Edexcel GCSE Maths Nov 2018 Paper 1F
📝 Non-Calculator Paper
This exam is non-calculator. All solutions show full arithmetic working methods.
Table of Contents
- Q1: Ordering Decimals
- Q2: Percentage Conversion
- Q3: Factors
- Q4: Rounding
- Q5: Order of Operations
- Q6: Fractions (Context)
- Q7: Pictogram
- Q8: Coordinates
- Q9: Substitution
- Q10: Prime Numbers
- Q11: Fractions
- Q12: Real-Life Graphs
- Q13: Ratio
- Q14: Geometric Proof
- Q15: Proportion (Recipes)
- Q16: Similarity
- Q17: Frequency Trees
- Q18: Best Value
- Q19: Transformations
- Q20: Indices
- Q21: Rearranging Formulae
- Q22: Percentages (Bonus)
- Q23: Inverse Proportion
- Q24: Speed Estimation
- Q25: Simultaneous Equations
- Q26: Area (Circle/Square)
- Q27: Probability Tree
- Q28: Polygons
Question 1 (1 mark)
Write the following numbers in order of size. Start with the smallest number.
\(0.4 \quad 0.02 \quad 0.37 \quad 0.152 \quad 0.2\)
Worked Solution
Strategy: Add placeholder zeros so all numbers have the same number of decimal places (3 places).
\(0.4 \rightarrow 0.400\)
\(0.02 \rightarrow 0.020\) (Smallest)
\(0.37 \rightarrow 0.370\)
\(0.152 \rightarrow 0.152\)
\(0.2 \rightarrow 0.200\)
Answer: \(0.02, 0.152, 0.2, 0.37, 0.4\)
✓ (B1)
Question 2 (1 mark)
Write \(0.6\) as a percentage.
Worked Solution
To convert a decimal to a percentage, multiply by 100.
\(0.6 \times 100 = 60\%\)
Answer: \(60\%\)
✓ (B1)
Question 3 (1 mark)
Here is a list of numbers: \(3, 5, 7, 12, 15, 18, 20\).
From the list, write down a factor of 10.
Worked Solution
A factor of 10 is a number that divides into 10 exactly (1, 2, 5, 10).
Looking at the list, the only number that is a factor of 10 is 5.
Answer: \(5\)
✓ (B1)
Question 4 (1 mark)
Write \(7829\) to the nearest 1000.
Worked Solution
Check the hundreds digit. If it’s 5 or more, round up.
In \(7\mathbf{8}29\), the hundreds digit is 8.
Since \(8 \geq 5\), we round up.
Answer: \(8000\)
✓ (B1)
Question 5 (3 marks)
(a) Work out \(3 \times 5 + 7\)
(b) Work out \(2^3\)
(c) Write brackets ( ) in this statement to make it correct: \(7 \times 2 + 3 = 35\)
Worked Solution
Part (a)
Using BIDMAS, multiply first:
\(3 \times 5 = 15\)
\(15 + 7 = 22\)
✓ (B1)
Part (b)
\(2^3 = 2 \times 2 \times 2 = 8\)
✓ (B1)
Part (c)
We need to multiply 7 by 5 to get 35.
\(7 \times (2 + 3) = 7 \times 5 = 35\)
✓ (B1)
Question 6 (3 marks)
Sue has 2 cats. Each cat eats \(\frac{1}{4}\) of a tin of cat food each day.
Sue buys 8 tins of cat food. Has Sue bought enough cat food to feed her 2 cats for 14 days? You must show how you get your answer.
Worked Solution
Step 1: Calculate daily consumption
2 cats each eat \(\frac{1}{4}\) tin.
Total per day = \(\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\) tin.
✓ (P1)
Step 2: Calculate total needed
For 14 days: \(14 \times \frac{1}{2} = 7\) tins.
✓ (P1)
She needs 7 tins. She has 8 tins.
Answer: Yes (since \(8 > 7\))
✓ (C1)
Question 7 (3 marks)
The pictogram shows numbers of trees in an orchard. There is a total of 30 trees.
Complete the pictogram for Plum trees.
Worked Solution
Step 1: Calculate existing trees
Key: 1 square = 4 trees.
