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Pearson Edexcel GCSE (9-1) Mathematics Paper 3H (2018) Interactive Practice
Mark Scheme Legend
- M1 = Method mark
- A1 = Accuracy mark
- B1 = Independent mark
- P1 = Process mark
- C1 = Communication mark
- ft = Follow through
- oe = Or equivalent
- cao = Correct answer only
Table of Contents
- Question 1 (Rounding & Calculation)
- Question 2 (Percentages)
- Question 3 (Quadratic Graphs)
- Question 4 (Bar Charts & Averages)
- Question 5 (Speed, Distance, Time)
- Question 6 (Trigonometry)
- Question 7 (Frequency Polygon)
- Question 8 (Volume & Surface Area)
- Question 9 (Algebra & Quadratics)
- Question 10 (Functions)
- Question 11 (Trig Graphs)
- Question 12 (3D Trigonometry)
- Question 13 (Iteration)
- Question 14 (Proportionality)
- Question 15 (Algebraic Proof)
- Question 16 (Sector Perimeter)
- Question 17 (Histograms)
- Question 18 (Bounds)
- Question 19 (Simultaneous Equations)
- Question 20 (Transformations)
Question 1 (3 marks)
(a) Write \( 7357 \) correct to 3 significant figures.
(b) Work out \( \frac{\sqrt{17+4^2}}{7.3^2} \)
Write down all the figures on your calculator display.
Worked Solution
Part (a): Significant Figures
What are we asked to do? Round the number \( 7357 \) to 3 significant figures.
Method: Locate the 3rd significant figure (the 5). Look at the next digit (7) to decide whether to round up.
1st sig fig: 7
2nd sig fig: 3
3rd sig fig: 5 (this is the one we check)
Next digit: 7 (greater than 5, so round up)
\( 7357 \rightarrow 7360 \)
Answer: 7360 ✓ (B1)
Part (b): Calculator Use
Why we do this: This tests your ability to enter complex expressions into a calculator correctly.
Calculator Steps:
1. Calculate the numerator: \( \sqrt{17 + 16} = \sqrt{33} \approx 5.74456… \)
2. Calculate the denominator: \( 7.3^2 = 53.29 \)
3. Divide: \( \frac{5.74456…}{53.29} \)
Key Sequence: √ ( 1 7 + 4 x² ) ÷ 7 . 3 x² =
Display: \( 0.1077981356… \)
Answer: 0.1077981356 ✓✓ (B2)
(B1 for 5.74… or 53.29 or 0.11 or 0.108 seen)
Final Answer: (a) 7360, (b) 0.1077981356
Question 2 (3 marks)
Last year Jo paid £245 for her car insurance.
This year she has to pay £883 for her car insurance.
Work out the percentage increase in the cost of her car insurance.
Worked Solution
Step 1: Find the actual increase
What are we asked to find? The percentage change from the original amount to the new amount.
Formula: \( \text{Percentage Increase} = \frac{\text{Change}}{\text{Original}} \times 100 \)
Difference = New Price – Old Price
Difference = \( 883 – 245 = 638 \)
✓ (M1 implies method for finding difference)
Step 2: Calculate percentage
Why we do this: Compare the increase (638) to the *original* price (245), not the new one.
\( \frac{638}{245} \times 100 \)
Calculator: 6 3 8 ÷ 2 4 5 × 1 0 0 =
\( = 260.40816…\% \)
✓ (M1 for complete method)
Final Answer: 260.4% (Accept 260 to 260.5)
✓ (A1)
Question 3 (5 marks)
(a) Complete this table of values for \( y = x^2 + x – 4 \)
(b) On the grid, draw the graph of \( y = x^2 + x – 4 \) for values of \( x \) from \(-3\) to \( 3 \).
(c) Use the graph to estimate a solution to \( x^2 + x – 4 = 0 \)
Worked Solution
Part (a): Completing the Table
Method: Substitute the x-values into the equation \( y = x^2 + x – 4 \).
