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ZIMSEC O Level · 4028/2 · J2011

Mathematics Paper 2 June 2011

Questions
58
Total marks
136
Time allowed
150 min
Syllabus code
4028/2

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Questions
58
Pass mark
35
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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]Fractions, Decimals and Percentages
Simplify (312−135)+123\left(3\tfrac{1}{2} - 1\tfrac{3}{5}\right) + 1\tfrac{2}{3}, giving your answer as a mixed number.

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Question 102

[2 marks]Fractions, Decimals and Percentages
Find the exact value of 10,03×0,1710,03 \times 0,17.

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Question 103

[2 marks]Fractions, Decimals and Percentages
Three farmers share 120 hectares of land in the ratio 3:4:53 : 4 : 5. Calculate, in hectares, the area of the largest piece of land.

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Question 104

[2 marks]Fractions, Decimals and Percentages
Find the Highest Common Factor of 27x2yz27x^2yz, 72xy3z272xy^3z^2 and 108xyz3108xyz^3.
  1. A9x2y3z39x^2y^3z^3
  2. B216xyz216xyz
  3. C3x2y3z33x^2y^3z^3
  4. D9xyz9xyz

Question 201

[2 marks]Algebraic Fractions
It is given that r=2q−5r = 2q - 5 and q=3p+2q = 3p + 2. Express rr in terms of pp, in its simplest form.

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Question 202

[2 marks]Algebraic Fractions
Express x3+x−45\dfrac{x}{3} + \dfrac{x - 4}{5} as a single fraction in its simplest form.
  1. A5x−1215\dfrac{5x - 12}{15}
  2. B8x−1215\dfrac{8x - 12}{15}
  3. C2x−48\dfrac{2x - 4}{8}
  4. D8x+1215\dfrac{8x + 12}{15}

Question 203

[3 marks]Algebraic Fractions
Solve the equation x3+x−45=4\dfrac{x}{3} + \dfrac{x - 4}{5} = 4.

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Question 204

[2 marks]Algebraic Fractions
Express 150 g as a percentage of 3 kg.

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Question 301

[2 marks]Circle Geometry
A, B, C and D lie on the circumference of a circle, and AC and BD meet at F. Given that AC^D=33∘A\hat{C}D = 33^\circ, calculate AB^DA\hat{B}D, in degrees.

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Question 302

[3 marks]Circle Geometry
A, B, C and D lie on the circumference of a circle, and CD is produced to E. Given that AC^D=33∘A\hat{C}D = 33^\circ and AD^E=87∘A\hat{D}E = 87^\circ, calculate CB^DC\hat{B}D, in degrees.

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Question 303

[3 marks]Circle Geometry
A, B, C and D lie on the circumference of a circle with BC=CDBC = CD, and CD is produced to E. Given that AC^D=33∘A\hat{C}D = 33^\circ and AD^E=87∘A\hat{D}E = 87^\circ, calculate AD^BA\hat{D}B, in degrees.

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Question 304

[2 marks]Circle Geometry
A sector POQ of a circle has centre O, with PO^Q=37∘P\hat{O}Q = 37^\circ and PO=8PO = 8 cm. PT is perpendicular to QO, with T lying on QO. Calculate PT, in centimetres, correct to three significant figures.

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Question 305

[3 marks]Circle Geometry
A sector POQ of a circle has centre O, with PO^Q=37∘P\hat{O}Q = 37^\circ and PO=8PO = 8 cm. Taking π\pi to be 3,142, calculate the area, in square centimetres, of the segment cut off by the chord PQ, correct to three significant figures.

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Question 401

[2 marks]Factorisation and Sets
Factorise completely 3np−6nq+ap−2aq3np - 6nq + ap - 2aq.
  1. A(3n+a)(p+2q)(3n + a)(p + 2q)
  2. B(n+3a)(3p−2q)(n + 3a)(3p - 2q)
  3. C(3n+a)(p−2q)(3n + a)(p - 2q)
  4. D(3n−a)(p+2q)(3n - a)(p + 2q)

Question 402

[2 marks]Factorisation and Sets
Factorise completely 12−4g−g212 - 4g - g^2.

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Question 403

[2 marks]Factorisation and Sets
In a survey of 72 girls, every girl watched at least one of the TV programmes Teen Scene or Fashion Show. Fifty girls watched Teen Scene and 62 girls watched Fashion Show. Find the number of girls who watched both programmes.

