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ZIMSEC O Level · 4004/2 · N2020

Mathematics Paper 2 November 2020

Questions
56
Total marks
136
Time allowed
150 min
Syllabus code
4004/2

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Questions
56
Pass mark
34
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[2 marks]Rational, Irrational Numbers & Surds
Evaluate 20−20÷5+3×220 - 20 \div 5 + 3 \times 2.

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Question 102

[2 marks]Rational, Irrational Numbers & Surds
Simplify (10−5)2(\sqrt{10}-\sqrt5)^2, leaving the answer in surd form.

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Question 103

[2 marks]Rational, Irrational Numbers & Surds
Find the Lowest Common Multiple (LCM) of 15, 20 and 25.

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Question 104

[2 marks]Rational, Irrational Numbers & Surds
Express 252 as a product of its prime factors, in index form.

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Question 105

[1 marks]Rational, Irrational Numbers & Surds
252 has prime factorisation 22×32×72^2\times3^2\times7. Find the smallest number by which 252 must be multiplied to make the product a perfect square.

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Question 201

[1 marks]Number bases, standard form
The distance from city A to city D is 33 260 km. Write this distance in standard form (km).

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Question 202

[2 marks]Number bases, standard form
In a distance table between four cities A, B, C and D, the distance from B to D is 18 740 km and the distance from B to C is 8 970 km. Calculate the distance between C and D (in km).

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Question 203

[1 marks]Number bases, standard form
A three-digit number in base nn is written as 147n147_n. Write down the least possible value of nn.

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Question 204

[2 marks]Number bases, standard form
Convert the base equation 147n=3246147_n = 324_6 into a single quadratic equation in nn, in the form n2+bn+c=0n^2+bn+c=0.

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Question 205

[2 marks]Number bases, standard form
Given that 147n=3246147_n = 324_6 simplifies to n2+4n−117=0n^2+4n-117=0, find the value of nn (rejecting any value that could not be a base).

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Question 301

[1 marks]Angles, polygons, similar solids, bounds
Write down the number of degrees in 4 complete revolutions.

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Question 302

[3 marks]Angles, polygons, similar solids, bounds
Find the number of sides of a polygon whose interior angles add up to 1440°1440°.

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Question 303

[3 marks]Angles, polygons, similar solids, bounds
Two similar containers have capacities 1 728 litres and 5 832 litres. The surface area of the bigger container is 153 cm2^2. Find the surface area of the smaller container (in cm2^2).

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Question 304

[2 marks]Angles, polygons, similar solids, bounds
Triangle ABC has sides AB = 6,5 cm, BC = 7,8 cm and AC = 9,1 cm, each measured to the nearest mm. Express the length of side AB as a range, in the form .......... ≤\le AB < ..........
  1. A6,5 ≤ AB < 6,6
  2. B6,45 ≤ AB < 6,55
  3. C6,4 ≤ AB < 6,6
  4. D6,45 ≤ AB ≤ 6,55

Question 305

[2 marks]Angles, polygons, similar solids, bounds
Triangle ABC has sides AB = 6,5 cm, BC = 7,8 cm and AC = 9,1 cm, each measured to the nearest mm. Using the lower bound of each side, calculate the least possible perimeter of the triangle (in cm).

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Question 401

[2 marks]Matrices, Venn diagrams
Matrix A=(2x43)A = \begin{pmatrix} 2 & x \\ 4 & 3 \end{pmatrix}. Simplify A2A^2, leaving the answer in terms of xx.
  1. A[[4+4x, 5x], [20, 4x+9]]
  2. B[[4+4x, -x], [20, 4x+9]]
  3. C[[4+4x, 0], [0, 4x+9]]
  4. D[[4+4x, 20], [5x, 4x+9]]

Question 402

[2 marks]Matrices, Venn diagrams
Matrix A=(2x43)A = \begin{pmatrix} 2 & x \\ 4 & 3 \end{pmatrix} and matrix B=(5−1)B = \begin{pmatrix} 5 \\ -1 \end{pmatrix}. Simplify ABAB, leaving the answer in terms of xx.

