GCSE Edexcel Higher Paper 3 (Nov 2018) Interactive Practice
Mark Scheme Legend
- M1 = Method mark (for a correct method or partial method)
- P1 = Process mark (for a correct process as part of a problem solving question)
- A1 = Accuracy mark (for a correct answer following a correct method)
- B1 = Unconditional accuracy mark (no method needed)
- C1 = Communication mark (fully correct statement with no contradiction)
Table of Contents
- Question 1 (Rounding & Calculation)
- Question 2 (Percentage Increase)
- Question 3 (Quadratic Graphs)
- Question 4 (Bar Charts & Percentages)
- Question 5 (Speed, Distance, Time)
- Question 6 (Right-angled Triangles)
- Question 7 (Frequency Polygon)
- Question 8 (Volume & Surface Area)
- Question 9 (Algebra: Expansion, Indices, Quadratics)
- Question 10 (Functions)
- Question 11 (Trigonometric Graphs)
- Question 12 (3D Trigonometry)
- Question 13 (Population Growth)
- Question 14 (Inverse Proportion)
- Question 15 (Algebraic Proof)
- Question 16 (Sector Area & 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
Step 1: Identify the 3rd significant figure.
The first significant figure is 7 (thousands). The second is 3 (hundreds). The third is 5 (tens).
We look at the next digit (the units column) to decide whether to round up.
Number: $735\underline{7}$
The digit after 5 is 7. Since 7 is 5 or greater, we round up the previous digit.
$5$ becomes $6$.
Replace digits after the rounding point with zeros to keep the place value.
$7357 \rightarrow 7360$
Answer: 7360 ✓ (B1)
Part (b): Calculator Use
Step 1: Calculate the numerator and denominator separately or use fraction button.
We need to be careful with the order of operations.
Numerator: $\sqrt{17 + 4^2} = \sqrt{17 + 16} = \sqrt{33} \approx 5.74456…$
Denominator: $7.3^2 = 53.29$
Calculation: $\frac{\sqrt{33}}{53.29}$
Calculator display: $0.1077981356…$
Note: The mark scheme requires at least the first 4 decimal places correct (rounded or truncated) to award the marks if the full display isn’t written.
Answer: 0.1077981356 ✓ (B2)
(Award B1 for 5.74… or 53.29 or 0.107 or 0.108 seen)
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 in cost
What are we being asked to find? The percentage increase, which is the change relative to the original amount.
$\text{Increase} = \text{New Price} – \text{Original Price}$
$\text{Increase} = 883 – 245 = 638$
Step 2: Calculate percentage increase
Method: Divide the increase by the original amount, then multiply by 100.
$\text{Percentage Increase} = \frac{\text{Increase}}{\text{Original Price}} \times 100$
$= \frac{638}{245} \times 100$
$= 2.60408… \times 100$
$= 260.408…\%$
Answer: 260.4% (accept 260 to 260.5) ✓ (A1)
(M1 for $883-245$, M1 for method to find percentage)
Question 3 (5 marks)
(a) Complete this table of values for $y = x^2 + x – 4$.
| $x$ | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| $y$ | -2 | -4 | -2 |
(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
Substitute $x$ values into $y = x^2 + x – 4$:
If $x = -3$: $y = (-3)^2 + (-3) – 4 = 9 – 3 – 4 = 2$
If $x = 0$: $y = (0)^2 + (0) – 4 = -4$
If $x = 2$: $y = (2)^2 + (2) – 4 = 4 + 2 – 4 = 2$
If $x = 3$: $y = (3)^2 + (3) – 4 = 9 + 3 – 4 = 8$
Values: 2, -4, 2, 8
Table Values: $y$: 2, -2, -4, -4, -2, 2, 8 ✓ (B2)
Part (b): Drawing the Graph
Graph: Smooth curve drawn through the points. ✓ (A1)
Part (c): Estimating Solutions
Method: The equation $x^2 + x – 4 = 0$ corresponds to where the graph $y = x^2 + x – 4$ crosses the line $y=0$ (the x-axis).
Read the x-coordinates where the curve intersects the x-axis.
The curve crosses near $x = -2.6$ and $x = 1.6$.
Answer: -2.6 or 1.6 ✓ (B1)
(Accept answers in range -2.6 to -2.5 and 1.5 to 1.6)
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. You must show all your working.
