GCSE Maths Nov 2018 – Paper 2 (Foundation) Interactive Practice
Mark Scheme Legend
- M1 = Method mark (correct method applied)
- A1 = Accuracy mark (correct answer)
- B1 = Independent mark (correct statement or value)
- P1 = Process mark (completing a process)
- C1 = Communication mark (clear explanation or reasoning)
Table of Contents
- Question 1 (Place Value)
- Question 2 (Square Numbers)
- Question 3 (Conversions)
- Question 4 (Roots)
- Question 5 (Fractions/Decimals)
- Question 6 (Ordering Fractions)
- Question 7 (Algebraic Simplification)
- Question 8 (Map Scales)
- Question 9 (Sequences)
- Question 10 (Number Machines)
- Question 11 (Percentages)
- Question 12 (Probability)
- Question 13 (Area & Perimeter)
- Question 14 (Probability Reasoning)
- Question 15 (Simple Interest)
- Question 16 (Transformations)
- Question 17 (Ratio & Proportion)
- Question 18 (Percentage Increase/Decrease)
- Question 19 (Solving & Factorising)
- Question 20 (Venn Diagrams)
- Question 21 (Scatter Graphs)
- Question 22 (Geometric Proof)
- Question 23 (Compound Interest)
- Question 24 (Area & Problem Solving)
- Question 25 (Gradients)
- Question 26 (Expanding & Factorising)
- Question 27 (Standard Form)
🧮 Calculator Paper
Calculator button sequences are shown for tricky calculations. You may use a calculator for all questions.
Question 1 (1 mark)
Write down the value of the 4 in the number 542.3
Worked Solution
Step 1: Identify Place Value
What are we looking for?
We need to determine the value of the digit 4 based on its position in the number.
Let’s look at the columns:
- 5 is in the Hundreds column
- 4 is in the Tens column
- 2 is in the Units column
- 3 is in the Tenths column
The 4 represents 4 tens.
Value = 40 or “tens”
✓ (B1) for 40 or tens
Final Answer:
40
Question 2 (1 mark)
Write down a square number that is also an odd number.
Worked Solution
Step 1: List Square Numbers
What is a square number?
A number obtained by multiplying an integer by itself (e.g., \(1 \times 1\), \(2 \times 2\)).
Let’s list the first few square numbers:
- \(1^2 = 1\) (Odd)
- \(2^2 = 4\) (Even)
- \(3^2 = 9\) (Odd)
- \(4^2 = 16\) (Even)
- \(5^2 = 25\) (Odd)
- \(6^2 = 36\) (Even)
- \(7^2 = 49\) (Odd)
- \(9^2 = 81\) (Odd)
Any odd square number is correct.
✓ (B1) for stating an odd square number (e.g., 1, 9, 25, 49, 81)
Final Answer:
9 (or 1, 25, 49, etc.)
Question 3 (2 marks)
(a) Change 4560 g into kg.
(b) Change 7.3 m into mm.
Worked Solution
Part (a): Grams to Kilograms
Conversion Fact: There are 1000 grams in 1 kilogram.
To change grams to kg, we divide by 1000.
✓ (B1) for 4.56
Part (b): Metres to Millimetres
Conversion Fact: There are 1000 millimetres in 1 metre (100 cm in a metre, 10 mm in a cm).
To change m to mm, we multiply by 1000.
✓ (B1) for 7300
Final Answer:
(a) 4.56 kg
(b) 7300 mm
Question 4 (1 mark)
Work out the cube root of 64
Worked Solution
Step 1: Understanding the Question
What is a cube root?
We are looking for a number that, when multiplied by itself three times, equals 64. (i.e., \(? \times ? \times ? = 64\))
Let’s test some numbers:
- \(2 \times 2 \times 2 = 8\)
- \(3 \times 3 \times 3 = 27\)
- \(4 \times 4 \times 4 = 16 \times 4 = 64\)
So, \(\sqrt[3]{64} = 4\).
Calculator Tip: Press SHIFT then √ (which usually gives \(\sqrt[3]{}\)), then type 64.
