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ZIMSEC O Level · 4028/2 · J2015

Mathematics Paper 2 June 2015

Questions
51
Total marks
136
Time allowed
150 min
Syllabus code
4028/2

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Questions
51
Pass mark
31
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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 Standard Form
Express 516\frac{5}{16} as a percentage.

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Question 102

[2 marks]Fractions, Decimals and Standard Form
Simplify 2723−(18)1327^{\frac{2}{3}} - \left(\frac{1}{8}\right)^{\frac{1}{3}}.

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Question 103

[3 marks]Fractions, Decimals and Standard Form
Simplify 345−(123+715)3\frac{4}{5} - \left(1\frac{2}{3} + \frac{7}{15}\right), giving the answer in its lowest terms.

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Question 104

[2 marks]Fractions, Decimals and Standard Form
Given that p=0,045p = 0,045 and r=2,513×10−4r = 2,513 \times 10^{-4}, evaluate prpr giving the answer in standard form correct to three significant figures.

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Question 201

[2 marks]Sets and Linear Equations
It is given that ξ={x:1≤x≤10,where x is an integer}\xi = \{x : 1 \le x \le 10, \text{where } x \text{ is an integer}\} and B={x:x is a factor of 20}B = \{x : x \text{ is a factor of } 20\}. List all the elements of B.
  1. A{1,2,4,5,10}\{1, 2, 4, 5, 10\}, the factors of 20 that lie inside the universal set
  2. B{1,2,4,5,10,20}\{1, 2, 4, 5, 10, 20\}, every factor of 20 including 20 itself
  3. C{2,4,6,8,10}\{2, 4, 6, 8, 10\}, the multiples of 2 that lie inside the universal set
  4. D{2,4,10}\{2, 4, 10\}, the even factors of 20 that lie inside the universal set

Question 202

[2 marks]Sets and Linear Equations
It is given that ξ={x:1≤x≤10,where x is an integer}\xi = \{x : 1 \le x \le 10, \text{where } x \text{ is an integer}\}, A={x:x is a prime number}A = \{x : x \text{ is a prime number}\} and B={x:x is a factor of 20}B = \{x : x \text{ is a factor of } 20\}. List all the elements of (A∪B)′(A \cup B)'.
  1. A{6,8,9}\{6, 8, 9\}, the members of ξ\xi that are in neither A nor B
  2. B{2,5}\{2, 5\}, the members of ξ\xi that belong to both A and B at once
  3. C{1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}, the members of ξ\xi that are not prime numbers
  4. D{3,6,7,8,9}\{3, 6, 7, 8, 9\}, the members of ξ\xi that are not factors of 20

Question 203

[2 marks]Sets and Linear Equations
It is given that ξ={x:1≤x≤10,where x is an integer}\xi = \{x : 1 \le x \le 10, \text{where } x \text{ is an integer}\}, A={x:x is a prime number}A = \{x : x \text{ is a prime number}\} and B={x:x is a factor of 20}B = \{x : x \text{ is a factor of } 20\}. List all the elements of A∩BA \cap B.

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Question 204

[3 marks]Sets and Linear Equations
Solve the equation 3x+13−x−45=12\frac{3x + 1}{3} - \frac{x - 4}{5} = \frac{1}{2}.

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Question 301

[2 marks]Ratio and Consumer Arithmetic
It is given that 300 cattle are to be shared in the ratio 12:10:812 : 10 : 8. Express the ratio in its simplest form.

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Question 302

[3 marks]Ratio and Consumer Arithmetic
It is given that 300 cattle are to be shared in the ratio 12:10:812 : 10 : 8. Calculate the difference between the largest and smallest shares.

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Question 303

[2 marks]Ratio and Consumer Arithmetic
A shop advertises 25% off all red shirts. The red shirts were originally marked at $25,00\$25,00 each. Calculate the amount that Joko paid for one red shirt.

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Question 304

[3 marks]Ratio and Consumer Arithmetic
Red shirts marked at $25,00\$25,00 each are on sale at 25% off. Tindo bought 10 such shirts at the sale price and sold them at $23,00\$23,00 each. Calculate the total profit Tindo made.

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Question 401

[2 marks]Circle Geometry and Number Bases
In the diagram A, B, C and D lie on a circle centre O, EAF is a tangent to the circle at A, and D, O and B lie on a straight line. Given that BA^F=40∘B\hat{A}F = 40^\circ and CA^D=30∘C\hat{A}D = 30^\circ, calculate AD^BA\hat{D}B.
  1. A50∘50^\circ, from the angles of triangle ADB after taking DA^BD\hat{A}B as 40∘40^\circ
  2. B40∘40^\circ, equal to the tangent-chord angle BA^FB\hat{A}F in the alternate segment
  3. C20∘20^\circ, taking the angle at the circumference as half of BA^FB\hat{A}F
  4. D80∘80^\circ, taking the angle at the circumference as twice BA^FB\hat{A}F

Question 402

[2 marks]Circle Geometry and Number Bases
In the diagram A, B, C and D lie on a circle centre O, EAF is a tangent to the circle at A, and D, O and B lie on a straight line that meets the tangent at F. Given that BA^F=40∘B\hat{A}F = 40^\circ, calculate AB^FA\hat{B}F in degrees.

