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Edexcel GCSE Mathematics – June 2018 Paper 2 (Higher)
Mark Scheme Legend
- (M1) – Method Mark: Awarded for a correct method or partial method.
- (P1) – Process Mark: Awarded for a correct process as part of a problem-solving question.
- (A1) – Accuracy Mark: Awarded after a correct method/process.
- (B1) – Unconditional Accuracy Mark: No method needed.
- (C1) – Communication Mark.
Table of Contents
- Question 1 (Algebraic Indices)
- Question 2 (LCM and HCF)
- Question 3 (Equation of a Line)
- Question 4 (Percentages and Ratio)
- Question 5 (Quadratic Graphs)
- Question 6 (Pressure and Force)
- Question 7 (Transformations)
- Question 8 (Probability from Categories)
- Question 9 (Compound Interest)
- Question 10 (Vectors)
- Question 11 (Functions)
- Question 12 (Proportionality Graphs)
- Question 13 (Circle Theorems)
- Question 14 (Distance-Time Graphs & Speed)
- Question 15 (Probability Trees)
- Question 16 (Circle Equation & Intersection)
- Question 17 (Histograms & Quartiles)
- Question 18 (3D Trigonometry)
- Question 19 (Volume & Surface Area)
- Question 20 (Rationalising Denominators)
- Question 21 (Density and Bounds)
Question 1 (5 marks)
(a) Simplify \(\quad m^{3} \times m^{4}\) (1)
(b) Simplify \(\quad (5np^{3})^{3}\) (2)
(c) Simplify \(\quad \frac{32q^{9}r^{4}}{4q^{3}r}\) (2)
Worked Solution
Part (a): Multiplying terms with the same base
Why we do this:
When we multiply algebraic terms that have the same base (like \(m\)), we apply the index law for multiplication: add the powers together. Mathematically, \(a^x \times a^y = a^{x+y}\).
Working:
\[ m^{3} \times m^{4} = m^{3+4} \] \[ = m^{7} \]✓ (B1) Correct answer only.
Part (b): Raising a product to a power
Why we do this:
When a term with multiple parts is raised to a power inside a bracket, everything inside the bracket gets raised to that power. This means we cube the number \(5\), cube the \(n\), and cube the \(p^3\). We use the power rule \((a^x)^y = a^{x \times y}\).
Working:
\[ (5np^{3})^{3} = 5^{3} \times n^{3} \times (p^{3})^{3} \]Calculate each part:
\[ 5^{3} = 5 \times 5 \times 5 = 125 \] \[ n^{3} = n^{3} \] \[ (p^{3})^{3} = p^{3 \times 3} = p^{9} \]Put it all back together:
\[ = 125n^{3}p^{9} \]✓ (B2) Correct answer only. (B1 is awarded if at least 2 of the 3 terms are correct in a single product).
Part (c): Dividing algebraic terms
Why we do this:
We divide this expression by handling the numbers, the \(q\)’s, and the \(r\)’s separately. For the letters, we use the index law for division: subtract the bottom power from the top power (\(a^x \div a^y = a^{x-y}\)). Remember that a letter on its own like \(r\) has an invisible power of 1 (\(r^1\)).
Working:
\[ \frac{32q^{9}r^{4}}{4q^{3}r^{1}} \]Divide the numbers: \(\frac{32}{4} = 8\)
Divide the \(q\) terms: \(\frac{q^{9}}{q^{3}} = q^{9-3} = q^{6}\)
Divide the \(r\) terms: \(\frac{r^{4}}{r^{1}} = r^{4-1} = r^{3}\)
Combine them all:
\[ = 8q^{6}r^{3} \]✓ (B2) Correct answer only. (B1 is awarded if at least 2 of the 3 terms are correct in a single product).
Final Answers:
(a) \(m^{7}\)
(b) \(125n^{3}p^{9}\)
(c) \(8q^{6}r^{3}\)
Total: 5 marks
Question 2 (3 marks)
(a) Find the lowest common multiple (LCM) of \(40\) and \(56\).
\(A = 2^{3} \times 3 \times 5\)
\(B = 2^{2} \times 3 \times 5^{2}\)
(b) Write down the highest common factor (HCF) of \(A\) and \(B\).
Worked Solution
Step 1: Finding the LCM of 40 and 56
Why we do this:
The lowest common multiple is the smallest number that both \(40\) and \(56\) will divide into evenly. We can solve this either by listing the times tables for both numbers until we find a match, or by using their prime factorisations.
Working (Listing Multiples Method):
Multiples of 40: \(40, 80, 120, 160, 200, 240, 280, 320\dots\)
Multiples of 56: \(56, 112, 168, 224, 280\dots\)
✓ (M1) For listing at least 3 multiples of both 40 and 56.
The smallest number in both lists is \(280\).
Alternative Method (Prime Factors):
\(40 = 8 \times 5 = 2^3 \times 5\)
\(56 = 8 \times 7 = 2^3 \times 7\)
The LCM takes the highest power of each prime factor present: \(2^3 \times 5 \times 7 = 8 \times 35 = 280\).
Step 2: Finding the HCF from Prime Factors
Why we do this:
To find the Highest Common Factor (HCF) when numbers are already written as products of their prime factors, we look for the factors that they share. For each prime number that appears in both \(A\) and \(B\), we take the lowest power.
Working:
\(A = 2^{3} \times 3 \times 5\)
\(B = 2^{2} \times 3 \times 5^{2}\)
Looking at the base of 2: they share \(2^2\).
Looking at the base of 3: they share \(3^1\) (which is just \(3\)).
Looking at the base of 5: they share \(5^1\) (which is just \(5\)).
Multiply these shared factors together:
\[ \text{HCF} = 2^{2} \times 3 \times 5 \] \[ \text{HCF} = 4 \times 3 \times 5 = 60 \]✓ (B1) For \(60\) or \(2^2 \times 3 \times 5\).
Final Answers:
(a) \(280\)
(b) \(60\)
Total: 3 marks
Question 3 (3 marks)
The line \(\mathbf{L}\) is shown on the grid.
Find an equation for \(\mathbf{L}\).
Worked Solution
Step 1: Understanding the Equation of a Straight Line
What are we being asked to find?
We need to find the equation of the line \(\mathbf{L}\) shown on the graph. The equation of a straight line is written in the form \(y = mx + c\), where:
- \(m\) is the gradient (steepness) of the line.
- \(c\) is the \(y\)-intercept (where the line crosses the vertical \(y\)-axis).
Step 2: Finding the y-intercept (\(c\))
Why we do this:
The \(y\)-intercept is usually the easiest part to find. We simply look at the graph to see exactly where the solid black line crosses the vertical \(y\)-axis.
Working:
Looking at the \(y\)-axis, the line crosses it at exactly \(-6\).
Therefore, \(c = -6\).
Step 3: Finding the gradient (\(m\))
Why we do this:
To find the gradient, we pick two easy-to-read points on the line and see how far up it goes for every step across. The formula is \(m = \frac{\text{Change in } y}{\text{Change in } x}\).
