M A T H E -

M AT I C S
 Introduction  
 
National  examinations  in  Malawi,  JCE  and  MSCE,  are  tough.  Students  as  
well   as   teachers   must   cover   syllabi   of   many   subjects   that   contain   a   large   number  
of   topics.   On   the   other   hand,   to   let   students   have   fun   of   Math   and  
Science   is   one   of   missions   for   us,   Japan   Overseas   Cooperation  
Volunteers.  However,  we  have  been  facing   challenges  that  we  can’t  secure  
the  time  to  carry  out  the  mission  because  of  the  coverage  of  the  syllabi.    

“How  can  we  cover  the  syllabi  efficiently?”  

“How  can  students  learn  effectively?”  
In   order   to   conquer   the   challenges,   we   have   produced   this   series,  
“JCE/MSCE  Past  Papers:  Questions,  Solutions  &  Analysis”.  
 
FEATURES:  
*  Questions  and  Answers  in  JCE  /  MSCE  from  2003  to  2012  
*  Classified  by  topic,  for  learning  /  teaching  effectively  
*  Graphs  that  show  which  topic  is  most  often  asked  
*  Actual  exams  of  2013  as  achievement  tests  
 
We   wish   that   students   can   learn   effectively   and   that   teachers   can   teach  
efficiently   when   these   are   used   correctly.   But   don’t   forget!   To   enjoy  
Maths  and  Science  is  more  important  than  just  to  get  high  
marks.  Learning  with  joy  and  wonder  will  finally  lead  you  to  the  bright  future.  
 
JICA  Math  &  Science  Teachers’  Committee  
 
*  There  are  no  solutions  of  MSCE  BIO  paper2  and  P/S  paper2  because  those  answers  depend  on    

environments,  conditions  and  materials.  

*  Actual  exams  of  2013  have  only  questions.  

 Contents ~JCE: Mathematics~

Introduction

How to use

Analysis

Preparation

Questions: Maths (2003~2012)

1. Basis of arithmetic 10. Linear equations

2. Indices and logarithms 11. Inequalities
3. Social and commercial 12. Linear simultaneous
arithmetic equations
4. Number patterns 13. Quadratic equations
5. Density and mixture 14. Coordinates
6. Proportions 15. Linear graph
7. Probability 16. Travel graph
8. Statistics 17. Sets
9. Algebraic expressions 18. Vectors
 19. Line and angles 24. Pythagoras theorem
20. Reflection and rotations 25. Quadrilaterals
21. Triangles 26. Mensuration
22. Convex polygons 27. Geometrical constructions
23. Similarity

Solutions: Maths (2003~2012)

2013 JCE Math Questions

(*Only questions)

Acknowledgements
 How  to  use  
 
**Analysis  /  Preparation**    
There   is   a   graph   made   by   analysing   past-­‐questions   from   2003   to   2012.  

You  will  know  which  topic  you  should  learn  /  teach  intensively.  And  you  will  

also  find  FUNDAMENTAL  but  IMPORTANT  TIPS  to  get  higher  marks.  
 
**Questions  /  Solutions**    
Questions   and   solutions   are   classified   and   re-­‐ordered   by   each   topic.  

Therefore,  you   can   learn   specific   topics,   or   topics   that   you   are   weak   in,  

selectively.   Think   deeply   on   your   own   before   you   check   the  

answer!  
 
**Questions  of  MSCE  BIO,  P/S  paper2**     (Only  MSCE  BIO,  P/S  books)  
Questions   are   sorted   by   year.   Although   there   are   no   solutions   because  

those   answers   depend   on   schools   and   their   conditions,   you   can   grasp   the  

trend.  Similar  questions  repeatedly  appear.  

 
**Actual  exams  of  2013**  
These   questions   aren’t   re-­‐ordered.   You   can   try   this   exam   to   check   your  

current  level  as  an  achievement  test.  You  must   follow   fixed   time   length  

when  you  try.  

 
 Preparation  
 
We  analysed  questions  from  2003  to  2012  and  made  this  graph.  This  graph  
shows  the  average  of  marks  of  each  topic.  It  means  it  also  shows  which  topic  is  
most   often   asked.   In   this   case,   the   length   of   “Indices   and   logarithms”   is  
longest:  MOST  ASKED  TOPIC.  The  length  of  “Coordinates”  is  shortest:  LEAST  
ASKED  TOPIC.  
 
    MOST  ASKED  TOPICS  IN  JCE  MATHEMATICS:  
1.  Indices  and  logarithms  
2.  Algebraic  expressions  
3.  Basis  of  arithmetic  
4.  Quadratic  equations  
5.  Geometrical  constructions  
 
Even   apart   from   daily   learning,   we   can   still   make   marks   better.   How   you  
write  exams  is  also  important.  
 
      TIPS  FOR  GETTING  HIGHER  MARKS:  
1.  Write  your  name  correctly  
2.  Write  your  answers  clearly  and  neatly  
3.  Try  to  solve  all  questions  
    (Time  designation  is  also  important)  
4.  Don’t  leave  blanks,  write  something  
    (Even  if  you  don’t  know  the  answer,  at  least,  you  can  mark  one  choice  in  multiple    
        questions!)  
      IN  ESSAY  QUESTIONS:  
5.  Don’t  itemise  and  number  the  sentences,  write  them  in  an  essay  form.  
6.  you  don’t  have  to  rewrite  “the  question”  in  your  answer  essay.  
 Analysis:  JCE  MATHEMATICS  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
J C E : M A T H S
 1 Basis of arithmetic
Section A

(−7)+5(−4)
2012-1. Simplify −3
(3 marks)

2012-11. a. Subtract 326 from 27810, giving your answer in base 10. (3 marks)

2010-10. a. Convert 910 to base 2. (3 marks)

2009-3. Simplify 21103 − 22023 + 10213, leaving your answer in base 3. (3 marks)

(−5)(−3)+6
2008-1. Simplify (3 marks)
(−7)

2008-6. Convert 12013 to base 10. (4 marks)

2007-2. Simplify 8(−5) + 2(3) − 5. (3 marks)

2006-1. Simplify (−10)(2) (−4). (3 marks)

2006-4. Convert 102410 to base 8. (3 marks)

2005-3. Convert 32112 to base 10. (4 marks)

2004-5. a. Evaluate 1123 + 1023 leaving your answer in base 3. (3 marks)

2003-1. Subtract -5 from -4. (3 marks)

2003-3. Convert 12647 to base 10. (4 marks)

Section B

2011-16. b. Convert 1405 to base 10. (3 marks)

1 Basis of arithmetic - Questions

 2 Indices and logarithms
Section A

2012-13.
1
a. Simplify 𝑚0 + 492 (3 marks)

2012-14.

b. Use logarithms to evaluate √47.89, leaving your answer correct to 2 decimal places. (5 marks)

2011-3.
1
Simplify (3𝑘)2 − (4𝑘 4 )2 (4 marks)

2011-14.
1391.0
b. Evaluate using logarithms leaving your answer correct to 2 significant figures. (4 marks)
17.7

2010-11.
Simplify 33 𝑒 4 ÷ 3𝑒 −4. (3 marks)

2009-9.
1
Use logarithms to evaluate (51 ∙ 8)3 giving your answer correct to one decimal place. (4 marks)

2009-10.
1
(𝑝2 𝑞 2 )2
a. Simplify 𝑞
. (3 marks)

2008-2.
Simplify 𝑦(𝑦 −2 )5. (3 marks)

2008-13.
Use logarithms to evaluate (4.56)3 and give your answer to 3 significant figures. (4 marks)

2 Indices and logarithms - Questions

2007-8.
Evaluate √667.9 and give your answer correct to 3 decimal places. (3 marks)

2007-9.
Simplify 2𝑚(3𝑚)2. (4 marks)

2006-11.
Evaluate (4 13)2 and give your answer correct to 3 significant figures. (3 marks)

2005-2.
1
Evaluate 643 + 22. (4 marks)

2003-12.