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Question 403

[2 marks]Circle Geometry and Number Bases
In the diagram A, B, C and D lie on a circle centre O, EAF is a tangent to the circle at A, and D, O and B lie on a straight line. Given that BA^F=40∘B\hat{A}F = 40^\circ and CA^D=30∘C\hat{A}D = 30^\circ, calculate CD^BC\hat{D}B in degrees.

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Question 404

[2 marks]Circle Geometry and Number Bases
Convert 651065_{10} to a number in base 3.

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Question 405

[3 marks]Circle Geometry and Number Bases
Simplify 31024+1110123102_4 + 11101_2, giving the answer in base 4.

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Question 501

[2 marks]Functional Notation and Change of Subject of Formula
Given that f(x)=3x2−7x+1f(x) = 3x^2 - 7x + 1, evaluate f(−1)f(-1).

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Question 502

[3 marks]Functional Notation and Change of Subject of Formula
Given that f(x)=3x2−7x+1f(x) = 3x^2 - 7x + 1, find the values of xx when f(x)=−1f(x) = -1.

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Question 503

[2 marks]Functional Notation and Change of Subject of Formula
Given that P=n2{2a+(n−1)d}P = \frac{n}{2}\{2a + (n - 1)d\}, express aa in terms of dd, nn and PP.
  1. Aa=Pn−(n−1)d2a = \frac{P}{n} - \frac{(n - 1)d}{2}, halving both terms after dividing by nn
  2. Ba=P−(n−1)d2na = \frac{P - (n - 1)d}{2n}, subtracting the (n−1)d(n - 1)d term before dividing by nn
  3. Ca=2Pn−(n−1)da = \frac{2P}{n} - (n - 1)d, dividing the whole bracket by 2 at the end
  4. Da=2Pn−(n−1)d2a = 2Pn - \frac{(n - 1)d}{2}, multiplying by nn instead of dividing

Question 504

[3 marks]Functional Notation and Change of Subject of Formula
Given that P=n2{2a+(n−1)d}P = \frac{n}{2}\{2a + (n - 1)d\}, find the value of aa when n=10n = 10, d=4d = 4 and P=20P = 20.

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Question 601

[2 marks]Constructions and Loci
A construction is to be drawn using a scale of 1 cm to represent 2 metres. Two points, M and N, are 10 metres apart on level ground. How far apart should M and N be drawn?
  1. A10 cm, by drawing one centimetre for each metre on the ground
  2. B5 cm, since each centimetre stands for 2 metres of ground
  3. C2 cm, by reading the scale as 2 cm to every 10 metres
  4. D20 cm, by multiplying the 10 metres by the 2 in the scale

Question 602

[3 marks]Constructions and Loci
From a point M on level ground the angle of elevation of a bird on top of a vertical pole is 30∘30^\circ. From a point N, 10 metres closer to the pole and on the same side, the angle of elevation of the same bird is 45∘45^\circ. Find the height of the pole in metres, correct to three significant figures.

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Question 603

[3 marks]Constructions and Loci
From a point M on level ground the angle of elevation of a bird on top of a vertical pole is 30∘30^\circ. From a point N, 10 metres closer to the pole and on the same side, the angle of elevation of the same bird is 45∘45^\circ. Find the distance of M from the bottom of the pole in metres, correct to three significant figures.

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Question 701

[3 marks]Variation and Mensuration
MM is directly proportional to (d−1)2(d - 1)^2. Given that M=12M = 12 when d=4d = 4, calculate MM when d=7d = 7.

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Question 702

[2 marks]Variation and Mensuration
ABCDE is the cross-section of a tobacco shed, with ABDE a rectangle and BCD a triangular roof. AB=BC=CD=DE=5AB = BC = CD = DE = 5 m and AE=8AE = 8 m. Calculate the perpendicular height of C above the side BD, in metres.

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Question 703

[2 marks]Variation and Mensuration
ABCDE is the cross-section of a tobacco shed, with ABDE a rectangle and BCD a triangular roof of perpendicular height 3 m. AB=DE=5AB = DE = 5 m and AE=BD=8AE = BD = 8 m. Calculate the area of the cross-section ABCDE, in square metres.

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Question 704

[2 marks]Variation and Mensuration
A tobacco shed 20 m long has a uniform cross-section of area 52 m2^2. Calculate the volume of the shed, in cubic metres.