Working:
Let’s pick two coordinates on the line that cross exactly at grid intersections:
Point 1: \((0, -6)\) (our \(y\)-intercept)
Point 2: \((2, 0)\) (where it crosses the \(x\)-axis)
Using the formula:
\[ m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{0 – (-6)}{2 – 0} \] \[ m = \frac{6}{2} = 3 \]✓ (M1) For a correct method to find the gradient, or stating \(m=3\).
✓ (M1) For substituting \(m=3\) and \(c=-6\) into a partial equation like \(y=3x+c\) or \(3x-6\).
What this tells us:
Our gradient is \(3\) and our intercept is \(-6\). We can combine these into our final equation.
Final Answer:
\(y = 3x – 6\)
Total: 3 marks
Question 4 (5 marks)
Raya buys a van for £\(8500\) plus VAT at \(20\%\).
Raya pays a deposit for the van.
She then pays the rest of the cost in \(12\) equal payments of £\(531.25\) each month.
Find the ratio of the deposit Raya pays to the total of the \(12\) equal payments.
Give your answer in its simplest form.
Worked Solution
Step 1: Understanding the Question
What are we being asked to find?
We need to build a ratio: \(\text{Deposit} : \text{Total Payments}\), and then simplify it. To do this, we need to figure out exactly how much the van cost altogether, how much she paid in monthly payments, and how much she paid upfront as a deposit.
Step 2: Find the total cost of the van
Why we do this:
The price is £\(8500\) plus VAT at \(20\%\). We need the total final cost including the tax.
Calculator Steps:
Find \(20\%\) of £\(8500\):
\[ 8500 \times 0.20 = 1700 \]Add the VAT to the original price:
\[ 8500 + 1700 = \text{£}10200 \](Alternatively, you can multiply by \(1.20\) directly: \(8500 \times 1.20 = 10200\))
✓ (P1) For the process to find \(20\%\) or \(120\%\) of the cost.
Step 3: Find the total of the 12 equal payments
Why we do this:
We need this number for the right side of our ratio, and we also need it to figure out what was left over to be paid as the deposit.
Working:
\[ 12 \times 531.25 = 6375 \]The total monthly payments sum to £\(6375\).
✓ (P1) For the process to find the total cost of payments.
Step 4: Find the deposit
Why we do this:
The total cost of the van is made up of the deposit and the monthly payments. If we subtract the monthly payments from the total cost, we are left with the deposit amount.
Working:
\[ 10200 – 6375 = 3825 \]The deposit was £\(3825\).
✓ (P1) For a complete process to find the value of the deposit.
Step 5: Create and simplify the ratio
Why we do this:
The question asks for the ratio of “deposit” to “total of the 12 equal payments”. We set it up in that order, and then divide both sides by common factors to simplify it as much as possible.
Working:
\(\text{Deposit} : \text{Payments}\)
\(3825 : 6375\)
✓ (P1) For finding a correct un-simplified ratio.
Let’s simplify. Both end in \(25\), so divide both sides by \(25\) (use your calculator!):
\[ (3825 \div 25) : (6375 \div 25) \] \[ 153 : 255 \]Can we simplify further? Check if they divide by \(3\) (sum of digits: \(1+5+3=9\) and \(2+5+5=12\), so yes they do):
\[ (153 \div 3) : (255 \div 3) \] \[ 51 : 85 \]Now check for prime factors. \(51 = 3 \times 17\). Does \(85\) divide by \(17\)? Yes, \(85 \div 17 = 5\).
Divide both sides by \(17\):
\[ 3 : 5 \]Final Answer:
\(3 : 5\)
✓ (A1) Correct simplified ratio.
Total: 5 marks
Question 5 (6 marks)
(a) Complete the table of values for \(y = x^{2} – x – 6\)
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| \(y\) | 6 | -6 |
(2)
(b) On the grid, draw the graph of \(y = x^{2} – x – 6\) for values of \(x\) from \(-3\) to \(3\).
[Grid provided in exam paper]
(2)
(c) Use your graph to find estimates of the solutions to the equation \(x^{2} – x – 6 = -2\)
(2)
Worked Solution
Step 1: Completing the table (Part a)
Why we do this:
We need to find the missing \(y\) values by substituting each missing \(x\) value into the equation \(y = x^{2} – x – 6\). Remember to be careful with negative numbers, especially when squaring them! \((- \times – = +)\)
Working:
When \(x = -2\):
\(y = (-2)^{2} – (-2) – 6\)
\(y = 4 + 2 – 6 = 0\)
When \(x = -1\):
\(y = (-1)^{2} – (-1) – 6\)
\(y = 1 + 1 – 6 = -4\)
When \(x = 1\):
\(y = (1)^{2} – (1) – 6\)
\(y = 1 – 1 – 6 = -6\)
When \(x = 2\):
\(y = (2)^{2} – (2) – 6\)
\(y = 4 – 2 – 6 = -4\)
When \(x = 3\):
\(y = (3)^{2} – (3) – 6\)
\(y = 9 – 3 – 6 = 0\)
✓ (B2) For all 5 correct figures (B1 for at least 2 correct).
Step 2: Drawing the graph (Part b)
Why we do this:
We plot all the coordinate pairs \((x, y)\) from our completed table onto the grid. A quadratic equation with a positive \(x^2\) term will always make a smooth U-shaped curve (a parabola).
Working:
Plot the points: \((-3, 6)\), \((-2, 0)\), \((-1, -4)\), \((0, -6)\), \((1, -6)\), \((2, -4)\), \((3, 0)\).
Join them with a smooth, continuous curve. Do not use a ruler between points.
✓ (M1) For plotting at least 5 points correctly from their table.
✓ (A1) For a fully correct, smooth curve.
Step 3: Estimating solutions (Part c)
Why we do this:
We are asked to solve \(x^{2} – x – 6 = -2\). Notice that the left side is exactly our equation for \(y\). So this is the same as asking: “What are the \(x\) values when \(y = -2\)?”
To find this graphically, we draw a horizontal line at \(y = -2\) and see where it crosses our curve.
Working:
Draw the line \(y = -2\) straight across the graph.
Read the \(x\) values by dropping down from the intersection points.
The intersections are roughly at \(x = -1.6\) and \(x = 2.6\).
✓ (M1) For drawing the line \(y = -2\) on the graph or showing the intersection clearly.
✓ (A1) For answers in the acceptable range: \(-1.5\) to \(-1.7\), and \(2.5\) to \(2.7\).
Final Answers:
(a) \(0, -4, -6, -4, 0\)
(b) Smooth curve drawn correctly.
(c) \(x = -1.6\) and \(x = 2.6\)
Total: 6 marks
Question 6 (3 marks)
A force of \(70\) newtons acts on an area of \(20\text{ cm}^{2}\).
The force is increased by \(10\) newtons.
The area is increased by \(10\text{ cm}^{2}\).
\[ \text{pressure} = \frac{\text{force}}{\text{area}} \]
Helen says,
“The pressure decreases by less than \(20\%\)”
Is Helen correct?
You must show how you get your answer.
Worked Solution
Step 1: Calculate the initial pressure
Why we do this:
To find out how much the pressure has decreased, we first need to know what the pressure was originally. We are given the formula and the starting values.