✓ (B1) for 4
Final Answer:
4
Question 5 (1 mark)
Write 0.31 as a fraction.
Worked Solution
Step 1: Convert Decimal to Fraction
Place Value:
The decimal 0.31 has two decimal places. The last digit is in the hundredths column.
So, we can write it directly over 100.
\[ 0.31 = \frac{31}{100} \]This cannot be simplified further as 31 is a prime number.
✓ (B1) for 31/100
Final Answer:
\(\frac{31}{100}\)
Question 6 (2 marks)
Here are four fractions.
\[ \frac{3}{4} \quad \frac{5}{7} \quad \frac{19}{25} \quad \frac{11}{15} \]Write the fractions in order of size.
Start with the smallest fraction.
Worked Solution
Step 1: Convert to Decimals
Why do this?
It’s much easier to compare decimals than fractions with different denominators. Use your calculator to divide the numerator by the denominator.
- \(\frac{3}{4} = 3 \div 4 = 0.75\)
- \(\frac{5}{7} = 5 \div 7 = 0.71428…\)
- \(\frac{19}{25} = 19 \div 25 = 0.76\)
- \(\frac{11}{15} = 11 \div 15 = 0.7333…\)
✓ (M1) for converting into decimals or percentages (at least two correct)
Step 2: Order the Decimals
Ordering from smallest to largest:
- \(0.714…\) (\(\frac{5}{7}\))
- \(0.733…\) (\(\frac{11}{15}\))
- \(0.75\) (\(\frac{3}{4}\))
- \(0.76\) (\(\frac{19}{25}\))
✓ (A1) for correct order
Final Answer:
\(\frac{5}{7}, \frac{11}{15}, \frac{3}{4}, \frac{19}{25}\)
Question 7 (2 marks)
(a) Simplify \( 3m – m – m + 3m \)
(b) Simplify \( 2 \times n \times p \times 4 \)
Worked Solution
Part (a): Collecting Like Terms
We work from left to right:
\[ 3m – m = 2m \] \[ 2m – m = 1m \] \[ 1m + 3m = 4m \]✓ (B1) for 4m
Part (b): Multiplying Terms
Multiply the numbers first, then the letters:
\[ 2 \times 4 = 8 \] \[ n \times p = np \]Combine them:
\[ 8np \]✓ (B1) for 8np
Final Answer:
(a) \(4m\)
(b) \(8np\)
Question 8 (2 marks)
A map has a scale of 1 cm to 14 km.
On the map, the distance between Manchester and London is 18.8 cm.
What is the real distance, in km, between Manchester and London?
Worked Solution
Step 1: Use the Scale
The scale tells us that every 1 cm on the map represents 14 km in real life.
We have 18.8 cm, so we need to multiply 18.8 by 14.
✓ (M1) for using the scale (14 × 18.8)
Step 2: Calculate
✓ (A1) for 263.2
Final Answer:
263.2 km
Question 9 (4 marks)
(a) The \(n\)th term of a sequence is \(3n + 4\).
Explain why 21 is not a term of this sequence.
(b) Here are the first three terms of a different sequence.
\[ 1 \quad 2 \quad 4 \]Write down two numbers that could be the 4th term and the 5th term of this sequence.
Give the rule you have used to get your numbers.
Worked Solution
Part (a): Checking if 21 is in the sequence
Method 1: Solve Equation
Set the \(n\)th term equal to 21 and see if \(n\) is a whole number.
Since \(n\) is not a whole number (integer), 21 is not in the sequence.
Method 2: List Terms
- \(n=1 \rightarrow 3(1)+4 = 7\)
- \(n=2 \rightarrow 3(2)+4 = 10\)
- …
- \(n=5 \rightarrow 3(5)+4 = 19\)
- \(n=6 \rightarrow 3(6)+4 = 22\)
The sequence goes …, 19, 22, … so it skips 21.
✓ (C2) for full explanation (e.g., solving to find non-integer n or showing terms 19 and 22)
Part (b): Extending the Sequence
We have terms 1, 2, 4. We need to find a pattern.