Worked Solution
Part (a): Percentage Calculation
Step 1: Read the chart.
Look at the bar for “20 to 24 books”. The height is 2.
Total students is given as 40.
Students buying 20+ books = 2
Percentage = $\frac{2}{40} \times 100$
$= \frac{1}{20} \times 100 = 5\%$
Answer: 5% ✓ (A1)
Part (b): Estimating the Mean
Step 1: Identify frequencies and midpoints.
Since the data is grouped, we use the midpoint of each group to represent the number of books.
| Interval | Frequency ($f$) | Midpoint ($x$) | $f \times x$ |
|---|---|---|---|
| 0 to 4 | 11 | $\frac{0+4}{2} = 2$ | $11 \times 2 = 22$ |
| 5 to 9 | 8 | $\frac{5+9}{2} = 7$ | $8 \times 7 = 56$ |
| 10 to 14 | 13 | $\frac{10+14}{2} = 12$ | $13 \times 12 = 156$ |
| 15 to 19 | 6 | $\frac{15+19}{2} = 17$ | $6 \times 17 = 102$ |
| 20 to 24 | 2 | $\frac{20+24}{2} = 22$ | $2 \times 22 = 44$ |
| Total | 40 | 380 |
Step 2: Calculate Mean.
$\text{Mean} = \frac{\text{Total } fx}{\text{Total Frequency}}$
$\text{Mean} = \frac{380}{40}$
$= 9.5$
Conclusion: $380 \div 40 = 9.5$. Shown. ✓ (C1)
(M1 for midpoints, M1 for $f \times x$, M1 for division by 40)
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.
(b) Lara’s average speed actually increases the further she goes.
How does this affect your answer to part (a)?
Worked Solution
Part (a): Rate Calculation
Step 1: Convert time to seconds to make calculation easier.
1 minute 54 seconds = $60 + 54 = 114$ seconds.
Step 2: Calculate Speed or Use Proportion.
We can find the time per meter or the speed.
Method 1 (Speed):
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{475}{114} \approx 4.166…$ m/s
New Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{700}{4.166…}$
Method 2 (Proportion – Easier):
$\frac{700}{475}$ is the ratio of distances.
Time = $114 \times \frac{700}{475} = 168$ seconds.
Step 3: Convert back to minutes and seconds.
$168 \div 60 = 2$ with remainder $48$.
Or $168 – 120 = 48$.
Time = 2 minutes 48 seconds.
Answer: 2 mins 48 secs ✓ (A1)
(P1 for converting time or finding speed, P1 for full process)
Part (b): Effect of Assumption
Reasoning: If speed increases, she travels faster.
Faster speed means less time taken.
Answer: It would take less time. ✓ (C1)
Question 6 (4 marks)
$ABC$ is a right-angled triangle.
$AC = 14$ cm.
Angle $C = 90^\circ$
size of angle $B$ : size of angle $A = 3 : 2$
Work out the length of $AB$.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Find the angles
Angles in a triangle sum to $180^\circ$.
Since $C = 90^\circ$, $A + B = 90^\circ$.
The ratio $B:A$ is $3:2$.
Total parts = $3 + 2 = 5$ parts.
Value of one part = $90^\circ \div 5 = 18^\circ$.
Angle $B = 3 \times 18 = 54^\circ$.
Angle $A = 2 \times 18 = 36^\circ$.
Step 2: Choose Trigonometric Ratio
We know Angle $A = 36^\circ$ and the Adjacent side $AC = 14$.
We want to find the hypotenuse $AB$.
SOH CAH TOA: We have Adjacent and want Hypotenuse $\rightarrow$ Cos.
$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
$\cos(36^\circ) = \frac{14}{AB}$
Step 3: Solve and Round
Rearrange to find $AB$:
$AB = \frac{14}{\cos(36^\circ)}$
Calculator: $AB = 17.3049…$
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 |
|---|---|
| $0 < s \leq 10$ | 14 |
| $10 < s \leq 20$ | 18 |
| $20 < s \leq 30$ | 26 |
| $30 < s \leq 40$ | 12 |
Draw a frequency polygon for this information.
Worked Solution
Step 1: Identify Plotting Points
For a frequency polygon, we plot the frequency against the midpoint of the class interval.
