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ZIMSEC O Level · 4028/2 · J2014

Mathematics Paper 2 June 2014

Questions
52
Total marks
136
Time allowed
150 min
Syllabus code
4028/2

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Questions
52
Pass mark
32
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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]Number, Algebra and Angles
Simplify 0,85−0,60,85 - 0,6, giving your answer as a common fraction in its lowest terms.

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Question 102

[2 marks]Number, Algebra and Angles
Evaluate 134÷125+1581\frac{3}{4} \div 1\frac{2}{5} + 1\frac{5}{8}, giving your answer as a mixed number.

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Question 103

[3 marks]Number, Algebra and Angles
Remove the brackets and simplify (a+2)(a−3)−3(a−5)(a+2)(a-3) - 3(a-5).

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Question 104

[2 marks]Number, Algebra and Angles
Giving your answer in standard form, find 25% of 3,168×10−43,168 \times 10^{-4}.
  1. A3,168×10−53,168 \times 10^{-5}, from dividing the index by 4 instead of the number
  2. B0,792×10−40,792 \times 10^{-4}, the right value but with a mantissa smaller than 1
  3. C7,92×10−57,92 \times 10^{-5}, a quarter of the number with a mantissa between 1 and 10
  4. D7,92×10−37,92 \times 10^{-3}, from moving the decimal point the wrong way when adjusting

Question 105

[2 marks]Number, Algebra and Angles
In the diagram, APX and BQY are parallel straight lines and AQ is the bisector of PQ^BP\hat{Q}B. Given that PA^Q=31∘P\hat{A}Q = 31^\circ, calculate AP^QA\hat{P}Q.
  1. A31∘31^\circ, taking AP^QA\hat{P}Q as alternate to PA^QP\hat{A}Q across the parallels
  2. B118∘118^\circ, from the three angles of triangle APQ adding to 180∘180^\circ
  3. C62∘62^\circ, taking AP^QA\hat{P}Q to be the whole of the bisected angle PQ^BP\hat{Q}B
  4. D149∘149^\circ, taking AP^QA\hat{P}Q as 180∘180^\circ less the single angle of 31∘31^\circ

Question 201

[2 marks]Algebra
Solve the equation 0,3x−1,7=1,8−0,4x0,3x - 1,7 = 1,8 - 0,4x.

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Question 202

[2 marks]Algebra
Solve the equation 3x=(−64)133x = (-64)^{\frac{1}{3}}.

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Question 203

[3 marks]Algebra
Factorise completely 6m2n2−mn−156m^2n^2 - mn - 15.
  1. A(6mn−5)(mn+3)(6mn - 5)(mn + 3), which expands to 6m2n2+13mn−156m^2n^2 + 13mn - 15
  2. B(6mn+5)(mn−3)(6mn + 5)(mn - 3), which expands to 6m2n2−13mn−156m^2n^2 - 13mn - 15
  3. C(3mn−5)(2mn+3)(3mn - 5)(2mn + 3), which expands to 6m2n2−mn−156m^2n^2 - mn - 15
  4. D(3mn+5)(2mn−3)(3mn + 5)(2mn - 3), which expands to 6m2n2+mn−156m^2n^2 + mn - 15

Question 204

[3 marks]Algebra
Express as a single fraction in its simplest form x−416−x2÷2x+4\dfrac{x-4}{16-x^2} \div \dfrac{2}{x+4}.

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Question 301

[2 marks]Sets
The Venn diagram shows information about the 52 families in a village. C is the set of 37 families with cattle, G the set of 24 with goats and S the set of 20 with sheep. Find, in terms of xx, the number of families with cattle only.

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Question 302

[2 marks]Sets
Using the Venn diagram, find, in terms of xx, the number of families with goats only. Set G holds 24 families in all.

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Question 303

[3 marks]Sets
In the Venn diagram the eight regions are 11+x11+x, x−1x-1, x−3x-3, 14−x14-x, 12−x12-x, 11−x11-x, xx and xx, and the village holds 52 families altogether. Find the value of xx.

