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ZIMSEC O Level · 4008/1, 4028/1 · J2001

Mathematics Paper 1 June 2001

Questions
56
Total marks
94
Time allowed
150 min
Syllabus code
4008/1, 4028/1

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Questions
56
Pass mark
34
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]Number
Giving your answer as a common fraction in its lowest terms, find the value of 725−6237\frac{2}{5} - 6\frac{2}{3}.

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Question 102

[1 marks]Number
Giving your answer as a common fraction in its lowest terms, find the value of 0,450,45 of 59\frac{5}{9}.

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Question 103

[1 marks]Number
Express 1516\frac{15}{16} as a terminating decimal.

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Question 201

[1 marks]Approximations & Estimations
Express 158,697158,697 correct to the nearest hundredth.

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Question 202

[1 marks]Approximations & Estimations
Express 158,697158,697 correct to two significant figures.

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Question 203

[1 marks]Number
Express 720 as a product of its prime factors.

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Question 301

[1 marks]Time
Find the sum of 2 days 15 hours and 3 days 21 hours, giving your answer in days and hours.

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Question 302

[2 marks]Number Bases
Express 34534_5 (34 base five) as a number in base two.

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Question 401

[3 marks]Algebra
Simplify a2−b2ab+a2÷ab−a22a3\dfrac{a^2-b^2}{ab+a^2} \div \dfrac{ab-a^2}{2a^3}.

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Question 501

[1 marks]Measures & Mensuration
Two similar cylinders have their volumes in the ratio 1:8. Write down the ratio of the areas of their circular surfaces.

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Question 502

[2 marks]Trigonometry, Bearing & Distances
AA, BB and CC are three points on level ground. The bearing of CC from BB is 138°138° and AB^C=92°A\hat{B}C = 92°. Calculate the bearing of BB from AA.

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Question 801

[1 marks]Laws of Indices
Evaluate (15)−2\left(\dfrac{1}{5}\right)^{-2}.

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Question 802

[2 marks]Laws of Indices
Evaluate −2723-27^{\frac{2}{3}}.

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Question 901

[2 marks]Measures & Mensuration
In the diagram, ABEABE and DCFDCF are parallel straight lines, ABCDABCD is a parallelogram, DC=7DC = 7 cm, EC=8EC = 8 cm and EC^F=65°E\hat{C}F = 65°. Using the information given below where necessary, find the area of △DEC\triangle DEC. [sin⁡65°=0,91\sin 65° = 0,91; cos⁡65°=0,42\cos 65° = 0,42; tan⁡65°=2,14\tan 65° = 2,14.]

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Question 902

[1 marks]Measures & Mensuration
In the diagram, ABEABE and DCFDCF are parallel straight lines, ABCDABCD is a parallelogram, DC=7DC = 7 cm, EC=8EC = 8 cm and EC^F=65°E\hat{C}F = 65°. Using the information given below where necessary, find the area of parallelogram ABCDABCD. [sin⁡65°=0,91\sin 65° = 0,91; cos⁡65°=0,42\cos 65° = 0,42; tan⁡65°=2,14\tan 65° = 2,14.]

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Question 1001

[1 marks]Functional Graphs
In the diagram, line l1l_1 and line l2l_2 are parallel. l1l_1 passes through (0,4)(0,4) and (3,0)(3,0). Find the gradient of line l1l_1.

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Question 1002

[2 marks]Functional Graphs
In the diagram, line l1l_1 and line l2l_2 are parallel. l1l_1 passes through (0,4)(0,4) and (3,0)(3,0), and l2l_2 passes through (5,4)(5,4). Find the equation of line l2l_2, giving your answer in the form ax+by=cax + by = c where aa, bb and cc are integers.

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Question 1101

[1 marks]Inequalities
The diagram shows an unshaded region bounded by the lines 2y=7x+142y = 7x+14, y+x=7y+x=7 and y=−4y=-4. Write down the inequality (involving only yy) that defines the unshaded region.

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Question 1102

[1 marks]Inequalities
The diagram shows an unshaded region bounded by the lines 2y=7x+142y = 7x+14, y+x=7y+x=7 and y=−4y=-4. Write down the inequality relating 2y2y and 7x+147x+14 that defines the unshaded region.

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Question 1103

[1 marks]Inequalities
The diagram shows an unshaded region bounded by the lines 2y=7x+142y = 7x+14, y+x=7y+x=7 and y=−4y=-4. Write down the inequality relating y+xy+x and 7 that defines the unshaded region.