41 672
Evaluate 161
using logarithms. (5 marks)

Section B

2010-18.
b. Use logarithms to evaluate 2 7 × 40 3, leaving your answer correct to two significant figures. (4
marks)

2007-16.
36.7
Use logarithm tables to evaluate (6 marks)
√2.8

2005-17.

Simplify 𝑎2 × 𝑎−4 ÷ 𝑎. (3 marks)

2005-19.
b. Evaluate √12.5 and give your answer correct to 2 decimal places. (3 marks)

2004-17.
123×27.3
Use logarithms to evaluate (6 marks)
2.631

2 Indices and logarithms - Questions

 3 Social and commercial arithmetic
Section A
2012-6. A lady bought a shirt marked K400.00 with 15% surtax and two pairs of socks at K120.00
per pair surtax free. Calculate the total amount of money she paid. (6 marks)

2010-4. The value of a bicycle depreciated by 7% in the first year. At the beginning of the second year
the value of the bicycle was K9300. Calculate the original value of the bicycle. (4 marks)

2009-5. A customer uses 192 units of electricity in a month. If the electricity supply company has a
fixed charge of K350.00 per month and K14.50 per unit of electricity, calculate the electricity bill for
the customer. (4 marks)

2008-7. Mr. Dzuwa works as an accountant and earns K 60 000 per month. He is allowed K 17 500
tax free per month. If he pays 15 % income tax on the rest, calculate the total income tax he pays in a
year. (5 marks)

2006-12. b. The surtax for a product at a market is 5%. If the cost of the product including the surtax
is K441.00, calculate the original price of the product. (4 marks)

2005-10. A school bought a car at K540,000. The value of the car depreciated by 5% after one year.
Calculate the value of the car at the beginning of the second year. (4 marks)

2004-8. The table below shows rate of income tax an employee pays at a company.
INCOME PER MONTH RATE OF TAX
First K4,800 0%
Next K9,600 10%
In excess of K14,400 25%
Calculate the income tax payable by a person who K24 400 per month. (6 marks)

Section B
2011-16. a. An agent got a commission of K2000 on sales of policies. If the rate of commission is 20%,
calculate the amount of policies sold. (3 marks)

2007-17. b. A lady is given 10% commission for every newspaper she sells. If she sold 1000
newspapers at K60 each, calculate her commission. (3 marks)

2003-19. b. A vendor reduced the price of a pocket radio from K800 to K720. Calculate the discount
percentage. (2 marks)

3 Social and commercial arithmetic - Questions

 4 Number patterns
Section A

2012-15.
b. The terms -8, P, 12, 22 are in a sequence. Calculate the value of P. (3 marks)

2010-12.
Given the sequence, 7, 11, 15, 19, 23, ... 283. Find the number of terms in the sequence. (5 marks)

2009-6.

Table 1 shows the positions of terms, the term and the connection between the position and the term
in a sequence.

Table 1

POSITION TERM CONNECTION BETWEEN

POSITION AND TERM

1 3 1 + (1 + 1)

2 5 2 + (2 + 1)

3 7 3 + (3 + 1)

Write down the nth term in its simplest form. (3 marks)

2008-10.
The terms −7, b, 17 are in arithmetic sequence. Calculate the value of b. (4 marks)

2007-13.
The nth term of a sequence is 2n − 1. If the last term is 11, calculate the number of terms of the
sequence. (4 marks)

4 Number patterns - Questions

2007-15.
b. Copy and complete the pattern of numbers in the table below.

NUMBER PATTERN TOTAL

1 1 1

2 1+2+1 4

3 1+2+3+2+1 9

4 1+2+3+4+3+2+1 16

(3 marks)

2006-2.
The nth term of a sequence is n2+n. Calculate the 5th term of the sequence. (3 marks)

2004-5.
b. Find the next term in the sequence 12, 32, 52 ,72, - -(1 mark)

2003-2.
a. study Figure 1 which shows triangular pattern of dots.

By drawing the 4th triangular pattern, find the number of dots which form it. (2 marks)

Section B

2011-19.

a. The nth term of a sequence is 3𝑛 − 2. Write down:

(i) the first three terms of the sequence. (3 marks)

(ii) the rule used to generate the sequence. (1 mark)

4 Number patterns - Questions

 5 Density and mixture
Section A
2012-11.
b. A cylinder of internal diameter 14 cm and height 10 cm is filled with a fluid of mass 3080g.
22
Calculate the density of the fluid. (Take 𝜋 = 7
). (4 marks)

2011-7.
A bottle weighs 77 g when empty, 101 g when full of water and 107 g when full of glycerine.
Find the relative density of glycerine. (4 marks)

2010-13.
A rectangular tank of volume 1800cm3 is filled with fuel. Given that the density of fuel is 0.7g/cm3,
calculate the mass of fuel in the tank. (3 marks)

2008-4.
A mathematical instrument box has a mass of 273 g.
If its relative density is 2.1, calculate the volume of the mathematical instrument box. (4 marks)

2007-10.
2 kg of white sugar costing K200 was mixed with 3 kg of brown sugar costing K270.
Calculate the cost of the mixture per kg. (4 marks)

2005-13.
A bottle weighs 20 g when empty, 55 g when full of petrol and 70 g when full of water.
Calculate the relative density of petrol. (4 marks)

2003-15.
A container Weighs 105 g when empty, 645 g When full of water and 915 g When full of liquid H.
Calculate the density of liquid H. (5 marks)

Section B
2011-17.
b. The Volume of a cylinder is 385 cm3. Given that its diameter is 7 cm, calculate the height of the
22
cylinder. (Take π = ). (3 marks)
7

2010-19.
b. In what ratio must white beans costing K146 per kg be mixed with red beans costing K170 per kg
in order for the mixture to cost K152 per kg? (4 marks)

5 Density and mixture - Questions

 6 Proportions
Section A

2011-10.
a. A family of four members has enough food for 30 days. If 2 visitors join them, how long would the
food last? (3 marks)

2009-10.
b. In a discount shop the price of an item was reduced from K700 to K616. Calculate the discount
percentage. (2 marks)

2008-14.
Figure 2 is a bar graph showing ways in which a person spends his/her monthly salary.

If the person’s monthly salary is K38000, calculate the monthly expenditure on food. (4 marks)

2007-11.
At a school 60% of the pupils are girls. If there are 48 girls, calculate the total number of pupils at the
school. (5 marks)

6 Proportions - Questions
2006-6.
A drum can hold 25 l of water. Calculate the number of bottles which can be used to fill the drum if
each bottle can contain 500 ml of Water. (3 marks)

2005-6. A family of 6 people uses 30 kg of flour in a month. If two visitors joined them for a month,
how many kilograms of flour would be required for that month? (3 marks)

2004-7. Figure 1 is a pie chart showing proportions of votes for candidates A, B and C.

Calculate the percentage of the votes that were for candidate C. (5 marks)

Section B

2005-18.
a. Chimwemwe and Chikondi share K540 so that Chikondi has K80 more than Chimwemwe. Find
their shares. (6 marks)

6 Proportions - Questions
 7 Probability
Section A

2012-8.
a. A die has its six faces marked 0, 1, 1, 1, 6, 6.
If it 1 is thrown once, find the probability that it will show a 6. (2 marks)

2011-14.
a. The letters of a word ABSTINENCE are written on identical pieces of paper and then put in a bag.
If one piece of paper is picked at random, find the probability that it has a letter N. (3 marks)

2010-6.
A learner has 6 red counters and 9 blue counters in a pocket.
If one counter is picked at random, find the probability that it is a blue counter. (3 marks)

2009-7.
A box contains 20 white beans, 30 black beans and 10 red beans.
Calculate the probability of picking a white bean. (3 marks)

2007-6.
Find the probability of choosing the letter A from the Word NYALA. (3 marks)

2003-5.
b. In order to decide which team begins playing football, a referee tosses a coin once.
If team A chooses head, what is the probability that team B begins playing? (2 marks)

7 Probability - Questions
 8 Statistics
Section A

2010-5.
The table below shows marks for 5 students A, B, C, D and E in a test.
Students A B C D E
Marks 7 11 6 7 9
Find the mean mark. (3 marks)

2009-2.
The mean mass of 3 packets of sugar is 400 g. If the mass of 2 packets is 900 g, calculate the mass of
the third packet. (3 marks)

2006-5.
a. Given the following list of numbers:
4, 1, 4, 2, 3, 7, 4, 3.
Calculate the median. (3 marks)

2005-4.
b. Figure 1 is a bar graph showing the number of textbooks according to subjects in a certain
classroom.