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Question 304

[3 marks]Sets
In the Venn diagram, x=4x = 4, so 3 families have goats only, 7 have goats and sheep only, 10 have cattle and goats only and 4 have all three animals. Find the number of families with goats but no cattle.
  1. A10, the goats-only families plus the goats-and-sheep-only families
  2. B24, every family in the set G, whether or not it also keeps cattle
  3. C3, only those families whose goats are the only animals they keep
  4. D14, the number of families that have both cattle and goats together

Question 401

[2 marks]Trigonometry and Bearings
P, Q and R are three points on level ground with cos⁡RP^Q=−13\cos R\hat{P}Q = -\dfrac{1}{3}. Calculate RP^QR\hat{P}Q, giving your answer in degrees correct to one decimal place.

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Question 402

[3 marks]Trigonometry and Bearings
In triangle PQR, PQ = 20 m, PR = 30 m and cos⁡RP^Q=−13\cos R\hat{P}Q = -\dfrac{1}{3}. Calculate the length of QR in metres, correct to 3 significant figures.

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Question 403

[3 marks]Trigonometry and Bearings
In triangle PQR, PQ = 20 m, QR = 41,2 m and RP^Q=109,5∘R\hat{P}Q = 109,5^\circ. Calculate PR^QP\hat{R}Q.
  1. A27,2∘27,2^\circ, from the sine rule using QR=41,2QR = 41,2 m and PQ=20PQ = 20 m
  2. B43,3∘43,3^\circ, which is the third angle PQ^RP\hat{Q}R rather than the angle at R
  3. C70,5∘70,5^\circ, from taking the acute angle whose cosine is +13+\frac{1}{3}
  4. D35,3∘35,3^\circ, from halving what is left when RP^QR\hat{P}Q is taken from 180∘180^\circ

Question 404

[3 marks]Trigonometry and Bearings
In triangle PQR, RP^Q=109,5∘R\hat{P}Q = 109,5^\circ, PR^Q=27,2∘P\hat{R}Q = 27,2^\circ and R is due south of Q, with P lying to the east of the line QR. Find the three-figure bearing of Q from P, correct to the nearest degree.
  1. A317∘317^\circ, the back bearing of the 136,7∘136,7^\circ bearing of P from Q
  2. B137∘137^\circ, which is the bearing of P from Q rather than of Q from P
  3. C043∘043^\circ, from measuring the angle at Q from north instead of from south
  4. D223∘223^\circ, the back bearing of 043∘043^\circ rather than the one of 137∘137^\circ

Question 501

[2 marks]Circle Geometry
In the diagram, ABCD is a cyclic quadrilateral and TA is a tangent at A. Given that TA^D=51∘T\hat{A}D = 51^\circ, calculate AC^DA\hat{C}D.
  1. A27∘27^\circ, taking AC^DA\hat{C}D to be the angle BD^CB\hat{D}C in the same segment
  2. B78∘78^\circ, taking AC^DA\hat{C}D as equal to the angle AP^DA\hat{P}D at the crossing
  3. C39∘39^\circ, taking the tangent to meet the chord AD at a right angle at A
  4. D51∘51^\circ, the angle AD subtends in the segment alternate to the tangent

Question 502

[3 marks]Circle Geometry
In the diagram, ABCD is a cyclic quadrilateral with AC and BD crossing at P. Given that AC^D=51∘A\hat{C}D = 51^\circ and AP^D=78∘A\hat{P}D = 78^\circ, calculate BA^CB\hat{A}C in degrees.

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Question 503

[2 marks]Circle Geometry
In triangle APD, AP^D=78∘A\hat{P}D = 78^\circ, the base angles at A and D are each 51∘51^\circ and AP = 5 cm. Calculate the length of AD in centimetres, correct to 3 significant figures.