Working:
Initial force \(= 70\text{ N}\)
Initial area \(= 20\text{ cm}^{2}\)
Step 2: Calculate the new pressure
Why we do this:
Next, we need the new pressure. We apply the increases given in the question to the original force and area, and then use the formula again.
Working:
New force \(= 70 + 10 = 80\text{ N}\)
New area \(= 20 + 10 = 30\text{ cm}^{2}\)
✓ (P1) For a correct process to calculate either the initial or new pressure.
Step 3: Find the percentage decrease
Why we do this:
Helen makes a claim about the percentage decrease. To check her claim, we calculate the actual percentage decrease. The formula for percentage change is: \(\frac{\text{Change}}{\text{Original}} \times 100\).
Working:
Decrease in pressure \(= 3.5 – 2.666… = 0.8333…\)
\[ \text{Percentage decrease} = \frac{0.8333…}{3.5} \times 100 \] \[ = 23.809… \% \]✓ (P1) For a complete process to make a comparison (e.g., finding the \(23.8\%\) or finding \(20\%\) of \(3.5\)).
What this tells us:
The actual decrease is approximately \(23.8\%\). Helen claimed it decreases by less than \(20\%\). Since \(23.8\%\) is greater than \(20\%\), Helen is incorrect.
(Alternative method: Find \(20\%\) of the original pressure. \(0.20 \times 3.5 = 0.7\). A \(20\%\) decrease would be \(3.5 – 0.7 = 2.8\). Because the new pressure is \(2.666…\), which is lower than \(2.8\), the pressure has decreased by more than \(20\%\).)
Final Answer:
No, Helen is incorrect. The pressure decreases by approximately \(24\%\) (which is greater than \(20\%\)).
✓ (A1) Correct conclusion supported by accurate figures.
Total: 3 marks
Question 7 (2 marks)
Enlarge shape A by scale factor \(\frac{1}{3}\) centre \((0,1)\)
Worked Solution
Step 1: Identifying coordinates and transformations
Why we do this:
When enlarging a shape from a specific centre, we measure the distance (horizontal and vertical steps) from the centre of enlargement to each corner of the original shape. We then multiply these distances by the scale factor to find the corners of our new shape.
Working:
Centre of Enlargement (CoE): \((0, 1)\)
Scale Factor (SF): \(\frac{1}{3}\)
Let’s map out the four vertices (corners) of Shape A:
- Vertex 1 at \((3, 4)\):
From \((0, 1)\) to \((3, 4)\) is \(3\) right, \(3\) up.
Apply SF \(\frac{1}{3}\): \(\frac{1}{3}\) of \(3\) right is \(1\) right, \(\frac{1}{3}\) of \(3\) up is \(1\) up.
From \((0, 1)\) go \(1\) right, \(1\) up \(\rightarrow\) New Vertex: \((1, 2)\) - Vertex 2 at \((6, 7)\):
From \((0, 1)\) to \((6, 7)\) is \(6\) right, \(6\) up.
Apply SF \(\frac{1}{3}\): \(2\) right, \(2\) up.
From \((0, 1)\) go \(2\) right, \(2\) up \(\rightarrow\) New Vertex: \((2, 3)\) - Vertex 3 at \((6, 10)\):
From \((0, 1)\) to \((6, 10)\) is \(6\) right, \(9\) up.
Apply SF \(\frac{1}{3}\): \(2\) right, \(3\) up.
From \((0, 1)\) go \(2\) right, \(3\) up \(\rightarrow\) New Vertex: \((2, 4)\) - Vertex 4 at \((3, 10)\):
From \((0, 1)\) to \((3, 10)\) is \(3\) right, \(9\) up.
Apply SF \(\frac{1}{3}\): \(1\) right, \(3\) up.
From \((0, 1)\) go \(1\) right, \(3\) up \(\rightarrow\) New Vertex: \((1, 4)\)
What this tells us:
Plot the new points \((1, 2)\), \((2, 3)\), \((2, 4)\), and \((1, 4)\) on the grid and join them up. Because the scale factor is fractional (\(1/3\)), the new shape is smaller than the original and closer to the centre of enlargement.
Final Answer:
Correct enlargement drawn with vertices at \((1,2)\), \((2,3)\), \((2,4)\), and \((1,4)\).
✓ (B2) For correct enlargement at \((1,2)\), \((2,3)\), \((2,4)\), \((1,4)\). (B1 for correct size and orientation in the wrong position).
Total: 2 marks
Question 8 (4 marks)
\(60\) people were asked if they prefer to go on holiday in Britain or in Spain or in Italy.
\(38\) of the people were male.
\(11\) of the \(32\) people who said Britain were female.
\(8\) males said Italy.
\(12\) people said Spain.
One of the females is chosen at random.
What is the probability that this female said Spain?
Worked Solution
Step 1: Organizing the information
Why we do this:
There are a lot of overlapping categories here (Gender and Country). The best way to organize this data so we can find the missing pieces is to draw a Two-Way Table.
Working:
Let’s set up a table and fill in the numbers given in the question.
| Britain | Spain | Italy | Total | |
|---|---|---|---|---|
| Male | 8 | 38 | ||
| Female | 11 | |||
| Total | 32 | 12 | 60 |
Step 2: Filling in the gaps
Why we do this:
We need to find the number of females who said Spain. To get there, we use simple addition and subtraction across the rows and columns.
Working:
1. Find total females: \(60 – 38 = 22\) females.
✓ (P1) For a process to find a first value (e.g., total females \(= 22\)).
2. Find males who said Britain: \(32 – 11 = 21\) males.
3. Find males who said Spain: We know total males = \(38\).
Britain Males + Spain Males + Italy Males \(= 38\)
\(21 + \text{Spain Males} + 8 = 38\)
\(29 + \text{Spain Males} = 38\)
Spain Males \(= 38 – 29 = 9\)
✓ (P1) For a process to find a secondary value (e.g., male/Spain \(= 9\)).
4. Find females who said Spain: We know total Spain = \(12\).
Spain Males + Spain Females \(= 12\)
\(9 + \text{Spain Females} = 12\)
Spain Females \(= 12 – 9 = 3\)
✓ (P1) For complete process to find female/Spain.
Step 3: Calculating the probability
Why we do this:
The question asks: “One of the females is chosen at random. What is the probability that this female said Spain?”
Because we are ONLY picking from the females, our denominator (bottom number) is the total number of females, NOT the total number of people.
Working:
\[ \text{Probability} = \frac{\text{Females who chose Spain}}{\text{Total Females}} \] \[ \text{Probability} = \frac{3}{22} \]Final Answer:
\(\frac{3}{22}\)
✓ (A1) For \(\frac{3}{22}\).
Total: 4 marks
Question 9 (4 marks)
Jean invests £\(12000\) in an account paying compound interest for \(2\) years.
In the first year the rate of interest is \(x\%\)
At the end of the first year the value of Jean’s investment is £\(12336\)
In the second year the rate of interest is \(\frac{x}{2}\%\)
What is the value of Jean’s investment at the end of \(2\) years?
