbf{a} \).
Step 2: Find Vectors to Points P and R
Point P: \( P \) lies on \( OX \) in ratio 1:2.
\( \overrightarrow{OP} = \frac{1}{3} \overrightarrow{OX} = \frac{1}{3}\mathbf{a} \)
Point R: \( R \) lies on \( OY \) in ratio 1:3.
\( \overrightarrow{OR} = \frac{1}{4} \overrightarrow{OY} = \frac{1}{4}\mathbf{b} \)
✓ (P1) for finding OP or OR
Step 3: Find Vectors \( \overrightarrow{ZP} \) and \( \overrightarrow{ZR} \)
Vector \( \overrightarrow{ZP} \):
\( \overrightarrow{ZP} = \overrightarrow{ZO} + \overrightarrow{OP} \)
\( \overrightarrow{ZO} = -(\mathbf{b} – \mathbf{a}) = \mathbf{a} – \mathbf{b} \)
\( \overrightarrow{ZP} = (\mathbf{a} – \mathbf{b}) + \frac{1}{3}\mathbf{a} = \frac{4}{3}\mathbf{a} – \mathbf{b} \)
Vector \( \overrightarrow{ZR} \):
\( \overrightarrow{ZR} = \overrightarrow{ZO} + \overrightarrow{OR} \)
\( \overrightarrow{ZR} = (\mathbf{a} – \mathbf{b}) + \frac{1}{4}\mathbf{b} = \mathbf{a} – \frac{3}{4}\mathbf{b} \)
✓ (P1) for finding ZP or ZR
Step 4: Compare Vectors
Let’s factorise both vectors to see the relationship.
\( \overrightarrow{ZP} = \frac{4}{3}\mathbf{a} – \mathbf{b} = \frac{1}{3}(4\mathbf{a} – 3\mathbf{b}) \)
\( \overrightarrow{ZR} = \mathbf{a} – \frac{3}{4}\mathbf{b} = \frac{1}{4}(4\mathbf{a} – 3\mathbf{b}) \)
Both vectors are multiples of \( (4\mathbf{a} – 3\mathbf{b}) \), which means \( Z, P, R \) are collinear.
Ratio ZP : ZR
\( \frac{1}{3} : \frac{1}{4} \)
Multiply by 12 to remove fractions:
\( 4 : 3 \)
✓ (P1) for writing vectors as multiples
✓ (A1) for 4:3
Final Answer:
4 : 3