GCSE Maths Foundation Paper 2 – June 2018
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 (Percentages)
- Question 2 (Rounding)
- Question 3 (Powers)
- Question 4 (Place Value)
- Question 5 (Unit Conversions)
- Question 6 (Number Properties)
- Question 7 (Combinations)
- Question 8 (Money/Best Buy)
- Question 9 (Speed/Distance/Time)
- Question 10 (Prime Numbers)
- Question 11 (Solving Algebra)
- Question 12 (Pie Charts)
- Question 13 (Probability)
- Question 14 (Frequency Tree)
- Question 15 (Geometry Reasoning)
- Question 16 (Averages/Tables)
- Question 17 (Ratio/Recipes)
- Question 18 (Transformations)
- Question 19 (Circles/Perimeter)
- Question 20 (Indices)
- Question 21 (LCM & HCF)
- Question 22 (Linear Graphs)
- Question 23 (Ratio/VAT)
- Question 24 (Quadratic Graphs)
- Question 25 (Force/Pressure)
- Question 26 (Volume of Prism)
Question 1 (1 mark)
Write \( \frac{4}{50} \) as a percentage.
Worked Solution
Step 1: Convert to a denominator of 100
Why we do this: “Percent” means “out of 100”. We need to find an equivalent fraction with 100 at the bottom.
Final Answer: 8%
✓ (B1)
Question 2 (1 mark)
Write 1.59 correct to 1 decimal place.
Worked Solution
Step 1: Check the critical digit
How to round: We want 1 decimal place, so we look at the 2nd decimal place (the 9). Since 9 is 5 or greater, we round the previous digit (5) up.
1.59 → Round up the 5 to 6.
1.6
Final Answer: 1.6
✓ (B1)
Question 3 (1 mark)
Work out the value of \( 3^5 \).
Worked Solution
Step 1: Expand the power
Meaning of power: The power of 5 means we multiply the base (3) by itself 5 times.
\( 3 \times 3 = 9 \)
\( 9 \times 3 = 27 \)
\( 27 \times 3 = 81 \)
\( 81 \times 3 = 243 \)
Final Answer: 243
✓ (B1)
Question 4 (1 mark)
Write down a 6 digit number that has 4 as its thousands digit.
You can only use the digit 4 once.
Worked Solution
Step 1: Understanding place value
Place value slots: [Hundred Thousands] [Ten Thousands] [Thousands] [Hundreds] [Tens] [Units]
We need to put a 4 in the Thousands slot and fill the rest with any other digits (except 4).
Example: 5 6 4 0 0 0
Other examples: 124,567 or 984,321
Final Answer: Any 6-digit number with 4 in the 4th position from right, e.g., 564,000
✓ (B1)
Question 5 (3 marks)
(a) Change 35 cm to mm.
(b) Change 7700 millilitres to litres.
(c) Change 0.32 kilograms to grams.
Worked Solution
Part (a): cm to mm
Conversion factor: 1 cm = 10 mm. To go from cm to mm, we multiply by 10.
✓ (B1)
Part (b): ml to litres
Conversion factor: 1000 ml = 1 litre. To go from ml to litres, we divide by 1000.
✓ (B1)
Part (c): kg to grams
Conversion factor: 1 kg = 1000 grams. To go from kg to grams, we multiply by 1000.
✓ (B1)
Final Answers: (a) 350 mm, (b) 7.7 litres, (c) 320 grams
Question 6 (3 marks)
Margaret is thinking of a number.
She says, “My number is odd. It is a factor of 36 and a multiple of 3.”
There are two possible numbers Margaret can be thinking of.
Write down these two numbers.
Worked Solution
Step 1: List factors of 36
What is a factor? Numbers that divide into 36 exactly.
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Step 2: Apply conditions
Condition 1: Must be ODD.
Condition 2: Must be a Multiple of 3.
Odd factors: 1, 3, 9
Which of these are multiples of 3?
- 1 is NOT a multiple of 3
- 3 IS a multiple of 3
- 9 IS a multiple of 3
Final Answer: 3 and 9
✓ (P1) for listing factors/multiples, ✓ (A2) for both correct
Question 7 (2 marks)
Mohsin, Yusuf and Luke are going to play a game.
At the end of the game, one of them will be in First place, one of them will be in Second place and one of them will be in Third place.
