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ZIMSEC O Level · 4008/1 · N2013

Mathematics Paper 1 November 2013

Questions
60
Total marks
100
Time allowed
150 min
Syllabus code
4008/1

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Questions
60
Pass mark
36
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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]Fractions, Decimals & Percentages
Evaluate 15+16\frac{1}{5} + \frac{1}{6}, giving the answer as a fraction in its lowest terms. [1]

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Question 102

[1 marks]Fractions, Decimals & Percentages
Evaluate 25÷4\frac{2}{5} \div 4, giving the answer as a fraction in its lowest terms. [1]

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Question 103

[1 marks]Fractions, Decimals & Percentages
Evaluate 34−14×23\frac{3}{4} - \frac{1}{4} \times \frac{2}{3}, giving the answer as a fraction in its lowest terms. [1]

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Question 201

[3 marks]Ordinary and Standard Form
Evaluate (0,3)3×0,020,0008\dfrac{(0,3)^3 \times 0,02}{0,0008}, giving your answer in standard form. [3]

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Question 301

[1 marks]Number
The temperature inside a freezer is −8°-8° C. During a power cut the temperature rose by 12°12° C. Find the temperature after the rise, in degrees Celsius. [1]

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Question 302

[2 marks]Number
Write down the next two terms in the following sequence: 1;12;14;18;…1; \frac{1}{2}; \frac{1}{4}; \frac{1}{8}; \ldots [2]

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Question 401

[1 marks]Change of Units
Write 3,35 minutes in minutes and seconds. [1]

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Question 402

[2 marks]Change of Units
If 1 kilometre is 58\frac{5}{8} of a mile, convert 75 miles to kilometres. [2]

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Question 501

[3 marks]Algebraic Expressions
A shopper spent $cd\$\frac{c}{d} on one item and half of that amount on each of three other items. Find how much she spent altogether, in dollars, in terms of cc and dd. [3]

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Question 601

[1 marks]Laws of Indices
Simplify (33)4273\dfrac{(3^3)^4}{27^3}. [1]

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Question 602

[1 marks]Laws of Indices
Simplify (4x2y6)12\left(4x^2y^6\right)^{\frac{1}{2}}. [1]

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Question 603

[1 marks]Laws of Indices
Simplify x0+x−2x^0 + x^{-2}. [1]

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Question 701

[1 marks]Fractions, Decimals & Percentages
Express 78\frac{7}{8} as a decimal fraction. [1]

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Question 702

[2 marks]Consumer Arithmetic
A car loses 55% of its value after four years. If it cost $8 500\$8\ 500 when new, find its value, in dollars, after the four years. [2]

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Question 801

[1 marks]Algebraic Expressions
Simplify 5m−2(x−3m)5m - 2(x - 3m). [1]

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Question 802

[2 marks]Linear Equations
Solve the equation x+57=32\dfrac{x+5}{7} = \dfrac{3}{2}. [2]

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Question 901

[1 marks]Change of Units
Express 200 km/h as a speed in km/min. [1]

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Question 902

[2 marks]Ratios, Rates & Proportions
Find the time taken for a racing driver to cover a 120 km race if he travels at a speed of 200 km/h, giving your answer in minutes. [2]

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Question 1001

[1 marks]Factorisation
Factorise completely 4y−44y - 4. [1]

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Question 1002

[2 marks]Factorisation
Factorise completely xy2−4x+2y2−8xy^2 - 4x + 2y^2 - 8. [2]

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Question 1101

[3 marks]Simultaneous Equations
Solve the simultaneous equations 5d−3e=−15d - 3e = -1 and 2d+3e=82d + 3e = 8, giving the values of dd and ee. [3]

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Question 1201

[3 marks]Quadratic Equations
Solve the equation (x+3)2=49(x + 3)^2 = 49. [3]

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Question 1202

[1 marks]Number
Write down the prime numbers between 20 and 30. [1]

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Question 1301

[3 marks]Algebraic Fractions
Simplify x2+3x+2x+2\dfrac{x^2 + 3x + 2}{x + 2}. [3]

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Question 1302

[1 marks]Polygons, Symmetry & Circles
Find the order of rotational symmetry of a right-angled isosceles triangle. [1]

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Question 1401

[1 marks]Sets
Given that n(A)=10n(\mathbf{A}) = 10 and n(B)=15n(\mathbf{B}) = 15, find the greatest possible value of n(A∪B)n(\mathbf{A} \cup \mathbf{B}). [1]