0 to 10: Midpoint = 5. Point: $(5, 14)$
10 to 20: Midpoint = 15. Point: $(15, 18)$
20 to 30: Midpoint = 25. Point: $(25, 26)$
30 to 40: Midpoint = 35. Point: $(35, 12)$
Step 2: Draw the Polygon
Graph: Points plotted at midpoints and joined with straight lines. ✓ (B2)
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 cm.
Work out the greatest number of these cubes that can be made.
Worked Solution
Step 1: Find dimensions of the cuboid
Let lengths be $x, y, z$. The areas are $xy=45$, $xz=27$, $yz=15$.
We need whole numbers that multiply to give these areas.
Look at factors:
$45 = 5 \times 9$
$27 = 3 \times 9$
$15 = 3 \times 5$
This works perfectly!
Dimensions are: 9 cm, 5 cm, 3 cm.
Step 2: Calculate Volume of Cuboid
$\text{Volume} = 9 \times 5 \times 3 = 135 \text{ cm}^3$
Alternatively: $\sqrt{45 \times 27 \times 15} = \sqrt{18225} = 135$
Step 3: Calculate Volume of Small Cube
Side length = 2.5 cm
$\text{Volume} = 2.5^3 = 15.625 \text{ cm}^3$
Step 4: Find Number of Cubes
Divide total volume by volume of one small cube.
$\frac{135}{15.625} = 8.64$
Since we need whole cubes, we truncate to 8.
Answer: 8 ✓ (A1)
Question 9 (8 marks)
(a) Expand and simplify $(x – 2)(2x + 3)(x + 1)$
$\frac{y^4 \times y^n}{y^2} = y^{-3}$
(b) Find the value of $n$.
(c) Solve $5x^2 – 4x – 3 = 0$
Give your solutions correct to 3 significant figures.
Worked Solution
Part (a): Expansion
Multiply two brackets first, then multiply the result by the third.
$(x-2)(2x+3) = 2x^2 + 3x – 4x – 6 = 2x^2 – x – 6$
Now multiply by $(x+1)$:
$(2x^2 – x – 6)(x + 1)$
$= 2x^3 – x^2 – 6x$ (multiplying by $x$)
$+ 2x^2 – x – 6$ (multiplying by $1$)
Combine like terms:
$2x^3 + (-x^2 + 2x^2) + (-6x – x) – 6$
$= 2x^3 + x^2 – 7x – 6$
Answer: $2x^3 + x^2 – 7x – 6$ ✓ (A1)
Part (b): Indices
Use index laws: $y^a \times y^b = y^{a+b}$ and $\frac{y^a}{y^b} = y^{a-b}$.
Numerator: $y^4 \times y^n = y^{4+n}$
Expression: $\frac{y^{4+n}}{y^2} = y^{4+n-2} = y^{n+2}$
We are told this equals $y^{-3}$.
So, $n + 2 = -3$
$n = -5$
Answer: $n = -5$ ✓ (A1)
Part (c): Quadratic Formula
For $ax^2 + bx + c = 0$, $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
$a=5, b=-4, c=-3$
$x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(5)(-3)}}{2(5)}$
$x = \frac{4 \pm \sqrt{16 – (-60)}}{10}$
$x = \frac{4 \pm \sqrt{76}}{10}$
Solution 1: $\frac{4 + \sqrt{76}}{10} \approx 1.2717…$
Solution 2: $\frac{4 – \sqrt{76}}{10} \approx -0.4717…$
Answer: 1.27 and -0.472 ✓ (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:
$x + 4 = 5$
$x = 1$
This is not fully correct.
(c) Explain why.
Worked Solution
Part (a): Evaluate Function
$f(23) = 4\sin(23)$
Calculator: $1.5629…$
Round to 3 s.f.: $1.56$
Answer: 1.56 ✓ (B1)
Part (b): Composite Function
$fg(34)$ means calculate $g(34)$ first, then put the result into $f$.
$g(34) = 2(34) – 3 = 68 – 3 = 65$
Now find $f(65)$:
$f(65) = 4\sin(65)$
Calculator: $3.6252…$
Answer: 3.63 ✓ (A1)
Part (c): Explanation
When taking the square root of 25, there are two answers: $+5$ and $-5$.
