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ZIMSEC O Level · 4004/2 · N2023

Mathematics Paper 2 November 2023

Questions
64
Total marks
136
Time allowed
150 min
Syllabus code
4004/2

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Questions
64
Pass mark
39
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]Fractions, Decimals & Percentages
The price of a packet of sweets was $1,20. The price increased to $1,50. Calculate the percentage increase.

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Question 102

[2 marks]Fractions, Decimals & Percentages
A new cereal packet contains 20% more cereal than the old packet. If the new packet has a mass of 264 g, calculate the mass, in grams, of the cereal in the old packet.

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Question 103

[2 marks]Consumer Arithmetic
A painter got a 7% discount after buying a large amount of paint. If the discount was $91, find the value, in dollars, of the paint he bought.

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Question 104

[3 marks]Circle Geometry
ABAB is a diameter of a circle centre O, and CC is a point on the circumference such that AC=7,5AC = 7,5 cm and BC=5BC = 5 cm. Calculate the length of ABAB, in centimetres.

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Question 201

[2 marks]Constructions & Loci
Triangle ABC has AB=6AB = 6 cm, AC=5,8AC = 5,8 cm and BA^C=135°B\hat{A}C = 135°. Calculate the length, in centimetres, of the perpendicular from C to BA produced.

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Question 202

[1 marks]Constructions & Loci
Describe fully the locus of points 4,5 cm from a fixed point B.

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Question 203

[2 marks]Constructions & Loci
In triangle ABC, AB=6AB = 6 cm, AC=5,8AC = 5,8 cm and BA^C=135°B\hat{A}C = 135°, and C is 4,1 cm from the line AB. Describe fully the locus represented by the line parallel to AB passing through C.

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Question 301

[1 marks]Sets
The universal set is ξ={x:x\xi = \{x : x is an integer and 30≤x≤40}30 \le x \le 40\} and P={x:xP = \{x : x is a prime number}\}. Find n(P)n(P).

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Question 302

[2 marks]Sets
The universal set is ξ={x:x\xi = \{x : x is an integer and 30≤x≤40}30 \le x \le 40\}, PP is the set of prime numbers in ξ\xi and QQ is the set of multiples of 4 in ξ\xi. List the members of Q∩P′Q \cap P'.

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Question 303

[3 marks]Trigonometry, Bearing & Distances
In triangle XYZ, XY=17XY = 17 cm, XZ=8XZ = 8 cm and YX^Z=65°30′Y\hat{X}Z = 65°30'. Calculate the area of triangle XYZ, in square centimetres.

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Question 304

[2 marks]Probability
There are 25 red sweets and xx blue sweets in a box. One sweet is selected at random and the probability that it is blue is 38\dfrac{3}{8}. Find xx.

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Question 305

[3 marks]Variation
P varies directly as q and inversely as the square of rr. Given that P=15P = 15 when q=9q = 9 and r=6r = 6, find PP when q=7q = 7 and r=2r = 2.

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Question 401

[2 marks]Algebraic Expressions
Remove the brackets and simplify 4(3−2p)−3(1−p)4(3 - 2p) - 3(1 - p).

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Question 402

[2 marks]Algebraic Expressions
Remove the brackets and simplify (3q−r)(q+2r)(3q - r)(q + 2r).

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Question 403

[1 marks]Factorisation, H.C.F & L.C.M
Factorise completely 8ab+6ab28ab + 6ab^2.

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Question 404

[2 marks]Factorisation, H.C.F & L.C.M
Factorise completely 18t2−218t^2 - 2.

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Question 405

[1 marks]Substitution
Given that y=18+3x2y = 18 + 3x^2, find the value of yy when x=−2x = -2.

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Question 406

[2 marks]Quadratic Equations
Given that y=18+3x2y = 18 + 3x^2, find the values of xx when y=93y = 93.

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Question 407

[2 marks]Change of Subject of Formula
Given that y=18+3x2y = 18 + 3x^2, express xx in terms of yy.

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Question 501

[3 marks]Algebraic Fractions
Express 4y−3−3y+4\dfrac{4}{y-3} - \dfrac{3}{y+4} as a single fraction in its simplest form.

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Question 502

[3 marks]Quadratic Equations
Solve the equation y−18=2y−1\dfrac{y-1}{8} = \dfrac{2}{y-1}.

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Question 503

[2 marks]Inequalities
Solve the inequality 21<4n−3≤2721 < 4n - 3 \le 27.

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Question 504

[1 marks]Co-ordinate Geometry
Points PP and QQ have coordinates (9;8)(9; 8) and (12;4)(12; 4) respectively. Find the gradient of QPQP.

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Question 505

[2 marks]Co-ordinate Geometry
The line PQ has gradient −43-\dfrac43 and the point RR is (4;−2)(4; -2). Find the equation of the line passing through RR and parallel to PQPQ.

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Question 601

[1 marks]Statistics & Probability
The ages of 40 learners are grouped as 8≤x<108 \le x < 10 (frequency density 3,5), 10≤x<1110 \le x < 11 (8), 11≤x<1211 \le x < 12 (6), 12≤x<1412 \le x < 14 (5), 14≤x<1614 \le x < 16 (1,5) and 16≤x<1916 \le x < 19 (2). State the modal class.