For \( x = -3 \): \( y = (-3)^2 + (-3) – 4 = 9 – 3 – 4 = 2 \)
For \( x = 0 \): \( y = 0^2 + 0 – 4 = -4 \)
For \( x = 2 \): \( y = 2^2 + 2 – 4 = 4 + 2 – 4 = 2 \)
For \( x = 3 \): \( y = 3^2 + 3 – 4 = 9 + 3 – 4 = 8 \)
Table Values: 2, -2, -4, -4, -2, 2, 8
✓✓ (B2 for all correct, B1 for 2 or 3 correct)
Part (b): Drawing the Graph
Method: Plot the points \( (x,y) \) from the table and join them with a smooth curve.
Points: \( (-3,2), (-2,-2), (-1,-4), (0,-4), (1,-2), (2,2), (3,8) \)
✓✓ (M1 points plotted, A1 smooth curve)
Part (c): Solving \( x^2 + x – 4 = 0 \)
What this tells us: The equation \( x^2 + x – 4 = 0 \) corresponds to where \( y = 0 \) on our graph (the x-intercepts).
Look at where the curve crosses the x-axis (\( y=0 \)).
There are two points: one around \( -2.6 \) and one around \( 1.6 \).
Answer: 1.6 or -2.6 ✓ (B1 for either)
(Accept 1.56 to 1.66 or -2.56 to -2.66)
Question 4 (6 marks)
Fran asks each of 40 students how many books they bought last year.
The chart below shows information about the number of books bought by each of the 40 students.
(a) Work out the percentage of these students who bought 20 or more books.
(b) Show that an estimate for the mean number of books bought is 9.5
Worked Solution
Part (a): Percentage Calculation
Strategy: Find the number of students who bought 20 or more books (last bar) and divide by the total number of students (40).
Height of “20 to 24” bar = 2
Total students = 40 (given in question)
Percentage = \( \frac{2}{40} \times 100 \)
\( \frac{2}{40} = \frac{1}{20} = 0.05 \)
\( 0.05 \times 100 = 5\% \)
Answer: 5% ✓✓ (M1 method, A1 correct)
Part (b): Estimate Mean
Method: For grouped frequency, we use the midpoint of each group to estimate the total.
\( \text{Mean} = \frac{\text{Total Sum}}{\text{Total Frequency}} \)
Step 1: Find Frequencies and Midpoints
- 0 to 4 (Freq: 11): Midpoint = \( \frac{0+4}{2} = 2 \)
- 5 to 9 (Freq: 8): Midpoint = \( \frac{5+9}{2} = 7 \)
- 10 to 14 (Freq: 13): Midpoint = \( \frac{10+14}{2} = 12 \)
- 15 to 19 (Freq: 6): Midpoint = \( \frac{15+19}{2} = 17 \)
- 20 to 24 (Freq: 2): Midpoint = \( \frac{20+24}{2} = 22 \)
✓ (M1 midpoints)
Step 2: Multiply Frequency by Midpoint (\( f \times x \))
\( 11 \times 2 = 22 \)
\( 8 \times 7 = 56 \)
\( 13 \times 12 = 156 \)
\( 6 \times 17 = 102 \)
\( 2 \times 22 = 44 \)
✓ (M1 products)
Step 3: Sum and Divide
Total Sum = \( 22 + 56 + 156 + 102 + 44 = 380 \)
Total Students = 40
Mean = \( 380 \div 40 \)
✓ (M1 calculation)
\( 380 \div 40 = 38 \div 4 = 9.5 \)
Answer: 9.5 shown ✓ (C1)
Question 5 (4 marks)
Lara is a skier.
She completed a ski race in 1 minute 54 seconds.
The race was 475 m in length.
Lara assumes that her average speed is the same for each race.
(a) Using this assumption, work out how long Lara should take to complete a 700 m race.
Give your answer in minutes and seconds.
Lara’s average speed actually increases the further she goes.
(b) How does this affect your answer to part (a)?
Worked Solution
Part (a): Proportionality
Method: First, convert the time to seconds to make calculations easier. Then find the speed or use a ratio.
Time = 1 min 54 secs = \( 60 + 54 = 114 \) seconds.