Answer: He missed the negative solution ($x+4=-5$). ✓ (C1)
Question 11 (2 marks)
Sketch the graph of $y = \tan x^\circ$ for $0 \leq x \leq 360$.
Worked Solution
Step 1: Identify Key Features of Tan Graph
The graph of $y = \tan x$ has specific properties:
- It starts at 0 ($ \tan 0 = 0 $).
- It is undefined (asymptote) at $90^\circ$ and $270^\circ$.
- It crosses the x-axis at $180^\circ$ and $360^\circ$.
- The shape is a repeating ‘S’ curve that goes towards $+\infty$ before an asymptote and comes from $-\infty$ after it.
Step 2: Draw the Sketch
Result: Graph showing asymptotes at 90 and 270, crossing x-axis at 0, 180, 360. ✓ (C2)
Question 12 (4 marks)
Here is a pyramid with a square base $ABCD$.
$AB = 5$ m.
The vertex $T$ is 12 m vertically above the midpoint of $AC$.
Calculate the size of angle $TAC$.
Worked Solution
Step 1: Understand the Geometry
We are looking for angle $TAC$.
Let $M$ be the midpoint of $AC$. The vertex $T$ is vertically above $M$.
This creates a right-angled triangle $TMA$ where:
- The angle at $M$ is $90^\circ$ (since $TM$ is vertical and $AM$ is horizontal).
- The height $TM = 12$ m.
- We need to find the length $AM$ to use trigonometry.
Step 2: Calculate Diagonal AC
The base $ABCD$ is a square with side 5 m.
Using Pythagoras in triangle $ABC$:
$AC^2 = 5^2 + 5^2$
$AC^2 = 25 + 25 = 50$
$AC = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
$AC \approx 7.071…$ m
Step 3: Calculate length AM
$M$ is the midpoint of $AC$, so $AM$ is half of $AC$.
$AM = \frac{5\sqrt{2}}{2} = 2.5\sqrt{2}$
$AM \approx 3.5355…$ m
Step 4: Calculate Angle TAC
In $\triangle TMA$, we have:
- Opposite ($TM$) = 12
- Adjacent ($AM$) = $3.5355…$
- We want angle $A$ (which is angle $TAC$).
Use TOA ($\tan$):
$\tan(A) = \frac{\text{Opp}}{\text{Adj}} = \frac{12}{3.5355…}$
$A = \tan^{-1}\left(\frac{12}{3.5355…}\right)$
$A = 73.583…^\circ$
Answer: 73.6$^\circ$ (1 d.p.) ✓ (A1)
(Accept 73.58 to 74.1)
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
Step 1: Understand the Iterative Formula
The formula tells us that to get the next year’s population ($P_{t+1}$), we multiply the current year’s population ($P_t$) by 1.01.
This represents a 1% increase each year.
Step 2: Calculate Year 2 and Year 3
Start of Year 1 ($P_1$): 400
Start of Year 2 ($P_2$):
$P_2 = 1.01 \times 400 = 404$
Start of Year 3 ($P_3$):
$P_3 = 1.01 \times 404 = 408.04$
Answer: 408 ✓ (A1)
(Mark scheme accepts 408 or 408.04. For population, integer is appropriate.)
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 Proportionality Equation
“Inverse proportion to $x^3$” means $y = \frac{k}{x^3}$ where $k$ is a constant.
$y = \frac{k}{x^3}$
Substitute the known values: $y=44$ when $x=a$.
$44 = \frac{k}{a^3}$
So, $k = 44a^3$.
Step 2: Calculate new y value
We want to find $y$ when $x = 2a$.
Substitute $k = 44a^3$ and $x = 2a$ into the general formula:
$y = \frac{44a^3}{(2a)^3}$
Expand $(2a)^3$:
$(2a)^3 = 2^3 \times a^3 = 8a^3$
So, $y = \frac{44a^3}{8a^3}$
Step 3: Simplify
The $a^3$ terms cancel out:
$y = \frac{44}{8}$
$y = 5.5$
Conclusion: Steps clearly show $y = 5.5$. ✓ (C1)
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 Odd Numbers
We need algebra to represent “any” consecutive odd numbers.
If $2n$ is an even number, then $2n+1$ is an odd number.
The next consecutive odd number would be $(2n+1) + 2 = 2n+3$.