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Question 1201

[3 marks]Trigonometry, Bearing & Distances
In the diagram, AB=11AB = 11 cm, BC=7BC = 7 cm and AC=6AC = 6 cm. Calculate the value of cos⁡AC^B\cos A\hat{C}B, giving your answer as a common fraction in its lowest terms.

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Question 1301

[3 marks]Equations
Solve the simultaneous equations 2x−3y=132x - 3y = 13, 3x+2y=03x + 2y = 0.

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Question 1401

[2 marks]Approximations & Estimations
Estimate, to the nearest whole number, the value of 7,9×80,61,8×3,1\dfrac{7,9 \times \sqrt{80,6}}{1,8 \times 3,1}.

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Question 1402

[2 marks]Approximations & Estimations
The length of a square is given as 10 cm correct to one significant figure. Find the smallest possible area of the square.

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Question 1501

[1 marks]Ordinary & Standard Form
Given that m=4,8×103m = 4,8 \times 10^3 and n=1,2×10−5n = 1,2 \times 10^{-5}, find 5m5m, giving your answer in standard form.

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Question 1502

[1 marks]Ordinary & Standard Form
Given that m=4,8×103m = 4,8 \times 10^3 and n=1,2×10−5n = 1,2 \times 10^{-5}, find n2n^2, giving your answer in standard form.

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Question 1503

[2 marks]Ordinary & Standard Form
Given that m=4,8×103m = 4,8 \times 10^3 and n=1,2×10−5n = 1,2 \times 10^{-5}, find nm\dfrac{n}{m}, giving your answer in standard form.

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Question 1601

[2 marks]Algebra
Expand and simplify (x+2)(5+2x−x2)(x + 2)(5 + 2x - x^2).

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Question 1602

[2 marks]Algebra
Factorise completely 3x2−6x−ax+2a3x^2 - 6x - ax + 2a.

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Question 1701

[1 marks]Circle Geometry
In the diagram, PAPA and PBPB are tangents to the circle centre OO. QQ is a point on the minor arc ABAB and AO^B=140°A\hat{O}B = 140°. Find OB^PO\hat{B}P.

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Question 1702

[1 marks]Circle Geometry
In the diagram, PAPA and PBPB are tangents to the circle centre OO. QQ is a point on the minor arc ABAB and AO^B=140°A\hat{O}B = 140°. Find AP^BA\hat{P}B.

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Question 1703

[2 marks]Circle Geometry
In the diagram, PAPA and PBPB are tangents to the circle centre OO. QQ is a point on the minor arc ABAB and AO^B=140°A\hat{O}B = 140°. Find AQ^BA\hat{Q}B.

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Question 1801

[2 marks]Matrices
A=(7−24−1)A = \begin{pmatrix} 7 & -2 \\ 4 & -1 \end{pmatrix}, B=(−12−47)B = \begin{pmatrix} -1 & 2 \\ -4 & 7 \end{pmatrix}. Find ABAB.

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Question 1802

[2 marks]Matrices
M=(−3−526)M = \begin{pmatrix} -3 & -5 \\ 2 & 6 \end{pmatrix}. Find the inverse of MM.

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Question 1901

[2 marks]Number
It is given that a=(11−7)\mathbf{a} = \begin{pmatrix} 11 \\ -7 \end{pmatrix}, 17=4,123\sqrt{17} = 4,123 and 1,7=1,304\sqrt{1,7} = 1,304. Find 0,17\sqrt{0,17}.

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Question 1902

[2 marks]Vector Geometry
It is given that a=(11−7)\mathbf{a} = \begin{pmatrix} 11 \\ -7 \end{pmatrix}, 17=4,123\sqrt{17} = 4,123 and 1,7=1,304\sqrt{1,7} = 1,304. Find the magnitude of a\mathbf{a} correct to two significant figures.

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Question 2001

[2 marks]Algebra
In a regular polygon of nn sides, the size of each interior angle x°x° is given by the formula x=180−360nx = 180 - \dfrac{360}{n}. Make nn the subject of the formula.

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Question 2002

[2 marks]Polygons, Symmetry & Circles
In a regular polygon of nn sides, the size of each interior angle x°x° is given by the formula x=180−360nx = 180 - \dfrac{360}{n}. Calculate the number of sides of a regular polygon whose interior angle is 150°150°.