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Question 602

[3 marks]Statistics & Probability
The ages of 40 learners are grouped as 8≤x<108 \le x < 10 (7 learners), 10≤x<1110 \le x < 11 (8), 11≤x<1211 \le x < 12 (6), 12≤x<1412 \le x < 14 (10), 14≤x<1614 \le x < 16 (3) and 16≤x<1916 \le x < 19 (6). Calculate an estimate of the mean age, in years.

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Question 603

[1 marks]Probability
In a group of 40 learners the youngest age class recorded is 8≤x<108 \le x < 10. One learner is chosen at random. Calculate the probability that the learner is below the age of 8 years.

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Question 604

[1 marks]Probability
In a group of 40 learners, 6 are in the class 16≤x<1916 \le x < 19 and the rest are younger than 16. One learner is chosen at random. Calculate the probability that the learner is below the age of 16 years.

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Question 605

[2 marks]Probability
Of 40 learners, 3 are in the age group 14≤x<1614 \le x < 16. If two learners are chosen at random, find the probability that they are both in that age group.

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Question 701

[2 marks]Functional Graphs
A table of values for y=6−x−x2y = 6 - x - x^2 gives y=py = p when x=−4x = -4 and y=qy = q when x=0x = 0. Find the value of pp and the value of qq.

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Question 702

[1 marks]Functional Graphs
The graph of y=6−x−x2y = 6 - x - x^2 is drawn for −4≤x≤3-4 \le x \le 3. Use it to find the maximum value of 6−x−x26 - x - x^2.

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Question 703

[1 marks]Functional Graphs
The graph of y=6−x−x2y = 6 - x - x^2 is drawn for −4≤x≤3-4 \le x \le 3. Write down the equation of its line of symmetry.

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Question 704

[2 marks]Functional Graphs
The graph of y=6−x−x2y = 6 - x - x^2 is drawn for −4≤x≤3-4 \le x \le 3. Use a tangent to find the gradient of the graph when x=−2x = -2.

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Question 705

[2 marks]Functional Graphs
The graph of y=6−x−x2y = 6 - x - x^2 passes through (−2;4)(-2; 4), (−1;6)(-1; 6) and (0;6)(0; 6). Estimate the area, in square units, bounded by the graph, the xx-axis, the yy-axis and the line x=−2x = -2.

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Question 801

[1 marks]Measures & Mensuration
A solid right cone has a base radius of 8 cm and a height of 15 cm. Calculate its slant height, in centimetres.

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Question 802

[2 marks]Measures & Mensuration
A solid right cone has a base radius of 8 cm and a height of 15 cm. Taking π\pi to be 227\frac{22}{7}, calculate the volume of the cone, in cubic centimetres.

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Question 803

[2 marks]Measures & Mensuration
A solid right cone has a base radius of 8 cm and a slant height of 17 cm. Taking π\pi to be 227\frac{22}{7}, calculate the curved surface area of the cone, in square centimetres.

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Question 804

[1 marks]Measures & Mensuration
A solid right cone has a base radius of 8 cm and a curved surface area of 29927 cm2\frac{2992}{7}\ \text{cm}^2. Taking π\pi to be 227\frac{22}{7}, calculate the total surface area of the cone, in square centimetres.

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Question 805

[2 marks]Measures & Mensuration
A square based cuboid of height 2 cm is made from 50 cm350\ \text{cm}^3 of clay. Find the length, in centimetres, of the side of the square base.

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Question 806

[2 marks]Measures & Mensuration
A pyramid with a base area of 15 cm215\ \text{cm}^2 is made from 50 cm350\ \text{cm}^3 of clay. Find the height of the pyramid, in centimetres.

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Question 807

[2 marks]Measures & Mensuration
A sphere is made from 50 cm350\ \text{cm}^3 of clay. Taking π\pi to be 227\frac{22}{7}, calculate the radius of the sphere, in centimetres. [Volume of a sphere =43πr3= \frac43\pi r^3]

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Question 901

[3 marks]Geometrical Transformation
Triangle A has vertices (2;2)(2;2), (4;2)(4;2) and (4;5)(4;5). It is mapped onto triangle C by an enlargement centre (0;0)(0;0), scale factor −2-2. Write down the vertices of triangle C.

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Question 902

[3 marks]Geometrical Transformation
Triangle A has vertices (2;2)(2;2), (4;2)(4;2) and (4;5)(4;5). Triangle D is its image under a clockwise rotation of 90° about the point (2;−2)(2;-2). Write down the vertices of triangle D.

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Question 903

[2 marks]Geometrical Transformation
Triangle A has vertices (2;2)(2;2), (4;2)(4;2), (4;5)(4;5) and triangle B has vertices (−4;3)(-4;3), (−2;3)(-2;3), (−2;6)(-2;6). Describe fully the single transformation that maps triangle A onto triangle B.

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Question 904

[2 marks]Geometrical Transformation
Describe fully the single transformation represented by the matrix (100−2)\begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix}.