Use the table below to list all the possible outcomes of the game.
Worked Solution
Step 1: List outcomes systematically
Strategy: Fix the person in First place, then swap the other two.
| First Place | Second Place | Third Place |
|---|---|---|
| Mohsin (M) | Yusuf (Y) | Luke (L) |
| Mohsin (M) | Luke (L) | Yusuf (Y) |
| Yusuf (Y) | Mohsin (M) | Luke (L) |
| Yusuf (Y) | Luke (L) | Mohsin (M) |
| Luke (L) | Mohsin (M) | Yusuf (Y) |
| Luke (L) | Yusuf (Y) | Mohsin (M) |
Final Answer: (M,Y,L), (M,L,Y), (Y,M,L), (Y,L,M), (L,M,Y), (L,Y,M)
✓ (M1) for at least 3 correct, ✓ (A1) for all 6 correct
Question 8 (5 marks)
Neil buys 30 pens, 30 pencils, 30 rulers and 30 pencil cases.
pens: 6 for 82p
pencils: 15 for 45p
rulers: 10 for £1.25
pencil cases: 37p each
What is the total amount of money Neil spends?
Worked Solution
Step 1: Calculate cost of Pens
Pack size: Pens come in packs of 6. We need 30 pens.
Number of packs = \( 30 \div 6 = 5 \) packs.
✓ (P1)
Step 2: Calculate cost of Pencils
Pack size: Pencils come in packs of 15.
Number of packs = \( 30 \div 15 = 2 \) packs.
✓ (P1)
Step 3: Calculate cost of Rulers
Pack size: Rulers come in packs of 10.
Number of packs = \( 30 \div 10 = 3 \) packs.
✓ (P1)
Step 4: Calculate cost of Pencil Cases
Individual items: These are sold singly.
Step 5: Total Sum
✓ (P1) for summation, ✓ (A1) for correct total
Final Answer: £19.85
Question 9 (4 marks)
Emily drives 186 miles in 3 hours.
(a) What is her average speed?
Sarah drives at an average speed of 58 mph for 4 hours.
(b) How many miles does Sarah drive?
Worked Solution
Part (a): Finding Speed
Formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
✓ (M1) for division, ✓ (A1) for 62
Part (b): Finding Distance
Formula: \( \text{Distance} = \text{Speed} \times \text{Time} \)
✓ (M1) for multiplication, ✓ (A1) for 232
Final Answer: (a) 62 mph, (b) 232 miles
Question 10 (3 marks)
(a) Write down all the prime numbers between 20 and 30.
Catherine says, “2 is the only even prime number.”
(b) Is Catherine right? You must give a reason for your answer.
Worked Solution
Part (a): Primes between 20 and 30
Definition: A prime number has exactly two factors: 1 and itself.
Let’s check numbers 20-30:
- 20, 22, 24, 26, 28, 30 (Even, not prime)
- 21 (Divisible by 3, 7)
- 25 (Divisible by 5)
- 27 (Divisible by 3, 9)
- 23 (Prime)
- 29 (Prime)
Answer: 23, 29
✓ (B2)
Part (b): Is 2 the only even prime?
Reasoning: Any even number greater than 2 can be divided by 2. Therefore, it will have more than two factors (1, 2, itself), making it NOT prime.
Answer: Yes, because all other even numbers are divisible by 2.
✓ (C1)
Question 11 (3 marks)
(a) Solve \( x + x + x = 51 \)
(b) Solve \( \frac{y}{4} = 3 \)
(c) Solve \( 2f + 7 = 18 \)
Worked Solution
Part (a)
✓ (B1)
Part (b)
Method: Multiply both sides by 4 to remove the fraction.
✓ (B1)
Part (c)
Step 1: Subtract 7 from both sides.
Step 2: Divide by 2.
✓ (B1)
Final Answers: (a) 17, (b) 12, (c) 5.5
Question 12 (3 marks)
A group of football fans were asked what their half time snack was.
| Snack | Number of fans |
|---|---|
| burger | 11 |
| pie | 17 |
| hot dog | 8 |
Draw an accurate pie chart for this information.