Option 1: Doubling Pattern
Rule: Multiply by 2 each time (Geometric Progression)
Sequence: 1, 2, 4, 8, 16
Option 2: Adding Pattern
Rule: Add 1, then add 2, then add 3… (Differences increase by 1)
Sequence: 1 (+1) → 2 (+2) → 4 (+3) → 7 (+4) → 11
Most common answer:
Terms: 8, 16
Rule: Double the previous term (or “Multiply by 2”)
✓ (B1) for stating two terms (e.g., 8, 16 or 7, 11)
✓ (C1) for explanation (e.g., “doubling” or “add 1 more each time”)
Question 10 (3 marks)
Here is a number machine.
(a) Work out the output when the input is 8
(b) Work out the input when the output is 28
Worked Solution
Part (a): Forward Calculation
Input = 8
Step 1: Multiply by 5 → \( 8 \times 5 = 40 \)
Step 2: Subtract 2 → \( 40 – 2 = 38 \)
✓ (B1) for 38
Part (b): Reverse Calculation
Reverse Operations: To find the input from the output, we go backwards and do the opposite operations.
- Opposite of “Subtract 2” is “Add 2”
- Opposite of “Multiply by 5” is “Divide by 5”
Output = 28
Step 1: Add 2 → \( 28 + 2 = 30 \)
Step 2: Divide by 5 → \( 30 \div 5 = 6 \)
✓ (M1) for inverse operations (+2 then ÷5)
✓ (A1) for 6
Final Answer:
(a) 38
(b) 6
Question 11 (3 marks)
Adam gets a bonus of 30% of £80
Katy gets a bonus of £28
Work out the difference between the bonus Adam gets and the bonus Katy gets.
Worked Solution
Step 1: Calculate Adam’s Bonus
We need to find 30% of £80. We can convert 30% to a decimal (0.30) and multiply.
So, Adam gets £24.
✓ (M1) for \(\frac{30}{100} \times 80\) or 24
Step 2: Calculate the Difference
Katy gets £28.
Adam gets £24.
Difference = \( 28 – 24 = 4 \)
✓ (M1) for finding the difference (28 – 24)
✓ (A1) for 4
Final Answer:
£4
Question 12 (2 marks)
There are 49 counters in a bag.
20 of the counters are red.
The rest of the counters are blue.
One of the counters is taken at random.
Find the probability that the counter is blue.
Worked Solution
Step 1: Find number of blue counters
✓ (P1) for finding number of blue counters (29)
Step 2: Write as a probability
✓ (A1) for \(\frac{29}{49}\) (or 0.59…)
Final Answer:
\(\frac{29}{49}\)
Question 13 (5 marks)
A square has an area of 81 cm\(^2\).
(a) Find the perimeter of the square.
The diagram shows a right-angled triangle and a parallelogram.
The area of the parallelogram is 5 times the area of the triangle.
The perpendicular height of the parallelogram is \(h\) cm.
(b) Find the value of \(h\).
Worked Solution
Part (a): Perimeter of Square
Step 1: Find the side length
Area of square = \(side \times side\). So, side = \(\sqrt{\text{Area}}\).
Step 2: Find Perimeter
\[ \text{Perimeter} = 9 \times 4 = 36 \text{ cm} \]✓ (P1) for finding side length 9
✓ (A1) for 36
Part (b): Finding h
Step 1: Area of Triangle
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ \text{Area} = \frac{1}{2} \times 9 \times 16 = 72 \text{ cm}^2 \]✓ (M1) for finding area of triangle (72)
Step 2: Area of Parallelogram
Area of parallelogram = \(5 \times \text{Area of Triangle}\)
\[ \text{Area} = 5 \times 72 = 360 \text{ cm}^2 \]Step 3: Find h
Area of parallelogram = \(\text{base} \times \text{perpendicular height}\)
\[ 360 = 30 \times h \] \[ h = \frac{360}{30} = 12 \]✓ (M1) for equation \(30 \times h = 360\) or \(h = 360 \div 30\)
✓ (A1) for 12
Final Answer:
(a) 36 cm
(b) \(h = 12\)
Question 14 (4 marks)
Victoria throws an ordinary fair 6-sided dice once.
