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GCSE Nov 2018 Edexcel Foundation Paper 3
Mark Scheme Legend
- M1 = Method mark
- A1 = Accuracy mark
- B1 = Independent mark
- C1 = Communication mark
- P1 = Process mark
- oe = or equivalent
- cao = correct answer only
- ft = follow through
Table of Contents
- Question 1 (Arithmetic)
- Question 2 (Fractions/Percentages)
- Question 3 (Square Roots)
- Question 4 (Fractions of Amount)
- Question 5 (3-D Shapes)
- Question 6 (Probability Scale)
- Question 7 (Unit Conversion)
- Question 8 (Money/Best Buy)
- Question 9 (Congruence)
- Question 10 (Reflection)
- Question 11 (Percentages)
- Question 12 (Pie Charts)
- Question 13 (Money Problems)
- Question 14 (Probability)
- Question 15 (Multiples)
- Question 16 (Ratio)
- Question 17 (Algebra/Perimeter)
- Question 18 (VAT & Interest)
- Question 19 (Inequalities)
- Question 20 (Calculator Use)
- Question 21 (Percentage Increase)
- Question 22 (Quadratic Graphs)
- Question 23 (Bar Charts & Mean)
- Question 24 (Speed, Distance, Time)
- Question 25 (Trigonometry)
- Question 26 (Sequences)
Question 1 (2 marks)
Write a number in each box to make the calculation correct.
(i) \( 56.3 + \square = 100 \)
(ii) \( \frac{2}{7} + \square = 1 \)
Worked Solution
Step 1: Part (i) – Missing Number
What we need to find: The number that adds to 56.3 to make 100. This is the difference between 100 and 56.3.
Calculation: \( 100 – 56.3 \)
Calculator: 100 - 56.3 = 43.7
Check: \( 56.3 + 43.7 = 100 \)
Step 2: Part (ii) – Completing the Fraction
Reasoning: A whole (1) is made up of 7 sevenths (\(\frac{7}{7}\)). We have \(\frac{2}{7}\), so we need to find how many more sevenths are needed.
\( 1 – \frac{2}{7} = \frac{7}{7} – \frac{2}{7} \)
\( = \frac{5}{7} \)
Final Answer:
(i) 43.7 ✓ (B1)
(ii) \(\frac{5}{7}\) ✓ (B1)
Question 2 (1 mark)
Write \(3\%\) as a fraction.
Worked Solution
Step 1: Understanding Percentages
Definition: “Percent” means “out of 100”.
\( 3\% = \frac{3}{100} \)
This fraction cannot be simplified further as 3 is a prime number and does not divide into 100.
Final Answer: \(\frac{3}{100}\) ✓ (B1)
Question 3 (1 mark)
Find \(\sqrt{1.44}\)
Worked Solution
Step 1: Calculate Square Root
Method: Use a calculator or recognize that \(12^2 = 144\), so \(\sqrt{144} = 12\), which helps estimate decimal placement.
Calculator: √ 1.44 =
Result: \(1.2\)
Check: \(1.2 \times 1.2 = 1.44\)
Final Answer: 1.2 ✓ (B1)
Question 4 (1 mark)
Work out \(\frac{1}{8}\) of \(720\)
Worked Solution
Step 1: Calculation
Meaning: Finding \(\frac{1}{8}\) of a number is the same as dividing the number by 8.
\( 720 \div 8 \)
We know \(72 \div 8 = 9\), so \(720 \div 8 = 90\).
Calculator: 720 ÷ 8 = 90
Final Answer: 90 ✓ (B1)
Question 5 (2 marks)
Here is a 3-D shape.
(a) Write down the name of this 3-D shape.
(b) Write down the number of edges of this 3-D shape.
Worked Solution
Step 1: Identifying the Shape
The shape has 6 rectangular faces. A 3-D box shape like this is called a cuboid.