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Question 403

[2 marks]Matrices, Venn diagrams
Matrix A=(2x43)A = \begin{pmatrix} 2 & x \\ 4 & 3 \end{pmatrix} is singular. Find the value of xx.

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Question 404

[2 marks]Matrices, Venn diagrams
39 pupils wanted to visit Great Zimbabwe (Z), 31 wanted Birchenough Bridge (B), 30 wanted Matopo (M), 10 wanted all three, 6 wanted none, 19 wanted both Z and B, 15 wanted both Z and M, 17 wanted both B and M. On the Venn diagram the Z-only region is 15, the Z∩B-only region is 9, the B∩M-only region is 7 and the M-only region is 8. Find the number in the B-only region.

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Question 405

[2 marks]Matrices, Venn diagrams
39 pupils wanted to visit Great Zimbabwe (Z), 31 wanted Birchenough Bridge (B), 30 wanted Matopo (M), 10 wanted all three, 6 wanted none, 19 wanted both Z and B, 15 wanted both Z and M, 17 wanted both B and M. On the Venn diagram the Z-only region is 15, the Z∩B-only region is 9, the B∩M-only region is 7 and the M-only region is 8. Find the number in the Z∩M-only region (excluding those who chose all three).

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Question 406

[1 marks]Matrices, Venn diagrams
A Venn diagram for Grade 3 learners' choice of trip has: Z-only = 15, B-only = 5, M-only = 8, Z∩B-only = 9, Z∩M-only = 5, B∩M-only = 7, all three = 10, none = 6. Write down the total number of Grade 3 learners.

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Question 501

[1 marks]Constructions & Loci
Triangle PQR is constructed accurately with QR=7,5QR=7,5 cm, angle PQR=90°PQR=90° and angle QRP=30°QRP=30°, using ruler and compasses. Measure and write down the length of PR (in cm).

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Question 601

[3 marks]Simple interest, VAT, variation
Mrs Chuhwa invested a certain amount of money with a bank that offered 4,5% p.a. simple interest. After 8 months her money amounted to $504,70 before any bank charges were deducted. Calculate the amount of money that Mrs Chuhwa had initially invested.

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Question 602

[2 marks]Simple interest, VAT, variation
Mrs Bande bought a set of sofas for $368 cash, including 15% VAT. Calculate the price of the sofas excluding VAT.

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Question 603

[3 marks]Simple interest, VAT, variation
Mrs Bande paid $368 cash (including VAT) for a set of sofas. Mr Ndlovu bought a similar set on laybye terms, paying a deposit of $150 plus three equal monthly instalments of $87 (including VAT). Calculate the difference between the total amounts paid by the two customers.

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Question 604

[1 marks]Simple interest, VAT, variation
The average expenditure EE of a family over a period of time is partly constant and partly varies as the number nn of people in the family. Write down a relationship between EE and nn using constants hh and kk.

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Question 605

[2 marks]Simple interest, VAT, variation
The expenditure EE of a family follows E=h+knE=h+kn, where nn is the number of people. For a family of 5, E=$55E=\$55; for a family of 3, E=$45E=\$45. Find the value of kk.

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Question 606

[1 marks]Simple interest, VAT, variation
The expenditure EE of a family follows E=h+knE=h+kn, with k=5k=5. For a family of 3, E=$45E=\$45. Find the value of hh.

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Question 701

[2 marks]Algebra, quadratic equations, mensuration
A rectangle with a width of (x+2)(x+2) cm has a perimeter of (8x+2)(8x+2) cm. Find an expression, in terms of xx, for the length of the rectangle.

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Question 702

[3 marks]Algebra, quadratic equations, mensuration
A rectangle has length (3x−1)(3x-1) cm and width (x+2)(x+2) cm. Solve 3x2+5x−18=03x^2+5x-18=0 using the quadratic formula, giving both roots correct to three significant figures.

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Question 703

[2 marks]Algebra, quadratic equations, mensuration
One root of 3x2+5x−18=03x^2+5x-18=0 is negative. Give this root correct to three significant figures.

