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ZIMSEC O Level · 4028/2 · N2015

Mathematics Paper 2 November 2015

Questions
59
Total marks
136
Time allowed
150 min
Syllabus code
4028/2

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Questions
59
Pass mark
36
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[3 marks]Fractions, Factorisation and Functional Notation
Find the value of (134+213)÷56\left(1\frac{3}{4} + 2\frac{1}{3}\right) \div \frac{5}{6}.

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Question 102

[2 marks]Fractions, Factorisation and Functional Notation
Factorise completely y2+10y−24y^2 + 10y - 24.

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Question 103

[2 marks]Fractions, Factorisation and Functional Notation
Factorise completely 27−3x227 - 3x^2.

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Question 104

[1 marks]Fractions, Factorisation and Functional Notation
It is given that f(x)=10+3x−x2f(x) = 10 + 3x - x^2. Find f(2)f(2).

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Question 105

[3 marks]Fractions, Factorisation and Functional Notation
It is given that f(x)=10+3x−x2f(x) = 10 + 3x - x^2. Find the values of xx for which f(x)=0f(x) = 0.

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Question 201

[2 marks]Trigonometry, Polygons and Percentages
In quadrilateral ABCD, BC=8BC = 8 cm, AC=12AC = 12 cm and AB^C=90∘A\hat{B}C = 90^\circ. Calculate AB, in centimetres, correct to 3 significant figures.

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Question 202

[2 marks]Trigonometry, Polygons and Percentages
In quadrilateral ABCD, AC=12AC = 12 cm, CA^D=46,5∘C\hat{A}D = 46,5^\circ and AC^D=90∘A\hat{C}D = 90^\circ. Calculate CD, in centimetres, correct to 3 significant figures.

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Question 203

[3 marks]Trigonometry, Polygons and Percentages
Quadrilateral ABCD has AB=8,94AB = 8,94 cm, BC=8BC = 8 cm, AC=12AC = 12 cm and CD=12,6CD = 12,6 cm, with right angles at B and at C. Calculate the area of ABCD, in square centimetres, correct to 3 significant figures.

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Question 204

[2 marks]Trigonometry, Polygons and Percentages
Four interior angles of a nonagon have a sum of 460∘460^\circ. The remaining interior angles are equal. Find the size of each of the equal angles.

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Question 205

[2 marks]Trigonometry, Polygons and Percentages
A shop sells a refrigerator at \$540. In the previous year the same type of refrigerator cost 8 % less. Calculate the cost price of the refrigerator in the previous year.
  1. A\$532,00, subtracting 8 dollars rather than 8 per cent of the price
  2. B\$496,80, since the previous year's price is 92 % of the \$540 charged now
  3. C\$586,96, taking \$540 as 92 % of the earlier price and dividing by 0,92
  4. D\$583,20, since 8 % of \$540 is added on to give the earlier price

Question 301

[2 marks]Matrices, Algebraic Fractions and Mensuration
Given M=(8−4−53)M = \begin{pmatrix} 8 & -4 \\ -5 & 3 \end{pmatrix} and N=(13)N = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, find MN.

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Question 302

[2 marks]Matrices, Algebraic Fractions and Mensuration
Find the inverse of M=(8−4−53)M = \begin{pmatrix} 8 & -4 \\ -5 & 3 \end{pmatrix}.
  1. A144(3458)\frac{1}{44}\begin{pmatrix} 3 & 4 \\ 5 & 8 \end{pmatrix}, taking the determinant as 24+2024 + 20 instead of 24−2024 - 20
  2. B14(3−4−58)\frac{1}{4}\begin{pmatrix} 3 & -4 \\ -5 & 8 \end{pmatrix}, swapping the leading diagonal but leaving the other two signs alone
  3. C14(845−3)\frac{1}{4}\begin{pmatrix} 8 & 4 \\ 5 & -3 \end{pmatrix}, changing the signs on the leading diagonal instead of swapping them
  4. D14(3458)\frac{1}{4}\begin{pmatrix} 3 & 4 \\ 5 & 8 \end{pmatrix}, with determinant 24−20=424 - 20 = 4 and both diagonals treated correctly

Question 303

[2 marks]Matrices, Algebraic Fractions and Mensuration
Given M=(8−4−53)M = \begin{pmatrix} 8 & -4 \\ -5 & 3 \end{pmatrix}, find M2M^2.