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Question 504

[3 marks]Circle Geometry
In the diagram, AC and BD cross at P, with BA^C=27∘B\hat{A}C = 27^\circ, AB^D=51∘A\hat{B}D = 51^\circ and AP = PD = 5 cm. Name, in the correct order, the triangle which is congruent to triangle ABP.
  1. ATriangle BPC, which has 51∘51^\circ at B, 78∘78^\circ at P and 51∘51^\circ at C
  2. BTriangle DCP, which has 27∘27^\circ at D, 51∘51^\circ at C and DP equal to AP
  3. CTriangle ADP, which has AP in common but 51∘51^\circ rather than 27∘27^\circ at A
  4. DTriangle DBC, which stands on the same chord but has no vertex at the point P

Question 601

[2 marks]Constructions and Loci
W and Y are two fixed points. Which of the following is the locus of points which are equidistant from W and Y?
  1. AThe circle whose centre is the mid-point of WY and whose radius is half of WY
  2. BThe bisector of the angle of the quadrilateral at the vertex W, drawn inwards
  3. CThe straight line drawn through W and Y and extended beyond both of them
  4. DThe perpendicular bisector of the line joining the two points W and Y

Question 602

[3 marks]Constructions and Loci
In quadrilateral WXYZ, WX = 5,3 cm, XY = 5 cm and WX^Y=120∘W\hat{X}Y = 120^\circ. Calculate the length of the diagonal WY in centimetres, correct to 3 significant figures.

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Question 603

[2 marks]Constructions and Loci
A point O inside quadrilateral WXYZ is found to be the same distance from each of W, X, Y and Z. What special name is given to quadrilateral WXYZ?
  1. AA cyclic quadrilateral, because its four vertices lie on one circle centre O
  2. BA rhombus, because the four sides are all constructed to the same length
  3. CA kite, because the two pairs of adjacent sides are constructed equal
  4. DA parallelogram, because both pairs of opposite sides come out parallel to each other

Question 701

[2 marks]Mensuration
Taking π=227\pi = \dfrac{22}{7}, calculate the volume of a copper ball of radius 3 cm, in cubic centimetres correct to 2 decimal places. Volume of a sphere =4πr33= \dfrac{4\pi r^3}{3}.

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Question 702

[2 marks]Mensuration
Taking π=227\pi = \dfrac{22}{7}, calculate the volume of a cylindrical rod of diameter 0,3 cm and length 15 cm, in cubic centimetres correct to 2 decimal places.

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Question 703

[3 marks]Mensuration
A copper ball of volume 7927\frac{792}{7} cubic centimetres is melted down and recast into rods each of volume 7,4257\frac{7,425}{7} cubic centimetres. Find the number of copper rods that can be made.
  1. A100, rounding the 106,67 rods' worth of copper down to the nearest hundred
  2. B107, rounding the 106,67 rods' worth of copper up to a whole number
  3. C113, dividing the ball's volume by the rod's length instead of its volume
  4. D106, since 106,67 rods' worth of copper only casts 106 complete rods

Question 704

[2 marks]Mensuration
A rod of length 15 cm is bent to form a circular bangle. Taking π=227\pi = \dfrac{22}{7}, calculate the radius of the bangle in centimetres, correct to 3 significant figures.

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Question 705

[3 marks]Mensuration
A rod of diameter 0,3 cm and length 15 cm is bent into a bangle, and 106 such bangles are made. Each is coated on its curved surface at 5c per square centimetre, with π=227\pi = \dfrac{22}{7}. Calculate the total cost in dollars, correct to the nearest cent.

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Question 801

[2 marks]Graphs of Quadratic Functions
A table of values is being drawn up for y=12+2x−x2y = 12 + 2x - x^2. The entry at x=−2x = -2 is written as pp. Find the value of pp.

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Question 802

[2 marks]Graphs of Quadratic Functions
In the table of values for y=12+2x−x2y = 12 + 2x - x^2, the entry at x=3x = 3 is written as qq. Find the value of qq.

