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ZIMSEC O Level · 4008/2 · J2012

Mathematics Paper 2 June 2012

Questions
49
Total marks
136
Time allowed
150 min
Syllabus code
4008/2

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Questions
49
Pass mark
30
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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]Algebraic Fractions
Express 334−2123\frac{3}{4} - 2\frac{1}{2} as a single fraction in its simplest form.

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Question 102

[3 marks]Algebraic Fractions
Express 2a−5a−4−12\frac{2a - 5}{a - 4} - \frac{1}{2} as a single fraction in its simplest form.
  1. A2a−6a−6\frac{2a - 6}{a - 6}
  2. B3a−142a−8\frac{3a - 14}{2a - 8}
  3. C3a−62a−8\frac{3a - 6}{2a - 8}
  4. Da−12a−8\frac{a - 1}{2a - 8}

Question 103

[2 marks]Algebraic Fractions
Find the numerical value of d(n−d2)d\left(n - d^2\right) when n=4n = 4 and d=12d = \frac{1}{2}.

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Question 104

[3 marks]Algebraic Fractions
By selling an article for 11,40 dollars, a shop owner made a loss of 5%. Calculate the cost price of the article, in dollars.

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Question 201

[2 marks]Algebraic Expressions
Factorise completely 2k2−7k−152k^2 - 7k - 15.
  1. A(2k+3)(k−5)(2k + 3)(k - 5)
  2. B(2k+5)(k−3)(2k + 5)(k - 3)
  3. C(2k−3)(k+5)(2k - 3)(k + 5)
  4. D(2k−5)(k+3)(2k - 5)(k + 3)

Question 202

[3 marks]Algebraic Expressions
Factorise completely 2am2−2an2−bm2+bn22am^2 - 2an^2 - bm^2 + bn^2.
  1. A(2a−b)(n−m)(n+m)(2a - b)(n - m)(n + m)
  2. B(2a−b)(m−n)(m−n)(2a - b)(m - n)(m - n)
  3. C(2a+b)(m−n)(m+n)(2a + b)(m - n)(m + n)
  4. D(2a−b)(m−n)(m+n)(2a - b)(m - n)(m + n)

Question 203

[2 marks]Algebraic Expressions
Mary has $(3x−4y)\$(3x - 4y) and Diana has $(2y−x)\$(2y - x). Write down, in terms of xx and yy, the simplified expression for the amount of money that Mary has more than Diana.

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Question 204

[3 marks]Algebraic Expressions
Mary has $(3x−4y)\$(3x - 4y) and Diana has $(2y−x)\$(2y - x). Given that Mary has $12\$12 and Diana has $8\$8, find the value of xx and the value of yy.
  1. Ax=12x = 12 and y=10y = 10
  2. Bx=28x = 28 and y=18y = 18
  3. Cx=18x = 18 and y=28y = 28
  4. Dx=20x = 20 and y=12y = 12

Question 301

[2 marks]Change of Subject of Formula
Given that T=11v6+20T = \frac{11v}{6} + 20, make vv the subject of the formula.
  1. Av=6T−2011v = \frac{6T - 20}{11}
  2. Bv=T−2066v = \frac{T - 20}{66}
  3. Cv=11(T−20)6v = \frac{11(T - 20)}{6}
  4. Dv=6(T−20)11v = \frac{6(T - 20)}{11}

Question 302

[3 marks]Change of Subject of Formula
Given that T=11v6+20T = \frac{11v}{6} + 20, find vv when T=vT = v.

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Question 303

[3 marks]Change of Subject of Formula
In triangle AMC, AM = 7 cm and MC = 8 cm, and the area of triangle AMC is 25 cm2^2. Calculate the acute angle AM^CA\hat{M}C, in degrees correct to one decimal place.

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Question 304

[2 marks]Change of Subject of Formula
Given that Q=(7346)\mathbf{Q} = \begin{pmatrix} 7 & 3 \\ 4 & 6 \end{pmatrix}, find the inverse of Q\mathbf{Q}.
  1. A130(7−3−46)\frac{1}{30}\begin{pmatrix} 7 & -3 \\ -4 & 6 \end{pmatrix}
  2. B130(6−3−47)\frac{1}{30}\begin{pmatrix} 6 & -3 \\ -4 & 7 \end{pmatrix}
  3. C130(6−4−37)\frac{1}{30}\begin{pmatrix} 6 & -4 \\ -3 & 7 \end{pmatrix}
  4. D154(6−3−47)\frac{1}{54}\begin{pmatrix} 6 & -3 \\ -4 & 7 \end{pmatrix}

Question 305

[2 marks]Change of Subject of Formula
It is given that P=(4−352)\mathbf{P} = \begin{pmatrix} 4 & -3 \\ 5 & 2 \end{pmatrix} and Q=(7346)\mathbf{Q} = \begin{pmatrix} 7 & 3 \\ 4 & 6 \end{pmatrix}. Find the matrix R\mathbf{R} such that P+R=Q\mathbf{P} + \mathbf{R} = \mathbf{Q}.