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Question 404

[2 marks]Factorisation and Sets
In a survey of 72 girls, every girl watched at least one of the TV programmes Teen Scene or Fashion Show. Fifty girls watched Teen Scene and 62 girls watched Fashion Show. Find the number of girls who watched Fashion Show only.

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Question 501

[2 marks]Number Bases and Logarithms
Convert 1123112_3 to a number in base 5.

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Question 502

[2 marks]Number Bases and Logarithms
Evaluate 11012+101121101_2 + 1011_2, giving your answer in base 2.

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Question 503

[3 marks]Number Bases and Logarithms
Express 2log⁡5(3x+2)−log⁡522\log_5(3x + 2) - \log_5 2 as a single logarithm.
  1. A(log⁡53x+22)2\left(\log_5\dfrac{3x + 2}{2}\right)^2
  2. Blog⁡5(6x+42)\log_5\left(\dfrac{6x + 4}{2}\right)
  3. Clog⁡5((3x+2)24)\log_5\left(\dfrac{(3x + 2)^2}{4}\right)
  4. Dlog⁡5((3x+2)22)\log_5\left(\dfrac{(3x + 2)^2}{2}\right)

Question 504

[3 marks]Number Bases and Logarithms
Solve the equation 3x2+4x−2=03x^2 + 4x - 2 = 0 and state its positive root, correct to two decimal places.

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Question 505

[2 marks]Number Bases and Logarithms
For the equation 3x2+4x−2=03x^2 + 4x - 2 = 0, find the value of b2−4acb^2 - 4ac.

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Question 601

[2 marks]Constructions and Loci
Shinga is 16 km from Tugwi. On a drawing made to a scale of 1 cm to represent 2 km, find the length, in centimetres, of the line joining Tugwi to Shinga.

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Question 602

[2 marks]Constructions and Loci
Hlangu is 15 km North-East of Shinga. State the bearing of Hlangu from Shinga, given as a three-figure bearing.
  1. A045∘045^\circ
  2. B135∘135^\circ
  3. C225∘225^\circ
  4. D315∘315^\circ

Question 603

[2 marks]Constructions and Loci
On a drawing made to a scale of 1 cm to represent 2 km, the locus of points 8 km from Shinga is a circle centred on Shinga. Find the radius, in centimetres, of that circle.

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Question 604

[2 marks]Constructions and Loci
Boterekwa is equidistant from Tugwi and from Hlangu. Name the locus of all points that are equidistant from Tugwi and Hlangu.
  1. AThe circle drawn with the line joining Tugwi and Hlangu as its diameter
  2. BThe bisector of the angle between the line to Tugwi and the line to Hlangu
  3. CThe arc of a circle centred on the midpoint of Tugwi and Hlangu
  4. DThe perpendicular bisector of the line joining Tugwi and Hlangu

Question 605

[3 marks]Constructions and Loci
Shinga is 16 km from Tugwi on a bearing of 150∘150^\circ, and Hlangu is 15 km North-East of Shinga. Boterekwa is 8 km from Shinga and equidistant from Tugwi and Hlangu. Taking the nearer of the two possible positions, find the distance, in kilometres to the nearest kilometre, from Boterekwa to Tugwi.

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Question 701

[3 marks]Matrices and Vector Geometry
Find the inverse of the matrix A=(3−41−2)\mathbf{A} = \begin{pmatrix} 3 & -4 \\ 1 & -2 \end{pmatrix}.
  1. A(120,51,5)\begin{pmatrix} 1 & 2 \\ 0,5 & 1,5 \end{pmatrix}
  2. B(1−20,5−1,5)\begin{pmatrix} 1 & -2 \\ 0,5 & -1,5 \end{pmatrix}
  3. C(−12−0,51,5)\begin{pmatrix} -1 & 2 \\ -0,5 & 1,5 \end{pmatrix}
  4. D(−34−12)\begin{pmatrix} -3 & 4 \\ -1 & 2 \end{pmatrix}

Question 702

[3 marks]Matrices and Vector Geometry
Solve the simultaneous equations 3x−4y=−33x - 4y = -3 and x−2y=−2x - 2y = -2, and state the value of yy.