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Question 304

[3 marks]Matrices, Algebraic Fractions and Mensuration
Express 2x−14−3x−512\frac{2x-1}{4} - \frac{3x-5}{12} as a single fraction in its simplest form.

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Question 305

[2 marks]Matrices, Algebraic Fractions and Mensuration
The area of a trapezium is 63 cm2^2 and the sum of the lengths of its two parallel sides is 22,5 cm. Calculate the perpendicular distance, in centimetres, between the two parallel sides.

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Question 401

[2 marks]Sets, Probability and Laws of Indices
In a class of 40 music students, 6 enjoyed both the mbira and the guitar, and 3 enjoyed all three of the mbira, the piano and the guitar. Find the value of ww, the number who enjoyed the mbira and the guitar but not the piano.

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Question 402

[3 marks]Sets, Probability and Laws of Indices
In a class of 40 music students, 20 enjoyed the guitar, 6 enjoyed the mbira and the guitar, 4 enjoyed the guitar and the piano, and 3 enjoyed all three instruments. Find the value of zz, the number who enjoyed the guitar only.

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Question 403

[2 marks]Sets, Probability and Laws of Indices
In a class of 40 music students, 20 enjoyed playing the guitar. Two students are selected at random from the class. Find the probability that both enjoyed playing the guitar.
  1. A1978\frac{19}{78}, since the second student comes from the 39 left, 19 of whom play guitar
  2. B1939\frac{19}{39}, using the second draw by itself and forgetting to count the first one
  3. C12\frac{1}{2}, which is the proportion of the whole class that enjoyed the guitar
  4. D14\frac{1}{4}, treating the two choices as independent with 20 out of 40 each time

Question 404

[2 marks]Sets, Probability and Laws of Indices
Solve the equation 9m−1=279^{m-1} = 27.

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Question 501

[3 marks]Change of Subject of Formula and Circle Geometry
Make xx the subject of the formula y=p−2xxy = \frac{p - 2x}{x}.

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Question 502

[1 marks]Change of Subject of Formula and Circle Geometry
ABCD is a circle with centre O and CD is a diameter. Given that AO^D=50∘A\hat{O}D = 50^\circ, find OC^AO\hat{C}A, in degrees.

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Question 503

[2 marks]Change of Subject of Formula and Circle Geometry
ABCD is a circle with centre O and AO^D=50∘A\hat{O}D = 50^\circ. Find OA^DO\hat{A}D, in degrees.

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Question 504

[2 marks]Change of Subject of Formula and Circle Geometry
ABCD is a cyclic quadrilateral in a circle with centre O, CD is a diameter and AD^C=65∘A\hat{D}C = 65^\circ. Find AB^CA\hat{B}C.
  1. A90∘90^\circ, treating CB as a diameter so that the angle at B stands in a semicircle
  2. B115∘115^\circ, since opposite angles of a cyclic quadrilateral add up to 180∘180^\circ
  3. C130∘130^\circ, doubling the 65∘65^\circ angle at D as though it stood at the centre of the circle
  4. D65∘65^\circ, taking AB^CA\hat{B}C to equal AD^CA\hat{D}C because they look like angles in the same segment

Question 505

[2 marks]Change of Subject of Formula and Circle Geometry
In circle ABCD with centre O, OA is parallel to CB, OA^C=25∘O\hat{A}C = 25^\circ and AB^C=115∘A\hat{B}C = 115^\circ. Find CA^BC\hat{A}B, in degrees.

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Question 601

[2 marks]Constructions and Loci
A parallelogram ABCD is constructed and the bisector of AB^CA\hat{B}C is drawn. Describe the locus that this bisector represents.
  1. AThe locus of points equidistant from the points A and C, the two ends of that diagonal
  2. BThe locus of points equidistant from the lines BA and BC, the two arms of the angle
  3. CThe locus of points a fixed distance from the vertex B, which is an arc of a circle
  4. DThe locus of points equidistant from the points B and C, a perpendicular bisector

Question 602

[2 marks]Constructions and Loci
In parallelogram ABCD, AB^C=120∘A\hat{B}C = 120^\circ. Write down the size of BC^DB\hat{C}D, in degrees.