10

6
No of
Books 4

Subject
Figure 1
Calculate the total number of books in the classroom. (3 marks)

8 Statistics - Questions
Section B

2012-17.
Table 1 shows the number of pupils and the months they were born.
Table 1

Month of Birth May June July August

Number of Pupils 5 6 9 4

Using a scale of 2 cm to represent 2 units on the vertical axis, draw a bar chart to represent the
information. (3 marks)

2011-17.
a. Figure 5 is a pie chart of students who play different games.

If the total number of students is 60, calculate the number of students who play football. (5 marks)

2008-19.
a. The ages of five people are 15, 20, 12, 10 and K. If their mean age is 14, calculate the age
represented by K. (3 marks)

2007-18.
a. The following are marks of 15 students:
6 8 7 6 9

9 6 8 7 10

7 7 9 9 9

Draw a bar chart on the graph paper provided to represent this information. (5 marks)

8 Statistics - Questions
 9 Algebraic expressions

Section A

2012-3. Given that 𝑒 = 4, 𝑓 = −2 and 𝑔 = −5, find the value of 2𝑒 − 𝑓 + 𝑔. (3 marks)

2012-5. Find the LCM of 30ef, 10e2f and 5f2. (3 marks)

2012-14. a. Make r the subject of the formula 𝐋 = 𝐫𝐚 + 𝐧 (3 marks)

2011-1. Simplify 2𝑘 − 7𝑓 − 4(2𝑘 − 3𝑓). (3 marks)

2011-5. Find the HCF of 4ar2, 6a2br and 8ac2r. (3 marks)

2010-1. Given that 𝑎 = −2, and 𝑏 = 5, find the value of 𝑎2 − 𝑏. (3 marks)

2𝑐−4
2010-7. a. Simplify 𝑐 2 −4. (3 marks)

2𝑝𝑟
2009-1. Given that 𝑝 = −1, 𝑡 = −2 and 𝑟 = 4, find the value of 𝑡
. (3 marks)

2009-4. b. Make g the subject of the formula 𝑕 = 𝑘 − (g + 𝑒). (3 marks)

2009-14. a. Simplify 𝑥 + 3𝑦 − (−𝑥) − 4𝑦. (2 marks)

2008-9. Factorise completely 𝑥𝑘 − 𝑥𝑚 − 𝑘𝑧 + 𝑚𝑧. (4 marks)

4𝑥 4𝑥−4
2008-11. Simplify ( ) ÷ ( ). (4 marks)
5 5

𝑥+𝑦
2007-3. Given that 𝑥 = 2 and 𝑦 = 4, find the value of 𝑥−𝑦
. (3 marks)

2007-4. Find the L.C.M. of 2𝑥, 12𝑥𝑦 and 18𝑥 2 𝑦. (4 marks)

2006-3. Factorise 𝑥 2 + 𝑥𝑦. (2 marks)

9 Algebraic expression
2005-1. If a = 3, b = 4, c = −2, calculate the value of 2b − ac. (3 marks)

2005-15. a. Expand −5(2𝑝 − 3𝑞). (3 marks)

2004-2. Expand (3𝑑 − 2)2 and simplify your answer. (4 marks)

2004-3. Find the HCF of 33 𝑥 2 𝑦𝑧 and 32 𝑥𝑧 3. (3 marks)

2 3
2004-4. Express 𝑥
− 𝑥−2 as a single fraction in its simplest form. (4 marks)

2004-11. Evaluate
a(b + c) when a = 4, b = −2 and c = −3. (3 marks)

2003-2. b. Factorise 4 -X2. (2 marks)

2003-5. a. Find the HCF of the following terms: 4 m2, 6 mn2 and 10 mn. (3 marks)

3𝑥𝑦 2
2003-6. If x=-5, y=-2 and m=2, find the value of (3 marks)
2

2003-13. b. Expand (2b +4)(d -3bd) and give your answer in the simplest form. (3 marks)

9 Algebraic expression
 10 Linear equations
Section A

2012-8.
𝑃+1 3
b. Solve the equation 5
+ 𝑃 = 5 (4 marks)

2011-2.
2
Solve the equation = 2. (4 marks)
r−1

2011-8.
A farmer has 50 bags of maize. He sells (𝑎 + 5) bags and shares (2𝑎 − 3) bags. If he has 30 bags
remaining, form an equation in a and solve it. (4 marks)

2009-15.
The width of a rectangular wire fence is 4 metres shorter than the length. If the total length of the
wire used is 36 metres. Calculate the width of the fence. (5 marks)

2008-12.
Given an equation 5𝑥 − 𝑦 = 8. Find:
(i) y−intercept
(ii) gradient of the equation (3 marks)

2005-4.
5𝑚
a. Solve the equation 3
= 𝑚 − 2. (4 marks)

10 Linear equations - Questions

Section B

2010-20.
r+2 1
a. Solve the equation 3
− 2 = 1. (4 marks)

2007-20.
b. Figure 5 shows a triangle ABC in which 𝐀𝐁 = (𝑥 + 2) om, 𝐁𝐂 = (𝑥 + 9)cm and 𝐀𝐂 = (𝑥 − 7)cm.

If the perimeter of the triangle is 55 cm, calculate the value of 𝑥.

10 Linear equations - Questions

 11 Inequalities
Section A

2012-2.
Figure 1 shows a number line where a graph of values of x are indicated.

Write down an inequality in x describing the graph. (3 marks)

2011-4.
Solve the inequality 𝑏 + 1 ≥ 13 − 3𝑏. (3 marks)

2010-3.
In a competition, team A scores n and team B scores (𝑛 + 5) goals. If the total number of goals
scored by teams A and B is less than 15, write down an inequality in n and solve it. (3 marks)

2009-8.
Solve the inequality 4𝑦 − 1 > 𝑦 + 11. (3 marks)

2007-7.
Solve the inequality 2𝑥 + 10 ≤ 4. (3 marks)

2006-5.
b. A student spent (8y + 40) kwacha on biscuits and 2y kwacha on sweets. If the total amount of
money spent was less than K80, formulate an inequality from this information. (3 marks)

2005-5.
Solve the inequality 4𝑥 − 2 > 𝑥 + 4. (4 marks)

2004-6.
Solve the inequality 3 − 5 (4 marks)

2003-7.
Solve the inequality − ≤ 0 (3 marks)

11 Inequalities - Questions
 12 Linear simultaneous equations
Section A
2012-15.
a. Solve the following simultaneous equations:
𝑥 + 3𝑦 = 11
5𝑥 + 4𝑦 = 22 (5 marks)

2008-15.
The graphs for 2𝑥 + 𝑦 = 5 and 𝑥 + 3𝑦 = 5 intersect at a point R. Without plotting the graphs,
calculate the coordinates of point R. (5 marks)

2004-9.
Solve the simultaneous equations:
𝑦 𝑥= 1
3𝑥 𝑦=5 (4 marks)

2003-14.
Solve the simultaneous equations:
𝑥+𝑦 =3
𝑥 2𝑦 = (5 marks)

Section B
2010-17.
Solve the following simultaneous equations:
2𝑚 𝑛=4
2𝑚 3𝑛 = 4 (5 marks)

2009-17.
b. The cost of two bananas and three mangoes is K28, and five bananas and one mango is K31.
Calculate the cost of each banana and each mango. (7 marks)

2006-18.
A bag contains 29 coins in which. some are 5 tambala coins and others are 2 tambala coins. Let the
number of 5 tambala coins be x and the number of 2 tambala coins be y.
a. Write down an equation involving x and y. (1 mark)
b. If the total amount of money in the bag is 100 tambala, write down another equation involving x
and y. (3 marks)
c. By solving the equations, find the number of coins of each type of coins. (2 marks)

12 Linear simultaneous equations - Questions

 13 Quadratic equations
Section A

2012-10.
A boy is 2 years younger than his sister. The product of their ages is 35. If the age of the boy is x years,
find the value of x. (6 marks)

2011-10.
b. Solve the equation 𝑘 2 = 10𝑘 − 25 (4 marks)

2010-14.
Solve the equation 𝑡 2 + 7𝑡 + 10 = 0. (4 marks)

2009-13.
Factorise 2𝑓 2 − 2 completely. (3 marks)

2006-14.
A woman is x2 years old while her son is x years old. If the sum of their ages is 42, calculate the age of
the woman. (7 marks)

Section B

2009-18.
b. Solve the equation 𝑦 2 + 4𝑦 − 32 = 0. (4 marks)

2007-20.
4
a. Solve the equation 9 − 4𝑡 2 = 0. (5 marks)

2006-16.
2
Solve the equation − = (8 marks)
2

13 Quadratic equations - Questions

2005-18.
b. Figure 8 shows an isosceles triangle ABC in which AB=AC.