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Question 705

[3 marks]Variation and Mensuration
A tobacco shed is 20 m long. Its cross-section ABDE below the roof line BD is a rectangle 8 m wide and 5 m high, with the roof above BD. Calculate the number of bales of tobacco that can be stored in the shed up to BD, given that each bale has a volume of 4 m3^3.

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Question 801

[2 marks]Inequalities and Linear Programming
A girl is given $6,00\$6,00 to buy fireworks. She buys xx rockets at 60 c each and yy crackers at 30 c each. Which inequality states that she does not spend more than she has?
  1. A60x+30y≥60060x + 30y \ge 600, requiring her to spend at least the whole amount
  2. B60x+30y≤660x + 30y \le 6, comparing the cost in cents with 6 dollars
  3. C30x+60y≤60030x + 60y \le 600, pricing the rockets at 30 c and the crackers at 60 c
  4. D60x+30y≤60060x + 30y \le 600, comparing the cost in cents with 600 c

Question 802

[2 marks]Inequalities and Linear Programming
A girl buys xx rockets and yy crackers. She wants at least 4 rockets, and the number of crackers should be more than or equal to twice the number of rockets. Write down two inequalities that satisfy these conditions.
  1. Ax>4x > 4 and y>2xy > 2x, since more than is what both of the conditions ask for here
  2. Bx≥4x \ge 4 and y≥2xy \ge 2x, taking at least and more than or equal as inclusive
  3. Cx≥4x \ge 4 and 2y≥x2y \ge x, taking twice as applying to the crackers
  4. Dx≤4x \le 4 and y≥2xy \ge 2x, reading at least 4 as a limit of 4 rockets

Question 803

[3 marks]Inequalities and Linear Programming
A girl has $6,00\$6,00 and buys xx rockets at 60 c each and yy crackers at 30 c each, subject to 2x+y≤202x + y \le 20, x≥4x \ge 4 and y≥2xy \ge 2x. Find the least amount of money, in dollars, that she can spend.

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Question 804

[3 marks]Inequalities and Linear Programming
A girl is given $6,00\$6,00 and buys 4 rockets at 60 c each and 8 crackers at 30 c each, which is the cheapest combination allowed. Find the change she would get.

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Question 805

[2 marks]Inequalities and Linear Programming
A girl spends the whole of her $6,00\$6,00 on xx rockets at 60 c each and yy crackers at 30 c each, so that 2x+y=202x + y = 20. Find the value of yy when she buys 5 rockets.

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Question 901

[2 marks]Geometrical Transformation
Quadrilateral ABCD has vertices at A(1; 0), B(2; 0), C(2; 2) and D(1; 2). State the special name given to quadrilateral ABCD.
  1. AA trapezium, because only one pair of opposite sides of the shape is parallel
  2. BA square, because all four sides are of equal length and all angles are right angles
  3. CA rectangle, because the angles are right angles and one side is 1 unit and the next 2
  4. DA rhombus, because opposite sides are parallel and the diagonals cross at right angles

Question 902

[3 marks]Geometrical Transformation
Quadrilateral ABCD has vertices A(1; 0), B(2; 0), C(2; 2), D(1; 2). Quadrilateral ABC1D1ABC_1D_1 has vertices A(1; 0), B(2; 0), C1C_1(6; 2), D1D_1(5; 2). Describe fully the single transformation that maps ABCD onto ABC1D1ABC_1D_1.
  1. AA translation of 4 units in the positive xx direction, applied to all four vertices
  2. BA shear with the xx-axis invariant and shear factor 2, moving C and D 4 units right
  3. CA stretch with the xx-axis invariant and stretch factor 3, taking C to (6;2)(6; 2)
  4. DA rotation about the origin through 90∘90^\circ, which fixes the points A and B

Question 903

[3 marks]Geometrical Transformation
Quadrilateral ABCD has vertices A(1; 0), B(2; 0), C(2; 2) and D(1; 2). It is mapped onto A2B2C2D2A_2B_2C_2D_2 by a reflection in the line y=x+2y = x + 2. Write down the coordinates of C2C_2, the image of C.

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Question 904

[3 marks]Geometrical Transformation
Quadrilateral ABCD has vertices A(1; 0), B(2; 0), C(2; 2) and D(1; 2). A3B3C3D3A_3B_3C_3D_3 is its image under an enlargement of scale factor −1-1 with (−1;−1)(-1; -1) as centre. Write down the coordinates of C3C_3, the image of C.

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Question 1001

[2 marks]Vector Geometry
The point M has coordinates (7;−3)(7; -3) and RM→=(64)\overrightarrow{RM} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}. Calculate the coordinates of R.