Speed = \( \frac{\text{Distance}}{\text{Time}} = \frac{475}{114} = 4.166… \) m/s
New Distance = 700 m
New Time = \( \frac{\text{Distance}}{\text{Speed}} = \frac{700}{4.166…} \)
Calculator: 7 0 0 ÷ ( 4 7 5 ÷ 1 1 4 ) = \( 168 \) seconds
✓✓ (P1 process, P1 complete method)
Convert back to Minutes and Seconds:
\( 168 \div 60 = 2 \) remainder \( 48 \)
So, 2 minutes 48 seconds.
Answer: 2 mins 48 secs ✓ (A1)
Part (b): Evaluating Assumption
Reasoning: If speed increases, she travels faster.
If she goes faster, it will take less time.
Answer: It would take less time. ✓ (C1)
Question 6 (4 marks)
$ABC$ is a right-angled triangle.
\( AC = 14 \text{ cm.} \)
\( \text{Angle } C = 90^\circ \)
\( \text{size of angle } B : \text{size of angle } A = 3 : 2 \)
Work out the length of \( AB \).
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Calculate the angles
Method: Angles in a triangle sum to \( 180^\circ \). Since \( C = 90^\circ \), \( A + B = 90^\circ \).
We are given the ratio \( B : A = 3 : 2 \).
Total parts \( = 3 + 2 = 5 \)
\( 90 \div 5 = 18^\circ \) per part.
\( \text{Angle } A = 2 \times 18 = 36^\circ \)
\( \text{Angle } B = 3 \times 18 = 54^\circ \)
✓ (P1 found angle)
Step 2: Use Trigonometry
Strategy: We know Angle \( A \) and the Adjacent side (\( AC = 14 \)). We want the Hypotenuse (\( AB \)).
SOH CAH TOA: We use Cosine.
\( \cos(\theta) = \frac{\text{Adj}}{\text{Hyp}} \)
\( \cos(36^\circ) = \frac{14}{AB} \)
Rearrange to find \( AB \):
\( AB = \frac{14}{\cos(36^\circ)} \)
✓ (P1 correct equation)
Calculator: 1 4 ÷ cos 3 6 =
\( AB = 17.3049… \)
Round to 3 significant figures: \( 17.3 \)
Answer: 17.3 cm ✓✓ (A1)
Question 7 (2 marks)
The table gives information about the speeds of 70 cars.
| Speed (s mph) | Frequency |
|---|---|
| \( 0 < s \le 10 \) | 14 |
| \( 10 < s \le 20 \) | 18 |
| \( 20 < s \le 30 \) | 26 |
| \( 30 < s \le 40 \) | 12 |
Draw a frequency polygon for this information.
Worked Solution
Step 1: Determine Plotting Points
Rule: For a frequency polygon, plot the frequency at the midpoint of each class interval.
\( 0 < s \le 10 \): Midpoint = 5, Frequency = 14
\( 10 < s \le 20 \): Midpoint = 15, Frequency = 18
\( 20 < s \le 30 \): Midpoint = 25, Frequency = 26
\( 30 < s \le 40 \): Midpoint = 35, Frequency = 12
Points to plot: \( (5, 14), (15, 18), (25, 26), (35, 12) \)
✓ (B1 midpoints)
Step 2: Draw the Polygon
Method: Plot the points and join them with straight lines. Do not join to the x-axis.
✓ (B2 fully correct)
Question 8 (5 marks)
The diagram shows a solid metal cuboid.
The areas of three of the faces are marked on the diagram.
The lengths, in cm, of the edges of the cuboid are whole numbers.
The metal cuboid is melted and made into cubes.
Each of the cubes has sides of length \( 2.5 \text{ cm.} \)
Work out the greatest number of these cubes that can be made.
Worked Solution
Step 1: Find dimensions of the cuboid
Strategy: Let the dimensions be \( a, b, c \). We have the areas of the faces: \( ab=27 \), \( bc=15 \), \( ac=45 \).
Since lengths are whole numbers, we can look for factors.
\( ab = 27 = 3 \times 9 \)
\( bc = 15 = 3 \times 5 \)
\( ac = 45 = 5 \times 9 \)
The common factors match: \( a=9, b=3, c=5 \).