Worked Solution
Step 1: Find the interest rate for the first year (\(x\%\))
Why we do this:
To find the value after 2 years, we must first know what the interest rate was. We know the starting amount (£\(12000\)) and the amount after 1 year (£\(12336\)). We can calculate the multiplier used, which tells us the percentage rate.
Working:
Multiplier = \(\frac{\text{End Value}}{\text{Start Value}}\)
\[ \text{Multiplier} = \frac{12336}{12000} = 1.028 \]✓ (P1) For start of process to find interest rate (e.g., \(12336 \div 12000\)).
A multiplier of \(1.028\) means the money is \(102.8\%\) of its original size. The interest rate is the portion above \(100\%\).
\[ x = 102.8\% – 100\% = 2.8\% \]✓ (P1) For complete process to find the interest rate for year 1 (\(x = 2.8\)).
(Alternative method: Interest earned = \(12336 – 12000 = 336\). Then calculate \(\frac{336}{12000} \times 100 = 2.8\%\).)
Step 2: Calculate the second year’s interest
Why we do this:
In the second year, the rate halves to \(\frac{x}{2}\%\). We need to apply this new rate to the amount in the account at the end of year 1 (£\(12336\)) because this is compound interest.
Working:
Second year rate = \(\frac{2.8\%}{2} = 1.4\%\)
A \(1.4\%\) increase gives a multiplier of \(1.014\). We multiply the Year 1 ending balance by this new multiplier.
\[ \text{Final Value} = 12336 \times 1.014 \]✓ (P1) For a complete process to find the value at the end of 2 years.
\[ \text{Final Value} = 12508.704 \]What this tells us:
Because this is money, we always round to 2 decimal places (the nearest penny) unless told otherwise. So, £\(12508.704\) becomes £\(12508.70\).
Final Answer:
£\(12508.70\)
✓ (A1) Accept \(12508.7\) to \(12508.71\) or \(12509\).
Total: 4 marks
Question 10 (3 marks)
The vector \(\mathbf{a}\) and the vector \(\mathbf{b}\) are shown on the grid.
(a) On the grid, draw and label vector \(-2\mathbf{a}\) (1)
(b) Work out \(\mathbf{a} + 2\mathbf{b}\) as a column vector. (2)
Worked Solution
Step 1: Drawing \(-2\mathbf{a}\) (Part a)
Why we do this:
First, we need to know what vector \(\mathbf{a}\) is in numbers. By looking at the grid, from the start of \(\mathbf{a}\) to the end, it goes \(1\) square right and \(2\) squares up. So, \(\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).
The vector \(-2\mathbf{a}\) means we multiply both directions by \(-2\). The negative sign reverses the direction, and the \(2\) doubles the length. So it goes \(2\) squares left and \(4\) squares down.
Working:
\[ -2\mathbf{a} = -2\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \end{pmatrix} \]We can draw this anywhere on the grid. Pick a starting point (like \((5, 5)\)), count \(2\) units left and \(4\) units down (to \((3, 1)\)), and draw an arrow pointing towards the end point.
✓ (B1) For correct vector drawn including arrow. May be drawn anywhere on the grid.
Step 2: Calculating \(\mathbf{a} + 2\mathbf{b}\) (Part b)
Why we do this:
We need to add vector \(\mathbf{a}\) to two lots of vector \(\mathbf{b}\). We already know \(\mathbf{a}\). Let’s find \(\mathbf{b}\) from the grid. From start to finish, \(\mathbf{b}\) goes \(1\) square right and \(3\) squares down. So, \(\mathbf{b} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}\).
Working:
\[ \mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -3 \end{pmatrix} \]Calculate \(2\mathbf{b}\):
\[ 2\mathbf{b} = 2 \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \end{pmatrix} \]Now add \(\mathbf{a}\) and \(2\mathbf{b}\) together by adding the top numbers and adding the bottom numbers:
\[ \mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ -6 \end{pmatrix} \] \[ = \begin{pmatrix} 1 + 2 \\ 2 + (-6) \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} \]✓ (M1) For writing \(\mathbf{a}\) and \(\mathbf{b}\) as column vectors and attempt to add \(\mathbf{a} + 2\mathbf{b}\).
Final Answers:
(a) Vector should be drawn \(2\) units left and \(4\) units down, with an arrow pointing towards the end.
(b) \( \begin{pmatrix} 3 \\ -4 \end{pmatrix} \)
✓ (A1) Correct column vector.
Total: 3 marks
Question 11 (3 marks)
\(f\) and \(g\) are functions such that
\[ f(x) = \frac{2}{x^{2}} \quad \text{and} \quad g(x) = 4x^{3} \](a) Find \(f(-5)\) (1)
(b) Find \(fg(1)\) (2)
Worked Solution
Step 1: Evaluating the function \(f(x)\) (Part a)
Why we do this:
The notation \(f(-5)\) means we need to take the function \(f(x)\) and substitute \(-5\) wherever we see an \(x\). We must be careful to put the \(-5\) in brackets before squaring it, because squaring a negative number gives a positive result.
Working:
\[ f(x) = \frac{2}{x^{2}} \] \[ f(-5) = \frac{2}{(-5)^{2}} \] \[ f(-5) = \frac{2}{25} \]✓ (B1) For correct answer \(\frac{2}{25}\) (or \(0.08\)).
Step 2: Evaluating the composite function \(fg(1)\) (Part b)
Why we do this:
The notation \(fg(1)\) represents a composite function. It means we work from the inside out: first we substitute \(1\) into the function \(g\), and then we take that resulting answer and substitute it into the function \(f\). It can also be written as \(f(g(1))\).
Working:
First, find \(g(1)\):
\[ g(x) = 4x^{3} \] \[ g(1) = 4(1)^{3} \] \[ g(1) = 4 \times 1 = 4 \]✓ (M1) For evaluating \(g(1) = 4\).
Now, substitute this result into \(f\):
We need to find \(f(4)\):
\[ f(x) = \frac{2}{x^{2}} \] \[ f(4) = \frac{2}{4^{2}} \] \[ f(4) = \frac{2}{16} \]Simplify the fraction:
\[ f(4) = \frac{1}{8} \]What this tells us:
The final value of \(fg(1)\) is \(\frac{1}{8}\) (or \(0.125\) as a decimal).
Final Answers:
(a) \(\frac{2}{25}\) (or \(0.08\))
(b) \(\frac{1}{8}\) (or \(0.125\))
✓ (A1) For the correct final answer.
Total: 3 marks
Question 12 (2 marks)
The graphs of \(y\) against \(x\) represent four different types of proportionality.
Graph A
Graph B
Graph C
Graph D
Match each type of proportionality in the table to the correct graph.
| Type of proportionality | Graph letter |
|---|---|
| \(y \propto x\) | |
| \(y \propto x^{2}\) | |
| \(y \propto \sqrt{x}\) | |
| \(y \propto \frac{1}{x}\) |
Worked Solution
Step 1: Identifying the shapes of proportionality graphs
Why we do this:
Each mathematical relationship produces a distinct curve shape when plotted. Let’s analyze them one by one.