Apple: 3 squares = \(3 \times 4 = 12\)
Cherry: 1 square + quarter square = \(4 + 1 = 5\)
Pear: 1 square + half square = \(4 + 2 = 6\)
Total so far = \(12 + 5 + 6 = 23\)
✓ (M1)
Step 2: Find Plum trees
Total trees = 30.
Plum = \(30 – 23 = 7\) trees.
✓ (M1)
Answer: 7 trees.
Draw: 1 full square (4) + three-quarters of a square (3).
✓ (C1)
Question 8 (3 marks)
(a) Write down the coordinates of point A.
(b) On the grid, mark with a cross (×) the point (2, 3). Label this point B.
(c) On the grid, draw the line with equation \(x = -4\).
Worked Solution
Part (a)
Point A is 2 units left and 1 unit down.
Answer: \((-2, -1)\) ✓ (B1)
Part (b)
Start at origin, go 2 right, 3 up.
✓ (B1) Correct point marked.
Part (c)
\(x = -4\) is a vertical line passing through -4 on the x-axis.
✓ (B1) Correct vertical line drawn.
Question 9 (2 marks)
\(g = 9\)
\(h = 4\)
Work out the value of \(2g + 3h\)
Worked Solution
Substitute the values:
\(2(9) + 3(4)\)
\(= 18 + 12\)
\(= 30\)
Answer: \(30\)
✓ (M1) subst
✓ (A1) cao
Question 10 (2 marks)
Write down two prime numbers that have a sum of 32.
Worked Solution
Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31…
Try subtracting primes from 32:
\(32 – 3 = 29\) (29 is prime). Pair: 3, 29.
\(32 – 13 = 19\) (19 is prime). Pair: 13, 19.
Answer: 3 and 29 (or 13 and 19)
✓ (M1) for identifying pair
✓ (A1) cao
Question 11 (3 marks)
Here are some fractions.
\(\frac{9}{12} \quad \frac{6}{8} \quad \frac{18}{24} \quad \frac{10}{16} \quad \frac{15}{20}\)
One of these fractions is not equivalent to \(\frac{3}{4}\).
(a) Which fraction?
(b) Work out \(\frac{1}{12} + \frac{5}{6}\)
Worked Solution
Part (a)
\(\frac{10}{16} = \frac{5}{8}\), which is not \(\frac{3}{4}\).
Answer: \(\frac{10}{16}\) ✓ (B1)
Part (b)
Make denominators same (12):
\(\frac{5}{6} = \frac{10}{12}\)
\(\frac{1}{12} + \frac{10}{12} = \frac{11}{12}\)
Answer: \(\frac{11}{12}\) ✓ (A1)
Question 12 (3 marks)
Tom uses his lorry to deliver bricks. You can use this graph to find the delivery cost.
(a) How much is the fixed charge?
(b) Work out the difference between the two delivery costs when one delivery is 20 miles longer than the other.
Worked Solution
Part (a)
Read graph at Distance = 0.
Answer: £20 ✓ (B1)
Part (b)
Read cost at 60 miles = £100. Read cost at 0 miles = £20.
Difference for 60 miles = £80.
Difference for 20 miles (1/3 of 60) = \(80 \div 3 \approx £26.67\).
Answer: ~£27 (Range 25-30 accepted) ✓ (A1)
Question 13 (2 marks)
Azmol, Ryan and Kim each played a game.
- Azmol’s score was four times Ryan’s score.
- Kim’s score was half of Azmol’s score.
Write down the ratio of Azmol’s score to Ryan’s score to Kim’s score.
Worked Solution
Let Ryan = 1.
Azmol = \(4 \times 1 = 4\).
Kim = \(\frac{1}{2} \times 4 = 2\).
Azmol : Ryan : Kim
4 : 1 : 2
Answer: 4 : 1 : 2
✓ (A1)
Question 14 (4 marks)
The diagram shows quadrilateral ABCD with each of its sides extended.
\(AB = AD\). Show that ABCD is a kite.
Worked Solution
1. Find angle ADC:
Angles on a straight line = \(180^\circ\).
\(180 – 75 = 105^\circ\).
2. Find angle ABC:
Interior angle at A = \(100^\circ\) (Vertically opposite).
Interior angle at C = \(50^\circ\) (Vertically opposite).
Sum of quad = \(360^\circ\).
\(360 – (100 + 105 + 50) = 360 – 255 = 105^\circ\).
