Edexcel – Higher – November 2018 – Paper 3

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

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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Speed (s mph)	Frequency
\( 0 < s \le 10 \)	14
\( 10 < s \le 20 \)	18
\( 20 < s \le 30 \)	26
\( 30 < s \le 40 \)	12

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

Pearson Edexcel GCSE (9-1) Mathematics Paper 3H (2018) Interactive Practice

Mark Scheme Legend

Table of Contents

Question 1 (3 marks)

Worked Solution

Part (a): Significant Figures

Part (b): Calculator Use

Question 2 (3 marks)

Worked Solution

Step 1: Find the actual increase

Step 2: Calculate percentage

Question 3 (5 marks)

Worked Solution

Part (a): Completing the Table

Part (b): Drawing the Graph

Part (c): Solving \( x^2 + x – 4 = 0 \)

Question 4 (6 marks)

Worked Solution

Part (a): Percentage Calculation

Part (b): Estimate Mean

Question 5 (4 marks)

Worked Solution

Part (a): Proportionality

Part (b): Evaluating Assumption

Question 6 (4 marks)

Worked Solution

Step 1: Calculate the angles

Step 2: Use Trigonometry

Question 7 (2 marks)

Worked Solution

Step 1: Determine Plotting Points

Step 2: Draw the Polygon

Question 8 (5 marks)

Worked Solution

Step 1: Find dimensions of the cuboid

Step 2: Calculate Volumes

Step 3: Divide to find number of cubes

Question 9 (8 marks)

Worked Solution

Part (a): Expansion

Part (b): Indices

Part (c): Quadratic Formula

Question 10 (4 marks)

Worked Solution

Part (a): Evaluate Function

Part (b): Composite Function

Part (c): Explanation

Question 11 (2 marks)

Worked Solution

Sketching Tan(x)

Question 12 (4 marks)

Worked Solution

Step 1: Calculate diagonal AC

Step 2: Find length AM

Step 3: Calculate Angle

Question 13 (2 marks)

Worked Solution

Calculate Year 2 and Year 3

Question 14 (3 marks)

Worked Solution

Step 1: Set up the equation

Step 2: Calculate new y

Question 15 (3 marks)

Worked Solution

Step 1: Define the numbers

Step 2: Difference of squares

Step 3: Factorise and Conclude

Question 16 (5 marks)

Worked Solution

Step 1: Calculate Length OD

Step 2: Calculate Arc Length ABC

Step 3: Calculate Total Perimeter

Question 17 (5 marks)

Worked Solution

Part (a): Calculate Frequency Density

Part (b): Estimate Median

Question 18 (5 marks)

Worked Solution

Step 1: Determine Bounds

Step 2: Calculate Max and Min Speed