If 𝐀𝐁 = (𝑥 + 20) cm and 𝐀𝐂 = 𝑥 2 cm, calculate the value of 𝑥. (6 marks)

2004-18.
Figure 6 shows a rectangle ABCD with AB = (x + 2) cm and BC = x cm.

If the area of the rectangle is 63 cm2, calculate the width of the rectangle. (7 marks)

2003-16.
Mavuto is, 12 years younger than his sister. The product of their ages is 108. How old is Mavuto?
(8 marks)

13 Quadratic equations - Questions

 14 Coordinates
Section A

2006-7.
Figure 1 shows a board used for playing a game.

If the position of X is (8,11) find the position of Y. (2 marks)

2004-14.
Figure 3 shows points Q and R on a plane.

Write down the co-ordinates of the points Q and R. (2 marks)

14 Coordinates - Questions
 15 Linear graph
Section A

2012-12.

Figure 4 shows a straight line graph AB.

Use the graph to find:

a. y coordinate when 𝒙 = −𝟏 (1 mark)
b. the gradient of the line AB. (3 marks)

2010-10.
b. The line 𝑦 = 𝑚𝑥 + 𝑐 has gradient 3 and passes through the point (1, 4). Find the value of c.
(3 marks)

15 Linear graph - Questions

2006-12.
a. Figure 3 shows a graph of a straight line AB which is parallel to the x-axis.

Find the equation of the line AB. (2 marks)

2005-9.
Figure 3 is a straight line graph which was drawn using a table of values.

𝑥 −3 3

𝑦 3

Using the graph, copy and complete the table of values of 𝑥 and 𝑦. (3 marks)

15 Linear graph - Questions

Section B

2011-18.
a. Complete the table of values for 𝑦 = 5 − 3𝑥. (1 mark)
𝒚 = 𝟐𝒙 − 𝟓
𝒙 -1 0 5

𝒚 -7 -5 5

𝒚 = 𝟓 − 𝟑𝒙
𝒙 0 2 5

𝒚 5 -10

b. Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 2 units on the vertical
axis, draw the graphs of 𝑦 = 2𝑥 − 5 and 𝑦 = 5 − 3𝑥. (4 marks)
c. Using the graphs in (b), solve the simultaneous equations
𝑦 = 5 − 3𝑥
𝑦 = 2𝑥 − 5 (4 marks)

2009-16.
Table 2 shows the values of 𝑥 and 𝑦 for the equation 𝑦 = 2𝑥 − 3.
Table 1
𝑥 0 1 2

𝑦 -1

a. Copy and complete the table. (2 marks)

b. Using a scale of 2cm to represent 1 unit on both axes, draw the graph of 𝑦 = 2𝑥 − 3. (4 marks)

2005-20.
a. Copy and complete the table of values for 𝑦 = 5 − 2𝑥 and 𝑦 = 𝑥 − 4.
𝒚 = 𝟓 − 𝟐𝒙
𝑥 −1 2 5

𝑦 7 −1

𝒚=𝒙−𝟒
𝑥 −1 2 5

𝑦 −5 −2

b. Using a scale of 2 cm to represent l unit on the 𝑥-axis and 2 cm to represent 1 unit on the 𝑦-axis,
draw on the same axes, the graphs of 𝑦 = 5 − 2𝑥 and 𝑦 = 𝑥 − 4. (6 marks)
c. Use the graph to solve the simultaneous equations: 𝑦 = 5 − 2𝑥 and 𝑦 = 𝑥 − 4. (2 marks)

15 Linear graph - Questions

 16 Travel graph

Section A

2012-4.
A minibus uses l0 litres of fuel in 70km. Calculate the distance it would cover with 7 litres of fuel.
(3 marks)

Section B

2010-16.
A cyclist left point A at 6:30 am for point C, which is 20 km away. At 7:00 am he stopped at point B for
1
1 2 hours having covered a distance of 12 km. He then continued at a speed of 16 km/h.

a. Using a scale of 2 cm to represent 4 km on the vertical axis and 2 cm to represent 30 minutes on the
horizontal axis, draw the distance-time graph for the cyclist. (5 marks)

b. Using your graph, find the time taken for the cyclist to reach his destination. (3 marks)

2008-20.
A minibus left city H for city Z, 200 km away, at 12:00 noon at a speed of 50 km/h. After traveling for
2 hours, there was a breakdown which took 1 hour to be repaired. The journey then continued at a
constant speed arriving at city Z at 17:30 hours.

a. Using a scale of 2 cm to present 1 hour on the horizontal axis and 2 cm to represent 25 km on the
vertical axis, draw a distance−time graph of the minibus. (5 marks)

b. How far from city H is the minibus at 16:54 hours. (1 mark)

c. Calculate the speed of the minibus after the breakdown. (4 marks)

16 Travel graph - Questions

2007-17.
a. Figure 3 shows a graph ABCD of Mr Phiri’s journey,

(i) How far is B from A? (1 mark)

(ii) Calculate the average speed of Mr Phiri’s journey from C to D. (3 marks)

2006-20.
Train A leaves station P at 6.00 am and travels at a Constant speed of 20 km per hour to station Q
which is 250 km away. Train B leaves station Q at 11.30 am for station P at a constant speed of 50
km/hr.
a. Using a scale of 1 cm to represent l hour on the horizontal axis and 2 cm to represent 50 km on the
vertical axis, draw the graphs on the same axes to show the journeys of the two trains. (8 marks)
b. At what time will the two trains meet? (2 marks)

2004-19.
a. Figure 7 shows a graph of a car travelling from Lilongwe to Blantyre. The car stopped in Dedza for
some time.

(i) What distance did the car travel? (1 mark)

(ii) For how much time was the car in motion? (4 marks)
16 Travel graph - Questions
2003-20.

Figure 8 shows graphs of a car and a bus travelling between two cities X and Y.

Using the graphs, calculate the difference between the speed of the car and the speed of the bus.
(8 marks)

16 Travel graph - Questions

 17 Sets
Section A

2011-11.
Given that
𝜉 = {fanta, coke. sprite, cherry plum, cocopina}
A = {fanta, sprite} and B = {cocopina, cherry plum}

a. Present the information using a venn diagram. (4 marks)

b. Find and state A ∩ B . (1 mark)

2010-9.
Given that ξ= {2, 3, 4, 5, 6, 7, 8, 9}, X = (Prime numbers) and Z = (Odd numbers)

x z

Using the given sets, copy the Venn diagram and fill in the elements. (5 marks)

2009-4. a. Figure 1 shows a venn diagram containing crops grown on farms R and D.

(i) Find the number of crops grown on farm D.