(Alternatively, use $2n-1$ and $2n+1$).
Let the two consecutive odd numbers be $2n+1$ and $2n+3$.
Step 2: Find the Difference of their Squares
Difference = $(2n+3)^2 – (2n+1)^2$
Expand brackets:
$(2n+3)(2n+3) = 4n^2 + 6n + 6n + 9 = 4n^2 + 12n + 9$
$(2n+1)(2n+1) = 4n^2 + 2n + 2n + 1 = 4n^2 + 4n + 1$
Subtract:
$(4n^2 + 12n + 9) – (4n^2 + 4n + 1)$
$= 4n^2 – 4n^2 + 12n – 4n + 9 – 1$
$= 8n + 8$
Step 3: Prove it is a Multiple of 8
$8n + 8 = 8(n + 1)$
Since $n$ is an integer, $(n+1)$ is an integer.
$8 \times \text{integer}$ is always a multiple of 8.
Proof Complete ✓ (C1)
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 cm.
$OCD$ is a straight line.
$AD = 14$ cm
Angle $AOD = 140^\circ$
Angle $OAD = 24^\circ$
Calculate the perimeter of the shape.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Identify Perimeter Components
The perimeter consists of:
- The straight line $AD$ (Given: 14 cm)
- The straight line $CD$ (Need to find)
- The major arc $ABC$ (Need to find)
Step 2: Calculate lengths in Triangle AOD
We know Angle $AOD = 140^\circ$ and Angle $OAD = 24^\circ$.
We can find the third angle $ADO$:
$180 – 140 – 24 = 16^\circ$
Now use the Sine Rule to find $OD$ and verify $OA$.
$\frac{OD}{\sin(24)} = \frac{14}{\sin(140)}$
$OD = \frac{14 \times \sin(24)}{\sin(140)} = \frac{14 \times 0.4067}{0.6427} \approx 8.858…$ cm
Check OA (Radius):
$\frac{OA}{\sin(16)} = \frac{14}{\sin(140)}$
$OA = \frac{14 \times \sin(16)}{\sin(140)} \approx 6.00…$ cm
This matches the given radius of 6 cm.
Step 3: Calculate Length CD
$OCD$ is a straight line.
$CD = OD – OC$
Since $C$ is on the circle, $OC = \text{radius} = 6$ cm.
$CD = 8.858… – 6 = 2.858…$ cm
Step 4: Calculate Arc Length ABC
The sector is the major sector (reflex angle).
Angle of sector = $360^\circ – 140^\circ = 220^\circ$.
$\text{Arc Length} = \frac{\theta}{360} \times 2\pi r$
$= \frac{220}{360} \times 2 \times \pi \times 6$
$= \frac{22}{36} \times 12\pi = \frac{11}{18} \times 12\pi = \frac{22\pi}{3}$
$\approx 23.038…$ cm
Step 5: Total Perimeter
$\text{Perimeter} = AD + CD + \text{Arc ABC}$
$= 14 + 2.858… + 23.038…$
$= 39.896…$
Round to 3 significant figures: 39.9 cm
Answer: 39.9 cm ✓ (A1)
(Accept 39.8 to 40)
Question 17 (5 marks)
The table shows information about the distances 570 students travelled to a university open day.
| Distance ($d$ miles) | Frequency |
|---|---|
| $0 < d \leq 20$ | 120 |
| $20 < d \leq 50$ | 90 |
| $50 < d \leq 80$ | 120 |
| $80 < d \leq 150$ | 140 |
| $150 < d \leq 200$ | 100 |
(a) Draw a histogram for the information in the table.
(b) Estimate the median distance.
Worked Solution
Part (a): Histogram Calculation
For a histogram, Frequency Density = Frequency ÷ Class Width.
| Interval | Width | Freq | FD |
|---|---|---|---|
| 0-20 | 20 | 120 | $120 \div 20 = 6$ |
| 20-50 | 30 | 90 | $90 \div 30 = 3$ |
| 50-80 | 30 | 120 | $120 \div 30 = 4$ |
| 80-150 | 70 | 140 | $140 \div 70 = 2$ |
| 150-200 | 50 | 100 | $100 \div 50 = 2$ |
Part (a): Drawing
Part (b): Estimating Median
Total frequency = 570.