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Question 703

[2 marks]Matrices and Vector Geometry
OABC is a trapezium in which OC is parallel to AB, with OA→=x\overrightarrow{OA} = x and OC→=y\overrightarrow{OC} = y. The diagonals OB and AC meet at D, where AD:DC=3:2AD : DC = 3 : 2. Express AD→\overrightarrow{AD} in terms of xx and yy.
  1. A35(y−x)\tfrac{3}{5}(y - x)
  2. B32(y−x)\tfrac{3}{2}(y - x)
  3. C35(x−y)\tfrac{3}{5}(x - y)
  4. D25(y−x)\tfrac{2}{5}(y - x)

Question 704

[2 marks]Matrices and Vector Geometry
OABC is a trapezium in which OC is parallel to AB, with OA→=x\overrightarrow{OA} = x and OC→=y\overrightarrow{OC} = y. The diagonals OB and AC meet at D, where AD:DC=3:2AD : DC = 3 : 2, and AB→=kOC→\overrightarrow{AB} = k\overrightarrow{OC} and OB→=hOD→\overrightarrow{OB} = h\overrightarrow{OD}. Find the value of hh.

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Question 705

[2 marks]Matrices and Vector Geometry
OABC is a trapezium in which OC is parallel to AB, with OA→=x\overrightarrow{OA} = x and OC→=y\overrightarrow{OC} = y. The diagonals OB and AC meet at D, where AD:DC=3:2AD : DC = 3 : 2, and AB→=kOC→\overrightarrow{AB} = k\overrightarrow{OC} and OB→=hOD→\overrightarrow{OB} = h\overrightarrow{OD}. Find the value of kk.

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Question 801

[2 marks]Geometrical Transformation
Triangle ABC has vertices A(2;1)A(2; 1), B(4;1)B(4; 1) and C(4;4)C(4; 4). Triangle A1B1C1A_1B_1C_1 is the reflection of triangle ABC in the line y=−2y = -2. Write down the coordinates of C1C_1.

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Question 802

[3 marks]Geometrical Transformation
Triangle ABC has vertices A(2;1)A(2; 1), B(4;1)B(4; 1) and C(4;4)C(4; 4), and triangle A2B2C2A_2B_2C_2 has vertices A2(6;4)A_2(6; 4), B2(10;4)B_2(10; 4) and C2(10;10)C_2(10; 10). Describe fully the single transformation that maps triangle ABC onto triangle A2B2C2A_2B_2C_2.
  1. AAn enlargement with centre (−2;−2)(-2; -2) and scale factor 2
  2. BAn enlargement with centre (0;0)(0; 0) and scale factor 2
  3. CAn enlargement with centre (2;1)(2; 1) and scale factor 3
  4. DAn enlargement with centre (−2;−2)(-2; -2) and scale factor 3

Question 803

[2 marks]Geometrical Transformation
Triangle ABC has vertices A(2;1)A(2; 1), B(4;1)B(4; 1) and C(4;4)C(4; 4). It is rotated through 90∘90^\circ anticlockwise about (0;0)(0; 0) onto triangle A3B3C3A_3B_3C_3. Write down the coordinates of A3A_3.

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Question 804

[2 marks]Geometrical Transformation
Write down the matrix that represents a rotation through 90∘90^\circ anticlockwise about the origin.
  1. A(−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
  2. B(100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
  3. C(0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
  4. D(01−10)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

Question 805

[3 marks]Geometrical Transformation
Triangle ABC has vertices A(2;1)A(2; 1), B(4;1)B(4; 1) and C(4;4)C(4; 4). Triangle A1B1C1A_1B_1C_1 is the reflection of triangle ABC in the line y=−2y = -2. Find the area, in square units, of triangle A1B1C1A_1B_1C_1.

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Question 901

[2 marks]Statistics and Probability
In a Mathematics examination the grades A, B, C, D, E and U were obtained by 10, 25, 40, 24, 21 and 30 candidates respectively. Calculate the number of candidates who wrote the examination.

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Question 902

[1 marks]Statistics and Probability
In a Mathematics examination the grades A, B, C, D, E and U were obtained by 10, 25, 40, 24, 21 and 30 candidates respectively. State the modal grade.
  1. AGrade A
  2. BGrade B
  3. CGrade C
  4. DGrade U

Question 903

[3 marks]Statistics and Probability
In a Mathematics examination the grades A, B, C, D, E and U were obtained by 10, 25, 40, 24, 21 and 30 candidates respectively. Given that the passing grades are A, B and C, find the probability that two candidates chosen at random both passed the examination.

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Question 904

[2 marks]Statistics and Probability
In a Mathematics examination the grades A, B, C, D, E and U were obtained by 10, 25, 40, 24, 21 and 30 candidates respectively. If a pie-chart is drawn for this information, calculate the angle, in degrees, of the sector that represents grade C.