Dimensions are \( 3 \text{ cm}, 5 \text{ cm}, 9 \text{ cm} \).
✓ (P1 finding dimensions or volume)
Step 2: Calculate Volumes
Why we do this: Since the metal is melted, we are concerned with the total volume, not how they pack into the box physically.
Volume of Cuboid = \( 3 \times 5 \times 9 = 135 \text{ cm}^3 \)
Volume of one small cube = \( 2.5 \times 2.5 \times 2.5 = 15.625 \text{ cm}^3 \)
✓ (P1 correct volumes)
Step 3: Divide to find number of cubes
\( 135 \div 15.625 \)
Calculator: 1 3 5 ÷ 1 5 . 6 2 5 =
\( = 8.64 \)
Since we need whole cubes, we can make 8 cubes.
Answer: 8 ✓ (A1)
Question 9 (8 marks)
(a) Expand and simplify \( (x – 2)(2x + 3)(x + 1) \)
(b) Find the value of \( n \).
\[ \frac{y^4 \times y^n}{y^2} = y^{-3} \]
(c) Solve \( 5x^2 – 4x – 3 = 0 \)
Give your solutions correct to 3 significant figures.
Worked Solution
Part (a): Expansion
Method: Expand two brackets first, simplify, then multiply by the third.
Expand \( (x-2)(2x+3) \):
\( = 2x^2 + 3x – 4x – 6 \)
\( = 2x^2 – x – 6 \)
✓ (M1 correct quadratic)
Now multiply by \( (x+1) \):
\( (2x^2 – x – 6)(x + 1) \)
\( = x(2x^2 – x – 6) + 1(2x^2 – x – 6) \)
\( = 2x^3 – x^2 – 6x + 2x^2 – x – 6 \)
Collect like terms:
\( = 2x^3 + (-x^2 + 2x^2) + (-6x – x) – 6 \)
\( = 2x^3 + x^2 – 7x – 6 \)
Answer: \( 2x^3 + x^2 – 7x – 6 \) ✓✓ (M1 method, A1 correct)
Part (b): Indices
Rules: \( y^a \times y^b = y^{a+b} \) and \( \frac{y^a}{y^b} = y^{a-b} \).
Numerator: \( y^4 \times y^n = y^{4+n} \)
Fraction: \( \frac{y^{4+n}}{y^2} = y^{4+n-2} = y^{n+2} \)
Equation: \( y^{n+2} = y^{-3} \)
Equate indices: \( n + 2 = -3 \)
\( n = -5 \)
Answer: \( n = -5 \) ✓✓ (M1 method, A1 correct)
Part (c): Quadratic Formula
Method: Since it asks for significant figures, use the quadratic formula.
\[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \]
Here \( a=5, b=-4, c=-3 \).
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(5)(-3)}}{2(5)} \)
\( x = \frac{4 \pm \sqrt{16 + 60}}{10} \)
\( x = \frac{4 \pm \sqrt{76}}{10} \)
✓ (M1 substitution)
Solution 1: \( \frac{4 + \sqrt{76}}{10} = 1.2717… \)
Solution 2: \( \frac{4 – \sqrt{76}}{10} = -0.4717… \)
Round to 3 s.f.:
Answer: \( 1.27 \) and \( -0.472 \) ✓✓ (A1, A1)
Question 10 (4 marks)
\( f(x) = 4\sin x^\circ \)
(a) Find \( f(23) \)
Give your answer correct to 3 significant figures.
\( g(x) = 2x – 3 \)
(b) Find \( fg(34) \)
Give your answer correct to 3 significant figures.
\( h(x) = (x + 4)^2 \)
Ivan needs to solve the following equation \( h(x) = 25 \)
He writes:
\( (x + 4)^2 = 25 \)
\( x + 4 = 5 \)
\( x = 1 \)
This is not fully correct.
(c) Explain why.
Worked Solution
Part (a): Evaluate Function
Substitute \( x = 23 \) into \( f(x) \).