The median is the $\frac{570}{2} = 285$th value.
Cumulative Frequencies:
0-20: 120
0-50: $120 + 90 = 210$
0-80: $210 + 120 = 330$
The 285th value lies in the 50 – 80 interval.
We need $(285 – 210) = 75$ more students from this group.
The group has 120 students and width 30.
Fraction into group = $\frac{75}{120}$
Distance into group = $\frac{75}{120} \times 30 = \frac{5}{8} \times 30 = 18.75$
Median = $50 + 18.75 = 68.75$
Answer: 68.75 miles ✓ (A1)
(Accept 66 to 71)
Question 18 (5 marks)
A high speed train travels a distance of 487 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 $\rightarrow \pm 0.5$ km.
LB = 486.5 km, UB = 487.5 km
Time (3 hours): Nearest minute.
3 hours = 180 minutes.
Nearest minute $\rightarrow \pm 0.5$ minutes.
LB = 179.5 min, UB = 180.5 min
Step 2: Calculate Speed Bounds
Speed = $\frac{\text{Distance}}{\text{Time}}$.
Lower Bound Speed = $\frac{\text{Lower Distance}}{\text{Upper Time}}$
Upper Bound Speed = $\frac{\text{Upper Distance}}{\text{Lower Time}}$
$\text{LB Speed} = \frac{486.5}{180.5} \approx 2.69529…$ km/min
$\text{UB Speed} = \frac{487.5}{179.5} \approx 2.71587…$ km/min
Step 3: Determine Suitable Degree of Accuracy
Compare the bounds:
LB: 2.695…
UB: 2.715…
To 1 decimal place: LB = 2.7, UB = 2.7. Match!
To 2 decimal places: LB = 2.70, UB = 2.72. Don’t match.
Answer: 2.7 km/minute ✓ (C1)
Reason: Both bounds round to 2.7 to 1 decimal place.
Question 19 (5 marks)
Solve algebraically the simultaneous equations
$2x^2 – y^2 = 17$
$x + 2y = 1$
Worked Solution
Step 1: Rearrange Linear Equation
Express $x$ in terms of $y$ because $x$ has no coefficient.
$x = 1 – 2y$
Step 2: Substitute into Quadratic
$2(1 – 2y)^2 – y^2 = 17$
Expand $(1-2y)^2 = 1 – 4y + 4y^2$
$2(1 – 4y + 4y^2) – y^2 = 17$
$2 – 8y + 8y^2 – y^2 = 17$
$7y^2 – 8y + 2 = 17$
$7y^2 – 8y – 15 = 0$
Step 3: Solve Quadratic
Factorise $(7y \dots)(y \dots) = 0$
We need factors of $-105$ ($7 \times -15$) that add to $-8$.
$-15$ and $7$.
$(7y – 15)(y + 1) = 0$
So, $y = \frac{15}{7}$ or $y = -1$.
Step 4: Find x values
When $y = -1$:
$x = 1 – 2(-1) = 1 + 2 = 3$
When $y = \frac{15}{7}$:
$x = 1 – 2(\frac{15}{7}) = 1 – \frac{30}{7} = \frac{7}{7} – \frac{30}{7} = -\frac{23}{7}$
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: Test Vertices
Transformations:
1. Rotation $180^\circ$ about $(-2,0)$. Formula: New $x = 2(-2) – x$, New $y = 2(0) – y$. i.e., $(-4-x, -y)$.
2. Translation $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$. Add $-3$ to $x$, $2$ to $y$.
Combined: $x’ = -4 – x – 3 = -7 – x$. $y’ = -y + 2$.
For invariance, $x’ = x$ and $y’ = y$.
Set $x = -7 – x \Rightarrow 2x = -7 \Rightarrow x = -3.5$
Set $y = -y + 2 \Rightarrow 2y = 2 \Rightarrow y = 1$
Step 2: Verify Point is on Triangle
The calculated point is $(-3.5, 1)$.
Is this point on Triangle A? The vertices are $(-4,1), (-3,1), (-3.5, 4)$.
The line $y=1$ connects $(-4,1)$ and $(-3,1)$.
$x = -3.5$ is the midpoint of this base.
So yes, $(-3.5, 1)$ is on the triangle.
Answer: $(-3.5, 1)$ ✓ (A1)