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Question 401

[2 marks]Constructions and Loci
A trapezium ABCD is constructed in which AB = 8,5 cm, AB^C=120∘A\hat{B}C = 120^\circ, BC = 6 cm, CD = 12 cm and AB is parallel to DC. Find the size of CB^DC\hat{B}D, in degrees.

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Question 402

[2 marks]Constructions and Loci
A circle is drawn having the 12 cm line CD as its diameter. Which description of that circle as a locus is the complete one?
  1. AThe locus of points at which CD subtends 90 degrees: centre the midpoint of CD, radius 6 cm
  2. BThe locus of points equidistant from C and D: the perpendicular bisector of CD, drawn as a circle
  3. CThe locus of points 12 cm from the midpoint of CD: a circle of centre that midpoint, radius 12 cm
  4. DThe locus of points at which CD subtends 45 degrees: centre the midpoint of CD, radius 12 cm

Question 501

[2 marks]Circle Geometry
K is 5 km due east of M, D is 8 km due south of K and C is 10 km due east of D. Calculate the length of the straight line MC, in kilometres.

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Question 502

[3 marks]Circle Geometry
K is 5 km due east of M, D is 8 km due south of K and C is 10 km due east of D. Calculate the bearing of C from M, correct to the nearest degree.

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Question 503

[2 marks]Circle Geometry
P, Q, R and S are points on a circle. AT is a tangent to the circle at P, and R lies in the segment alternate to the angle QP^TQ\hat{P}T. Given that PR^Q=32∘P\hat{R}Q = 32^\circ, find QP^TQ\hat{P}T.
  1. A16∘16^\circ
  2. B32∘32^\circ
  3. C58∘58^\circ
  4. D64∘64^\circ

Question 504

[2 marks]Circle Geometry
P, Q, R and S are points on a circle with centre O, and the line PS passes through O. Given that PR^Q=32∘P\hat{R}Q = 32^\circ, find QP^SQ\hat{P}S, in degrees.

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Question 505

[2 marks]Circle Geometry
P, Q, R and S are points on a circle with centre O, the line PS passes through O, and QO is parallel to RS. Given that QP^S=58∘Q\hat{P}S = 58^\circ, find PS^RP\hat{S}R.
  1. A64∘64^\circ
  2. B116∘116^\circ
  3. C32∘32^\circ
  4. D58∘58^\circ

Question 601

[2 marks]Similarity and Congruency
Find the equation of the straight line which passes through P(3; −4)P(3;\ -4) and Q(−1; 2)Q(-1;\ 2).
  1. A3x+2y=−63x + 2y = -6
  2. B2x+3y=−62x + 3y = -6
  3. C3x−2y=−73x - 2y = -7
  4. D3x+2y=13x + 2y = 1

Question 602

[3 marks]Similarity and Congruency
Solve the equation 32x−5−4x−3=0\frac{3}{2x - 5} - \frac{4}{x - 3} = 0.

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Question 603

[2 marks]Similarity and Congruency
ABCD is a cyclic quadrilateral in which the diagonals AC and BD intersect at P. Name, in correct order, the triangle that is similar to triangle APD.
  1. ATriangle DPC, since A corresponds to D, P to P and D to C
  2. BTriangle CPB, since A corresponds to C, P to P and D to B
  3. CTriangle BPC, since A corresponds to B, P to P and D to C
  4. DTriangle BPA, since A corresponds to B, P to P and D to A

Question 604

[3 marks]Similarity and Congruency
ABCD is a cyclic quadrilateral in which the diagonals AC and BD intersect at P, with AP = 3 cm, PC = 4 cm and BP = 6 cm. Given that the area of triangle CPD is 8 cm2^2, calculate the area of the quadrilateral ABCD, in cm2^2.

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Question 701

[2 marks]Trigonometry, Bearing & Distances
The points H, G and D lie in a straight line on level ground. DE is a tree 6 m high standing at D, and the angle of elevation of E from G is 52∘52^\circ. HF is a tower standing at H, and the angle of depression of E from F is 24∘24^\circ. Find GE^FG\hat{E}F.
  1. A76∘76^\circ
  2. B104∘104^\circ
  3. C28∘28^\circ
  4. D52∘52^\circ

Question 702

[3 marks]Trigonometry, Bearing & Distances
The points H, G and D lie in a straight line on level ground. DE is a tree 6 m high standing at D, and the angle of elevation of E from G is 52∘52^\circ. Calculate the length of GE, in metres, correct to three significant figures.