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Question 803

[3 marks]Graphs of Quadratic Functions
Find the positive root of the equation 12+2x−x2=012 + 2x - x^2 = 0, correct to 1 decimal place.

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Question 804

[2 marks]Graphs of Quadratic Functions
Find the gradient of the curve y=12+2x−x2y = 12 + 2x - x^2 at the point where x=2x = 2.

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Question 805

[3 marks]Graphs of Quadratic Functions
Find the value of the integer kk for which the equation 12+2x−x2=k12 + 2x - x^2 = k has roots −1-1 and 3.
  1. Ak=3k = 3, taking the product of the two roots −1-1 and 3 to be +3+3 here
  2. Bk=15k = 15, adding 3 to 12 rather than subtracting when comparing terms
  3. Ck=−3k = -3, taking the constant term of x2−2x−3=0x^2 - 2x - 3 = 0 as kk itself
  4. Dk=9k = 9, since k−12k - 12 must equal the −3-3 in x2−2x−3=0x^2 - 2x - 3 = 0

Question 901

[2 marks]Variation
A saleswoman's salary P is partly constant and her commission partly varies as N, the number of cars she sells in a month. Which expression gives P in terms of N and the constants h and k?
  1. AP=h+kN2P = h + kN^2, a fixed part plus a part varying as the square of NN
  2. BP=h+kNP = h + kN, a fixed part hh plus a part varying directly as NN
  3. CP=hkNP = hkN, the whole salary varying directly as the number of cars sold
  4. DP=h+kNP = h + \frac{k}{N}, a fixed part plus a part varying inversely as NN

Question 902

[3 marks]Variation
A saleswoman's salary is P=h+kNP = h + kN dollars, where N is the number of cars sold in a month. She earns 675 dollars for 7 cars and 900 dollars for 10 cars. Find the value of k.

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Question 903

[2 marks]Variation
A saleswoman's salary is P=h+kNP = h + kN dollars with k=75k = 75, and she earns 675 dollars in a month when she sells 7 cars. Find her salary in dollars for a month in which she sells no cars.

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Question 904

[3 marks]Variation
A saleswoman's salary in dollars is given by P=150+75NP = 150 + 75N, where N is the number of cars she sells in one month. Calculate her salary in dollars for a month in which she sells 9 cars.

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Question 905

[2 marks]Variation
A saleswoman earns a commission of 75 dollars on each car she sells, and that commission is 212%2\frac{1}{2}\% of the price of one car. Calculate the price of each car in dollars.

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Question 1001

[3 marks]Quadratic Equations and Statistics
Solve the equation 3x2−5x−15=03x^2 - 5x - 15 = 0 and give the positive root correct to 2 decimal places.

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Question 1002

[2 marks]Quadratic Equations and Statistics
Solve the equation 3x2−5x−15=03x^2 - 5x - 15 = 0 and give the negative root correct to 2 decimal places.

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Question 1003

[3 marks]Quadratic Equations and Statistics
A survey of the masses of 80 pupils gives these cumulative frequencies: 2 pupils at 45 kg or less, 10 at 50 kg, 21 at 55 kg, 41 at 60 kg, 60 at 65 kg, 72 at 70 kg, 78 at 75 kg and 80 at 80 kg. Find the median mass in kilograms.

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Question 1004

[3 marks]Quadratic Equations and Statistics
For the 80 pupils, the cumulative frequencies are 60 at 65 kg, 72 at 70 kg, 78 at 75 kg and 80 at 80 kg. Find the number of pupils whose masses are more than 72 kg.
  1. A8, the 80 pupils less the 72 who are 70 kg or lighter on the table
  2. B12, the difference between the cumulative frequencies at 70 kg and 65 kg
  3. C6, the 80 pupils less the 74 who are 72 kg or lighter on the curve
  4. D74, the cumulative frequency read off the curve at a mass of 72 kg

Question 1101

[2 marks]Vectors
In quadrilateral OXYZ, P is a point on OZ with OP→=(−1−2)\overrightarrow{OP} = \begin{pmatrix} -1 \\ -2 \end{pmatrix} and OX→=(50)\overrightarrow{OX} = \begin{pmatrix} 5 \\ 0 \end{pmatrix}. Find XP→\overrightarrow{XP}, writing the column vector as (top; bottom).