She says, “The probability of getting a 3 is half the probability of getting a 6”
(a) Is Victoria correct? You must explain your answer.
Andy throws the dice twice.
He says, “The probability of getting a 6 on both throws is \(\frac{2}{6}\)”
(b) Is Andy correct? You must explain your answer.
Indre throws the dice once. She also throws a coin to get Heads or Tails.
(c) List all the possible outcomes she can get.
Worked Solution
Part (a)
No.
On a fair dice, the probability of getting a 3 is \(\frac{1}{6}\).
The probability of getting a 6 is also \(\frac{1}{6}\).
They are equal, not half.
✓ (C1) for “No” with correct explanation
Part (b)
No.
When events happen together (independent events), we multiply the probabilities, not add them.
\[ P(6 \text{ and } 6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \]Andy has incorrectly added them (\(\frac{1}{6} + \frac{1}{6} = \frac{2}{6}\)).
✓ (C1) for “No” with correct explanation
Part (c): Sample Space
We combine the Dice outcomes (1,2,3,4,5,6) with the Coin outcomes (H, T).
Possible outcomes:
- 1H, 2H, 3H, 4H, 5H, 6H
- 1T, 2T, 3T, 4T, 5T, 6T
✓ (B2) for all 12 correct outcomes
Question 15 (3 marks)
Remi invests £600 for 5 years in a savings account.
By the end of the 5 years he has received a total of £75 simple interest.
Work out the annual rate of simple interest.
Worked Solution
Step 1: Interest per year
Since it is simple interest, the same amount is paid each year. The total is £75 over 5 years.
✓ (M1) for 75 ÷ 5 or 15
Step 2: Calculate Percentage Rate
We need to find what percentage £15 is of the original £600.
✓ (M1) for method to find percentage (15/600 * 100)
✓ (A1) for 2.5%
Final Answer:
2.5%
Question 16 (2 marks)
Describe fully the single transformation that maps shape A onto shape B.
Worked Solution
Step 1: Identify Transformation Type
The shape has flipped over. It hasn’t turned (rotation) or slid (translation) or changed size (enlargement).
This is a Reflection.
Step 2: Identify Mirror Line
The mirror line is exactly halfway between Shape A and Shape B.
Shape A is above the x-axis, Shape B is below it, at the same distance.
The mirror line is the x-axis (or the line \(y = 0\)).
✓ (B1) for Reflection
✓ (B1) for in the x-axis (or y=0)
Final Answer:
Reflection in the x-axis
Question 17 (3 marks)
Adrian is going to make concrete. He is going to use:
- 180 kg of cement
- 375 kg of sand
- 1080 kg of stone
Cement, sand and stone are sold in bags.
| 1 bag cement | 1 bag sand | 1 bag stone |
|---|---|---|
| 25 kg | 22.5 kg | 50 kg |
Adrian already has:
- 10 bags of cement
- 20 bags of sand
- 20 bags of stone
Work out what bags he needs to buy to make the concrete.
Worked Solution
Step 1: Calculate amount already owned
Cement: \( 10 \text{ bags} \times 25 \text{ kg} = 250 \text{ kg} \)
Sand: \( 20 \text{ bags} \times 22.5 \text{ kg} = 450 \text{ kg} \)
Stone: \( 20 \text{ bags} \times 50 \text{ kg} = 1000 \text{ kg} \)
✓ (P1) for calculating total weight he has (or bags needed)
Step 2: Compare with what is needed
Cement: Needs 180 kg. Has 250 kg. (Enough)
Sand: Needs 375 kg. Has 450 kg. (Enough)
Stone: Needs 1080 kg. Has 1000 kg.
He needs \( 1080 – 1000 = 80 \text{ kg} \) more stone.
Step 3: Calculate bags to buy
He needs 80 kg of stone.