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Question 603

[2 marks]Constructions and Loci
State what has to be constructed to show the locus of points equidistant from the two points B and C.
  1. AThe bisector of the angle at B, the line that splits AB^CA\hat{B}C into two equal parts
  2. BThe line BC itself, extended equally far beyond B and beyond C in both directions
  3. CThe perpendicular bisector of BC, the line through the midpoint of BC at right angles to it
  4. DA circle with BC as its diameter, drawn so that it passes through both B and C

Question 604

[2 marks]Constructions and Loci
In parallelogram ABCD, AB=8AB = 8 cm and BC=10BC = 10 cm. Calculate the perimeter of the parallelogram, in centimetres.

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Question 701

[1 marks]Functional Graphs
A table of values for y=12x(5−x)y = \frac{1}{2}x(5 - x) gives the value of yy at x=3x = 3 as pp. Calculate the value of pp.

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Question 702

[2 marks]Functional Graphs
State the maximum value of yy for the function y=12x(5−x)y = \frac{1}{2}x(5 - x).

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Question 703

[2 marks]Functional Graphs
Write down the range of values of xx for which y=12x(5−x)y = \frac{1}{2}x(5 - x) is positive.

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Question 704

[2 marks]Functional Graphs
State the equation of the straight line that must be drawn together with the curve y=12x(5−x)y = \frac{1}{2}x(5 - x) in order to solve the equation 12x(5−x)=x−1\frac{1}{2}x(5 - x) = x - 1.

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Question 705

[3 marks]Functional Graphs
Solve the equation 12x(5−x)=x−1\frac{1}{2}x(5 - x) = x - 1, giving the answers correct to 2 decimal places.
  1. Ax=0x = 0 or x=5x = 5, the two points at which the curve itself crosses the xx-axis
  2. Bx=3x = 3 or x=−1x = -1, the roots of x2−2x−3=0x^2 - 2x - 3 = 0 found by dropping the fraction
  3. Cx=3,56x = 3,56 or x=−0,56x = -0,56, the two roots of x2−3x−2=0x^2 - 3x - 2 = 0 to 2 decimal places
  4. Dx=2x = 2 or x=−1x = -1, read off where the line y=x−1y = x - 1 meets the two axes

Question 706

[2 marks]Functional Graphs
For y=12x(5−x)y = \frac{1}{2}x(5 - x) the values of yy at x=2x = 2, x=3x = 3 and x=4x = 4 are 3, 3 and 2. Using the trapezium rule with these three ordinates, estimate the area bounded by the curve, the xx-axis and the lines x=2x = 2 and x=4x = 4, in square units.

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Question 801

[2 marks]Quadratic Equations and Linear Programming
The equation 21−x−4x=3\frac{2}{1-x} - \frac{4}{x} = 3 reduces to 3x2+3x−4=03x^2 + 3x - 4 = 0. Calculate the value of the discriminant b2−4acb^2 - 4ac of that quadratic.

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Question 802

[3 marks]Quadratic Equations and Linear Programming
Solve the equation 3x2+3x−4=03x^2 + 3x - 4 = 0, giving the answers correct to 3 significant figures.

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Question 803

[2 marks]Quadratic Equations and Linear Programming
The unshaded region R is defined by three inequalities, one of which is y≥4−xy \geq 4 - x. Its other two boundaries are a vertical line through x=4x = 4 and a line joining (0;4)(0; 4) to (8;0)(8; 0), with the shading outside R in each case. Write down the other two inequalities.
  1. Ax≤4x \leq 4 and x+2y≤8x + 2y \leq 8, the vertical boundary and the shallower slanting boundary
  2. Bx≥4x \geq 4 and x+2y≥8x + 2y \geq 8, taking the shaded side of each boundary as the wanted one
  3. Cx≤4x \leq 4 and y≤4−2xy \leq 4 - 2x, reading the shallow line's gradient as −2-2 and not −12-\frac{1}{2}
  4. Dy≤4y \leq 4 and x+2y≤8x + 2y \leq 8, using a horizontal boundary in place of the vertical one

Question 804

[3 marks]Quadratic Equations and Linear Programming
The region R has vertices at (0;4)(0; 4), (4;0)(4; 0) and (4;2)(4; 2). Find the maximum value of 3y+x3y + x for points in R.