3. Conclusion:
We have one pair of equal opposite angles (\(105^\circ\)) and two adjacent equal sides (\(AB=AD\)).
This defines a Kite.
✓ (C1) for complete proof
Question 15 (4 marks)
Mix 5 parts orange juice with 2 parts lemonade.
(a) Shahid has 300ml orange and 200ml lemonade. What is the greatest amount of drink he can make?
(b) If he only has 160ml lemonade, does this affect the amount? Explain.
Worked Solution
Part (a)
Orange limit: \(300 \div 5 = 60\) ml per part.
Lemonade limit: \(200 \div 2 = 100\) ml per part.
Smallest limit determines max volume. Use 60 ml/part.
Total parts = \(5+2=7\).
Total volume = \(7 \times 60 = 420\) ml.
Answer: 420 ml ✓ (A1)
Part (b)
New lemonade limit: \(160 \div 2 = 80\) ml per part.
Orange limit is still 60 ml per part.
Since \(60 < 80\), the orange juice is still the limiting factor.
Answer: No, he can still make 420ml. ✓ (C1)
Question 16 (1 mark)
Rectangle A: 8cm x 6cm. Rectangle B: 12cm x 10cm.
Jim says they are similar because \(8+4=12\) and \(6+4=10\).
Is Jim correct? Explain.
Worked Solution
Similar shapes relate by a scale factor (multiplication), not addition.
Scale for width: \(12 \div 8 = 1.5\).
Scale for height: \(10 \div 6 = 1.66…\)
Scale factors are different.
Answer: No, you multiply for similarity, not add. ✓ (C1)
Question 17 (5 marks)
80 people. 48 are women. 61 like coffee. 8 men do not like coffee.
(a) Complete the frequency tree.
(b) One person who likes coffee is chosen. Probability they are a woman?
Worked Solution
Part (a)
Men = \(80 – 48 = 32\).
Men Like Coffee = \(32 – 8 = 24\).
Women Like Coffee = \(61 (\text{Total Like}) – 24 (\text{Men Like}) = 37\).
Women Not Like = \(48 – 37 = 11\).
✓ (C2) for correct tree values.
Part (b)
Denominator = People who like coffee = 61.
Numerator = Women who like coffee = 37.
Answer: \(\frac{37}{61}\) ✓ (M1 A1)
Question 18 (4 marks)
Food Mart: 400g. 20% off normal price of £5.
Jan’s Store: 400g + 30% extra. Price £5.
Which is better value?
Worked Solution
Food Mart:
Price: \(£5 – 20\% = £5 – £1 = £4\).
Value: \(400\text{g for } £4 \rightarrow 100\text{g for } £1\).
Jan’s Store:
Weight: \(400 + 30\% = 400 + 120 = 520\text{g}\).
Value: \(520\text{g for } £5 \rightarrow 104\text{g for } £1\).
Conclusion: Jan’s Store gives you more grams per pound (104g vs 100g).
Answer: Jan’s Store ✓ (C1)
Question 19 (2 marks)
Rotate shape A (vertices at (4,-3), (5,-4), (4,-5)) \(180^\circ\) about \((1, 0)\).
Worked Solution
Centre of rotation is \((1, 0)\).
Point \((4, -3)\): Move 3 right, 3 down from center. Rotate \(\rightarrow\) 3 left, 3 up from center \(\rightarrow (1-3, 0+3) = (-2, 3)\).
Repeat for others.
New vertices: \((-2, 3), (-3, 4), (-2, 5)\).
✓ (B2) Correct shape drawn.
Question 20 (2 marks)
Work out the value of \(\frac{3^7 \times 3^{-2}}{3^3}\).
Worked Solution
Numerator: \(3^7 \times 3^{-2} = 3^{7 + (-2)} = 3^5\).
Divide: \(\frac{3^5}{3^3} = 3^{5-3} = 3^2\).
\(3^2 = 9\).
Answer: 9 ✓ (A1)
Question 21 (4 marks)
\(v^2 = u^2 + 2as\)
\(u = 12, a = -3, s = 18\).
(a) Work out v.
(b) Make s the subject.
Worked Solution
Part (a)
\(v^2 = 12^2 + 2(-3)(18)\)
\(v^2 = 144 – 108 = 36\)
\(v = \sqrt{36} = 6\) (or -6).