Worked Solution
Step 1: Calculate Angles
Total fans: \( 11 + 17 + 8 = 36 \)
Degrees per person: \( 360^\circ \div 36 = 10^\circ \)
Burger: \( 11 \times 10^\circ = 110^\circ \)
Pie: \( 17 \times 10^\circ = 170^\circ \)
Hot dog: \( 8 \times 10^\circ = 80^\circ \)
Check: \( 110 + 170 + 80 = 360^\circ \) ✓
✓ (M1)
Step 2: Draw Pie Chart
✓ (A2) for fully correct labelled chart
Question 13 (2 marks)
A scout group has a raffle to raise money for charity. There is 1 prize to be won.
Laura buys 12 raffle tickets.
A total of 350 raffle tickets are sold.
Find the probability that Laura does not win the prize.
Worked Solution
Step 1: Probability of NOT winning
Method: If there is 1 winning ticket, all other tickets are losing tickets.
Total tickets = 350.
Winning tickets owned by Laura = 12? No, Laura buys 12. If she wins, one of her 12 is the winner.
Wait, “Probability Laura does not win”. This means NONE of her 12 tickets is the winner.
Or simply: Probability = (Total tickets NOT owned by Laura) / Total.
Tickets NOT owned by Laura = \( 350 – 12 = 338 \)
Probability = \( \frac{338}{350} \)
✓ (M1)
Final Answer: \( \frac{338}{350} \) (or 0.966…)
✓ (A1)
Question 14 (3 marks)
Each worker in a factory is either left-handed or right-handed.
22 of the 45 workers are male.
16 of the 34 right-handed workers are female.
Complete the frequency tree for this information.
Worked Solution
Step 1: Fill in known values
Total workers = 45
Male = 22
Therefore, Female = \( 45 – 22 = 23 \)
Step 2: Work backwards from “Right-handed” info
Total Right-handed = 34 (This isn’t directly on the tree, but we know parts sum to it? No, usually the tree splits gender then hand, or hand then gender? The diagram shows 45 -> Male/Female -> Left/Right).
Let’s trace the diagram structure provided in the exam paper:
45 -> (Male, Female)
Male -> (Left, Right)
Female -> (Left, Right)
Given:
1. 16 of the 34 right-handed workers are female. -> So Female Right-handed = 16.
2. Total Right-handed = 34. So Male Right-handed = \( 34 – 16 = 18 \).
3. Total Male = 22. We know Male Right = 18. So Male Left-handed = \( 22 – 18 = 4 \).
4. Total Female = 23. We know Female Right = 16. So Female Left-handed = \( 23 – 16 = 7 \).
✓ (C3) for fully correct tree
Completed Frequency Tree
Question 15 (2 marks)
(a) Mary needs to work out the size of angle \( x \) in a triangle with base angles at A and B, and vertex at C. Angle B = 63°.
She writes: \( x = 63^\circ \) because base angles of an isosceles triangle are equal.
Mary is wrong. Explain why.
(b) William needs to work out angle \( y \) where a transversal crosses two parallel lines. Angle 57° and angle \( y \) form a ‘Z’ shape.
William writes: \( y = 180^\circ – 57^\circ = 123^\circ \) because angles on a straight line add up to 180°.
One of William’s reasons is wrong. Write down the correct reason.
Worked Solution
Part (a)
Analysis: In the diagram (which shows tick marks on AC and BC), AC = BC. This means the angles opposite these sides are equal. So Angle A = Angle B = 63°.
\( x \) is the angle at C (the top), not a base angle.
Answer: \( x \) is not a base angle. (Wait, strictly, we calculate \( x = 180 – 63 – 63 = 54^\circ \)).
✓ (C1)
Part (b)
Analysis: The angles form a ‘Z’ shape between parallel lines. These are Alternate Angles.
Alternate angles are equal.
Answer: Alternate angles are equal. (So \( y = 57^\circ \))
✓ (C1)
Question 16 (3 marks)
Marla buys some bags of buttons.
There are 19 buttons or 20 buttons or 21 buttons or 22 buttons in each bag.
The table gives some information about the number of buttons in each bag.
The total number of buttons is 320. Complete the table.
| Number of buttons | Frequency |
|---|---|
| 19 | … |
| 20 | 7 |
| 21 | 3 |
| 22 | 1 |
Worked Solution
Step 1: Calculate buttons from known bags
\( 20 \times 7 = 140 \) buttons
\( 21 \times 3 = 63 \) buttons
\( 22 \times 1 = 22 \) buttons
✓ (P1)
Step 2: Find buttons remaining
Sum so far: \( 140 + 63 + 22 = 225 \)
Total needed: 320
Buttons remaining for the ’19’ bags: \( 320 – 225 = 95 \)
✓ (P1)
Step 3: Calculate missing frequency
Each bag holds 19 buttons.