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Question 1002

[2 marks]Vector Geometry
The point M has coordinates (7;−3)(7; -3) and RM→=(64)\overrightarrow{RM} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}. Find MR→\overrightarrow{MR}.
  1. A(−64)\begin{pmatrix} -6 \\ 4 \end{pmatrix}, reversing only the horizontal component of RM→\overrightarrow{RM}
  2. B(1−7)\begin{pmatrix} 1 \\ -7 \end{pmatrix}, the position vector of the point R found earlier
  3. C(−6−4)\begin{pmatrix} -6 \\ -4 \end{pmatrix}, the reverse of RM→\overrightarrow{RM}, so both components change sign
  4. D(64)\begin{pmatrix} 6 \\ 4 \end{pmatrix}, the same as RM→\overrightarrow{RM} because M and R are the same pair of points

Question 1003

[2 marks]Vector Geometry
ORST is a quadrilateral in which OR→=u\overrightarrow{OR} = \boldsymbol{u}, OT→=2v\overrightarrow{OT} = 2\boldsymbol{v} and TS→=2u+v\overrightarrow{TS} = 2\boldsymbol{u} + \boldsymbol{v}. Express OS→\overrightarrow{OS} in terms of u\boldsymbol{u} and v\boldsymbol{v}.

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Question 1004

[3 marks]Vector Geometry
ORST is a quadrilateral in which OR→=u\overrightarrow{OR} = \boldsymbol{u} and OT→=2v\overrightarrow{OT} = 2\boldsymbol{v}. Express RT→\overrightarrow{RT} in terms of u\boldsymbol{u} and v\boldsymbol{v}.

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Question 1005

[3 marks]Vector Geometry
In quadrilateral ORST the diagonals OS and RT meet at P, where RP→=(2k−1)u+3kv\overrightarrow{RP} = (2k - 1)\boldsymbol{u} + 3k\boldsymbol{v} and also RP→=hRT→\overrightarrow{RP} = h\overrightarrow{RT} with RT→=−u+2v\overrightarrow{RT} = -\boldsymbol{u} + 2\boldsymbol{v}. Calculate the value of kk.

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Question 1101

[3 marks]Simultaneous Equations, Bearings and Trigonometry
Solve the simultaneous equations 3x−2y=83x - 2y = 8 and 5x−4y=125x - 4y = 12.

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Question 1102

[3 marks]Simultaneous Equations, Bearings and Trigonometry
P is 4 km north of Q, and R is 6 km from P on a bearing of 073∘073^\circ. Calculate QR in km, correct to three significant figures.

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Question 1103

[3 marks]Simultaneous Equations, Bearings and Trigonometry
P is 4 km north of Q, and R is 6 km from P on a bearing of 073∘073^\circ, with QR=8,13QR = 8,13 km. Calculate PQ^RP\hat{Q}R in degrees, correct to one decimal place.

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Question 1104

[3 marks]Simultaneous Equations, Bearings and Trigonometry
P is 4 km north of Q, and R is 6 km from P on a bearing of 073∘073^\circ. Given that PQ^R=44,9∘P\hat{Q}R = 44,9^\circ, find the bearing of R from Q to the nearest degree.

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Question 1201

[3 marks]Statistics and Probability
The cumulative frequency of Mathematics marks for 500 students is: 10 students scored x≤10x \le 10, 50 scored x≤20x \le 20, 150 scored x≤30x \le 30, 245 scored x≤40x \le 40, 325 scored x≤50x \le 50, 400 scored x≤60x \le 60, 465 scored x≤70x \le 70, 490 scored x≤80x \le 80, 495 scored x≤90x \le 90 and 500 scored x≤100x \le 100. Find the median mark, correct to the nearest whole mark.

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Question 1202

[3 marks]Statistics and Probability
For 500 students the cumulative frequency of Mathematics marks is: 10 at x≤10x \le 10, 50 at x≤20x \le 20, 150 at x≤30x \le 30, 245 at x≤40x \le 40, 325 at x≤50x \le 50, 400 at x≤60x \le 60, 465 at x≤70x \le 70, 490 at x≤80x \le 80, 495 at x≤90x \le 90 and 500 at x≤100x \le 100. Find the inter-quartile range, correct to the nearest whole mark.

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Question 1203

[3 marks]Statistics and Probability
Of 500 students in a Mathematics examination, 325 scored marks less than or equal to 50. Two students are chosen at random, one after the other and without replacement. Find the probability that both students got marks less than or equal to 50, correct to three significant figures.

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Question 1204

[2 marks]Statistics and Probability
In a Mathematics examination taken by 500 students, 325 of them scored marks less than or equal to 50. How many students scored more than 50 marks?
  1. A75, the difference between the 325 students and the 250 halfway mark
  2. B250, taking half of the 500 students who sat the examination
  3. C325, the same count as the students who scored 50 marks or less
  4. D175, the 500 students less the 325 who scored 50 marks or less

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