1 bag of stone = 50 kg.
1 bag = 50 kg (Not enough)
2 bags = 100 kg (Enough)
He needs to buy 2 bags of stone.
✓ (P1) for complete process
✓ (C1) for 2 bags of stone
Final Answer:
2 bags of stone
Question 18 (2 marks)
Bill wants to increase 150 by 3%
He writes down:
\(150 \times 1.3 = 195\)
Bill’s method is wrong.
(a) Explain why.
Sally wants to decrease 150 by 3%
(b) Complete this statement to show how Sally can decrease 150 by 3%
\(150 \times \dots\dots\dots = \dots\dots\dots\)
Worked Solution
Part (a): Explanation
Bill used 1.3, which represents an increase of 30% (since \(1.3 = 130\%\)).
For 3%, the decimal should be 0.03, so the multiplier should be \(1 + 0.03 = 1.03\).
✓ (C1) for explanation (e.g., 1.3 is 30% or should use 1.03)
Part (b): Percentage Decrease
A 3% decrease means we start with 100% and subtract 3%.
\[ 100\% – 3\% = 97\% \]As a decimal multiplier, \(97\% = 0.97\).
\[ 150 \times 0.97 = 145.5 \]✓ (B1) for 0.97 and 145.5
Final Answer:
(a) 1.3 increases by 30%, not 3%.
(b) \(150 \times 0.97 = 145.5\)
Question 19 (4 marks)
(a) Solve \( 3(x – 4) = 12 \)
(b) Factorise fully \( 9b – 3b^2 \)
Worked Solution
Part (a): Solving
Method 1: Expand brackets first
\[ 3x – 12 = 12 \] \[ 3x = 12 + 12 \] \[ 3x = 24 \] \[ x = 8 \]Method 2: Divide first
Divide both sides by 3:
\[ x – 4 = \frac{12}{3} \] \[ x – 4 = 4 \] \[ x = 4 + 4 \] \[ x = 8 \]✓ (M1) for correct first step
✓ (A1) for 8
Part (b): Factorising
We need to find the Highest Common Factor (HCF) of \(9b\) and \(3b^2\).
Numbers: HCF of 9 and 3 is 3.
Letters: HCF of \(b\) and \(b^2\) is \(b\).
So, we take \(3b\) outside the bracket.
✓ (M1) for partial factorisation e.g., \(3(3b – b^2)\) or \(b(9 – 3b)\)
✓ (A1) for \(3b(3 – b)\)
Final Answer:
(a) \(x = 8\)
(b) \(3b(3 – b)\)
Question 20 (6 marks)
\(\mathscr{E} = \{ \text{even numbers between 1 and 25} \}\)
\(A = \{2, 8, 10, 14\}\)
\(B = \{6, 8, 20\}\)
\(C = \{8, 18, 20, 22\}\)
(a) Complete the Venn diagram for this information.
A number is chosen at random from \(\mathscr{E}\).
(b) Find the probability that the number is a member of \(A \cap B\).
Worked Solution
Part (a): Venn Diagram
First, list the Universal Set \(\mathscr{E}\):
Evens 1-25: {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
Now place the numbers:
- Intersection of all three (A ∩ B ∩ C): Is there a number in all lists? Yes, 8.
- Intersection A ∩ B (excluding C): Numbers in A and B but not C? None.
- Intersection B ∩ C (excluding A): Numbers in B and C but not A? 20.
- Intersection A ∩ C (excluding B): Numbers in A and C but not B? None.
- A only: Remaining in A: 2, 10, 14.
- B only: Remaining in B: 6.
- C only: Remaining in C: 18, 22.
- Outside circles: Numbers in \(\mathscr{E}\) not yet used: 4, 12, 16, 24.
Correct Placement:
- Centre (All 3): 8
- B ∩ C only: 20
- A only: 2, 10, 14
- B only: 6
- C only: 18, 22
- Outside: 4, 12, 16, 24
✓ (C4) for fully correct diagram
Part (b): Probability
\(A \cap B\) means numbers in BOTH A and B.