(ii) Write down crops grown on both farms. (3 marks)

2008-3. Given that M and N are sets containing fruits where

M = {tomato, pumpkin, mango, paw-paw, peach}
N = {guava, tomato, banana, mango, paw-paw}
Draw a venn diagram to present this information. (4 marks)

17 Sets - Questions
2007-5. Given Set A = {2,3,4,5,6,7} and Set B = {2,4,6,8} Find A ∩ B. (3 marks)

2006-10. Given set A = {3, 4}, list any three subsets of the set. (4 marks)

2005-8. Given that X and Y are sets containing birds where

set ={ , , , } and
set ={ , , , , , }.
How many elements are in the union X and Y? (2 marks)

2004-10. Given that ε = {3, 4, 5, 6, 7, 9}

A = {4, 6, 8}
B = {3, 6, 9}
Draw venn diagram to represent this information. (4 marks)

2003-13. a. Figure 5 shows a venn diagram illustrating the relationship between sets P and Q.

Write down:
(i) all the elements in P.
(ii) P∩Q (2 marks)

Section B

2012-18.
b. Given that:
P = {2,4,6,8}
Q = {1,2,3,4,5,6,7}.
List down elements in:

(i) P ∩ Q (2 marks)
(ii) P ∪ Q (2 marks)

17 Sets - Questions
 18 Vectors
Section A

2009-11.

−4
Given that ⃗⃗⃗⃗⃗
𝐴𝐵 = ( ) and ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ , find ⃗⃗⃗⃗⃗
𝐶𝐷 = −3𝐴𝐵 𝐶𝐷. (3 marks)
2

2008-8.
𝑥
⃗⃗⃗⃗⃗ = (6) , ⃗⃗⃗⃗⃗
Given that 𝑃𝑄 𝑃𝑅 = (𝑦) and ⃗⃗⃗⃗⃗
1
𝑅𝑄 = ( ) . If ⃗⃗⃗⃗⃗
𝑃𝑄 = ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ , find ⃗⃗⃗⃗⃗
𝑃𝑅 + 𝑅𝑄 𝑃𝑅 . (4 marks)
5 3

2006-9.
If A=(2,3) and B=(5,-2), find
⃗⃗⃗⃗⃗
(i) 𝐴𝐵
⃗⃗⃗⃗⃗ , hence show that 𝐴𝐵
(ii) 𝐵𝐴 ⃗⃗⃗⃗⃗ − 𝐵𝐴
⃗⃗⃗⃗⃗ = 2𝐴𝐵
⃗⃗⃗⃗⃗ (8 marks)

2005-7.
In Figure 2, H and K are points on a vector drawn on a graph paper.

a. Find points H and K in column form.

b. Find the column vector of HK. (5 marks)

2 2
2003-9. Given that vector ⃗⃗⃗⃗⃗
𝐴𝐵 = ( ) and vector ⃗⃗⃗⃗⃗
𝐷𝐶 = ( ) , calculate ⃗⃗⃗⃗⃗ + 𝐷𝐶
(𝐴𝐵 ⃗⃗⃗⃗⃗ ). (3 marks)
3 5

18 Vectors - Questions
Section B

2012-18.

2 −1
a. Given that 𝑎 = ( ) and 𝑏 = ( ), find 𝑎 − 3𝑏. (4 marks)
−3 2

2011-19.
⃗⃗⃗⃗⃗ = (−4) and 𝐸𝐹
b. Given that 𝐶𝐷 ⃗⃗⃗⃗⃗ = 𝐸𝐹
⃗⃗⃗⃗⃗ = ( 𝑎 ). If 2𝐶𝐷 ⃗⃗⃗⃗⃗ , find the value of a. (3 marks)
5 0

2010-18.
a. Figure 4 shows vectors KL and LM.
y
6

4
L M
2

–3 –2 –1 0 1 2 3 4 x

–2

K –4

Figure 4
⃗⃗⃗⃗⃗ + 𝐿𝑀
Find 𝐾𝐿 ⃗⃗⃗⃗⃗⃗ . (3 marks)

2007-19.

⃗⃗⃗⃗⃗ = (1) and Vector 𝐵𝐶

b. Vector 𝐴𝐵 ⃗⃗⃗⃗⃗ = (3). Using a scale of 2 cm to represent 1 unit on both 𝑥 and 𝑦
3 0
⃗⃗⃗⃗⃗ and 𝐵𝐶
axes, draw the vectors 𝐴𝐵 ⃗⃗⃗⃗⃗ on the same axes. (4 marks)

2004-19.

2 2 −3 3
b. If 𝑎 = ( ), 𝑏⃗ = ( ) using these vectors, show that 𝑎 − ( ) = 𝑏⃗ + ( ). (4 marks)
4 −2 1 5

18 Vectors - Questions
 19 Line and angles
Section A

2012-7.
In Figure 2, ABC is parallel to DEF and GH. Angle GHB = 125° and angle BEF = 105°.

Calculate angle DBE. (4 marks)

2010-2.
In Figure 1, straight lines AMC and DMB intersect at M, angle BMC = 130˚ and angle AMD = 5y˚.

A
B

M
5y°
130˚

D
C
Figure 1

Calculate the value of y˚. (2 marks)

19 Line and angles - Questions

2008-5.
In Figure 1 WX and YZ are parallel lines. Angle WAB = 147° and angle CBZ = 62°.

Calculate the values of angles marked m and n. (5 marks)

2007-1.
In Figure 1, ABC is a straight line.

Figure 1
If angle DBC = 80°, calculate the value of 𝑥. (3 marks)

2006-8.
In Figure 2, AB, CD and EF are parallel to one another. Angle HIB =140° and angle GEF = 110°.

Calculate the angles labelled u and w. (7 marks)

19 Line and angles - Questions

2003-8. In Figure3, angle ABF = 78°, angle ACE =2xo angle ECD =xo and BF is parallel to CE.

Calculate angle ACD. (4 marks)

Section B

2005-19.
a. In Figure 9, ABC is a straight line, angle ABH = 54°, angle HBK= 𝑥°, angle KBC= 2𝑥° and angle BCP
= 84°.

(i) Calculate the value of 𝑥.

(ii) Show that lines BK and PC are parallel. (7 marks)

19 Line and angles - Questions

 20 Reflection and rotations
Section A

2010-15.
Figure 3 shows line AB with coordinates A(2, 4) and B(4, 6) rotated through 90˚ angle clockwise to
line A´B´which has coordinates A´(8, 2) and B´(10, 0).

y B
6
A
4 A

2 B
x

–2
Figure 3
a. Copy the figure on the graph paper provided. (2 marks)
b. Find the centre of rotation. (2 marks)

2006-13. Figure 4 shows a nail AB and its image A'B'.

a. Copy the figure on the graph paper provided. (1 mark)

b. Draw the position of the mirror line. (2 marks)

20 Reflection and rotations

2005-11.

Figure 4 is a diagram of an E-shaped object.

On the answer sheet provided, draw the object to show how it will appear after rotating it through an
angle of 270° anti-clockwise. (2 marks)

2004-15.
Figure 4 shows an arrow AB with coordinates A (1, 3) and B (4, 3).

If the arrow is rotated 90° clockwise about Point A, find the new coordinates of B. (3 marks)

20 Reflection and rotations

 Section B

2012-19. Figure 7 shows a quadriìateral ABCD on a graph paper.

On the same graph paper, draw and label the reflection of the quadrilateral ABCD if the mirror line is
MN. (4 marks)

2009-18.
a. Figure 5 shows a triangle ABC on the Cartesian plane.

(i)Copy the figure on graph paper provided.

(ii) If triangle ABC is rotated through 90° clockwise about the point (0,0), draw its image. (4 marks)

20 Reflection and rotations

2008-16.

Figure 3 shows an L−shaped object on a graph.

a. Copy the figure on the graph paper provided. (1 mark)

b. On the same graph, draw and label the reflection of the object if the mirror line is the y−axis.
(3 marks)
c. What are the coordinates of C’ in the reflection? (1 mark)

20 Reflection and rotations

 21 Triangles
Section A

2011-12.
In Figure 2 PQRS is a rectangle.

If SN = RN, show that triangle PSN is congruent to triangle QRN. (4 marks)

2011-13.
In Figure 3 UQ is parallel to SQ, QT = QR, angle PQU = 68° and angle TRS = 18°.

Calculate angle QTV. (5 marks)

2009-12.
In Figure 2 HC and DG are straight lines. Angle CBF = 92°, angle ABE = 35° and AB is parallel to DG.