Name: Cuboid
Step 2: Counting Edges
Strategy: Count the edges in groups:
- Bottom edges: 4
- Top edges: 4
- Vertical edges connecting top and bottom: 4
Total = \( 4 + 4 + 4 = 12 \)
Final Answer:
(a) Cuboid ✓ (B1)
(b) 12 ✓ (B1)
Question 6 (2 marks)
An ordinary fair dice is thrown once.
(a) On the probability scale below, mark with a cross (\(\times\)) the probability that the dice lands on an odd number.
(b) Write down the probability that the dice lands on a number greater than 4.
Worked Solution
Step 1: Part (a) – Odd Numbers
Reasoning: A fair dice has numbers 1, 2, 3, 4, 5, 6.
The odd numbers are: 1, 3, 5 (Three numbers).
Total outcomes: 6.
Probability = \(\frac{3}{6} = \frac{1}{2}\)
We mark a cross exactly at \(\frac{1}{2}\) on the scale.
Step 2: Part (b) – Number greater than 4
Reasoning: We need numbers strictly greater than 4.
Numbers: 5, 6 (Two numbers).
Probability = \(\frac{2}{6}\)
This simplifies to \(\frac{1}{3}\), but \(\frac{2}{6}\) is correct.
Final Answer:
(a) Cross marked at \(\frac{1}{2}\) ✓ (B1)
(b) \(\frac{2}{6}\) (or \(\frac{1}{3}\)) ✓ (B1)
Question 7 (2 marks)
Shaun is \(1.88\) m tall.
David is \(6\) cm taller than Shaun.
How tall is David?
Worked Solution
Step 1: Unify Units
Problem: Shaun’s height is in metres, the difference is in centimetres.
We must work in the same units. Let’s convert metres to centimetres.
\(1 \text{ m} = 100 \text{ cm}\)
\(1.88 \text{ m} = 1.88 \times 100 = 188 \text{ cm}\)
Step 2: Add the Height Difference
David is 6 cm taller, so we add 6 to Shaun’s height.
\(188 \text{ cm} + 6 \text{ cm} = 194 \text{ cm}\)
Step 3: Convert Back (Optional)
\(194 \text{ cm} = 1.94 \text{ m}\)
Final Answer: 1.94 m (or 194 cm) ✓ (M1 A1)
Question 8 (4 marks)
2 pens cost £2.38
5 folders cost £5.60
Ben wants to buy 20 pens and 20 folders.
He only has £50.
Does Ben have enough money to buy 20 pens and 20 folders?
You must show how you get your answer.
Worked Solution
Step 1: Calculate Cost of Pens
Strategy: We know the cost of 2 pens. We need 20 pens.
How many sets of 2 fit into 20? \(20 \div 2 = 10\) sets.
Cost of 20 pens = \(10 \times £2.38\)
Calculator: 10 × 2.38 = 23.80
Cost of pens = £23.80
Step 2: Calculate Cost of Folders
Strategy: We know the cost of 5 folders. We need 20 folders.
How many sets of 5 fit into 20? \(20 \div 5 = 4\) sets.
Cost of 20 folders = \(4 \times £5.60\)
Calculator: 4 × 5.60 = 22.40
Cost of folders = £22.40
Step 3: Total Cost and Comparison
Add the costs together and compare with £50.
Total Cost = £23.80 + £22.40 = £46.20
Compare: £46.20 < £50.00
Final Answer: Yes, he has enough money. (Total cost is £46.20) ✓ (P1 P1 P1 C1)
Question 9 (2 marks)
The diagram shows five shapes on a centimetre grid.
(a) Write down the name of shape E.
Two of the shapes are congruent.
(b) Write down the letters of these two shapes.
Worked Solution
Step 1: Identifying Shape E
Shape E has 4 sides. One pair of opposite sides is parallel (the top horizontal line and the bottom horizontal line). A quadrilateral with one pair of parallel sides is called a Trapezium.