Looking at the diagram, the overlap of A and B contains {8}.
There is only 1 number in \(A \cap B\).
The total numbers in \(\mathscr{E}\) is 12.
\[ P(A \cap B) = \frac{1}{12} \]✓ (M1) for identifying 1 or 12
✓ (A1) for 1/12
Final Answer:
(a) See diagram above.
(b) \(\frac{1}{12}\)
Question 21 (2 marks)
Sean has information about the height, in cm, and the weight, in kg, of each of ten rugby players.
He is asked to draw a scatter graph and a line of best fit for this information.
Sean has plotted the points accurately.
Write down two things that are wrong with his answer.
Worked Solution
Identifying Errors
Looking at the graph, we can see two main issues:
- Line of Best Fit: The line forces its way through the bottom-left corner (140, 85). A line of best fit should go through the middle of the data points, following the trend. This line is too steep and leaves most points below it.
- X-Axis Scale (Height): Look at the numbers on the bottom axis. It goes 140, 160, 170…
- The gap from 140 to 160 is 20 units.
- The gap from 160 to 170 is 10 units.
- The number 150 is missing from the scale, or the scale is inconsistent (non-linear).
✓ (C1) for line of best fit error
✓ (C1) for scale error (150 missing)
Question 22 (4 marks)
BEG is a triangle.
ABC and DEF are parallel lines.
Work out the size of angle \(x\).
Give a reason for each stage of your working.
Worked Solution
Step 1: Find Angle ABE or CBE
We have parallel lines ABC and DEF with transversal BE.
Angle DEB = 110°.
Angle CBE and Angle DEB are alternate angles (Z-angles), so they are equal.
Reason: Alternate angles are equal.
✓ (M1) for finding related angle (e.g., ABE=70 or CBE=110)
Step 2: Find Angle EBG inside the triangle
Angle CBE is made up of Angle CBG and Angle EBG.
\[ 110^\circ = 35^\circ + \text{Angle EBG} \] \[ \text{Angle EBG} = 110^\circ – 35^\circ = 75^\circ \]Step 3: Find Angle BEG inside the triangle
Angles on a straight line add up to 180°.
\[ \text{Angle BEG} = 180^\circ – 110^\circ – 25^\circ = 45^\circ \]Step 4: Find Angle x (BGE)
Angles in a triangle add up to 180°.
\[ x = 180^\circ – 75^\circ – 45^\circ \] \[ x = 60^\circ \]✓ (A1) for 60
✓ (C1) for complete reasons (Alternate angles, Angles on straight line, Angles in triangle)
Final Answer:
\(x = 60^\circ\)
Question 23 (5 marks)
Northern Bank has two types of account. Both accounts pay compound interest.
- Cash savings account: Interest 2.5% per annum
- Shares account: Interest 3.5% per annum
Ali invests £2000 in the cash savings account.
Ben invests £1600 in the shares account.
(a) Work out who will get the most interest by the end of 3 years. You must show all your working.
In the 3rd year the rate of interest for the shares account is changed to 4% per annum.
(b) Does this affect who will get the most interest by the end of 3 years? Give a reason for your answer.
Worked Solution
Part (a): Compound Interest Calculation
Formula: \( \text{Amount} = \text{Principal} \times (\text{Multiplier})^n \)
Ali (2.5%): Multiplier = 1.025
Ben (3.5%): Multiplier = 1.035
Ali’s Total:
\[ 2000 \times 1.025^3 = 2153.78 \]Interest = \( 2153.78 – 2000 = £153.78 \)
Ben’s Total:
\[ 1600 \times 1.035^3 = 1773.95 \]Interest = \( 1773.95 – 1600 = £173.95 \)
Comparison:
£173.95 > £153.78
Ben gets the most interest.
✓ (C1) Ben supported by correct values
Part (b): Changing Rate
If Ben’s rate increases to 4% in the 3rd year, he will get even more interest.
Since he already had more interest than Ali, increasing his rate will only widen the gap.