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Question 1402

[1 marks]Sets
Given that n(A)=10n(\mathbf{A}) = 10 and n(B)=15n(\mathbf{B}) = 15, find the greatest possible value of n(A∩B)n(\mathbf{A} \cap \mathbf{B}). [1]

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Question 1403

[2 marks]Sets
In the Venn diagram, the whole of set R is shaded, together with the part of P∩Q\mathbf{P} \cap \mathbf{Q} that lies outside R. Use set notation to describe the shaded region in terms of sets P\mathbf{P}, Q\mathbf{Q} and R\mathbf{R}. [2]

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Question 1501

[2 marks]Matrices
Given that A=(−2−162)\mathbf{A} = \begin{pmatrix} -2 & -1 \\ 6 & 2 \end{pmatrix} and B=(0−143)\mathbf{B} = \begin{pmatrix} 0 & -1 \\ 4 & 3 \end{pmatrix}, find 3A−B3\mathbf{A} - \mathbf{B}. [2]

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Question 1502

[2 marks]Matrices
Given that B=(0−143)\mathbf{B} = \begin{pmatrix} 0 & -1 \\ 4 & 3 \end{pmatrix}, find B2\mathbf{B}^2. [2]

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Question 1601

[1 marks]Number Bases
Write down the largest four-digit number in base eight. [1]

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Question 1602

[2 marks]Number Bases
Convert 1118111_8 to a number in base two. [2]

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Question 1603

[1 marks]Number Bases
Find the sum of 4445444_5 and 21521_5, giving your answer in base five. [1]

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Question 1701

[1 marks]Statistics
A rugby team scored the following points in 12 matches: 21; 18; 3; 12; 15; 18; 42; 18; 24; 6; 12; 3. Find the mode. [1]

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Question 1702

[2 marks]Statistics
A rugby team scored the following points in 12 matches: 21; 18; 3; 12; 15; 18; 42; 18; 24; 6; 12; 3. Find the mean. [2]

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Question 1703

[1 marks]Statistics
A rugby team scored the following points in 12 matches: 21; 18; 3; 12; 15; 18; 42; 18; 24; 6; 12; 3. In the next match, the team scored 55 points. Write down the median score for the 13 matches. [1]

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Question 1801

[2 marks]Scales & Simple Map Problems
The scale of a map is 1:10 000. Two hills are 4,5 cm apart on the map. Find the actual distance between the hills, giving your answer in kilometres. [2]

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Question 1802

[2 marks]Scales & Simple Map Problems
The scale of a map is 1:10 000. Two towns are 80 km apart. Find the distance between them on the map, giving your answer in centimetres. [2]

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Question 1901

[1 marks]Substitution
The temperature T°T° C at a height of HH metres above sea level is given by the formula T=20−H150T = 20 - \dfrac{H}{150}. Calculate the temperature, in degrees Celsius, at 4 500 metres. [1]

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Question 1902

[2 marks]Change of Subject of Formula
The temperature T°T° C at a height of HH metres above sea level is given by the formula T=20−H150T = 20 - \dfrac{H}{150}. Make HH the subject of the formula. [2]

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Question 1903

[1 marks]Change of Subject of Formula
The temperature T°T° C at a height of HH metres above sea level is given by the formula T=20−H150T = 20 - \dfrac{H}{150}. Find the height, in metres, at which the temperature is 12°12° C. [1]

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Question 2001

[1 marks]Variation
The number of revolutions, nn, of a wheel over a fixed distance varies inversely as the circumference, CC cm, of the wheel. Write down an equation involving nn, CC and a constant kk. [1]

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Question 2002

[3 marks]Variation
The number of revolutions, nn, of a wheel over a fixed distance varies inversely as the circumference, CC cm, of the wheel. If a wheel of circumference 80 cm makes 10 revolutions, find the number of revolutions made by a wheel of circumference 200 cm. [3]

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Question 2101

[2 marks]Similarity & Congruency
In the diagram, XY is parallel to QR, PY = 2 cm, YR = 3 cm and XY = 4 cm. Find the length of QR, in centimetres. [2]

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Question 2102

[2 marks]Similarity & Congruency
In the diagram, XY is parallel to QR, PY = 2 cm, YR = 3 cm and XY = 4 cm. Find the ratio area △PXYarea △PQR\dfrac{\text{area } \triangle PXY}{\text{area } \triangle PQR}. [2]

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Question 2201

[1 marks]Probability
Tendai and Vimbai take a driving test. The probability that Tendai will pass is 35\frac{3}{5} and the probability that Vimbai will pass is 23\frac{2}{3}. State which one of them is more likely to pass. [1]