Step 2: Finding Congruent Shapes
Definition: Congruent shapes are identical in size and shape (they can be rotated or reflected).
Let’s check dimensions:
- Shape C is a right-angled triangle with base 2 and height 2 units (approximately). Actually, checking the grid: Shape D is 2 units wide and 2 units high. Shape C is a triangle 2 units wide and 2 units high.
- Check C and D: Both are right-angled triangles with legs of the same length.
Shape C and Shape D are congruent.
Final Answer:
(a) Trapezium ✓ (B1)
(b) C and D ✓ (B1)
Question 10 (1 mark)
On the grid, reflect the shaded shape in the mirror line.
Worked Solution
Step 1: Reflection Principle
How to reflect: Every point on the original shape must be the same distance from the mirror line but on the opposite side.
The base of the L-shape is on the line, so it stays on the line.
The top of the vertical part is 2 squares up (60 units), so it reflects to 2 squares down.
The horizontal arm is 2 squares up, reflecting to 2 squares down.
Final Answer: See diagram above. ✓ (C1)
Question 11 (3 marks)
There are men and women at a meeting.
There are 28 women.
\(30\%\) of the people at the meeting are men.
Work out the total number of people at the meeting.
Worked Solution
Step 1: Determine Percentage of Women
Reasoning: The total percentage of people is \(100\%\). If \(30\%\) are men, the rest are women.
\(100\% – 30\% = 70\%\)
So, \(70\%\) of the people are women.
Step 2: Use the Known Number
We know that \(70\%\) corresponds to 28 women. We want to find \(100\%\) (the total).
\(70\% = 28\)
Divide by 7 to find \(10\%\):
\(10\% = 28 \div 7 = 4\)
Multiply by 10 to find \(100\%\):
\(100\% = 4 \times 10 = 40\)
Alternatively: \(\frac{28}{0.7} = 40\)
Final Answer: 40 people ✓ (P1 P1 A1)
Question 12 (3 marks)
Joan asked each of 60 people to name their favourite vegetable.
Here are her results.
| Vegetable | Frequency |
|---|---|
| Peas | 24 |
| Carrots | 16 |
| Mushrooms | 20 |
Draw an accurate pie chart for her results.
Worked Solution
Step 1: Calculate Angles
Method: A full circle is \(360^\circ\). The total frequency is 60.
Degrees per person = \(360 \div 60 = 6^\circ\).
Multiply each frequency by 6 to get the angle.
Peas: \(24 \times 6 = 144^\circ\)
Carrots: \(16 \times 6 = 96^\circ\)
Mushrooms: \(20 \times 6 = 120^\circ\)
Check: \(144 + 96 + 120 = 360^\circ\)
Step 2: Draw the Chart
Draw a circle, start from a vertical radius, and measure the angles.
Final Answer: Pie chart with angles: Peas \(144^\circ\), Carrots \(96^\circ\), Mushrooms \(120^\circ\). ✓ (M1 A1 A1)
Question 13 (4 marks)
Annie sold:
- 45 books at £1.20 each
- 34 candles at £1.50 each
- some calendars at 90p each
She got a total of £150.
Work out how many calendars Annie sold.
Worked Solution
Step 1: Calculate Income from Books and Candles
Books: \(45 \times 1.20 = 54.00\)
Candles: \(34 \times 1.50 = 51.00\)
Total known income: \(54 + 51 = 105\)
Step 2: Find Income from Calendars
Subtract the known income from the total to see how much came from calendars.
Calendar Income = Total – Known
\(£150 – £105 = £45\)
Step 3: Calculate Number of Calendars
Caution: The price is in pence (90p), but our total is in pounds (£45). Convert units.
£45 = 4500p, OR 90p = £0.90.
Number of calendars = \(45 \div 0.90\)
Calculator: 45 ÷ 0.9 = 50
Final Answer: 50 calendars ✓ (P1 P1 P1 A1)
Question 14 (4 marks)
Here is a 4-sided spinner.