Working:
- \(y \propto x\) (Direct Proportion): This means \(y = kx\). This is the equation of a straight line passing through the origin \((0,0)\) with a constant positive gradient. This matches Graph B.
- \(y \propto x^{2}\): This means \(y = kx^2\). This is a quadratic curve (a parabola) passing through the origin. As \(x\) gets larger, \(y\) increases at an accelerating (faster) rate, making the curve bend upwards. This matches Graph D.
- \(y \propto \sqrt{x}\): This means \(y = k\sqrt{x}\). This curve passes through the origin. As \(x\) gets larger, \(y\) increases but at a decelerating (slower) rate, meaning it flattens out as it goes to the right. This matches Graph A.
- \(y \propto \frac{1}{x}\) (Inverse Proportion): This means \(y = \frac{k}{x}\). This is a reciprocal curve. As \(x\) gets bigger, \(y\) gets smaller, approaching zero but never touching the axes. This matches Graph C.
What this tells us:
The correct order from top to bottom is B, D, A, C.
Final Answers:
\(y \propto x\) : B
\(y \propto x^{2}\) : D
\(y \propto \sqrt{x}\) : A
\(y \propto \frac{1}{x}\) : C
✓ (B2) For all 4 correct. (B1 for at least 2 correct).
Total: 2 marks
Question 13 (4 marks)
\(A\), \(B\), \(C\) and \(D\) are points on the circumference of a circle, centre \(O\).
\(FDE\) is a tangent to the circle.
(a) Show that \(y – x = 90\)
You must give a reason for each stage of your working.
(3)
Dylan was asked to give some possible values for \(x\) and \(y\).
He said,
“\(y\) could be \(200\) and \(x\) could be \(110\), because \(200 – 110 = 90\)”
(b) Is Dylan correct?
You must give a reason for your answer.
(1)
Worked Solution
Step 1: Using circle theorems to connect \(x\) and \(y\) (Part a)
Why we do this:
We are asked to prove a geometric relationship involving \(x\) and \(y\). We must identify relevant circle theorems that connect the angles inside the circle to the tangent outside the circle.
Two key features in the diagram are:
- A tangent (\(FDE\)) meeting a radius (\(OD\)).
- A triangle inside a circle (\(ABD\)) connecting to a tangent (which suggests the Alternate Segment Theorem).
Working:
1. The line \(OD\) is a radius and \(FDE\) is a tangent. The angle between a tangent and a radius is always \(90^\circ\).
\[ \text{Angle } ODE = 90^\circ \]Reason: A radius and a tangent meet at right angles (\(90^\circ\)).
✓ (M1) For finding one missing angle, e.g., stating \(\text{Angle } ODE = 90^\circ\).
2. Look at the whole angle \(\text{Angle } BDE\). It is made up of two smaller angles added together:
\[ \text{Angle } BDE = \text{Angle } BDO + \text{Angle } ODE \] \[ \text{Angle } BDE = x + 90^\circ \]3. Now use the Alternate Segment Theorem. The angle between a chord (\(BD\)) and a tangent (\(FDE\)) is equal to the angle subtended by that chord in the alternate segment (which is Angle \(BAD\), or \(y\)).
\[ \text{Angle } BDE = \text{Angle } BAD \] \[ \text{Angle } BDE = y \]Reason: Alternate Segment Theorem.
✓ (C1) For all correct circle theorems given appropriate for their working.
4. Since we have two different expressions for \(\text{Angle } BDE\), we can set them equal to each other:
\[ y = x + 90 \]Subtract \(x\) from both sides to rearrange:
\[ y – x = 90 \]✓ (A1) For a complete correct method leading to \(y – x = 90\).
Step 2: Evaluating Dylan’s claim (Part b)
Why we do this:
Even though \(200 – 110 = 90\), we must check if \(y = 200^\circ\) is actually possible within the context of the diagram. \(y\) is an interior angle of the cyclic quadrilateral \(ABCD\) (and also an angle inside triangle \(ABD\)).
Working:
\(y\) is the angle \(\text{Angle } BAD\). Angles inside a triangle add up to \(180^\circ\). Therefore, a single interior angle of a convex polygon or a triangle cannot be \(200^\circ\) (it would be a reflex angle, curving ‘outwards’).
✓ (C1) For correct explanation.
Final Answers:
(a) Proof completed correctly with appropriate reasons.
(b) No, Dylan is incorrect because \(y\) is an angle inside a triangle (or cyclic quadrilateral) and therefore must be less than \(180^\circ\).
Total: 4 marks
Question 14 (3 marks)
The distance-time graph shows information about part of a car journey.
Use the graph to estimate the speed of the car at time \(5\) seconds.
Worked Solution
Step 1: Finding speed from a distance-time graph
Why we do this:
On a distance-time graph, the speed is represented by the gradient (steepness) of the line. Because the line is curved (meaning the speed is changing), to find the speed at a specific instant (\(t = 5\)), we must draw a straight tangent line that just touches the curve at exactly \(5\) seconds.
Working:
1. Locate \(5\) seconds on the horizontal axis and go straight up to the curve.
2. Draw a tangent line at this point. A tangent should touch the curve at exactly one point without crossing through it, following the slope of the curve at that exact moment.
✓ (P1) For drawing a tangent to the curve at time = 5.
Step 2: Calculating the gradient of the tangent
Why we do this:
The gradient of our tangent line is our estimated speed. We calculate gradient using the formula: \(\frac{\text{change in vertical } (y)}{\text{change in horizontal } (x)}\).
Note: Because everyone draws tangents slightly differently by hand, the examiners accept a range of answers. The example below shows one possible valid tangent.
Working:
Pick two easy-to-read points on your drawn tangent line. For example, your tangent might pass through \((0, -35)\) and \((5, 40)\), or you might read a triangle with base from \(2\) to \(7\) seconds.
Let’s say our tangent creates a right-angled triangle where the vertical change is \(75\text{ m}\) and horizontal change is \(5\text{ s}\).
\[ \text{Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}} \] \[ \text{Speed} = \frac{75}{5} = 15 \]✓ (P1) For a process to find the gradient using their drawn tangent.
What this tells us:
The estimated speed is \(15\text{ m/s}\). According to the mark scheme, any calculated value between \(11\) and \(19\text{ m/s}\) is fully correct, as long as it comes from the gradient of a drawn tangent.
Final Answer:
\(15\text{ m/s}\) (Accept any value between \(11\) and \(19\))
✓ (A1) For answer in the range \(11 – 19\text{ m/s}\) derived from a tangent.
Total: 3 marks
Question 15 (5 marks)
A darts team is going to play a match on Saturday and on Sunday.
The probability that the team will win on Saturday is \(0.45\)
If they win on Saturday, the probability that they will win on Sunday is \(0.67\)
If they do not win on Saturday, the probability that they will win on Sunday is \(0.35\)
(a) Complete the probability tree diagram.
(2)
(b) Find the probability that the team will win exactly one of the two matches.
(3)
Worked Solution
Step 1: Completing the Probability Tree (Part a)
Why we do this:
In a probability tree, the sum of probabilities on any set of branches coming from the same point must always add up to \(1\) (because it covers \(100\%\) of the possible outcomes). If they win or they don’t win, those are the only two options.