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Question 1001

[2 marks]Matrices
Given that A=(−34−122)A = \begin{pmatrix} -3 & 4 \\ -\frac12 & 2 \end{pmatrix} and B=(−2261)B = \begin{pmatrix} -2 & 2 \\ 6 & 1 \end{pmatrix}, find A+BA + B.

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Question 1002

[2 marks]Matrices
Given that A=(−34−122)A = \begin{pmatrix} -3 & 4 \\ -\frac12 & 2 \end{pmatrix}, find A−1A^{-1}, the inverse of matrix A.

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Question 1003

[3 marks]Matrices
The determinant of the matrix (k−1413k)\begin{pmatrix} k-1 & 4 \\ 1 & 3k \end{pmatrix} is 14. Find the possible values of kk.

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Question 1004

[1 marks]Circle Geometry
In the diagram P, Q, R and S lie on a circle centre O, PS is a diameter and SP^R=43°S\hat{P}R = 43°. Find PS^RP\hat{S}R, in degrees.

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Question 1005

[1 marks]Circle Geometry
In the diagram P, Q, R and S lie on a circle centre O and PO^Q=68°P\hat{O}Q = 68°. Find PR^QP\hat{R}Q, in degrees.

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Question 1006

[1 marks]Circle Geometry
In the diagram the tangent to the circle at Q meets SP produced at T, and TO^Q=68°T\hat{O}Q = 68°. Find PT^QP\hat{T}Q, in degrees.

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Question 1007

[1 marks]Circle Geometry
In the diagram P, Q, R and S lie on a circle centre O, PS is a diameter, TO^Q=68°T\hat{O}Q = 68° and SP^R=43°S\hat{P}R = 43°. Find RQ^OR\hat{Q}O, in degrees.

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Question 1101

[1 marks]Vector Geometry
In the diagram OA⃗=a\vec{OA} = a and OB⃗=b\vec{OB} = b. Express AB⃗\vec{AB} in terms of aa and bb.

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Question 1102

[1 marks]Vector Geometry
In the diagram OA⃗=a\vec{OA} = a, OB⃗=b\vec{OB} = b and AX:XB=2:3AX : XB = 2 : 3, with X on AB. Express BX⃗\vec{BX} in terms of aa and bb.

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Question 1103

[2 marks]Vector Geometry
In the diagram OA⃗=a\vec{OA} = a, OB⃗=b\vec{OB} = b and AX:XB=2:3AX : XB = 2 : 3, with X on AB. Express OX⃗\vec{OX} in terms of aa and bb.

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Question 1104

[2 marks]Vector Geometry
In the diagram OA⃗=a\vec{OA} = a, OB⃗=b\vec{OB} = b and BC⃗=kOA⃗\vec{BC} = k\vec{OA}. Express AC⃗\vec{AC} in terms of kk, aa and bb.

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Question 1105

[1 marks]Vector Geometry
In the diagram OA⃗=a\vec{OA} = a, OB⃗=b\vec{OB} = b and BC⃗=kOA⃗\vec{BC} = k\vec{OA}. Express OC⃗\vec{OC} in terms of kk, aa and bb.

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Question 1106

[1 marks]Vector Geometry
Given that OX⃗=35a+25b\vec{OX} = \frac35 a + \frac25 b and OC⃗=hOX⃗\vec{OC} = h\vec{OX}, express OC⃗\vec{OC} in terms of hh, aa and bb.

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Question 1107

[3 marks]Vector Geometry
Given that OC⃗=ka+b\vec{OC} = ka + b and also OC⃗=3h5a+2h5b\vec{OC} = \frac{3h}{5}a + \frac{2h}{5}b, find the values of kk and hh.

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Question 1108

[1 marks]Vector Geometry
Given that AC⃗=(k−1)a+b\vec{AC} = (k-1)a + b and k=1,5k = 1,5, express AC⃗\vec{AC} in terms of aa and bb.

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Question 1201

[3 marks]Trigonometry, Bearing & Distances
Points X, Y and Z lie on horizontal ground with YX^Z=85°Y\hat{X}Z = 85°, XY=6,5XY = 6,5 m and XZ=7,2XZ = 7,2 m. Calculate YZ, in metres.

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Question 1202

[3 marks]Trigonometry, Bearing & Distances
In triangle XYZ, YX^Z=85°Y\hat{X}Z = 85°, XZ=7,2XZ = 7,2 m and YZ=9,27YZ = 9,27 m. Calculate XY^ZX\hat{Y}Z, in degrees correct to one decimal place.

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Question 1203

[3 marks]Trigonometry, Bearing & Distances
The bearing of Z from X is 250°, YX^Z=85°Y\hat{X}Z = 85° with Y on the northern side of XZ, and XY^Z=50,7°X\hat{Y}Z = 50,7°. Calculate the bearing of Z from Y, correct to the nearest degree.

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Question 1204

[2 marks]Trigonometry, Bearing & Distances
In triangle XYZ, XY=6,5XY = 6,5 m, XZ=7,2XZ = 7,2 m, YX^Z=85°Y\hat{X}Z = 85° and YZ=9,27YZ = 9,27 m. Calculate the shortest distance from X to YZ, in metres.

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