The table shows the probabilities that when the spinner is spun it will land on 1, on 3 and on 4.
| Number | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Probability | 0.2 | 0.4 | 0.1 |
The spinner is spun once.
(a) Work out the probability that the spinner will land on 2.
(b) Which number is the spinner least likely to land on?
Jake is going to spin the spinner 60 times.
(c) Work out an estimate for the number of times the spinner will land on 1.
Worked Solution
Step 1: Part (a) – Probability of 2
Rule: Probabilities of all possible outcomes sum to 1.
Sum of knowns: \(0.2 + 0.4 + 0.1 = 0.7\)
\(P(2) = 1 – 0.7 = 0.3\)
Step 2: Part (b) – Least Likely
Compare the probabilities: 0.2, 0.3, 0.4, 0.1.
The smallest probability is 0.1, which corresponds to number 4.
Answer: 4
Step 3: Part (c) – Expected Frequency
Formula: Expected times = Probability \(\times\) Total Trials.
Probability of 1 is \(0.2\).
\(0.2 \times 60 = 12\)
Final Answer:
(a) 0.3 ✓ (B1)
(b) 4 ✓ (B1)
(c) 12 ✓ (M1 A1)
Question 15 (3 marks)
Bert has 100 cards. There is a whole number from 1 to 100 on each card. No cards have the same number.
Bert puts a star on every card that has a multiple of 3 on it.
He puts a circle on every card that has a multiple of 5 on it.
Work out how many cards have both a star and a circle on them.
Worked Solution
Step 1: Understanding the Condition
Reasoning: A card has both a star (multiple of 3) and a circle (multiple of 5) if the number is a multiple of both 3 and 5.
The Lowest Common Multiple (LCM) of 3 and 5 is 15.
So, we are looking for multiples of 15.
Step 2: List or Calculate Multiples
We need multiples of 15 up to 100.
List: 15, 30, 45, 60, 75, 90.
Next is 105, which is > 100.
Count: There are 6 numbers.
Alternative: \(100 \div 15 = 6.66…\), so 6 complete multiples.
Final Answer: 6 ✓ (P1 P1 A1)
Question 16 (2 marks)
Write down the ratio of 450 grams to 15 grams.
Give your answer in its simplest form.
Worked Solution
Step 1: Set up the Ratio
\(450 : 15\)
Step 2: Simplify
Divide both sides by the smaller number (15), or common factors.
Divide by 5: \(90 : 3\)
Divide by 3: \(30 : 1\)
Alternative: \(450 \div 15 = 30\), so \(30:1\).
Final Answer: 30:1 ✓ (M1 A1)
Question 17 (5 marks)
The diagram shows a pentagon. The pentagon has one line of symmetry.
\(AE = 4x\)
\(AB = 2x + 1\)
\(BC = x + 2\)
All these measurements are given in centimetres.
The perimeter of the pentagon is 18 cm.
(a) Show that \(10x + 6 = 18\)
(b) Find the value of \(x\).
Worked Solution
Step 1: Use Symmetry Properties
Reasoning: Since the pentagon has a vertical line of symmetry:
- \(ED = AB = 2x + 1\)
- \(CD = BC = x + 2\)
Step 2: Form Perimeter Expression
Perimeter is the sum of all 5 sides.
\(P = AE + AB + BC + CD + DE\)
\(P = (4x) + (2x+1) + (x+2) + (x+2) + (2x+1)\)
Step 3: Simplify and Equate
Collect x terms: \(4x + 2x + x + x + 2x = 10x\)
Collect number terms: \(1 + 2 + 2 + 1 = 6\)
Total Perimeter: \(10x + 6\)
Given Perimeter is 18, so: \(10x + 6 = 18\)
Part (a) shown. ✓ (M1 M1 C1)
Step 4: Solve for x (Part b)
\(10x + 6 = 18\)
Subtract 6: \(10x = 12\)
Divide by 10: \(x = 1.2\)
Final Answer (b): \(x = 1.2\) ✓ (M1 A1)
Question 18 (4 marks)
Trevor buys a boat.
