November 2022
Paper 1 – Non-Calculator
Complete question-by-question practice with worked solutions and reconstructed diagrams.
Official Exam Materials
Download the original examination materials for November 2022 Paper 1H directly from Pearson.
Paper information
- Answer all 24 questions.
- Show all stages of your working.
- Diagrams on this page have been redrawn for learning and may not be to scale.
- Use the Reveal solution button after attempting each question.
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Worked Solutions — All 24 Questions
Question 1
Prime factorisation · 3 marks3 marksReveal solution
- 500 = 2 × 2 × 5 × 5 × 5
- Collect equal prime factors: 22 × 53.
Question 2
Fractions and mixed numbers · 4 marks4 marks(a) Work out 1 ⅗ + 2 ¼. Give your answer as a mixed number.
(b) Show that 2 ⅔ ÷ 6 = 4/9.
Reveal solution
(a) Convert to improper fractions:
1⅗ + 2¼ = 8/5 + 9/4 = 32/20 + 45/20 = 77/20 = 3 17/20
(b)
2⅔ ÷ 6 = 8/3 × 1/6 = 8/18 = 4/9
Question 3
Index laws · 2 marks2 marksReveal solution
(2−5 × 28)2 = (23)2 = 26
Question 4
Decimal multiplication · 2 marks2 marksReveal solution
4 × 32 = 128
There are 3 + 2 = 5 decimal places altogether, so
0.004 × 0.32 = 0.00128
Question 5
Proportion and scaling · 2 marks2 marksA car factory is going to make four different car models A, B, C and D. 80 people are asked which model they would be most likely to buy.
| Car model | Number of people |
|---|---|
| A | 23 |
| B | 15 |
| C | 30 |
| D | 12 |
The factory is going to make 40,000 cars next year. Work out how many model B cars it should make.
Reveal solution
Model B represents 15/80 = 3/16 of the people.
40,000 × 3/16 = 2,500 × 3 = 7,500
Question 6
Ratio · 6 marks6 marksRizwan writes down three numbers a, b and c.
a : b = 1 : 3 b : c = 6 : 5
(a)(i) Find a : b : c.
(a)(ii) Express a as a fraction of the total of the three numbers.
Emma writes down three numbers m, n and p.
n = 2m p = 5n
(b) Find m : p.
Reveal solution
(a)(i) Make the b parts equal: 1:3 = 2:6. Therefore a:b:c = 2:6:5.
(a)(ii) Total parts = 2+6+5=13, so the fraction is 2/13.
(b) Since n=2m and p=5n, then p=10m. Hence m:p=1:10.
Question 7
Compound measures · 2 marks2 markspressure = force / area
A storage tank exerts a force of 10,000 newtons on the ground. The base of the tank in contact with the ground is a 4 m by 2 m rectangle. Work out the pressure on the ground due to the tank.
Reveal solution
Area of base = 4 × 2 = 8 m².
Pressure = force ÷ area = 10,000 ÷ 8 = 1,250 N/m²
Question 8
HCF and multiples · 2 marks2 marksReveal solution
Choose m=7×5=35 and n=7×2=14. Their HCF is 7, 35 is a multiple of 5 and 14 is even.
Question 9
Cubic graphs · 4 marks4 marks(a) Complete the table of values for y = 6x − x³.
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 9 | ? | ? | ? | ? | 4 | −9 |
(b) On the grid, draw the graph for values of x from −3 to 3.
The curve is shown only for −3 ≤ x ≤ 3. It follows y = 6x − x³, with the left-hand section below the x-axis between x = −2 and x = 0.
Reveal solution
Substitute each value of x into y=6x−x³:
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 9 | −4 | −5 | 0 | 5 | 4 | −9 |
Question 10
Probability · 4 marks4 marksLina spins a biased 5-sided spinner 40 times. Here are her results.
| Score | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency | 6 | 8 | 9 | 7 | 10 |
(a) Lina spins the spinner another two times. Work out an estimate for the probability that she gets a score of 5 both times.
(b) Derek is going to spin the spinner a large number of times. Work out an estimate for the percentage of times Derek can expect to get a score of 1.
Reveal solution
(a) Estimated probability of a 5 is 10/40=1/4. For two spins:
1/4 × 1/4 = 1/16
(b) Estimated probability of a 1 is 6/40=3/20=0.15, so the percentage is 15%.
Question 11
Enlargement · 2 marks2 marksSchematic reconstruction of the coordinate diagram; the transformation description is based on the paper's plotted shapes.
Reveal solution
The corresponding lengths in Q are one third of those in P, and the shapes have the same orientation.
Question 12
Simultaneous equations · 4 marks4 marks5x + 2y = 11
4x + 3y = 6
Reveal solution
Multiply the first equation by 3 and the second by 2:
15x + 6y = 33
8x + 6y = 12
Subtract to eliminate y: 7x=21, so x=3.
Substitute into the first equation: 15+2y=11, so 2y=−4 and y=−2.
Question 13
Inverse proportion · 3 marks3 marksp is inversely proportional to t. Complete the table.
| t | 100 | 25 | ? | 2 |
|---|---|---|---|---|
| p | 1 | ? | 5 | ? |
Reveal solution
Inverse proportion means p=k/t. Using p=1 when t=100, we get k=100. Therefore p=100/t.
When t=25, p=4. When p=5, t=20. When t=2, p=50.
| t | 100 | 25 | 20 | 2 |
|---|---|---|---|---|
| p | 1 | 4 | 5 | 50 |
Question 14
Histograms · 3 marks3 marksThe table shows information about the weights, in grams, of some potatoes. Draw a histogram.
| Weight (w grams) | Number of potatoes |
|---|---|
| 50 < w ≤ 70 | 20 |
| 70 < w ≤ 80 | 50 |
| 80 < w ≤ 90 | 60 |
| 90 < w ≤ 110 | 30 |
Histogram bar heights are frequency densities, not frequencies.