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Question 901

[2 marks]Vector Geometry
If v=(8u)v = \begin{pmatrix} 8 \\ u \end{pmatrix} and ∣v∣=17|v| = 17, find the two possible values of uu.

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Question 902

[1 marks]Vector Geometry
In triangle XYZ, XZ→=p\overrightarrow{XZ} = p and XY→=q\overrightarrow{XY} = q. Express YZ→\overrightarrow{YZ} in terms of pp and qq.

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Question 903

[2 marks]Vector Geometry
In triangle XYZ, XZ→=p\overrightarrow{XZ} = p, XY→=q\overrightarrow{XY} = q and M lies on YZ with 3YM=YZ3YM = YZ. Express XM→\overrightarrow{XM} in terms of pp and qq.

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Question 904

[2 marks]Vector Geometry
In triangle XYZ, XM→=13p+23q\overrightarrow{XM} = \frac{1}{3}p + \frac{2}{3}q, where pp and qq are not parallel. Given also that XM→=hp+kq\overrightarrow{XM} = hp + kq, find the value of hh and the value of kk.
  1. Ah=23h = \frac{2}{3} and k=13k = \frac{1}{3}, matching the two coefficients the other way round
  2. Bh=13h = \frac{1}{3} and k=13k = \frac{1}{3}, taking M as the midpoint of YZ rather than one third along
  3. Ch=13h = \frac{1}{3} and k=23k = \frac{2}{3}, matching each coefficient to the vector it multiplies
  4. Dh=3h = 3 and k=3k = 3, reading 3YM=YZ3YM = YZ as multiplying by 3 rather than dividing by 3

Question 905

[2 marks]Vector Geometry
N lies on XZ with XN→=13XZ→\overrightarrow{XN} = \frac{1}{3}\overrightarrow{XZ}. Write down the numerical value of XNNZ\frac{XN}{NZ}.

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Question 906

[2 marks]Vector Geometry
N lies on XZ with XN=13XZXN = \frac{1}{3}XZ, and Y is the third vertex of triangle XYZ. Write the ratio area of triangle XYN : area of triangle XYZ in its simplest form.

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Question 1001

[2 marks]Geometrical Transformation
Triangle LMN, with L (3; 1), M (2; 2) and N (0; 1), is reflected in the line y=1y = 1. Two of the three vertices are unmoved by this reflection. Write down the letters of those two vertices.

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Question 1002

[3 marks]Geometrical Transformation
Triangle LMN, with L (3; 1), M (2; 2) and N (0; 1), is mapped onto triangle L2M2N2L_2M_2N_2 by a rotation through 180∘180^\circ about the point (−1;0)(-1; 0). Find the coordinates of L2L_2, M2M_2 and N2N_2.
  1. AL2(−3;−1)L_2(-3; -1), M2(−2;−2)M_2(-2; -2), N2(0;−1)N_2(0; -1), rotating about the origin instead of (−1;0)(-1; 0)
  2. BL2(−5;−1)L_2(-5; -1), M2(−4;−2)M_2(-4; -2), N2(−2;−1)N_2(-2; -1), using the rule (x;y)→(−2−x;−y)(x; y) \to (-2 - x; -y)
  3. CL2(−5;1)L_2(-5; 1), M2(−4;2)M_2(-4; 2), N2(−2;1)N_2(-2; 1), reflecting in the line x=−1x = -1 rather than rotating
  4. DL2(5;1)L_2(5; 1), M2(4;2)M_2(4; 2), N2(2;1)N_2(2; 1), changing the sign of xx only and leaving yy untouched

Question 1003

[2 marks]Geometrical Transformation
The point L (3; 1) is mapped by the matrix (−2001)\begin{pmatrix} -2 & 0 \\ 0 & 1 \end{pmatrix} onto the point L3L_3. Write down the xx-coordinate of L3L_3.