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Question 703

[3 marks]Trigonometry, Bearing & Distances
In triangle FEG, EF = 9 m, GE = 7,61 m and GE^F=76∘G\hat{E}F = 76^\circ. Calculate the length of FG, in metres, correct to three significant figures.

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Question 704

[3 marks]Trigonometry, Bearing & Distances
The points H, G and D lie in a straight line on level ground, F is the top of a tower standing at H and E is the top of a tree standing at D. In triangle FEG, EF = 9 m, GE = 7,61 m and GE^F=76∘G\hat{E}F = 76^\circ, and the angle of depression of E from F is 24∘24^\circ. Find the angle of depression of G from F, in degrees correct to one decimal place.

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Question 801

[2 marks]Geometrical Transformation
Triangle A has vertices at (3; 1)(3;\ 1), (1; 2)(1;\ 2) and (2; 4)(2;\ 4). Triangle B is the image of triangle A under an anticlockwise rotation of 90∘90^\circ about (−2; 2)(-2;\ 2). Write down the coordinates of the image of the vertex (3; 1)(3;\ 1).

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Question 802

[3 marks]Geometrical Transformation
A single transformation PP maps triangle A, with vertices (3; 1)(3;\ 1), (1; 2)(1;\ 2) and (2; 4)(2;\ 4), onto triangle C, with vertices (9; 1)(9;\ 1), (3; 2)(3;\ 2) and (6; 4)(6;\ 4). Find the matrix which represents PP.
  1. A(0310)\begin{pmatrix} 0 & 3 \\ 1 & 0 \end{pmatrix}
  2. B(1003)\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}
  3. C(3003)\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}
  4. D(3001)\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}

Question 803

[3 marks]Geometrical Transformation
A single transformation PP maps triangle A, with vertices (3; 1)(3;\ 1), (1; 2)(1;\ 2) and (2; 4)(2;\ 4), onto triangle C, with vertices (9; 1)(9;\ 1), (3; 2)(3;\ 2) and (6; 4)(6;\ 4). Describe fully the single transformation PP.
  1. AA stretch parallel to the yy-axis, scale factor 3, invariant line the xx-axis
  2. BA shear parallel to the xx-axis with the yy-axis invariant and shear factor 3
  3. CA stretch parallel to the xx-axis, scale factor 3, invariant line the yy-axis
  4. DAn enlargement of scale factor 3 with the origin as centre, which trebles both coordinates

Question 804

[2 marks]Geometrical Transformation
Triangle A has vertices at (3; 1)(3;\ 1), (1; 2)(1;\ 2) and (2; 4)(2;\ 4). It is mapped onto triangle D by an enlargement of scale factor −2-2 with the origin as the centre. Write down the coordinates of the image of the vertex (2; 4)(2;\ 4).

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Question 901

[3 marks]Vector Geometry
The resistance, RR newtons, to a train travelling at vv km/h is given by R=c+dv2R = c + dv^2, where cc and dd are constants. Given that R=4R = 4 when v=20v = 20 and that R=10R = 10 when v=40v = 40, find the value of cc and the value of dd.
  1. Ac=0,005c = 0,005 and d=2d = 2
  2. Bc=2c = 2 and d=0,005d = 0,005
  3. Cc=3c = 3 and d=0,005d = 0,005
  4. Dc=2c = 2 and d=0,05d = 0,05

Question 902

[2 marks]Vector Geometry
The resistance, RR newtons, to a train travelling at vv km/h is given by R=c+dv2R = c + dv^2, where c=2c = 2 and d=0,005d = 0,005. Find vv when R=3R = 3.

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Question 903

[2 marks]Vector Geometry
OAB is a triangle in which K is the point on BA with BK:KA=1:2BK : KA = 1 : 2. Given that OA→=3a\overrightarrow{OA} = 3\mathbf{a} and OB→=3b\overrightarrow{OB} = 3\mathbf{b}, express BK→\overrightarrow{BK} in terms of a\mathbf{a} and b\mathbf{b}.

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Question 904

[2 marks]Vector Geometry
OAB is a triangle in which H lies on OA with OH:HA=1:2OH : HA = 1 : 2 and K lies on BA with BK:KA=1:2BK : KA = 1 : 2. Given that OA→=3a\overrightarrow{OA} = 3\mathbf{a} and OB→=3b\overrightarrow{OB} = 3\mathbf{b}, express HK→\overrightarrow{HK} in terms of a\mathbf{a} and b\mathbf{b}.
  1. A2b2\mathbf{b}
  2. B2a2\mathbf{a}
  3. Ca+b\mathbf{a} + \mathbf{b}
  4. D2a−2b2\mathbf{a} - 2\mathbf{b}

Question 905

[2 marks]Vector Geometry
In triangle OAB, HK→=2b\overrightarrow{HK} = 2\mathbf{b} and OB→=3b\overrightarrow{OB} = 3\mathbf{b}. Write down the ratio HKOB\frac{HK}{OB}.