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Question 2101

[1 marks]Geometrical Transformation
The diagram shows △ABC\triangle ABC with A(−3,−1)A(-3,-1), B(−1,−1)B(-1,-1), C(−1,−3)C(-1,-3). Find the coordinates of PP, QQ and RR, the images of AA, BB and CC respectively under a translation of vector (74)\begin{pmatrix} 7 \\ 4 \end{pmatrix}.

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Question 2102

[3 marks]Geometrical Transformation
The diagram shows △ABC\triangle ABC and △DEF\triangle DEF. Describe fully the single transformation that maps △ABC\triangle ABC onto △DEF\triangle DEF.

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Question 2201

[2 marks]Scales & Simple Map Problems
The scale of a map is given as 1:50 000. Calculate the length of a line on the map that represents a road 6,4 km long.

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Question 2202

[3 marks]Scales & Simple Map Problems
The scale of a map is given as 1:50 000. Calculate the area, in hectares, of a piece of land represented on the map by an area of 3 cm². [1 hectare = 10 000 m².]

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Question 2301

[1 marks]Variation
mm is directly proportional to n2n^2 and m=28m = 28 when n=2n = 2. Write down an expression for mm in terms of nn and a constant kk.

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Question 2302

[2 marks]Variation
mm is directly proportional to n2n^2 and m=28m = 28 when n=2n = 2, so that m=kn2m = kn^2. Calculate the value of kk.

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Question 2303

[2 marks]Variation
mm is directly proportional to n2n^2, so that m=7n2m = 7n^2. Calculate the values of nn when m=63m = 63.

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Question 2401

[1 marks]Logarithms
Given that log⁡216=2,334\log 216 = 2,334 and log⁡3=0,477\log 3 = 0,477, find log⁡2,16\log 2,16.

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Question 2402

[2 marks]Logarithms
Given that log⁡216=2,334\log 216 = 2,334 and log⁡3=0,477\log 3 = 0,477, find log⁡6\log 6.

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Question 2403

[2 marks]Logarithms
Given that log⁡216=2,334\log 216 = 2,334 and log⁡3=0,477\log 3 = 0,477, find log⁡72\log 72.

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Question 2501

[1 marks]Simultaneous Equations
The graph of x+2y+p=0x + 2y + p = 0 and the graph of 3x+qy=93x + qy = 9 intersect at the point (−2,3)(-2, 3). Find the value of pp.

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Question 2502

[1 marks]Simultaneous Equations
The graph of x+2y+p=0x + 2y + p = 0 and the graph of 3x+qy=93x + qy = 9 intersect at the point (−2,3)(-2, 3). Find the value of qq.

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Question 2503

[2 marks]Algebra
Factorise 4x2−12x+94x^2 - 12x + 9.

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Question 2504

[2 marks]Algebra
Given that 4x2−12x+9=(2x−3)24x^2 - 12x + 9 = (2x-3)^2, solve the equation 4x2−12x+9=164x^2 - 12x + 9 = 16.

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Question 2601

[1 marks]Functional Graphs
The diagram shows the velocity-time graph of a car during a period of 20 minutes: velocity falls from 90 km/h to 30 km/h during the first 5 minutes, stays at 30 km/h from 5 to 15 minutes, then rises back to 90 km/h from 15 to 20 minutes. Find the retardation, in km/h per minute, during the first 5 minutes.

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Question 2602

[2 marks]Functional Graphs
The diagram shows the velocity-time graph of a car during a period of 20 minutes: velocity falls from 90 km/h to 30 km/h during the first 5 minutes, stays at 30 km/h from 5 to 15 minutes, then rises back to 90 km/h from 15 to 20 minutes. Find the distance, in kilometres, travelled at constant velocity.

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Question 2603

[2 marks]Functional Graphs
The diagram shows the velocity-time graph of a car during a period of 20 minutes: velocity falls from 90 km/h to 30 km/h during the first 5 minutes, stays at 30 km/h from 5 to 15 minutes, then rises back to 90 km/h from 15 to 20 minutes. Find the total distance travelled, in kilometres.

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Question 2604

[3 marks]Functional Graphs
The diagram shows the velocity-time graph of a car during a period of 20 minutes: velocity falls from 90 km/h to 30 km/h during the first 5 minutes, stays at 30 km/h from 5 to 15 minutes, then rises back to 90 km/h from 15 to 20 minutes. Find the average velocity, in km/h, during the 20 minutes.

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