So, No, it does not change who gets the most interest.
✓ (C1) for correct conclusion with reason
Question 24 (5 marks)
The diagram shows a floor in the shape of a trapezium.
John is going to paint the floor.
Each 5 litre tin of paint costs £16.99
1 litre of paint covers an area of 2 m\(^2\)
John has £160 to spend on paint.
Has John got enough money to buy all the paint he needs? You must show how you get your answer.
Worked Solution
Step 1: Calculate Area of Floor
Area of Trapezium = \(\frac{a + b}{2} \times h\)
✓ (P1) for calculating area (91)
Step 2: Calculate Paint Needed
1 litre covers 2 m\(^2\).
\[ \text{Litres needed} = 91 \div 2 = 45.5 \text{ litres} \]Step 3: Calculate Tins Needed
Each tin is 5 litres.
\[ \text{Tins} = 45.5 \div 5 = 9.1 \text{ tins} \]You can’t buy 0.1 of a tin, so he must buy 10 tins.
✓ (P1) for finding integer number of tins (10)
Step 4: Calculate Cost
✓ (P1) for finding total cost
Step 5: Conclusion
He has £160.
Cost is £169.90.
£169.90 > £160.
No, he does not have enough money.
✓ (C1) for correct conclusion supported by figures
Question 25 (3 marks)
\(A\) is the point with coordinates \((5, 9)\)
\(B\) is the point with coordinates \((d, 15)\)
The gradient of the line \(AB\) is 3
Work out the value of \(d\).
Worked Solution
Step 1: Use Gradient Formula
Gradient \( m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 – y_1}{x_2 – x_1} \)
✓ (P1) for setting up the equation
Step 2: Solve for d
Multiply both sides by \((d – 5)\):
\[ 3(d – 5) = 6 \]Divide by 3:
\[ d – 5 = 2 \]Add 5:
\[ d = 7 \]✓ (A1) for 7
Final Answer:
\(d = 7\)
Question 26 (4 marks)
(a) Expand and simplify \( (5x + 2)(2x – 3) \)
(b) Factorise \( x^2 + 4x + 3 \)
Worked Solution
Part (a): Expand Double Brackets
Use FOIL (First, Outside, Inside, Last):
- First: \(5x \times 2x = 10x^2\)
- Outside: \(5x \times -3 = -15x\)
- Inside: \(2 \times 2x = 4x\)
- Last: \(2 \times -3 = -6\)
Collect like terms (\(-15x + 4x\)):
\[ 10x^2 – 11x – 6 \]✓ (A1) for correct expansion
Part (b): Factorise Quadratic
We need two numbers that multiply to make 3 and add to make 4.
Factors of 3: 1 and 3.
\(1 + 3 = 4\). This works.
✓ (A1) for correct factorisation
Final Answer:
(a) \(10x^2 – 11x – 6\)
(b) \((x + 1)(x + 3)\)
Question 27 (4 marks)
(a) Write the number 0.000 075 47 in standard form.
(b) Write \( 3.42 \times 10^4 \) as an ordinary number.
(c) Work out \( \frac{2.3 \times 10^4 \times 6.7 \times 10^3}{5 \times 10^{-8}} \)
Worked Solution
Part (a)
Move the decimal point 5 places to the right to get 7.547.
\[ 7.547 \times 10^{-5} \]✓ (B1) for \(7.547 \times 10^{-5}\)
Part (b)
Move the decimal point 4 places to the right.
\[ 3.42 \rightarrow 34200 \]✓ (B1) for 34200
Part (c)
Step 1: Calculate Numerator
\[ (2.3 \times 6.7) \times (10^4 \times 10^3) \] \[ 15.41 \times 10^7 \]Step 2: Divide by Denominator
\[ \frac{15.41}{5} \times \frac{10^7}{10^{-8}} \] \[ 3.082 \times 10^{7 – (-8)} \] \[ 3.082 \times 10^{15} \]✓ (M1) for correct process or one correct calculation
✓ (A1) for \(3.082 \times 10^{15}\)