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Question 1102

[3 marks]Vectors
In quadrilateral OXYZ, OP→=(−1−2)\overrightarrow{OP} = \begin{pmatrix} -1 \\ -2 \end{pmatrix}, OX→=(50)\overrightarrow{OX} = \begin{pmatrix} 5 \\ 0 \end{pmatrix}, OZ→=3OP→\overrightarrow{OZ} = 3\overrightarrow{OP} and ZY→=2OX→\overrightarrow{ZY} = 2\overrightarrow{OX}. Find OY→\overrightarrow{OY}, writing the column vector as (top; bottom).

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Question 1103

[3 marks]Vectors
R lies on XP with OR→=(5−6h−2h)\overrightarrow{OR} = \begin{pmatrix} 5-6h \\ -2h \end{pmatrix}, and R also lies on OY with OR→=k(7−6)\overrightarrow{OR} = k\begin{pmatrix} 7 \\ -6 \end{pmatrix}. Find the value of h.
  1. Ah=15h = \frac{1}{5}, which is the value of kk rather than the value of hh
  2. Bh=25h = \frac{2}{5}, which is the fraction of XP making up RP and not XR
  3. Ch=35h = \frac{3}{5}, from h=3kh = 3k used together with 5−18k=7k5 - 18k = 7k
  4. Dh=53h = \frac{5}{3}, from inverting the correct ratio of XR to XP here

Question 1104

[2 marks]Vectors
R lies on XP with OR→=(5−6h−2h)\overrightarrow{OR} = \begin{pmatrix} 5-6h \\ -2h \end{pmatrix}, and R also lies on OY with OR→=k(7−6)\overrightarrow{OR} = k\begin{pmatrix} 7 \\ -6 \end{pmatrix}. Find the value of k.

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Question 1105

[2 marks]Vectors
R lies on XP with XR→=35XP→\overrightarrow{XR} = \frac{3}{5}\overrightarrow{XP}. Write down the numerical value of the ratio XRRP\dfrac{XR}{RP}.

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Question 1201

[3 marks]Transformations
Triangle W has a vertex at (1;−1)(1;-1). Find the image of this vertex under a reflection in the line y=x+2y = x + 2, writing your answer as (x; y).

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Question 1202

[3 marks]Transformations
Which matrix represents an enlargement of scale factor −12-\frac{1}{2} with the origin as centre?
  1. A(−1200−12)\begin{pmatrix} -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}, scale factor −12-\frac{1}{2} on both axes
  2. B(−200−2)\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}, an enlargement of scale factor −2-2 instead
  3. C(−120012)\begin{pmatrix} -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}, a stretch reversing only the xx direction
  4. D(120012)\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}, an enlargement of scale factor +12+\frac{1}{2}

Question 1203

[3 marks]Transformations
Find the image of the point (4;4)(4;4) under an enlargement of scale factor −12-\frac{1}{2} with the origin as centre, writing your answer as (x; y).

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Question 1204

[3 marks]Transformations
Triangle W has vertices (1;−1)(1;-1), (7;−1)(7;-1) and (4;4)(4;4). Triangle Z has vertices (1;−3)(1;-3), (1;−9)(1;-9) and (6;−6)(6;-6). Describe fully the single transformation which maps triangle W onto triangle Z.
  1. AA rotation of 180∘180^\circ about the centre (0;−2)(0;-2), a half turn
  2. BA rotation of 90∘90^\circ clockwise about the centre (0;−2)(0;-2)
  3. CA rotation of 90∘90^\circ anticlockwise about the centre (0;−2)(0;-2)
  4. DA rotation of 90∘90^\circ clockwise about the origin (0;0)(0;0)

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