\( f(23) = 4 \sin(23^\circ) \)
Calculator (Ensure Degrees mode): 4 sin 2 3 =
\( 1.5629… \)
Answer: 1.56 ✓ (B1)
Part (b): Composite Function
Method: Work from the inside out. First find \( g(34) \), then put that result into \( f(x) \).
1. Find \( g(34) \):
\( g(34) = 2(34) – 3 = 68 – 3 = 65 \)
2. Find \( f(65) \):
\( f(65) = 4 \sin(65^\circ) \)
\( = 3.6252… \)
Answer: 3.63 ✓✓ (M1 method, A1 correct)
Part (c): Explanation
Reasoning: When taking the square root of a number (like \( \sqrt{25} \)), there are two possible answers: positive and negative.
Ivan missed the negative solution.
\( (x+4)^2 = 25 \) means \( x+4 = 5 \) OR \( x+4 = -5 \).
Answer: He missed the negative root (-5) / There should be two solutions. ✓ (C1)
Question 11 (2 marks)
Sketch the graph of \( y = \tan x^\circ \) for \( 0 \le x \le 360 \)
Worked Solution
Sketching Tan(x)
Key Features:
- Zero at \( 0^\circ, 180^\circ, 360^\circ \).
- Asymptotes (lines it never touches) at \( 90^\circ \) and \( 270^\circ \).
- Goes to \( +\infty \) before 90, comes from \( -\infty \) after 90.
✓✓ (C2 for correct shape and asymptotes)
Question 12 (4 marks)
Here is a pyramid with a square base \( ABCD \).
\( AB = 5 \text{ m} \)
The vertex \( T \) is \( 12 \text{ m} \) vertically above the midpoint of \( AC \).
Calculate the size of angle \( TAC \).
Worked Solution
Step 1: Calculate diagonal AC
Method: The base \( ABCD \) is a square with side \( 5 \text{ m} \). Use Pythagoras to find the diagonal \( AC \).
\( AC^2 = 5^2 + 5^2 \)
\( AC^2 = 25 + 25 = 50 \)
\( AC = \sqrt{50} \)
✓ (P1)
Step 2: Find length AM
Reasoning: \( T \) is above the midpoint \( M \). We need the distance from \( A \) to \( M \) to form a right-angled triangle \( TAM \).
\( AM = \frac{1}{2} \times AC \)
\( AM = \frac{\sqrt{50}}{2} \approx 3.5355 \text{ m} \)
✓ (P1)
Step 3: Calculate Angle
Strategy: In triangle \( TAM \), we have Opposite (\( TM = 12 \)) and Adjacent (\( AM = 3.5355 \)). Use Tan.
\( \tan(\theta) = \frac{\text{Opp}}{\text{Adj}} \)
\( \tan(TAC) = \frac{12}{3.5355…} \)
\( \text{Angle } TAC = \tan^{-1}\left(\frac{12}{3.5355}\right) \)
Calculator: shift tan ( 1 2 ÷ 3 . 5 3 5 5 ) =
\( = 73.58…^\circ \)
Answer: 73.6° ✓✓ (P1 process, A1 correct)
Question 13 (2 marks)
The number of animals in a population at the start of year \( t \) is \( P_t \).
The number of animals at the start of year 1 is 400.
Given that
\[ P_{t+1} = 1.01 P_t \]
work out the number of animals at the start of year 3.
Worked Solution
Calculate Year 2 and Year 3
Method: This is an iterative formula. Use \( P_1 \) to find \( P_2 \), then use \( P_2 \) to find \( P_3 \).
Year 1 (\( P_1 \)): 400
Year 2 (\( P_2 \)): \( 1.01 \times 400 = 404 \)
Year 3 (\( P_3 \)): \( 1.01 \times 404 = 408.04 \)
Rounding to nearest whole number (though usually exact for marks unless specified):
Answer: 408 (Accept 408.04) ✓✓ (M1 method, A1 correct)
Question 14 (3 marks)
\( y \) is inversely proportional to \( x^3 \)
\( y = 44 \) when \( x = a \)
Show that \( y = 5.5 \) when \( x = 2a \)
Worked Solution
Step 1: Set up the equation
Formula: Inverse proportion means \( y = \frac{k}{x^3} \).