The cost of the boat is £14 200 plus VAT at 20%.
Trevor pays a deposit of £5000.
He pays the rest of the cost in 10 equal payments.
Work out the amount of each of the 10 payments.
Worked Solution
Step 1: Calculate Total Cost with VAT
Method: Add 20% to the base price.
\(20\% \text{ of } 14200 = 0.2 \times 14200 = 2840\)
Total = \(14200 + 2840\).
Or use multiplier \(1.2\).
\(14200 \times 1.2 = 17040\)
Total Cost = £17,040
Step 2: Subtract Deposit
\(17040 – 5000 = 12040\)
Amount left to pay = £12,040
Step 3: Calculate Installments
Divide by 10:
\(12040 \div 10 = 1204\)
Final Answer: £1204 ✓ (P2 P1 A1)
Question 19 (7 marks)
(a) On the number line, show the inequality \(x < 4\)
\(3 < y \leqslant 7\) where \(y\) is an integer.
(b) Write down all the possible values of \(y\).
(c) Solve \(3x + 5 \geqslant x + 17\)
Worked Solution
Step 1: Part (a) – Number Line
Rules:
- \( < \) means an open circle (not included).
- Arrow points to the left (less than).
Step 2: Part (b) – Integer Values
\(3 < y \leqslant 7\)
\(y\) is strictly greater than 3, so starts at 4.
\(y\) is less than or equal to 7, so includes 7.
Values: 4, 5, 6, 7 ✓ (B2)
Step 3: Part (c) – Solving Inequality
\(3x + 5 \geqslant x + 17\)
Subtract \(x\) from both sides:
\(2x + 5 \geqslant 17\)
Subtract 5 from both sides:
\(2x \geqslant 12\)
Divide by 2:
\(x \geqslant 6\)
Final Answer (c): \(x \geqslant 6\) ✓ (M1 M1 A1)
Question 20 (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
Step 1: Part (a) – Significant Figures
Count 3 digits from the start: 7, 3, 5.
The next digit is 7, so we round up the 5 to 6.
Replace following digits with zeros to maintain magnitude.
Answer: 7360 ✓ (B1)
Step 2: Part (b) – Calculation
Numerator: \( \sqrt{17 + 16} = \sqrt{33} \approx 5.74456 \)
Denominator: \( 7.3^2 = 53.29 \)
Calculation: \( 5.74456… \div 53.29 \)
Calculator display: 0.1077981356…
Final Answer (b): 0.1077981356 ✓ (B2)
Question 21 (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 Actual Increase
\( \text{Increase} = \text{New Price} – \text{Old Price} \)
\( 883 – 245 = 638 \)
Step 2: Calculate Percentage
Divide the increase by the original amount.
\( \frac{638}{245} \times 100 \)
\( 2.60408… \times 100 = 260.408… \)
Final Answer: 260.4% (accept 260 to 260.5) ✓ (M1 M1 A1)
Question 22 (5 marks)
(a) Complete this table of values for \(y = x^2 + x – 4\)
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| \(y\) | 2 | -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
Step 1: Part (a) – Table Values
When \(x = 0\): \(0^2 + 0 – 4 = -4\)
When \(x = 2\): \(2^2 + 2 – 4 = 4 + 2 – 4 = 2\)
When \(x = 3\): \(3^2 + 3 – 4 = 9 + 3 – 4 = 8\)
Completed values: 2, -2, -4, -4, -2, 2, 8.
Step 2: Part (b) – Plotting
Plot points: (-3, 2), (-2, -2), (-1, -4), (0, -4), (1, -2), (2, 2), (3, 8).
Join with a smooth curve.