Calculate angles marked x and y. (6 marks)

21 Triangle - Questions
2009-14.
b. In Figure 3 N lies on a side of the square EGXW. EM and XL are perpendicular to WN. Angle EWM
= angle WXL.

Show that triangle WME and XLW are congruent. (4 marks)

2007-14.
b. In Figure 2, PQRS is a parallelogram. QT meets SR produced at T.

If SR=RT, prove that triangle PSR is congruent to triangle QRT. (4 marks)

2004-1. In an acute angled triangle PQR, angle P = 65° and angle Q = 50°. Find the size of angle R.
(4 marks)

21 Triangle - Questions
 Section B

2006-17. In Figure 5, LK is parallel to MS and LM=MS

If KM and LS are straight lines, angle KLS=30° and angle KML=70°,

a. calculate the value of the angle KMS. (5 marks)
b. what type of triangle is KLM? (3 marks)

2005-16. Figure 7 shows a rectangular floor AB CD in which N is the mid-point of AB.

If ND = NC and angle DNC = 70°, calculate the value of the angle AND. (5 marks)

2003-18. a. In Figure 6, QL= PM, LM= MN and angle QLM = angle NML = 90°.

Prove that triangles LMQ and MNP are congruent. (4 marks)

21 Triangle - Questions
 22 Convex polygons
Section A

2011-6.
The sum of the interior angles of a regular convex polygon is 1440°. Calculate the size of
each-exterior angle of the polygon. (5 marks)

2010-7.
b. The sum of interior angles of a regular polygon is 1980°. Calculate the number of sides of the
polygon. (4 marks)

2007-12.
The sum of the interior angles of a regular convex polygon is 2340°. Calculate the exterior angle of the
polygon. (6 marks)

2006-15.
The sum of the interior angles of a regular convex polygon is l620°. Calculate the number of sides of
the polygon. (5 marks)

2005-14.
The interior angle of a regular convex polygon is 168°. Calculate the number of sides of the polygon.
(4 marks)

2004-13.
A convex polygon has 21 sides. Find the sum of the interior angles of the polygon. (4 marks)

2003-4.
In Figure 2, ABCD is a quadrilateral in which angle ABC I 96°, angle BCD = 105° and angle BAD=I 73°.
'

Calculate the size of angle ADC. (3 marks)

22 Convex polygons - Questions

Section B

2012-16.
Figure 6 shows a pentagon ABCDE in which angle ABC =3y°, angle BCD = 100°, angle CDE = 114°,
angle DEA = y° and angle EAB = 126°.

Calculate the size of angle DEA. (5 marks)

2008-17.
In Figure 4, ABCDE is a convex pentagon in which angle BCD = 85°, angle CDE = 115° and angle DEA
= 120°.

a. If EA and CB are produced to meet at F, calculate angle AFB. (3 marks)

b. Given that angle ABC = 110°, what type of triangle is AFB? (3 marks)

2003-18.
b. Each interior angle of a regular convex polygon is two times the exterior angle. Calculate the
number of sides of the polygon. (4 marks)

22 Convex polygons - Questions

 23 Similarity
Section A

2012-13. b. In Figure 5, ABCD is a square.

If angle BEA = Angle DFA, Show that triangle BEA is similar to triangle DFA. (3 marks)

2011-9. In Figure 1 triangle AEB is similar to triangle CED.

𝐀𝐂 = 𝑥 cm, 𝐂𝐄 = 4 cm, 𝐁𝐃 = 6 cm and 𝐃𝐄 = 9 cm.

Calculate the value of 𝑥. (5 marks)

2010-8. In Figure 2, ABDE is a rectangle, angle AFE = angle BCD.

A B

F E D C
Figure 2

Show that triangle AFE is similar to triangle BCD. (4 marks)

23 Similarity - Questions
2005-15.
b. Figure 6 shows two similar triangles DCE and NCM. DC=10cm, CE=12 cm and CM=30cm.

Calculate the length of CN. (4 marks)

2004-12.
In Figure 2, triangle PQT is similar to triangle PRS in which RS 6 cm,QT =4 cm and PT=3 cm.

Calculate the length of PS. (4 marks)

Section B
2008-18. b. In Figure 5 triangle ABC is similar to triangle BDC. Angle BAC = angle DBC, BD = 9 cm,
BC = 10 cm and DC = 6 cm.

Calculate the length of AB. (4 marks)

23 Similarity - Questions
2007-19. a. Figure4 shows two similar triangles ABC and ZYX. Angle BAC = angle XZY, angle ACB =
angle YXZ, AB = 3 cm, AC=7 cm and YZ = 15 cm.

Calculate the length of ZX. (4 marks)

2003-19.
a. In Figure 7, angle BAC= angle CBD, AB = 8 cm, BD = 6 cm, AD= 4 cm, and BC=7 cm.

(i) Prove that triangles CAB and CBD are similar.

(ii) Calculate, the length of CD. (6 marks)

23 Similarity - Questions
 24 Pythagoras’s theorem
Section A
2011-15.
Figure 4 is a right angled triangle PQR in which PQ = d cm, QR = 8 cm and PR = (d + 4) cm.

Calculate the value of d. (4 marks)

2003-11.
In a triangle ABC, AB=7cm, AC= 3 cm and angle ACB=90°. Calculate the length of BC giving your
answer correct to 2 significant figures. (5 marks)

Section B

2009-17.
a. Figure 4 shows a parallelogram BCFD and triangle AEF. AC = 10 cm, BD = 5 cm, EF = 3 cm and
angle AEF= 90°.

Calculate AE. (5 marks)

2007-18.
b. In a triangle XYZ, 𝐗𝐘 = 3 cm, 𝐘𝐙 = 4 cm and 𝐗𝐙 = 5 cm. Show that triangle XYZ is right angled.
(5 marks)

24 Pythagoras’s theorem - Questions

 25 Quadrilaterals

Section A

2012-9.
Figure 3 shows a parallelogram RSTU in which angle RUT = (2𝑥 + 30)° and angle RST = (7𝑥 − 25)°.

Calculate the size of angle RUT. (5 marks)

2005-12.
Figure 5 shows a trapezium GHJK in which GH is parallel to KJ, HJ=HR and angle HJR= angle GKR.

Prove that GHRK is a parallelogram. (4 marks)

25 Quadrilaterals - Questions
Section B

2010-20.
b. The diagonals of a rhombus ABCD are 12cm and 16cm. Calculate the area of the rhombus.
(5 marks)

2008-19.
b. In Figure 6 WXYZ is a parallelogram in which WX is parallel to ZY and ZW is parallel to YX. Angle
WZY= (𝑎2 − 2𝑎)°and WXY = 63°.

Calculate the values of a. (6 marks)

2004-20.
b. The diagonals of a rhombus are 10 cm and 24 cm. Calculate the length of a side of the rhombus.
(5 marks)

25 Quadrilaterals - Questions
 26 Mensuration

Section A

2007-14.
22
a. Calculate the volume of a cylinder whose radius is 7 cm and height 10 cm. (Take π = 7
). (3 marks)

2003-10.
Figure 4 shows a cylinder whose diameter is 14 cm and height is 10 cm.

22
Taking π to be , calculate the volume of the cylinder. (3 marks)
7

Section B

2009-19.

Figure 6 shows a closed cylinder with diameter 7 cm and height 20 cm.

22
Calculate the total surface area of the cylinder. (Take π = 7
). (6 marks)

26 Mensuration - Questions
2004-16.

Figure 5 shows a right pyramid with a rectangular base ABCD measuring 8 cm by 6 cm.
The heights of triangles VBC and VAB are 12 cm and 14 cm respectively.