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Question 1001

[3 marks]Measures and Mensuration
Solve the equation 3q2−5q−5=03q^2 - 5q - 5 = 0, giving the positive root correct to two decimal places.

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Question 1002

[2 marks]Measures and Mensuration
Solve the equation 3q2−5q−5=03q^2 - 5q - 5 = 0, giving the negative root correct to two decimal places.

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Question 1003

[3 marks]Measures and Mensuration
A composite solid is made up of a cylinder of height 2,2 m with a cone standing on it, both having the same radius rr metres. The slant height of the cone is 4,81 m and the total height of the solid is 5,5 m. Calculate rr, in metres.

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Question 1004

[2 marks]Measures and Mensuration
A composite solid is made up of a cylinder of height 2,2 m with a cone standing on it, both of radius 3,5 m. The slant height of the cone is 4,81 m. Taking π=227\pi = \frac{22}{7}, find the surface area of the solid excluding its base.
  1. A5353 m2^2
  2. B101101 m2^2
  3. C140140 m2^2
  4. D4848 m2^2

Question 1005

[2 marks]Measures and Mensuration
A composite solid is made up of a cylinder of height 2,2 m with a cone of height 3,3 m standing on it, both of radius 3,5 m. Taking π=227\pi = \frac{22}{7}, calculate the volume of the solid, in m3^3, correct to three significant figures.

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Question 1101

[2 marks]Functional Graphs
A table of values for y=x3−5x+3y = x^3 - 5x + 3 is drawn up for x=−3,−2,−1,0,1,2,3x = -3, -2, -1, 0, 1, 2, 3. The entry at x=−2x = -2 is mm and the entry at x=3x = 3 is nn. Find mm and nn.
  1. Am=5m = 5 and n=9n = 9
  2. Bm=21m = 21 and n=15n = 15
  3. Cm=5m = 5 and n=15n = 15
  4. Dm=−5m = -5 and n=15n = 15

Question 1102

[3 marks]Functional Graphs
The curve y=x3−5x+3y = x^3 - 5x + 3 and the line y=x+3y = x + 3 are drawn on the same axes for −3≤x≤3-3 \le x \le 3. Write down the roots of the equation x3−5x+3=x+3x^3 - 5x + 3 = x + 3.
  1. Ax=0x = 0, x=2,45x = 2,45 and x=−2,45x = -2,45
  2. Bx=3x = 3, x=2,45x = 2,45 and x=−2,45x = -2,45
  3. Cx=0x = 0, x=3x = 3 and x=−3x = -3
  4. Dx=0x = 0, x=6x = 6 and x=−6x = -6

Question 1103

[2 marks]Functional Graphs
Estimate the gradient of the curve y=x3−5x+3y = x^3 - 5x + 3 at the point where x=−2x = -2.

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Question 1201

[2 marks]Statistics and Probability
The heights of 100 pupils, in centimetres, have frequencies 2, 14, 24, 30, 16, 10, 3 and 1 in the eight classes from 110<h≤115110 < h \le 115 up to 145<h≤150145 < h \le 150, each 5 cm wide. In the cumulative frequency table, pp is the entry for h≤125h \le 125, qq is the entry for h≤135h \le 135 and rr is the entry for h≤140h \le 140. Find pp, qq and rr.
  1. Ap=40p = 40, q=70q = 70 and r=86r = 86
  2. Bp=38p = 38, q=84q = 84 and r=94r = 94
  3. Cp=40p = 40, q=86q = 86 and r=96r = 96
  4. Dp=24p = 24, q=16q = 16 and r=10r = 10

Question 1202

[2 marks]Statistics and Probability
For the heights of 100 pupils, the cumulative frequency is 40 at 125 cm and 70 at 130 cm. Estimate the median height, in centimetres, to the nearest centimetre.

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Question 1203

[3 marks]Statistics and Probability
The heights hh cm of 100 pupils have frequencies 2, 14, 24, 30, 16, 10, 3 and 1 in the eight classes 110<h≤115110 < h \le 115, 115<h≤120115 < h \le 120, 120<h≤125120 < h \le 125, 125<h≤130125 < h \le 130, 130<h≤135130 < h \le 135, 135<h≤140135 < h \le 140, 140<h≤145140 < h \le 145 and 145<h≤150145 < h \le 150. Calculate the approximate mean height, in centimetres, correct to the nearest centimetre.

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Question 1204

[3 marks]Statistics and Probability
Of 100 pupils, 30 have heights hh cm with 125<h≤130125 < h \le 130 and 16 have 130<h≤135130 < h \le 135. Two pupils are chosen at random from the group. Find the probability that both of them have a height of more than 125 cm but not more than 135 cm.

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