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Question 905

[2 marks]Statistics and Probability
In a Mathematics examination the grades A, B, C, D, E and U were obtained by 10, 25, 40, 24, 21 and 30 candidates respectively. Calculate the percentage of the candidates who obtained grade U.

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Question 906

[2 marks]Statistics and Probability
In a Mathematics examination the grades A, B, C, D, E and U were obtained by 10, 25, 40, 24, 21 and 30 candidates respectively, and the passing grades are A, B and C. Find the probability that one candidate chosen at random failed the examination.

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Question 1001

[2 marks]Variation, Inequalities and Linear Programming
The length, ll, of a rectangle of constant area varies inversely as bb, the width of the rectangle. State the relationship between ll, bb and a constant kk.
  1. Al=kbl = \dfrac{k}{b}
  2. Bl=bkl = \dfrac{b}{k}
  3. Cl=k+bl = k + b
  4. Dl=kbl = kb

Question 1002

[3 marks]Variation, Inequalities and Linear Programming
The length, ll, of a rectangle of constant area varies inversely as bb, its width. Given that l=8,5l = 8,5 cm when b=6b = 6 cm, find bb, in centimetres, when l=10,2l = 10,2 cm.

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Question 1003

[2 marks]Variation, Inequalities and Linear Programming
Find the coordinates of the point where the line x+2y=8x + 2y = 8 crosses the line y=−2y = -2.

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Question 1004

[3 marks]Variation, Inequalities and Linear Programming
A region is defined by x≥3x \ge 3, y≥−2y \ge -2 and x+2y≤8x + 2y \le 8. Find the maximum value of 3x−2y3x - 2y over that region.

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Question 1005

[2 marks]Variation, Inequalities and Linear Programming
A region is defined by x≥3x \ge 3, y≥−2y \ge -2 and x+2y≤8x + 2y \le 8. Find the coordinates of the corner of that region which lies on both x=3x = 3 and x+2y=8x + 2y = 8.

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Question 1101

[2 marks]Consumer Arithmetic
A set with 1 element has 2 subsets, a set with 2 elements has 4 subsets and a set with 3 elements has 8 subsets. Find the number of subsets of a set with 5 elements.

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Question 1102

[2 marks]Consumer Arithmetic
The number of subsets of a set with mm elements is 2m2^m. A certain set has 128 subsets. Find the number of elements in that set.

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Question 1103

[2 marks]Consumer Arithmetic
For a set of nn elements the number of subsets is rr, where a set of 1 element has 2 subsets, a set of 2 elements has 4 and a set of 3 elements has 8. Express rr in terms of nn.
  1. Ar=2n−1r = 2^{n - 1}
  2. Br=2nr = 2^n
  3. Cr=2nr = 2n
  4. Dr=n2r = n^2

Question 1104

[3 marks]Consumer Arithmetic
Mbudzi Investments borrowed 6 000 dollars from a bank. At the end of the first month the bank charged interest of 80 dollars on that balance of 6 000 dollars. Find the rate of simple interest per annum, as a percentage.

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Question 1105

[3 marks]Consumer Arithmetic
A loan is charged interest at the end of each month at 113%1\tfrac{1}{3}\% of the balance standing at that time. Find the interest, in dollars correct to the nearest cent, charged on a balance of 5 878,40 dollars.

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Question 1201

[2 marks]Travel Graphs
The velocity of a particle moving along a straight line is given by v=15+7t−2t2v = 15 + 7t - 2t^2, where vv is in m/s and tt is in seconds. Find the value of vv when t=1t = 1.

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Question 1202

[2 marks]Travel Graphs
The velocity of a particle moving along a straight line is given by v=15+7t−2t2v = 15 + 7t - 2t^2, where vv is in m/s and tt is in seconds. Find the value of vv when t=4t = 4.

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Question 1203

[2 marks]Travel Graphs
The velocity of a particle moving along a straight line is given by v=15+7t−2t2v = 15 + 7t - 2t^2, where vv is in m/s and tt is in seconds. Find the value of tt, in seconds, at which vv is greatest.

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Question 1204

[3 marks]Travel Graphs
The velocity of a particle moving along a straight line is given by v=15+7t−2t2v = 15 + 7t - 2t^2, where vv is in m/s and tt is in seconds. Find the maximum value of vv, in m/s, correct to three significant figures.

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Question 1205

[3 marks]Travel Graphs
The velocity of a particle moving along a straight line is given by v=15+7t−2t2v = 15 + 7t - 2t^2, where vv is in m/s and tt is in seconds. Find the acceleration of the particle, in metres per second per second, when t=3t = 3.

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