Answer: 6 ✓ (A1)
Part (b)
\(v^2 = u^2 + 2as\)
\(v^2 – u^2 = 2as\)
\(s = \frac{v^2 – u^2}{2a}\)
Answer: \(s = \frac{v^2 – u^2}{2a}\) ✓ (A1)
Question 22 (5 marks)
Bonus £2100. 10 people. 3 managers get 40% equally. 7 salesmen get the rest equally.
Salesman says: “If shared equally between all 10, I get 25% more.” Is he correct?
Worked Solution
1. Current Salesman Share:
Managers: \(40\%\) of \(2100 = £840\).
Salesmen Total: \(2100 – 840 = £1260\).
Per Salesman: \(1260 \div 7 = £180\).
2. Equal Share Scenario:
\(2100 \div 10 = £210\).
3. Compare with 25% increase:
\(180 \times 1.25\) (or \(180 + 45\)) \(= £225\).
He gets £210, not £225.
Answer: No, he would need £225 to be correct. ✓ (A1)
Question 23 (3 marks)
120 minutes to fill pool with 5 taps.
(a) How many minutes with 3 taps?
(b) State one assumption.
Worked Solution
Part (a)
Total work = \(5 \times 120 = 600\) tap-minutes.
With 3 taps: \(600 \div 3 = 200\) minutes.
Answer: 200 ✓ (A1)
Part (b)
Assumption: All taps flow at the same rate.
Question 24 (4 marks)
Plane speed 213 mph.
(a) Estimate seconds to travel 1 mile.
(b) Is it an underestimate or overestimate?
Worked Solution
Part (a)
Round 213 to 200.
200 miles in 1 hour (3600 seconds).
1 mile = \(3600 \div 200 = 18\) seconds.
Answer: 18 ✓ (A1)
Part (b)
We divided by a smaller number (200 instead of 213).
Dividing by a smaller number gives a larger answer.
Answer: Overestimate ✓ (C1)
Question 25 (3 marks)
Solve:
\(5x + y = 21\)
\(x – 3y = 9\)
Worked Solution
Multiply eqn (1) by 3:
\(15x + 3y = 63\) (3)
Add eqn (2) and (3):
\(16x = 72\)
\(x = 72 \div 16 = 4.5\)
Substitute back:
\(4.5 – 3y = 9 \rightarrow -3y = 4.5 \rightarrow y = -1.5\)
Answer: \(x=4.5, y=-1.5\) ✓ (A1)
Question 26 (4 marks)
Square ABCD side 20cm. AB is diameter of semicircle. AC is arc of circle centre B.
Show that \(\frac{\text{area shaded}}{\text{area square}} = \frac{\pi}{8}\).
Worked Solution
Area Square = \(20 \times 20 = 400\).
Area Quarter Circle (Radius 20) = \(\frac{\pi \times 20^2}{4} = 100\pi\).
Area Semicircle (Radius 10) = \(\frac{\pi \times 10^2}{2} = 50\pi\).
Shaded Area = Quarter – Semicircle = \(100\pi – 50\pi = 50\pi\).
Fraction = \(\frac{50\pi}{400} = \frac{5\pi}{40} = \frac{\pi}{8}\).
✓ (C1) Complete proof
Question 27 (4 marks)
Bag 1: 3 Red, 7 Green. Bag 2: 5 Red, 4 Green.
(a) Complete tree diagram.
(b) Work out probability of taking two red balls.
Worked Solution
Part (a)
Bag 1: P(Red) = 3/10, P(Green) = 7/10.
Bag 2: P(Red) = 5/9, P(Green) = 4/9.
✓ (B2) Correct tree
Part (b)
P(Red and Red) = \(\frac{3}{10} \times \frac{5}{9} = \frac{15}{90}\).
Simplify to \(\frac{1}{6}\).
Answer: \(\frac{15}{90}\) ✓ (A1)
Question 28 (3 marks)
The size of each interior angle of a regular polygon is 11 times the size of each exterior angle.
Work out how many sides the polygon has.
Worked Solution
Let Exterior Angle = \(x\).
Interior Angle = \(11x\).
Interior + Exterior = \(180^\circ\).
\(11x + x = 180 \rightarrow 12x = 180\).
\(x = 15^\circ\) (Exterior angle).
Number of sides = \(360 \div 15 = 24\).
Answer: 24 ✓ (A1)