Step 3: Part (c) – Estimate Solutions
\(x^2 + x – 4 = 0\) is where the graph crosses the x-axis (\(y=0\)).
Looking at the graph, it crosses between 1 and 2, and between -2 and -3.
Values: approx \(1.6\) and \(-2.6\). (Accept 1.56 or -2.56) ✓ (B1)
Question 23 (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.
(a) Work out the percentage of these students who bought 20 or more books.
(b) Show that an estimate for the mean number of books bought is 9.5.
Worked Solution
Step 1: Part (a) – Reading the Chart
Read frequencies from the chart:
- 0-4: 11
- 5-9: 8
- 10-14: 13
- 15-19: 6
- 20-24: 2
Students buying 20 or more = The last bar (2 students).
Total students = 40.
\(\frac{2}{40} \times 100 = 5\%\)
Answer (a): 5% ✓ (M1 A1)
Step 2: Part (b) – Estimate Mean
Method: Use the midpoint of each interval multiplied by the frequency.
Midpoints: 2, 7, 12, 17, 22.
\(11 \times 2 = 22\)
\(8 \times 7 = 56\)
\(13 \times 12 = 156\)
\(6 \times 17 = 102\)
\(2 \times 22 = 44\)
Sum of products: \(22 + 56 + 156 + 102 + 44 = 380\)
Mean = \(\frac{380}{40} = 9.5\)
Shown correctly. ✓ (M1 M1 M1 C1)
Question 24 (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
Step 1: Convert Time to Seconds
1 minute 54 seconds = \(60 + 54 = 114\) seconds.
Step 2: Calculate Speed or Use Proportion
Method: Ratio of distances equals ratio of times.
Time per metre = \( \frac{114}{475} = 0.24 \) seconds/metre.
Time for 700m = \( 700 \times 0.24 = 168 \) seconds.
Step 3: Convert Back to Minutes and Seconds
\(168 \div 60 = 2\) remainder \(48\).
2 minutes 48 seconds.
Final Answer (a): 2 mins 48 secs ✓ (P1 P1 A1)
Step 4: Part (b) – Effect of increasing speed
If speed increases, she travels faster. Faster speed means less time.
Answer (b): It would take less time. ✓ (C1)
Question 25 (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: Calculate Angles
Angles in a triangle sum to \(180^\circ\). Angle \(C = 90^\circ\).
So \(A + B = 90^\circ\).
Ratio \(B:A = 3:2\). Total parts = 5.
1 part = \(90 \div 5 = 18^\circ\).
Angle \(A = 2 \times 18 = 36^\circ\).
Angle \(B = 3 \times 18 = 54^\circ\).
Step 2: Use Trigonometry (SOH CAH TOA)
We know Angle \(A = 36^\circ\).
We know Adjacent side \(AC = 14\).
We want Hypotenuse \(AB\).
Use Cosine: \(\cos(A) = \frac{\text{Adj}}{\text{Hyp}}\).
\(\cos(36^\circ) = \frac{14}{AB}\)
\(AB = \frac{14}{\cos(36^\circ)}\)
Step 3: Calculate
\(AB = 14 \div 0.8090… = 17.3049…\)
Final Answer: 17.3 cm ✓ (P1 P1 P1 A1)
Question 26 (2 marks)
Here are the first four terms of an arithmetic sequence.
5 11 17 23
Write down an expression, in terms of \(n\), for the \(n\)th term of the sequence.
Worked Solution
Step 1: Find the Difference
Check the gap between terms.
\(11 – 5 = 6\)
\(17 – 11 = 6\)
Common difference is 6. So the expression starts with \(6n\).
Step 2: Adjust the Zero Term
If we use \(6n\): 6, 12, 18…
We want: 5, 11, 17…
We need to subtract 1 from each term.
Formula: \(6n – 1\)
Final Answer: \(6n – 1\) ✓ (M1 A1)