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Question 2202

[2 marks]Probability
Tendai and Vimbai take a driving test. The probability that Tendai will pass is 35\frac{3}{5} and the probability that Vimbai will pass is 23\frac{2}{3}. Calculate the probability that they both fail. [2]

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Question 2203

[2 marks]Probability
Tendai and Vimbai take a driving test. The probability that Tendai will pass is 35\frac{3}{5} and the probability that Vimbai will pass is 23\frac{2}{3}. Calculate the probability that only one of them will pass. [2]

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Question 2301

[1 marks]Vector Geometry
In the diagram PQ→=p\overrightarrow{PQ} = \mathbf{p}, QR→=q\overrightarrow{QR} = \mathbf{q} and RS→=r\overrightarrow{RS} = \mathbf{r}, and triangles PQT, QTR and TRS are equilateral. Express PS→\overrightarrow{PS} in terms of p\mathbf{p}, q\mathbf{q} and r\mathbf{r}. [1]

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Question 2302

[2 marks]Vector Geometry
In the diagram QR→=q\overrightarrow{QR} = \mathbf{q}, and triangles PQT, QTR and TRS are equilateral. Express PS→\overrightarrow{PS} in terms of q\mathbf{q} only. [2]

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Question 2303

[2 marks]Vector Geometry
In the diagram PQ→=p\overrightarrow{PQ} = \mathbf{p}, QR→=q\overrightarrow{QR} = \mathbf{q} and RS→=r\overrightarrow{RS} = \mathbf{r}, and triangles PQT, QTR and TRS are equilateral, so that PS→=2q\overrightarrow{PS} = 2\mathbf{q}. Express p\mathbf{p} in terms of q\mathbf{q} and r\mathbf{r}. [2]

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Question 2401

[1 marks]Co-ordinate Geometry
The graph of y=mx+cy = mx + c passes through the points A (0;2)(0; 2) and B (5;−3)(5; -3). Find the value of cc. [1]

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Question 2402

[2 marks]Co-ordinate Geometry
The graph of y=mx+cy = mx + c passes through the points A (0;2)(0; 2) and B (5;−3)(5; -3). Find the value of mm. [2]

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Question 2403

[2 marks]Co-ordinate Geometry
The graph of y=mx+cy = mx + c passes through the points A (0;2)(0; 2) and B (5;−3)(5; -3). Calculate the length of AB, leaving your answer in surd form. [2]

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Question 2501

[2 marks]Travel Graphs
The diagram is the speed-time graph of a sprinter during an athletics training session. The sprinter accelerates uniformly from rest to 10 m/s in the first 4 seconds, runs at 10 m/s until t=10t = 10, accelerates to a speed V at t=12t = 12, then decelerates uniformly to rest at t=17t = 17. Calculate the distance, in metres, the sprinter covers during the first 10 seconds. [2]

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Question 2502

[2 marks]Travel Graphs
The diagram is the speed-time graph of a sprinter during an athletics training session. The sprinter is running at 10 m/s at t=10t = 10 and reaches a speed V at t=12t = 12. Given that the acceleration during the time interval from t=10t = 10 to t=12t = 12 is 5 m/s2^2, find the value of V, in m/s. [2]

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Question 2503

[2 marks]Travel Graphs
The diagram is the speed-time graph of a sprinter during an athletics training session. The sprinter reaches 20 m/s at t=12t = 12 and comes to rest at t=17t = 17. Calculate the deceleration of the sprinter, in m/s2^2, from t=12t = 12 to the time the sprinter stops running. [2]

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Question 2601

[2 marks]Circle Geometry
The points P, Q and R lie on the circumference of a circle, centre O. PQ = 5 cm, PR = 8 cm and QP^R=80°Q\hat{P}R = 80°. Calculate the area of triangle PQR, in cm2^2. [sin 80° = 0.985; cos 80° = 0.174; tan 80° = 5.67] [2]

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Question 2602

[2 marks]Circle Geometry
The points P, Q and R lie on the circumference of a circle, centre O. PQ = 5 cm, PR = 8 cm and QP^R=80°Q\hat{P}R = 80°. Calculate the value of QR2^2. [sin 80° = 0.985; cos 80° = 0.174; tan 80° = 5.67] [2]

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Question 2603

[2 marks]Circle Geometry
The points P, Q and R lie on the circumference of a circle, centre O. PQ = 5 cm, PR = 8 cm and QP^R=80°Q\hat{P}R = 80°. Find the reflex QO^RQ\hat{O}R, in degrees. [2]

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