Calculate the total surface area of the right pyramid. (6 marks)

26 Mensuration - Questions
 27 Geometrical constructions
Section A

2007-15.
a. Using a ruler and a pair of compasses only, construct triangle PQR in which PQ= 11cm, QR =
9 cm and PR = 7 cm. (3 marks)

Section B

2012-20.
a. Using a ruler and a pair of compasses only, Construct in the same diagram:
(i) a triangle XYZ in which XY = 9 cm, YZ = 11 cm and ZX = 7 cm.
(ii) the inscribed circle of the triangle XYZ. (5 marks)
b. Measure and state the length of the radius of the circle. (1 mark)

2011-20.
Using a pair of compasses and a ruler only, construct in the same diagram:
(i) a line AB which is 10 cm.
(ii) point C which is equidistant from point A and point B and is also 6 cm from AB.
(iii) measure and state angle BAC. (5 marks)

2010-19.
a. Using a ruler and a pair of compasses only, construct in the same diagram:
(i) a triangle PQR in which PQ = QR = PR = 5cm.
(ii) construct the circumscribed circle of the triangle PQR. (7 marks)

2009-20.
Using a ruler and a pair of compasses only, construct in the same diagram:
a. A triangle UVW in which angle WUV = 90°, UV = 5 cm and UW = 6 cm. (3 marks)
b. Produce line UV to point X, and line UW to point Y. Bisect angles YWV and XVW. The
bisectors should meet at Z. (4 marks)
c. Measure and state the length of line UZ. (1 mark)

27 Geometrical constructions - Questions

2008-18.
a. Using a pair of compasses and a ruler only, construct in the same diagram:
(i) lines AB and AC such that AB = 8 cm, AC = 6 cm and angle BAC = 60°.
(ii) the locus of a point equidistant from lines AB and AC. (6 marks)

2006-19.
Using a ruler and a pair of compasses only, construct in the same diagram:
a. a line AB = 10 cm (1 mark)
b. a locus of points equidistant from points A and B. Let P be a point on the locus such that
AP=8 cm (3 marks)
c. measure and state angle APB. (2 marks)

2004-20.
a. Using a ruler and a pair of compasses only, construct triangle ABC in which AB = 9.0 cm,
angle BAC = 150° and CA = 4.5 cm. Measure and state angle ABC. (6 marks)

2003-17.
Using a ruler and a pair of compasses only, construct in the same diagram:
a. triangle PQR in which PQ = 8 cm, QR= 7 cm and PR= 6 cm. (4 marks)
b. a locus of points equidistant from PR and PQ. If point S is on the locus such that PS=10 cm,
measure and state the length of QS. (4 marks)

27 Geometrical constructions - Questions

SOLUTIONS
 1 Basis of arithmetic
2007-2.
Section A 8(−5) + 2(3) − 5 = −40 + 6 − 5
2012-1. = −39…Answer
(−7) + 5(−4) (−7) + (−20)
=
−3 −3 2006-1.
−27 (−10)(2) (−4)
= = 9 … Answer
−3 = (−20) (−4) = 5…Answer

2012-11. 2006-4.
a.
8 ) 1024
326 = 3 × 61 + 3 × 60
8 ) 128 …0
= 18 + 3 × 1 = 18 + 3 = 2110 Read the remainders
8 ) 16 …0
∴ 27810 − 2110 = 25710…Answer from down to up.
8)2 …0
0 …2
2010-10.
a. ∴ 102410 = 20008…Answer
2)9
2 ) 4 …1 2005-3.
Read the remainders 32112 = 3 × 122 + 2 × 121 + 1 × 120
2 ) 2 …0
from down to up.
2 ) 1 …0 = 432 + 24 + 1 = 457…Answer
0 …1
2004-5.
∴ 10012…Answer a. 1123 + 1023 = 2213…Answer

2009-3. 2003-1.
21103 − 22023 + 10213 (−4) − (−5) = −4 + 5 = 1…Answer
= 21103 + 10213 − 22023
2110 10201 2003-3.
+ 1021 －2202 12647 = 1 × 73 + 2 × 72 + 6 × 71 + 4 × 70
10201 222 ∴ 2223 …Answer = 343 + 98 + 42 + 4 = 48710…Answer

2008-1.
(−5) × (−3) + 6 15 + 6 21 3 Section B
= = = − = −3
(−7) −7 −7 1
2011-16.
2008-6. b. 1405 = 1 × 52 + 4 × 51 + 0 × 50
33(3×3×3) 32(3×3) 31(3) 30(1) = 25 + 20 + 0
1 2 0 1 = 4510…Answer
12013 = 1 × 30 + 0 × 31 + 2 × 32 + 1 × 33
= 1 × 1 + 0 × 3 + 2 × 9 + 1 × 27
= 4610 … Answer

1 Basis of arithmetic - Answers

 2 Indices and logarithms
2010-11.
Section A
3
32 𝑒 4 ÷ 3𝑒 ;4 = 9𝑒 4 ÷
2012-13. 𝑒4
a. 𝑒4
= 9𝑒 4 × = 3𝑒 8 … Answer
1 3
𝑚0 + 492 = 1 + √49 = 1 + 7

= 8…Answer 2009-9.
1 1
log(51.8)3 = × log(51.8)
2012-14. 3
b. log 51.8 = 1.7143
log 47.89 = 1.6802 (from the table ”LOGARITHMS)
∴ log √47.89 = (log 4.789) ÷ 2 1 1
log(51.8)3 = × 1.7143 = 0.571433
= 1.6802 ÷ 2 = 0.8401 3
∴ √47.89 = 10log √47.89 = 100.8401 = 0.6…Answer
= 6.9203 ≈ 6.92(2d.p.)…Answer
2009-10.
2011-3. a.
1 1 1 1
(3𝑘)2 − (4𝑘 4 )2 = 9𝑘 2 − 2𝑘 2 (𝑝2 𝑞2)2 𝑝2×2 𝑞2×2 𝑝𝑞
= =
𝑞 𝑞 𝑞
= 7𝑘 2…Answer
= 𝑝 … Answer
2011-14.
b. 2008-2.
1391.0 𝑦(𝑦 ;2 )5 = 𝑦(𝑦 ;2×5 ) = 𝑦(𝑦 ;10 )
log = log 1391.0 − log 17.7
17.7 = 𝑦 × 𝑦 ;10 = 𝑦 1 × 𝑦 ;10
In logarithm tables, = 𝑦 ;9…Answer
log 13.91 = 1.1433
∴ log 1391.0 = log(100 × 13.91) 2008-13.
= 2 + 1.1433 = 3.1433 log(4.56)3 = 3log(4.56).
log 17.7 = 1.2480 According to the log table 45.6 is 6.590.
∴ log 1391.0 − log 17.7 = 3.1433 − 1.2480 log(45.6) = 6590
= 1.8953 log(4.56) = 0.6590
log 78.57 = 1.8949 + 0.0004 = 1.8953 log(4.56)3 = 3 × 0.6590 = 1.977…Answer
1391.0
∴ = 101.8956 = 78.63
17.7 2007-8.
= 79(to 2s. f)…Answer log 66.79 = 1.8247
∴ log √667.9 = (log 667.9) ÷ 2
= (1 + 1.8247) ÷ 2 = 1.41235
∴ √667.9 = 101.41235 = 10 × 100.41235
= 25.84…Answer