Working:
Saturday:
Win = \(0.45\)
Not Win = \(1 – 0.45 = 0.55\)
✓ (B1) For \(0.55\) in the correct position.
Sunday (If won on Saturday):
Win = \(0.67\) (Given in question)
Not Win = \(1 – 0.67 = 0.33\)
Sunday (If did NOT win on Saturday):
Win = \(0.35\) (Given in question)
Not Win = \(1 – 0.35 = 0.65\)
✓ (B1) For the branches for the second game correct.
Step 2: Finding “exactly one win” (Part b)
Why we do this:
Winning “exactly one” match can happen in two different ways:
- They WIN on Saturday AND NOT WIN on Sunday.
- They NOT WIN on Saturday AND WIN on Sunday.
To find the probability of a specific path along the tree (“AND”), we multiply the probabilities. To find the total probability of multiple different successful paths (“OR”), we add them together.
Working:
Path 1: P(Win, Not Win) = \(0.45 \times 0.33 = 0.1485\)
✓ (M1) For one correct product on a relevant branch path.
Path 2: P(Not Win, Win) = \(0.55 \times 0.35 = 0.1925\)
Total Probability = P(Path 1) + P(Path 2)
\[ \text{Total} = 0.1485 + 0.1925 = 0.341 \]✓ (M1) For a correct method to sum the two relevant paths: \((0.45 \times 0.33) + (0.55 \times 0.35)\).
What this tells us:
The total probability that they win exactly one of the two matches is \(0.341\).
Final Answers:
(a) Tree completed: Sat Not Win = 0.55; Sun Top Branch = 0.67, 0.33; Sun Bottom Branch = 0.35, 0.65.
(b) \(0.341\)
✓ (A1) For the exact answer \(0.341\).
Total: 5 marks
Question 16 (5 marks)
(a) On the grid, draw the graph of \(\quad x^{2} + y^{2} = 12.25\)
(2)
(b) Hence find estimates for the solutions of the simultaneous equations
\(x^{2} + y^{2} = 12.25\)
\(2x + y = 1\)
(3)
Worked Solution
Step 1: Drawing the circle (Part a)
Why we do this:
An equation in the format \(x^{2} + y^{2} = r^{2}\) represents a circle centered at the origin \((0,0)\) with a radius of \(r\). We need to find the radius by square-rooting the number on the right, and then draw it.
Working:
\[ r^{2} = 12.25 \] \[ r = \sqrt{12.25} = 3.5 \]We draw a circle with its centre at \((0,0)\) and a radius of \(3.5\) units. It should pass exactly halfway between the \(3\) and the \(4\) on all four axes.
✓ (B2) For a correct circle drawn with radius \(3.5\) and centre \((0,0)\).
Step 2: Solving the simultaneous equations graphically (Part b)
Why we do this:
The word “Hence” means we must use our previous work (the graph). We have the circle graph for the first equation. If we draw the graph for the second equation (\(2x + y = 1\)), the points where the line and the circle cross (intersect) are the solutions to both equations simultaneously.
Working:
Rearrange the linear equation to make it easy to draw:
\[ 2x + y = 1 \implies y = -2x + 1 \]Plot a few points to draw this line:
- If \(x = 0\), \(y = 1\) \(\rightarrow (0, 1)\)
- If \(x = 2\), \(y = -2(2) + 1 = -3\) \(\rightarrow (2, -3)\)
- If \(x = -1\), \(y = -2(-1) + 1 = 3\) \(\rightarrow (-1, 3)\)
Draw a straight line through these points. Then, read the coordinates where this line crosses the circle.
✓ (M1) For \(2x + y = 1\) drawn, or correctly eliminating one variable algebraically.
Step 3: Reading the intersection points
What this tells us:
By looking at the graph, we find the \(x\) and \(y\) values for both intersection points. Because we are reading from a hand-drawn graph, small variations (estimates) are accepted in the final answers.
Working:
Intersection 1 (Top Left): \(x \approx -1.2\), \(y \approx 3.3\)
Intersection 2 (Bottom Right): \(x \approx 2.0\), \(y \approx -2.9\)
✓ (A1) For one correct pair of \(x\) and \(y\) values.
✓ (A1) For both correct pairs of \(x\) and \(y\) values, unambiguously matched.
Final Answers:
(a) Correct circle drawn (radius \(3.5\))
(b) \(x = 2.0\), \(y = -2.9\) and \(x = -1.2\), \(y = 3.3\)
Total: 5 marks
Question 17 (4 marks)
The histogram shows information about the times taken by some students to finish a puzzle.
(a) Complete the frequency table for this information.
| Time taken (\(t\) minutes) | Frequency |
|---|---|
| \(0 < t \leq 5\) | 4 |
| \(5 < t \leq 15\) | |
| \(15 < t \leq 25\) | |
| \(25 < t \leq 30\) | |
| \(30 < t \leq 50\) |
(2)
(b) Find an estimate for the lower quartile of the times taken to finish the puzzle.
(2)
Worked Solution
Step 1: Finding frequencies from the histogram (Part a)
Why we do this:
In a histogram, the area of each bar represents the frequency. The formula is: \(\text{Frequency} = \text{Frequency Density} \times \text{Class Width}\). We can read the frequency density from the y-axis and the class width from the x-axis for each bar.
Working:
Let’s check the first bar to confirm our method: Width = \(5\), Height (Density) = \(0.8\). Area = \(5 \times 0.8 = 4\). This matches the table!
Bar 2 (\(5 < t \leq 15\)):
Width = \(15 – 5 = 10\). Height = \(0.4\).
Frequency = \(10 \times 0.4 = 6\)
Bar 3 (\(15 < t \leq 25\)):
Width = \(25 – 15 = 10\). Height = \(0.6\).
Frequency = \(10 \times 0.6 = 6\)
Bar 4 (\(25 < t \leq 30\)):
Width = \(30 – 25 = 5\). Height = \(1.0\).
Frequency = \(5 \times 1.0 = 5\)
Bar 5 (\(30 < t \leq 50\)):
Width = \(50 – 30 = 20\). Height = \(0.2\).
Frequency = \(20 \times 0.2 = 4\)
✓ (M1) For a correct method to find at least 2 frequencies from bars of different widths.
✓ (A1) For all correct frequencies: \(6, 6, 5, 4\).
Step 2: Finding the Lower Quartile position (Part b)
Why we do this:
The lower quartile (LQ) is the value a quarter (\(1/4\)) of the way through the data. First, we need the total number of students (Total Frequency).
Working:
Total Frequency \((n) = 4 + 6 + 6 + 5 + 4 = 25\)
Wait, let me recalculate:
Bar 1: 4
Bar 2: 10 * 0.4 = 4 (Oh, I made a mistake above! \(10 \times 0.4 = 4\), not 6!)
Bar 3: 10 * 0.6 = 6
Bar 4: 5 * 1.0 = 5
Bar 5: 20 * 0.2 = 4
Total Frequency = \(4 + 4 + 6 + 5 + 4 = 23\).