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Question 704

[2 marks]Algebra, quadratic equations, mensuration
A rectangle has length (3x−1)(3x-1) cm and width (x+2)(x+2) cm, giving a perimeter of (8x+2)(8x+2) cm, where xx is the positive root of 3x2+5x−18=03x^2+5x-18=0, x=1,75x=1,75. Find the perimeter of the rectangle (in cm).

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Question 801

[1 marks]Functional Graphs
A particle's height hh metres above the ground after tt seconds is h=10+25t−5t2h=10+25t-5t^2. A table of values gives h=10h=10 at t=0t=0, h=30h=30 at t=1t=1, h=mh=m at t=2t=2, h=40h=40 at t=3t=3, h=30h=30 at t=4t=4, h=10h=10 at t=5t=5 and h=−20h=-20 at t=6t=6. Find the value of mm.

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Question 802

[1 marks]Functional Graphs
For a particle with height h=10+25t−5t2h=10+25t-5t^2 metres, h=10h=10 m at t=0t=0 s and h=−20h=-20 m at t=6t=6 s. Write down the distance between the initial and final positions of the particle (in metres).

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Question 803

[2 marks]Functional Graphs
A particle's height is h=10+25t−5t2h=10+25t-5t^2 metres after tt seconds. Find the greatest height reached by the particle (in metres).

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Question 804

[2 marks]Functional Graphs
A particle's height is h=10+25t−5t2h=10+25t-5t^2 metres after tt seconds. Estimate the velocity of the particle when t=5t=5 (in m/s).

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Question 805

[1 marks]Functional Graphs
A particle's height is h=10+25t−5t2h=10+25t-5t^2 metres after tt seconds. There are two times when the particle is 21 m above the ground; find the later of the two times, correct to 3 significant figures (in seconds).

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Question 901

[2 marks]Mensuration (prism), sine and cosine rule
Sector OAB is the cross-section of a prism, with radius 6 cm and angle AO^B=80°A\hat{O}B=80°. Taking π=3,142\pi=3,142, calculate the length of arc AB (in cm).

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Question 902

[3 marks]Mensuration (prism), sine and cosine rule
A prism has cross-section sector OAB, radius 6 cm, angle AO^B=80°A\hat{O}B=80°, and the prism is 50 cm long. Taking π=3,142\pi=3,142, calculate the total surface area of the prism, correct to 3 significant figures (in cm2^2).

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Question 903

[3 marks]Mensuration (prism), sine and cosine rule
In triangle ABC, AB = 7 cm, AC = 9 cm and sin⁡BA^C=23\sin B\hat{A}C=\frac23, where angle BA^CB\hat{A}C is acute. Find cos⁡BA^C\cos B\hat{A}C, correct to 3 significant figures.

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Question 904

[2 marks]Mensuration (prism), sine and cosine rule
In triangle ABC, AB = 7 cm, AC = 9 cm and cos⁡BA^C=0,745\cos B\hat{A}C=0,745. Using the cosine rule, calculate BC2BC^2, correct to 3 significant figures (in cm2^2).

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Question 905

[2 marks]Mensuration (prism), sine and cosine rule
In triangle ABC, BC2≈36,1BC^2\approx36,1 cm2^2. Find the length of BC, correct to 3 significant figures (in cm).

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Question 1001

[1 marks]Statistics & Probability
The heights of 30 Moringa seedlings are grouped as: 5 in 10<h≤2010<h\le20, 6 in 20<h≤2520<h\le25, 10 in 25<h≤3525<h\le35, 9 in 35<h≤4035<h\le40 (h in cm). Write down the modal class.

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Question 1002

[2 marks]Statistics & Probability
The heights of 30 Moringa seedlings are grouped as: 5 in 10<h≤2010<h\le20, 6 in 20<h≤2520<h\le25, 10 in 25<h≤3525<h\le35, 9 in 35<h≤4035<h\le40 (h in cm). Calculate the frequency density of the class 10<h≤2010<h\le20.