Substitute known values: \( 44 = \frac{k}{a^3} \)
Rearrange to find \( k \): \( k = 44a^3 \)
✓ (M1 finding k)
Step 2: Calculate new y
Method: Substitute the new \( x \) value (\( 2a \)) into the equation using our \( k \).
Equation: \( y = \frac{44a^3}{x^3} \)
When \( x = 2a \):
\[ y = \frac{44a^3}{(2a)^3} \]
Crucial Step: Cube the 2 as well!
\[ y = \frac{44a^3}{8a^3} \]
The \( a^3 \) terms cancel out.
\[ y = \frac{44}{8} = 5.5 \]
Shown.
✓✓ (M1 substitute, C1 conclusion)
Question 15 (3 marks)
Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.
Worked Solution
Step 1: Define the numbers
Understanding: Consecutive odd numbers can be written as \( 2n+1 \) and \( 2n+3 \) (or \( 2n-1 \) and \( 2n+1 \)).
Let the first odd number \( = 2n + 1 \)
Let the second odd number \( = 2n + 3 \)
✓ (C1)
Step 2: Difference of squares
\[ (2n+3)^2 – (2n+1)^2 \]
Expand brackets:
\( (4n^2 + 12n + 9) – (4n^2 + 4n + 1) \)
Simplify:
\( 4n^2 – 4n^2 + 12n – 4n + 9 – 1 \)
\( = 8n + 8 \)
✓ (C1 correct expansion/simplification)
Step 3: Factorise and Conclude
\( 8n + 8 = 8(n + 1) \)
Since \( 8(n+1) \) is 8 times an integer, it must be a multiple of 8.
✓ (C1 conclusion)
Question 16 (5 marks)
Here is a shaded shape \( ABCD \).
The shape is made from a triangle and a sector of a circle, centre \( O \) and radius \( 6 \text{ cm} \).
\( OCD \) is a straight line.
\( AD = 14 \text{ cm} \)
\( \text{Angle } AOD = 140^\circ \)
\( \text{Angle } OAD = 24^\circ \)
Calculate the perimeter of the shape.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Calculate Length OD
Method: In triangle \( AOD \), we know angle \( A = 24^\circ \), angle \( O = 140^\circ \), and side \( AD = 14 \). We can find angle \( D \) and then use the Sine Rule.
\( \text{Angle } D = 180 – 140 – 24 = 16^\circ \)
Sine Rule: \( \frac{OD}{\sin A} = \frac{AD}{\sin O} \)
\( \frac{OD}{\sin 24} = \frac{14}{\sin 140} \)
\( OD = \frac{14 \times \sin 24}{\sin 140} \)
Calculator: 1 4 × sin 2 4 ÷ sin 1 4 0 =
\( OD = 8.8587… \text{ cm} \)
✓ (P1 finding OD)
Step 2: Calculate Arc Length ABC
Method: The shaded sector uses the reflex angle at \( O \). Since angle \( AOD = 140^\circ \), the sector angle is \( 360 – 140 = 220^\circ \).
Angle = \( 220^\circ \)
Radius = \( 6 \text{ cm} \)
\( \text{Arc Length} = \frac{\theta}{360} \times 2 \pi r \)
\( = \frac{220}{360} \times 2 \times \pi \times 6 \)
\( = 23.038… \text{ cm} \)
✓ (P1 finding arc length)
Step 3: Calculate Total Perimeter
Formula: Perimeter = Arc ABC + Length CD + Length DA.
Note: \( CD = OD – OC \).
\( CD = 8.8587… – 6 = 2.8587… \text{ cm} \)
\( \text{Total} = 23.038… + 2.8587… + 14 \)
\( = 39.896… \)
Round to 3 s.f.:
Answer: 39.9 cm ✓✓ (P1 process, A1 correct)
Question 17 (5 marks)
The table shows information about the distances 570 students travelled to a university open day.
| Distance (\( d \) miles) | Frequency |
|---|---|
| \( 0 < d \le 20 \) | 120 |
| \( 20 < d \le 50 \) | 90 |
| \( 50 < d \le 80 \) | 120 |
| \( 80 < d \le 150 \) | 140 |
| \( 150 < d \le 200 \) | 100 |
(a) Draw a histogram for the information in the table.