\( 95 \div 19 = 5 \)
Final Answer: 5
✓ (A1)
Question 17 (3 marks)
Here is the list of ingredients for making 30 biscuits:
- 225 g butter
- 110 g caster sugar
- 275 g plain flour
- 75 g chocolate chips
Lucas has:
- 900 g butter
- 1000 g caster sugar
- 1000 g plain flour
- 225 g chocolate chips
What is the greatest number of biscuits Lucas can make?
Worked Solution
Step 1: Calculate maximum batches for each ingredient
Strategy: Divide the amount Lucas has by the amount needed for one batch (30 biscuits) to see how many batches he can make.
Butter: \( 900 \div 225 = 4 \) batches
Sugar: \( 1000 \div 110 = 9.09… \) batches
Flour: \( 1000 \div 275 = 3.63… \) batches
Chips: \( 225 \div 75 = 3 \) batches
✓ (P1)
Step 2: Identify limiting ingredient
The ingredient that runs out first limits the production. Here, it’s the Chocolate Chips (only 3 batches).
Max batches = 3
Total biscuits = \( 3 \times 30 = 90 \)
✓ (P1)
Final Answer: 90 biscuits
✓ (A1)
Question 18 (2 marks)
Describe fully the single transformation that maps shape A onto shape B.
Worked Solution
Step 1: Identify the type of transformation
The shape has been flipped over. This is a Reflection.
Step 2: Find the mirror line
Shape A is 3 units right of the y-axis.
Shape B is 3 units left of the y-axis.
The mirror line is exactly in the middle, which is the y-axis (or line \( x = 0 \)).
Final Answer: Reflection in the y-axis (or x = 0)
✓ (B2)
Question 19 (5 marks)
A farmer has a field in the shape of a semicircle of diameter 50 m.
The farmer asks Jim to build a fence around the edge of the field.
Jim tells him how much it will cost.
Total cost = £29.86 per metre of fence plus £180 for each day’s work
Jim takes three days to build the fence.
Work out the total cost.
Worked Solution
Step 1: Calculate the Perimeter
The perimeter consists of the curved part (half circumference) plus the straight edge (diameter).
Circumference \( C = \pi d \)
Curved part = \( \frac{1}{2} \times \pi \times 50 = 25\pi \approx 78.54 \) m
Straight part = 50 m
Total Perimeter = \( 78.54 + 50 = 128.54 \) m
✓ (P1)
Step 2: Calculate Fence Material Cost
\( 128.54 \text{ m} \times \text{£}29.86 = \text{£}3838.20 \)
✓ (P1)
Step 3: Calculate Labour Cost
\( 3 \text{ days} \times \text{£}180 = \text{£}540 \)
✓ (P1)
Step 4: Total Cost
\( 3838.20 + 540 = 4378.20 \)
Final Answer: £4378.20
✓ (A1)
Question 20 (5 marks)
(a) Simplify \( m^3 \times m^4 \)
(b) Simplify \( (5np^3)^3 \)
(c) Simplify \( \frac{32q^9r^4}{4q^3r} \)
Worked Solution
Part (a)
Rule: When multiplying terms with the same base, add the powers.
✓ (B1)
Part (b)
Rule: Apply the power of 3 to EVERYTHING inside the bracket.
\( 5^3 = 125 \)
\( n^3 = n^3 \)
\( (p^3)^3 = p^{3 \times 3} = p^9 \)
Result: \( 125n^3p^9 \)
✓ (B2)
Part (c)
Rule: Divide numbers, subtract powers.
Numbers: \( 32 \div 4 = 8 \)
q terms: \( q^9 \div q^3 = q^{9-3} = q^6 \)
r terms: \( r^4 \div r^1 = r^{4-1} = r^3 \)
Result: \( 8q^6r^3 \)
✓ (B2)
Question 21 (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
Part (a)
Method: List multiples or use Prime Factors.