Lower Quartile position = \(\frac{n}{4} = \frac{23}{4} = 5.75\)
(Or \(\frac{n+1}{4} = \frac{24}{4} = 6th\) position. Both are acceptable).
✓ (M1) For \(\frac{23+1}{4} = 6\) or \(\frac{23}{4} = 5.75\).
Step 3: Estimating the Lower Quartile value
Why we do this:
We know the LQ is the \(5.75th\) (or \(6th\)) student. We count through the bars to find where this student is, and assume the students are spread evenly throughout that bar.
Working:
The first bar (\(0\) to \(5\) mins) contains the first \(4\) students.
We need the \(5.75th\) student, so we need to go \(1.75\) students into the second bar.
The second bar (\(5\) to \(15\) mins) contains \(4\) students and has a width of \(10\) mins.
We want the fraction \(\frac{1.75}{4}\) of this class width:
\[ \text{Extra time} = \frac{1.75}{4} \times 10 = 0.4375 \times 10 = 4.375 \text{ mins} \]Add this to the start of the class interval (\(5\) mins):
\[ \text{Lower Quartile} = 5 + 4.375 = 9.375 \text{ mins} \](If using the 6th position: we need \(2\) students into the second bar. \(\frac{2}{4} \times 10 = 5\). Add to 5 mins = \(10\) mins.)
Final Answers:
(a) Frequencies: \(4, 6, 5, 4\)
(b) \(10\) minutes (or \(9.375\) minutes)
✓ (A1) For \(10\) or \(9.375\).
Total: 4 marks
Question 18 (4 marks)
\(ABCDEFGH\) is a cuboid.
\(AB = 7.3\text{ cm}\)
\(CH = 8.1\text{ cm}\)
Angle \(BCA = 48^{\circ}\)
Find the size of the angle between \(AH\) and the plane \(ABCD\).
Give your answer correct to \(1\) decimal place.
Worked Solution
Step 1: Visualizing the 3D geometry
What are we being asked to find?
The “angle between a line and a plane” means the angle between the line (\(AH\)) and its “shadow” on the plane (\(AC\)). So we are looking for the angle \(HAC\) inside the right-angled triangle \(AHC\).
To find angle \(HAC\) using trigonometry (SOH CAH TOA), we need two sides of the triangle \(AHC\). We know the height \(CH = 8.1\text{ cm}\), but we don’t know the base \(AC\) yet.
Step 2: Finding the length of AC
Why we do this:
Before we can solve the 3D triangle, we must solve a 2D triangle on the base. Look at the rectangular base \(ABCD\). Triangle \(ABC\) is a right-angled triangle (the corner \(B\) of a cuboid is \(90^\circ\)). We know \(AB = 7.3\) and angle \(BCA = 48^\circ\). We can use SOH CAH TOA to find the hypotenuse \(AC\).
Working:
In right-angled triangle \(ABC\):
- Angle \(= 48^\circ\)
- Opposite side (\(AB\)) \(= 7.3\)
- Hypotenuse (\(AC\)) \(= ?\)
We use Sine (SOH): \(\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
\[ \sin(48^{\circ}) = \frac{7.3}{AC} \]✓ (P1) For a correct trigonometric statement.
Rearrange to find \(AC\):
\[ AC = \frac{7.3}{\sin(48^{\circ})} \] \[ AC = 9.823… \text{ cm} \]✓ (P1) For a complete correct process to find \(AC\).
(Keep this exact value in your calculator for the next step!)
Step 3: Calculating the 3D angle
Why we do this:
Now we go back to our vertical 3D triangle \(AHC\). It has a right angle at \(C\). We want to find angle \(HAC\). We know the Opposite side (\(CH = 8.1\)) and the Adjacent side (\(AC = 9.823…\)).
Working:
Use Tangent (TOA): \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)
\[ \tan(\text{Angle } CAH) = \frac{8.1}{9.823…} \]✓ (P1) For a correct statement using angle \(CAH\).
\[ \tan(\text{Angle } CAH) = 0.8245… \]To find the angle, use inverse tangent (\(\tan^{-1}\)):
\[ \text{Angle } CAH = \tan^{-1}(0.8245…) \] \[ \text{Angle } CAH = 39.506…^{\circ} \]What this tells us:
The angle is \(39.506…^\circ\). The question asks for \(1\) decimal place, so we round it to \(39.5^\circ\).
Final Answer:
\(39.5^{\circ}\)
✓ (A1) For answer in the range \(39.5 – 39.51\).
Total: 4 marks
Question 19 (5 marks)
Shape \(S\) is one quarter of a solid sphere, centre \(O\).
The volume of \(S\) is \(576\pi\text{ cm}^{3}\).
Find the surface area of \(S\).
Give your answer correct to \(3\) significant figures.
You must show your working.
Volume of sphere \(= \frac{4}{3}\pi r^{3}\)
Surface area of sphere \(= 4\pi r^{2}\)
Worked Solution
Step 1: Finding the radius of the sphere
Why we do this:
We cannot find the surface area without knowing the radius (\(r\)). We are told the volume of this quarter-sphere is \(576\pi\). This means a full sphere would have a volume four times as large. We can set up an equation to find \(r\).
Working:
Volume of full sphere \(= 4 \times 576\pi = 2304\pi\)
Set this equal to the volume formula:
\[ \frac{4}{3}\pi r^{3} = 2304\pi \]✓ (P1) For correct use of the volume formula (e.g. \(\frac{1}{4} \times \frac{4}{3} \times \pi \times r^{3} = 576\pi\)).
Cancel \(\pi\) from both sides and multiply by \(3\), then divide by \(4\):
\[ \frac{4}{3}r^{3} = 2304 \] \[ r^{3} = \frac{2304 \times 3}{4} \] \[ r^{3} = 1728 \]Take the cube root to find \(r\):
\[ r = \sqrt[3]{1728} = 12\text{ cm} \]✓ (P1) For a complete correct process to find \(r\).
Step 2: Calculating the surface area of the shape
Why we do this:
A common mistake is to just find the surface area of a full sphere and divide it by \(4\). However, when you cut a sphere, you expose new flat surfaces. Shape \(S\) has three faces:
- The curved outer surface (which is \(\frac{1}{4}\) of a full sphere’s surface).
- Two flat semicircular faces that meet at the centre (like slices of an orange). Two semicircles make one full circle of area \(\pi r^2\).
We need to add all these areas together.
Working:
1. Area of curved surface:
\[ \text{Curved Area} = \frac{1}{4} \times (4\pi r^{2}) = \pi r^{2} \]✓ (P1) For a process to find the curved surface area OR the flat surfaces.
2. Area of two flat semicircles:
\[ \text{Flat Area} = 2 \times \left(\frac{1}{2}\pi r^{2}\right) = \pi r^{2} \]3. Total Surface Area:
\[ \text{Total SA} = \pi r^{2} + \pi r^{2} = 2\pi r^{2} \]✓ (P1) For a process to find the complete surface area.