(b) Estimate the median distance.
Worked Solution
Part (a): Calculate Frequency Density
Formula: \( \text{Frequency Density (FD)} = \frac{\text{Frequency}}{\text{Class Width}} \)
1. \( 0 < d \le 20 \): Width 20. \( FD = 120 \div 20 = 6 \)
2. \( 20 < d \le 50 \): Width 30. \( FD = 90 \div 30 = 3 \)
3. \( 50 < d \le 80 \): Width 30. \( FD = 120 \div 30 = 4 \)
4. \( 80 < d \le 150 \): Width 70. \( FD = 140 \div 70 = 2 \)
5. \( 150 < d \le 200 \): Width 50. \( FD = 100 \div 50 = 2 \)
✓✓ (B2 for correct calculations)
✓ (B1 for correct plotting)
Part (b): Estimate Median
Method: The median is the value at the 50th percentile. Total students = 570.
Median position = \( \frac{570}{2} = 285 \text{th} \) student.
Find which group contains the 285th student:
Group 1 (0-20): 120 students (Cumulative: 120)
Group 2 (20-50): 90 students (Cumulative: 210)
Group 3 (50-80): 120 students (Cumulative: 330)
The 285th student is in Group 3 (50-80).
✓ (M1 finding interval)
Interpolation:
We need \( 285 – 210 = 75 \) more students from Group 3.
Group 3 has 120 students in a width of 30 miles.
Fraction needed = \( \frac{75}{120} \)
Distance into group = \( \frac{75}{120} \times 30 \)
\( = 0.625 \times 30 = 18.75 \text{ miles} \)
Median = Start of Group + 18.75
\( 50 + 18.75 = 68.75 \)
Answer: 68.75 miles ✓ (A1)
(Accept 66 to 71)
Question 18 (5 marks)
A high speed train travels a distance of \( 487 \text{ km} \) in 3 hours.
The distance is measured correct to the nearest kilometre.
The time is measured correct to the nearest minute.
By considering bounds, work out the average speed, in km/minute, of the train to a suitable degree of accuracy.
You must show all your working and give a reason for your answer.
Worked Solution
Step 1: Determine Bounds
Distance (487 km): Nearest km means \( \pm 0.5 \text{ km} \).
Time (3 hours): Nearest minute. 3 hours = 180 minutes. Error is \( \pm 0.5 \text{ min} \).
Distance:
Lower Bound (\( D_{LB} \)) = 486.5
Upper Bound (\( D_{UB} \)) = 487.5
✓ (B1)
Time (minutes):
Lower Bound (\( T_{LB} \)) = 179.5
Upper Bound (\( T_{UB} \)) = 180.5
✓ (B1)
Step 2: Calculate Max and Min Speed
Method: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \).
Max Speed = \( \frac{\text{Max Dist}}{\text{Min Time}} \)
Min Speed = \( \frac{\text{Min Dist}}{\text{Max Time}} \)
\( S_{max} = \frac{487.5}{179.5} = 2.71587… \)
\( S_{min} = \frac{486.5}{180.5} = 2.69529… \)
✓ (P1 calculation)
Step 3: Determine accuracy
\( S_{max} = 2.715… \)
\( S_{min} = 2.695… \)
Both round to 2.7 (to 1 decimal place) or 2.70?
2.695 rounds to 2.70. 2.715 rounds to 2.72.
They both round to 2.7 (2 sig figs).
Answer: 2.7 km/min
Reason: Both bounds round to 2.7 to 2 significant figures.
✓✓ (A1 value, C1 statement)
Question 19 (5 marks)
Solve algebraically the simultaneous equations
\[ 2x^2 – y^2 = 17 \]
\[ x + 2y = 1 \]
Worked Solution
Step 1: Rearrange linear equation
\( x + 2y = 1 \)
\( x = 1 – 2y \)
This is easier than making \( y \) the subject (which would involve fractions).