\( 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5 \)
\( 56 = 8 \times 7 = 2^3 \times 7 \)
LCM must contain highest power of each prime factor:
\( 2^3 \times 5 \times 7 \)
\( 8 \times 35 = 280 \)
✓ (A1)
Part (b)
Method: For HCF, take the LOWEST power of common factors.
2s: \( 2^2 \) (lowest)
3s: \( 3^1 \) (lowest)
5s: \( 5^1 \) (lowest)
HCF = \( 2^2 \times 3 \times 5 = 4 \times 15 = 60 \)
✓ (B1)
Question 22 (3 marks)
The line L is shown on the grid. Find an equation for L.
Worked Solution
Step 1: Find the y-intercept (c)
Where does the line cross the vertical y-axis?
It crosses at -6. So \( c = -6 \).
✓ (M1)
Step 2: Find the Gradient (m)
Formula: \( m = \frac{\text{change in } y}{\text{change in } x} \)
Pick two points: \( (0, -6) \) and \( (2, 0) \).
Rise: from -6 to 0 is +6.
Run: from 0 to 2 is +2.
\( m = \frac{6}{2} = 3 \)
✓ (M1)
Final Answer: \( y = 3x – 6 \)
✓ (A1)
Question 23 (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: Calculate Total Cost of Van
Net price = £8500
VAT = \( 20\% \text{ of } 8500 = 0.2 \times 8500 = 1700 \)
Total Cost = \( 8500 + 1700 = \text{£}10,200 \)
✓ (P1)
Step 2: Calculate Total of Monthly Payments
\( 12 \times 531.25 = \text{£}6375 \)
✓ (P1)
Step 3: Calculate Deposit
Deposit = Total Cost – Monthly Payments
Deposit = \( 10200 – 6375 = \text{£}3825 \)
✓ (P1)
Step 4: Form and Simplify Ratio
Ratio = Deposit : Payments
\( 3825 : 6375 \)
Divide by 25: \( 153 : 255 \)
Divide by 51: \( 3 : 5 \)
✓ (A1)
Final Answer: 3 : 5
Question 24 (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 |
(b) On the grid, draw the graph of \( y = x^2 – x – 6 \) for values of x from -3 to 3.
(c) Use your graph to find estimates of the solutions to the equation \( x^2 – x – 6 = -2 \)
Worked Solution
Part (a)
\( x = -2: (-2)^2 – (-2) – 6 = 4 + 2 – 6 = 0 \)
\( x = -1: (-1)^2 – (-1) – 6 = 1 + 1 – 6 = -4 \)
\( x = 1: 1^2 – 1 – 6 = -6 \)
\( x = 2: 2^2 – 2 – 6 = 4 – 2 – 6 = -4 \)
\( x = 3: 3^2 – 3 – 6 = 9 – 3 – 6 = 0 \)
Missing values: 0, -4, -6, -4, 0
✓ (B2)
Part (c)
Method: Draw the line \( y = -2 \) across the graph. Find the x-values where this line crosses the curve.
Answer: \( x = -1.6 \) and \( x = 2.6 \) (approx)
✓ (A1)
Question 25 (3 marks)
A force of 70 newtons acts on an area of 20 cm².
\( \text{pressure} = \frac{\text{force}}{\text{area}} \)
The force is increased by 10 newtons.
The area is increased by 10 cm².
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 Original Pressure
Step 2: Calculate New Pressure
New Force = \( 70 + 10 = 80 \)
New Area = \( 20 + 10 = 30 \)
\[ P_2 = \frac{80}{30} = 2.66… \]✓ (P1)
Step 3: Calculate Percentage Decrease
Decrease = \( 3.5 – 2.66… = 0.833… \)
Percentage Decrease = \( \frac{0.833…}{3.5} \times 100 \)
\( = 23.8\% \)
✓ (P1)
Conclusion: The decrease is ~23.8%, which is MORE than 20%.
Is Helen correct? No.
✓ (A1)
Question 26 (5 marks)
Here is a triangular prism.
Work out the volume of the prism.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Find the base of the triangular face
The front face is a right-angled triangle. We know the height (7.2) and hypotenuse (8.4). Use Pythagoras.
✓ (P1)
Step 2: Calculate Area of Triangle
✓ (P1)
Step 3: Calculate Volume
✓ (P1)
Final Answer: 280 cm³ (to 3 s.f.)
✓ (A1)