Now substitute \(r = 12\):
\[ \text{Total SA} = 2 \times \pi \times (12)^{2} \] \[ \text{Total SA} = 2 \times \pi \times 144 = 288\pi \] \[ \text{Total SA} = 904.7786… \]What this tells us:
The total surface area is \(904.778…\text{ cm}^2\). The question asks for \(3\) significant figures. The third significant figure is the \(4\), and the next digit is \(7\) (so we round up).
Final Answer:
\(905\text{ cm}^{2}\)
✓ (A1) For answer in the range \(904.7 – 905\) or \(288\pi\).
Total: 5 marks
Question 20 (2 marks)
Martin did this question.
Rationalise the denominator of \(\quad \frac{14}{2 + \sqrt{3}}\)
Here is how he answered the question.
Martin’s answer is wrong.
(a) Find Martin’s mistake.
(1)
Sian did this question.
Rationalise the denominator of \(\quad \frac{5}{\sqrt{12}}\)
Here is how she answered the question.
Sian’s answer is wrong.
(b) Find Sian’s mistake.
(1)
Worked Solution
Part (a): Analyzing Martin’s Work
Why we do this:
We need to go through Martin’s algebraic expansion step-by-step to find the mathematical error.
Working:
Look at his expansion of the denominator:
\[ (2 + \sqrt{3})(2 – \sqrt{3}) \]When expanding this difference of two squares, the last term comes from multiplying \((+\sqrt{3})\) by \((-\sqrt{3})\).
\[ \sqrt{3} \times -\sqrt{3} = -3 \]However, Martin wrote \(+3\) in his second line of working: \(4 + 2\sqrt{3} – 2\sqrt{3} + 3\).
Because of this, his denominator became \(4 + 3 = 7\), instead of \(4 – 3 = 1\).
Part (b): Analyzing Sian’s Work
Why we do this:
We track Sian’s steps to see where her logic breaks down, specifically when she simplifies the surd \(\sqrt{12}\).
Working:
Look at her second line of working. She claims that \(\sqrt{12} = 3\sqrt{2}\).
Let’s simplify \(\sqrt{12}\) ourselves by finding the largest square number that goes into \(12\):
\[ \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \]She incorrectly wrote \(3\sqrt{2}\) (which is actually \(\sqrt{18}\)) instead of \(2\sqrt{3}\).
Final Answers:
(a) Martin incorrectly calculated \(\sqrt{3} \times -\sqrt{3}\) as \(+3\). It should be \(-3\).
✓ (C1) For a correct explanation.
(b) Sian incorrectly simplified the surd \(\sqrt{12}\). She wrote \(3\sqrt{2}\) but it should be \(2\sqrt{3}\).
✓ (C1) For a correct explanation.
Total: 2 marks
Question 21 (5 marks)
Jackson is trying to find the density, in \(\text{g/cm}^{3}\), of a block of wood.
The block of wood is in the shape of a cuboid.
He measures
the length as \(13.2\text{ cm}\), correct to the nearest mm
the width as \(16.0\text{ cm}\), correct to the nearest mm
the height as \(21.7\text{ cm}\), correct to the nearest mm
He measures the mass as \(1970\text{ g}\), correct to the nearest \(5\text{ g}\).
By considering bounds, work out the density of the wood.
Give your answer to a suitable degree of accuracy.
You must show all your working and give a reason for your final answer.
Worked Solution
Step 1: Finding the Upper and Lower Bounds
Why we do this:
Because the measurements have been rounded, the true values could be slightly higher or lower. We must find the Lower Bound (LB) and Upper Bound (UB) for each measurement.
– “Nearest mm” means the nearest \(0.1\text{ cm}\). The error interval is half of this: \(\pm 0.05\text{ cm}\).
– “Nearest \(5\text{ g}\)” has an error interval of half of \(5\): \(\pm 2.5\text{ g}\).
Working:
- Length: \(13.2 \pm 0.05\) → LB = \(13.15\text{ cm}\), UB = \(13.25\text{ cm}\)
- Width: \(16.0 \pm 0.05\) → LB = \(15.95\text{ cm}\), UB = \(16.05\text{ cm}\)
- Height: \(21.7 \pm 0.05\) → LB = \(21.65\text{ cm}\), UB = \(21.75\text{ cm}\)
- Mass: \(1970 \pm 2.5\) → LB = \(1967.5\text{ g}\), UB = \(1972.5\text{ g}\)
✓ (B1) For identifying at least one correct bound for mass or length.
Step 2: Calculating Bounds for the Volume
Why we do this:
Density is \(\frac{\text{Mass}}{\text{Volume}}\). So, we first need to find the maximum possible volume and the minimum possible volume of the cuboid. Volume = Length \(\times\) Width \(\times\) Height.
Working:
To get the smallest volume (Lower Bound), multiply all the smallest lengths together:
\[ \text{Volume LB} = 13.15 \times 15.95 \times 21.65 = 454.0925125\text{ cm}^{3} \]To get the largest volume (Upper Bound), multiply all the largest lengths together:
\[ \text{Volume UB} = 13.25 \times 16.05 \times 21.75 = 462.5409375\text{ cm}^{3} \]✓ (P1) For a correct process to find a bound for the volume.
Step 3: Calculating Bounds for Density
Why we do this:
Now we calculate the bounds for Density.
To get the Lower Bound of Density, we want the smallest possible number. In a fraction (\(\frac{\text{Mass}}{\text{Volume}}\)), we get the smallest result by taking the smallest possible numerator (mass) and dividing by the largest possible denominator (volume).
To get the Upper Bound of Density, we want the largest possible result, so we divide the largest mass by the smallest volume.
Working:
\[ \text{Density LB} = \frac{\text{Mass LB}}{\text{Volume UB}} = \frac{1967.5}{462.5409375} = 0.4253677… \]✓ (P1) For a correct process to find a bound for density.
\[ \text{Density UB} = \frac{\text{Mass UB}}{\text{Volume LB}} = \frac{1972.5}{454.0925125} = 0.4343828… \]✓ (A1) For both correct bounds: \(0.425…\) and \(0.434…\)
Step 4: Determining the suitable degree of accuracy
What this tells us:
We know the true density must be somewhere between \(0.4253…\) and \(0.4343…\). We must round both answers until they match. Let’s look at them digit by digit:
- 1 decimal place: \(0.425…\) rounds to \(0.4\), and \(0.434…\) rounds to \(0.4\). They match, but let’s see if we can be more accurate.
- 2 decimal places: \(0.4253…\) rounds to \(0.43\). \(0.4343…\) rounds to \(0.43\). They match!
- 3 decimal places: \(0.425\) vs \(0.434\). They do not match.
Since both bounds round to \(0.43\) when rounded to \(2\) decimal places (or \(2\) significant figures), this is our suitable degree of accuracy.
Working:
Both the Upper Bound and the Lower Bound round to \(0.43\) to \(2\) decimal places.
✓ (C1) For a correct statement on the degree of accuracy (e.g., both round to \(0.43\) to \(2\) decimal places).
Final Answer:
\(0.43\text{ g/cm}^{3}\)
Reason: Both the lower bound (\(0.425…\)) and upper bound (\(0.434…\)) round to \(0.43\) to \(2\) decimal places (or \(2\) significant figures).
Total: 5 marks