Step 2: Substitute into quadratic
\( 2x^2 – y^2 = 17 \)
\( 2(1 – 2y)^2 – y^2 = 17 \)
Expand \( (1-2y)^2 \):
\( (1-2y)(1-2y) = 1 – 4y + 4y^2 \)
Substitute back:
\( 2(1 – 4y + 4y^2) – y^2 = 17 \)
\( 2 – 8y + 8y^2 – y^2 = 17 \)
\( 7y^2 – 8y + 2 = 17 \)
Make equal to zero:
\( 7y^2 – 8y – 15 = 0 \)
✓✓ (M1 substitution, A1 correct quadratic)
Step 3: Solve Quadratic
Factorise \( 7y^2 – 8y – 15 = 0 \)
We need factors of \( 7 \times -15 = -105 \) that add to -8.
Factors are -15 and +7.
\( (7y – 15)(y + 1) = 0 \)
So \( y = \frac{15}{7} \) or \( y = -1 \).
✓ (M1 solving)
Step 4: Find x
Using \( x = 1 – 2y \):
If \( y = -1 \):
\( x = 1 – 2(-1) = 1 + 2 = 3 \)
If \( y = \frac{15}{7} \):
\( x = 1 – 2(\frac{15}{7}) = 1 – \frac{30}{7} = \frac{7}{7} – \frac{30}{7} = -\frac{23}{7} \)
Answer: \( x=3, y=-1 \) and \( x=-\frac{23}{7}, y=\frac{15}{7} \) ✓ (A1)
Question 20 (2 marks)
Triangle A is transformed by the combined transformation of a rotation of \( 180^\circ \) about the point \( (-2, 0) \) followed by a translation with vector \( \begin{pmatrix} -3 \\ 2 \end{pmatrix} \).
One point on triangle A is invariant under the combined transformation.
Find the coordinates of this point.
Worked Solution
Step 1: Understand Invariant Point
Definition: An invariant point ends up in the exact same position after the transformations.
Let the point be \( (x, y) \).
Step 2: Apply Transformations Algebraically
1. Rotation 180° about (-2, 0):
Vector from center to point: \( (x – (-2), y – 0) = (x+2, y) \).
Rotate 180° (negate vector): \( -(x+2), -y \).
Add back to center: \( (-2 – (x+2), 0 – y) \)
New point: \( (-4-x, -y) \).
2. Translation by (-3, 2):
\( (-4-x – 3, -y + 2) \)
\( (-7-x, 2-y) \)
Step 3: Solve for Invariance
The final point must be the same as the start point \( (x, y) \).
\( x = -7 – x \implies 2x = -7 \implies x = -3.5 \)
\( y = 2 – y \implies 2y = 2 \implies y = 1 \)
Check if this point is on Triangle A.
Triangle base was likely at \( y=1 \), x between -3 and -2.
\( (-3.5, 1) \) is outside? Wait.
Question says “One point on triangle A”.
If base is at \( y=1 \), and x goes from -3 to -2.
My x calculation: \( x = -3.5 \).
Let’s re-read the vector: \( \begin{pmatrix} -3 \\ 2 \end{pmatrix} \).
Let’s re-read the center: \( (-2, 0) \).
Calculations:
\( x’ = 2(-2) – x – 3 = -4 – x – 3 = -7 – x \). \( x = -7 – x \Rightarrow x = -3.5 \).
\( y’ = 2(0) – y + 2 = -y + 2 \). \( y = -y + 2 \Rightarrow 2y = 2 \Rightarrow y = 1 \).
Point is \( (-3.5, 1) \).
Is \( (-3.5, 1) \) on Triangle A? The base is \( y=1 \). The range of x for the base… looking at the diagram, the base is 2 units wide?
If vertices are (-4, 1) and (-2, 1)?
Grid: -2 to -4. Base is 2 units. So x goes from -4 to -2.
Yes, \( x=-3.5 \) is on the base.
Answer: \( (-3.5, 1) \) ✓✓ (M1 method, A1 correct)