Reveal solution
Frequency density = frequency ÷ class width.
- 50 < w ≤ 70: 20 ÷ 20 = 1
- 70 < w ≤ 80: 50 ÷ 10 = 5
- 80 < w ≤ 90: 60 ÷ 10 = 6
- 90 < w ≤ 110: 30 ÷ 20 = 1.5
Question 15
Arc length · 3 marks3 marksThe diagram shows a sector of a circle of radius 18 cm. The length of the arc is 4π cm. Work out the value of x.
Reveal solution
Use the arc length formula:
4π = (x/360) × 2π × 18
4π = xπ/10 ⇒ x=40
Question 16
Algebraic proof · 4 marks4 marks(a) Prove that
(2m + 1)² − (2n − 1)² = 4(m + n)(m − n + 1)
(b) Sophia says that the result in part (a) shows that the difference of the squares of any two odd numbers must be a multiple of 4. Is Sophia correct? Give reasons.
Reveal solution
(a) Use the difference of two squares:
[(2m+1)−(2n−1)][(2m+1)+(2n−1)]
= (2m−2n+2)(2m+2n)
= 2(m−n+1) × 2(m+n)
= 4(m+n)(m−n+1)
This is the required result.
(b) Yes. Any two odd integers can be written in the forms 2m+1 and 2n−1. The expression is 4 multiplied by integers, so it is divisible by 4.
Question 17
Fractional indices · 2 marks2 marksReveal solution
(8/27)4/3 = [(8/27)1/3]4 = (2/3)4 = 16/81
Question 18
Circle theorems · 3 marks3 marksA and B are points on a circle, centre O. DBC is the tangent to the circle at B. Angle AOB = x°. Show that angle ABC = ½x°. Give a reason for each stage of your working.
Reveal solution
- Radius OB is perpendicular to the tangent at B, so ∠OBC=90°.
- OA=OB (radii), so triangle OAB is isosceles. Each base angle is (180−x)/2 = 90−x/2.
- Angles on a straight line at B add to 180°: ∠ABC = 180−90−(90−x/2)=x/2.
Question 19
Algebraic fractions and quadratics · 5 marks5 marksSolve
1/x − 1/(x+1) = 4
Give your answer in the form a ± b√2 where a and b are fractions.
Reveal solution
Combine the fractions:
[(x+1)−x]/[x(x+1)] = 4
1 = 4x(x+1) ⇒ 4x²+4x−1=0
Using the quadratic formula:
x = [−4 ± √(16+16)]/8 = [−4 ± 4√2]/8
x = −1/2 ± (1/2)√2
Question 20
Probability · 3 marks3 marksAlfie has 11 cards: 3 blue cards, 7 green cards and 1 white card. Alfie takes at random 2 of these cards. Work out the probability that he takes cards of different colours.
Reveal solution
It is easier to find the probability of getting two cards of the same colour, then subtract from 1.
P(two blue) = 3/11 × 2/10 = 6/110
P(two green) = 7/11 × 6/10 = 42/110
P(two white) = 0
P(same colour) = 48/110
P(different colours) = 1−48/110 = 62/110 = 31/55
Question 21
Trigonometric graphs · 2 marks2 marksThe diagram shows a sketch of part of the curve with equation y = cos x°. P is a minimum point on the curve. Write down the coordinates of P.
Reveal solution
For y=cos x°, the minimum value is −1, first reached at x=180°.
Question 22
Sine rule · 4 marks4 marksHere is a triangle ABC. Work out the value of sin ABC. Give your answer in the form m/n where m and n are integers.
Reveal solution
Using the sine rule:
sin B / 6.5 = sin 30° / 10.7
sin B = (6.5 × 1/2) / 10.7
= (13/2 × 1/2) × 10/107 = 65/214
Question 23
Geometric sequences · 5 marks5 marksHere are the first five terms of a geometric sequence:
√5 10 20√5 200 400√5
(a) Work out the next term of the sequence.
The 4th term of a different geometric sequence is 5√2/4. The 6th term is 5√2/8. Given that the terms of this sequence are all positive, (b) work out the first term. Show all your working.
Reveal solution
(a) Common ratio:
r = 10/√5 = 2√5
Next term = 400√5 × 2√5 = 400 × 10 = 4000
(b) Let the common ratio be r. Then
r² = (5√2/8) ÷ (5√2/4) = 1/2
All terms are positive, so r=1/√2. If the first term is a, then
a(1/√2)³ = 5√2/4
a = (5√2/4) × (2√2) = 5
Question 24
Volume and surface area · 6 marks6 marksHere is a solid sphere and a solid cone. All measurements are in cm.
Volume of sphere = 4/3 πr³ Volume of cone = 1/3 πr²h
The volume of the sphere is equal to the volume of the cone.
(a) Find r : h. Give your answer in its simplest form.
Here is a different solid sphere and a different solid cone. Surface area of sphere = 4πr²; curved area of cone = πrl. The surface area of the sphere is equal to the total surface area of the cone. (b) Find r:h. Give your answer in the form 1:n.
Reveal solution
(a) Equate the volumes:
4/3 πr³ = 1/3 πr²h
Multiply by 3 and divide by πr²: 4r=h.
(b) The total surface area of the cone is its curved surface area plus its circular base:
4πr² = πrl + πr²
3πr² = πrl ⇒ l=3r
Using Pythagoras in the cone's cross-section:
h² = l²−r² = 9r²−r² = 8r²
h=√8 r
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