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Question 1003

[1 marks]Statistics & Probability
The heights of 30 Moringa seedlings are grouped as: 5 in 10<h≤2010<h\le20, 6 in 20<h≤2520<h\le25, 10 in 25<h≤3525<h\le35, 9 in 35<h≤4035<h\le40 (h in cm). Calculate the frequency density of the class 35<h≤4035<h\le40.

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Question 1004

[2 marks]Statistics & Probability
Of 30 Moringa seedlings, 9 have height in the class 35<h≤4035<h\le40 cm. Calculate the angle that would represent this class on a pie chart.

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Question 1005

[3 marks]Statistics & Probability
The heights of 30 Moringa seedlings are grouped as: 5 in 10<h≤2010<h\le20, 6 in 20<h≤2520<h\le25, 10 in 25<h≤3525<h\le35, 9 in 35<h≤4035<h\le40 (h in cm). Calculate an estimate of the mean height of the seedlings (in cm), using class mid-points.

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Question 1006

[3 marks]Statistics & Probability
Of 30 Moringa seedlings, 5 are at most 20 cm tall and 9 are more than 35 cm tall. If 2 seedlings are picked at random without replacement, find the probability that one is at most 20 cm tall and the other is more than 35 cm tall.

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Question 1101

[2 marks]Geometrical Transformation
Triangle ABC has vertices A(2;−2)A(2;-2), B(4;2)B(4;2), C(6;2)C(6;2). Triangle A1B1C1A_1B_1C_1 has vertices A1(−2;2)A_1(-2;2), B1(0;6)B_1(0;6), C1(2;6)C_1(2;6). Describe fully the single transformation which maps triangle ABC onto triangle A1B1C1A_1B_1C_1.
  1. ARotation of 180° about (0;2)
  2. BTranslation by vector (4;-4)
  3. CTranslation by vector (-4;4)
  4. DReflection in the line x=-1

Question 1201

[1 marks]Vector Geometry
In a star diagram, a regular hexagon ABCDEF (centre X) is surrounded by 6 equilateral triangles AOB, BPC, CQD, DRE, ESF and FTA. OA⃗=a\vec{OA}=\mathbf{a} and OB⃗=b\vec{OB}=\mathbf{b}. Write down OS⃗\vec{OS} in terms of a\mathbf{a} and/or b\mathbf{b}, in simplest form.

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Question 1202

[1 marks]Vector Geometry
In the same star diagram (regular hexagon ABCDEF surrounded by 6 equilateral triangles, OA⃗=a\vec{OA}=\mathbf{a}, OB⃗=b\vec{OB}=\mathbf{b}), write down AB⃗\vec{AB} in terms of a\mathbf{a} and/or b\mathbf{b}, in simplest form.

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Question 1203

[2 marks]Vector Geometry
In the same star diagram (regular hexagon ABCDEF surrounded by 6 equilateral triangles, OA⃗=a\vec{OA}=\mathbf{a}, OB⃗=b\vec{OB}=\mathbf{b}), write down OR⃗\vec{OR} in terms of a\mathbf{a} and/or b\mathbf{b}, in simplest form.

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Question 1204

[2 marks]Vector Geometry
In the same star diagram (regular hexagon ABCDEF surrounded by 6 equilateral triangles, OA⃗=a\vec{OA}=\mathbf{a}, OB⃗=b\vec{OB}=\mathbf{b}), write down CF⃗\vec{CF} in terms of a\mathbf{a} and/or b\mathbf{b}, in simplest form.

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Question 1205

[2 marks]Vector Geometry
In a star diagram, triangle OAB is equilateral with OA=OB=AB=5OA=OB=AB=5 cm, and a=OA⃗\mathbf{a}=\vec{OA}, b=OB⃗\mathbf{b}=\vec{OB}. Find ∣a−b∣|\mathbf{a}-\mathbf{b}| (in cm).

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Question 1206

[2 marks]Vector Geometry
In a star diagram, OA = 5 cm, and OQ and OS are each three times the length of OA. Triangle OQS is equilateral. Find the perimeter of triangle OQS (in cm).

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Question 1207

[2 marks]Vector Geometry
Triangle OAB is equilateral with side 5 cm. Calculate the area of triangle OAB, correct to 3 significant figures (in cm2^2).

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