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Question 1004

[3 marks]Geometrical Transformation
Describe fully the single transformation P represented by the matrix (−2001)\begin{pmatrix} -2 & 0 \\ 0 & 1 \end{pmatrix}.
  1. AAn enlargement centred at the origin with scale factor −2-2, doubling both coordinates
  2. BA reflection in the yy-axis on its own, since the xx-coordinates simply change sign
  3. CA one-way stretch parallel to the yy-axis, with the xx-axis invariant, scale factor −2-2
  4. DA one-way stretch parallel to the xx-axis, with the yy-axis invariant, scale factor −2-2

Question 1005

[2 marks]Geometrical Transformation
The transformation represented by the matrix (−2001)\begin{pmatrix} -2 & 0 \\ 0 & 1 \end{pmatrix} leaves one line fixed point by point. Write down the equation of that line.

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Question 1101

[3 marks]Statistics
The marks of 80 students are grouped in classes with mid-values 10, 25, 35, 45, 55, 65, 75, 85 and 95 and frequencies 0, 5, 19, 18, 16, 14, 4, 2 and 2. Estimate the mean mark, correct to 3 significant figures.

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Question 1102

[1 marks]Statistics
A cumulative frequency distribution for 80 marks reads 0, 5, 24, 42, 58, 72, q, 78 and 80 for m≤20m \leq 20, 30, 40, 50, 60, 70, 80, 90 and 100. The class 70<m≤8070 < m \leq 80 has frequency 4. Find the value of q.

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Question 1103

[3 marks]Statistics
For 80 students the cumulative frequency is 24 at a mark of 40 % and 42 at a mark of 50 %. Estimate the median mark, in per cent, to the nearest whole number.

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Question 1104

[3 marks]Statistics
Reading from a cumulative frequency curve for 80 students, about 33 students scored 45 % or less and about 74 scored 75 % or less. Estimate the number of students whose marks were more than 45 % but less than 75 %.

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Question 1105

[2 marks]Statistics
A grouped distribution has the class 30<m≤4030 < m \leq 40 with frequency 19, and the running total up to m≤40m \leq 40 is 24. State the point that must be plotted for this class when a cumulative frequency curve is drawn.
  1. A(40;24)(40; 24), the upper class boundary plotted against the running total up to it
  2. B(35;24)(35; 24), the mid-value of the class plotted against the running total up to it
  3. C(40;19)(40; 19), the upper class boundary plotted against that one class's frequency
  4. D(30;24)(30; 24), the lower class boundary plotted against the total up to the upper one

Question 1201

[2 marks]Measures and Mensuration
A right cone has a base radius of 8 cm and a slant height of 10 cm. Calculate the perpendicular height of the cone, in centimetres.

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Question 1202

[2 marks]Measures and Mensuration
A right cone has a base radius of 8 cm and a slant height of 10 cm. Taking π\pi to be 227\frac{22}{7}, calculate the curved surface area of the cone, in square centimetres, correct to 3 significant figures.

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Question 1203

[2 marks]Measures and Mensuration
A right cone has a base radius of 8 cm and a perpendicular height of 6 cm. Taking π\pi to be 227\frac{22}{7}, calculate the volume of the cone, in cubic centimetres, correct to 3 significant figures.

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Question 1204

[3 marks]Measures and Mensuration
A cone of base radius 8 cm and slant height 10 cm is cut open to make a sector ABC of a circle centre O, where the shaded sector is the major one. Taking π\pi to be 227\frac{22}{7}, calculate reflex AO^CA\hat{O}C, in degrees.

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Question 1205

[3 marks]Measures and Mensuration
A cone PQR is similar to a right cone of base radius 8 cm and slant height 10 cm, and has a slant height of 18 cm. Taking π\pi to be 227\frac{22}{7}, calculate the base area of the cone PQR.
  1. A362362 cm2^2, scaling the base area by 1,81,8 instead of by 1,81,8 squared
  2. B1 1731\,173 cm2^2, scaling the base area by 1,81,8 cubed as though it were a volume
  3. C652652 cm2^2, since lengths scale by 1,81,8, so the base radius becomes 14,414,4 cm
  4. D201201 cm2^2, the base area of the original cone, left unscaled altogether

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