2 Indices and logarithms - Answers

2007-9. 2007-16.
2 2 3 36.7
2𝑚(3𝑚) = 2𝑚 × 9𝑚 = 18𝑚 …Answer 1
log = log 36.7 − log 2.82
√2.8
2006-11. 1
= log 36.7 − log 2.8
log 41.3 = 1.6160 2
∴ log(4.13)2 = log(41.3 − 1) × 2 = 1.232 log 36.7 = 1.5674, log 28.0 = 1.4314
(4.13)2 = 101.232
= 10 × 10 0.232
1
∴ log 2.8 = log ( × 28.0)
= 10 × 1.706 = 17.06 10
≈ 17.1(to 3 s. f. )…Answer 1
= log + log 28.0 = log 10;1 + log 28.0
10
2005-2. = −1 + 1.4472 = 0.4472
1 1 1 1
643 + 22 = (26 )3 + 22 ∴ log 36.7 − log 2.8 = 1.5674 − × 0.4472
2 2
1 = 1.5674 − 0.2236 = 1.3438
= 26×3 + 22 = 22 + 22 36.7
∴ = 101.3438 = 22.0191 …
= 4 + 4 = 8…Answer √2.8
≒ 22…Answer
2003-12.
41 × 672 2005-17.
log10 ( )
161 𝑎2 × 𝑎;4 ÷ 𝑎 = 𝑎2:(;4);1
= log10 41 × log10 672 − log10 161 = 𝑎;3…Answer
= log10 41 × 2log10 67 − log10 161
=(log10 10 + log10 4.1) + 2(log10 10 + log10 6.7) 2005-19.
− (log10 100 + log10 1.61) b.
=1.6128 + 2 × 1.8261 − 2.2068 = 3.0582 log 1.25 = 0.0969
Antilogarithm of 0.0582 = 1.144 1
log √12.5 = log 12.5
Antilogarithm of 3.0582 = 1144…Answer 2
1
= × (1 + 0.0969) = 0.54845
2
Section B
∴ √12.5 = 100.54845
2010-18. = 3.535 ≈ 3.54(2d. p. )..Answer
log(2.7 × 40.3) = log 2.7 + log 40.3
From logarithm tables, log 27 = 1.4314 and 2004-17.
log 40.3 = 1.6053 Using the logarithm table
∴ log 2.7 = 0.4314 log 1.23 = 0.0899 ∴ log 123 = 2.0899
log 2.7 + log 40.3 = 0.4314 + 1.6053 = 2.0367 log 2.73 = 0.4362 ∴ log 27.3 = 1.4362
2.0367 log 2.631 = 0.4202
∴ 2.7 × 40.3 = 10
log 10.88 = 1.0334 + 0.0033 = 1.0367 123 × 27.3
log = log 123 + log 27.3 − log 2.631
∴ log 108.8 = 2.0367 2.631
∴ 2.7 × 40.3 = 102.0367 = 108.8 = 2.0899 + 1.4362 − 0.4202 =3.1059
= 110 (to 2 s. f. ) Using the antilogarithm table
100.1059 = 1.274 + 0.003 = 1.277
123×27.3
∴ = 103.1059 = 1277…Answer
2.631

2 Indices and logarithms - Answers

 3 Social and commercial arithmetic
2005-10.
Section A The value of the car at the beginning of the first
2012-6. year = 95% of K540,000 = K513,000
15 The value of the car at the beginning of the
surtax of a shirt = K400 × = K60
100 second year = 95% of K513,000
total amount of money
= K487,350…Answer
= K400 + K60 + 2 × K120
= K460 + K240 = K700…Answer
2004-8.
K24400 = K4800 + K9600 + K10000
2010-4.
0 of K4800 = K0
Depreciation 100% − 7% = 93%
10 of K9600 = K960
Let the original value of the bicycle be 𝑥
25 of K10000 = K2500
93
𝑥× = 9300 the total income tax = K960 + K2500
100
= K3460…Answer
93 100 100
𝑥× × = 9300 ×
100 93 93
𝑥 = 10000 Section B
∴ K10,000…Answer
2011-16.
a.
2009-5.
Let the amount of policies sold be 𝑥.
electricity bill = unit cost × number of units +
20
fixed charge 𝑥× = K2000
100
= K14.50×192 + K350.00
100
= K2784.00 + K350.00= K3134.00…Answer 𝑥 = K2000 × = K10000
20
The amount of policies sold is K10000…Answer
2008-7.
The K 17 500 of K 60 000 is not taxable.
2007-17.
Taxed income = K 60 000 − K 17 500 = K 42 500
b.
Income tax = rate × taxable salary
otal sell = K60 × 1000 = K60,000
15
15 % × K 42 500 = × K42500 Commission = K60,000 × 10%
100
10
= 15 × 425 = 6375 = K60,000 × = K6,000 … ns er
100
K 6375 per month
K 6375 × 12 month = K 76 500…Answer
2003-19.
b.
2006-12.
K800 − K720 = K80
b. If K441.00 includes tax at 5%, then
80
105% of basic cost = K441.00 he discount percentage = × 100
800
∴ he original price of the product
= 10%...Answer
100
= K441.00 × = K420.00 … ns er
105

3 Social and commercial arithmetic - Answers

 4 Number patterns
2007-15.
Section A
NUMBER PATTERN TOTAL
2012-15. 1 1 1
b. 2 1+2+1 4
P − (−8) = 12 − P 3 1+2+3+2+1 9
P + 8 = 12 − P 4 1+2+3+4+3+2+1 16
P + P = 12 − 8 5 1+2+3+4+5+4+3+2+1 25
2P = 4
P = 2…Answer
2006-2.
𝑎5 = 52 + 5
2010-12.
= 25 + 5 = 30…Answer
This sequence is an AP with the common
difference (+4).
2004-5.
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑, where
b. 92
𝑎1 = 7, 𝑎𝑛 = 283 and 𝑑 = 4
∴ 283 = 7 + (𝑛 − 1)4
2003-2.
−4𝑛 = −280
a.
𝑛 = 70
∴ The number of terms is 70…Answer

2009-6.
1st term = 1 + (1 + 1)
2nd term= 2 + (2 + 1)
3rd term= 3 + (3 + 1)
The number of dots of the 4th triangle is
nth term= 𝑛 + (𝑛 + 1) = 2𝑛 + 1…Answer
15…Answer

2008-10.
17 − (−7) = 17 + 7 = 24 Section B
24 2011-19.
7+ = −7 + 12 = 5
2
a.
∴ 𝑏 = 5…Answer
(i)
𝑎1 = 3 × 1 − 2 = 1
2007-13.
𝑎2 = 3 × 2 − 2 = 4
2n − 1 = 11
𝑎3 = 3 × 3 − 2 = 7
2n = 12
∴ The first three terms are 1,4,7…Answer
n=6
(ii)
∴ The number of terms is 6…Answer
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑
The nth term of a sequence is 3𝑛 − 2…Answer

4 Number patterns - Answers

 5 Density and mixture
2005-13.
Section A The weight of petrol = 55 − 20 = 35g
2012-11. The weight of water = 70 − 20 = 50g
2
b.Volume of cylinder = π𝑟 ℎ 35
The relative density = = 0.7 … Answer
22 14 14 50
=( × × × 10) cm3
7 2 2
= (22 × 7 × 10)cm3 2003-15.
3 The volume of the container = 645 − 105
= 1540cm
Mass = 540cm3
Density =
volume The mass of liquid H = 915 − 105
3080 g = 810g
= = 2 g/cm3 … Answer
1540 cm3 810g
The Density of liquid H =
540cm3
2011-7. = 1.5g⁄cm3…Answer
101g − 77g = 24g is the mass of the water.
107g − 77g = 30g is the mass of the glycerin.
The relative density of the glycerin is
Section B
30g 2011-17.
= 1.25 … Answer
24g b.
Let the height of the cylinder be 𝑥.
2010-13. 7 2
( ) × π × 𝑥 = 385
Density 2
Mass = = 0.7 × 1800
Volume 49 22
× × 𝑥 = 385
= 1260g…Answer 4 7
77
𝑥 = 385
2008-4. 2
Relative density = Mass of substance/Volume of 2
𝑥 = 385 × = 10
substance 77
Volume of mathematical instrument box: x cm3 ∴ The height is 10cm …Answer
273
2.1 =
𝑥 2010-19.
273 b.
𝑥= = 130
2.1 Let white beans is 𝑥 kg and red beans is
∴130cm3…Answer (1 − 𝑥)kg.
146 × 𝑥 + 170 × (1 − 𝑥) = 152
2007-10. −24𝑥 = −18
White suger cost + Brown suger cost 3 3 1
∴ 𝑥 = ,1 − 𝑥 = 1 − =
= K200 + K270 = K470 4 4 4
The cost of the mixture per kg 3 1
∴ white: red = : = 3: 1 … Answer
K470 4 4
= = 𝐾94⁄kg … Answer
5kg
5 Density and mixture - Answers
 6 Proportions
2006-6.
Section A 25 litre = 25000ml
2011-10. The number of bottles which can be used to fill
a. the drum is
4 × 30 120 25000
= = 20days … Answer = 500 … Answer
6 6 500

2009-10. 2005-6.
b. 30kg ÷ 6 people = 5 kg⁄person
percentage change 5 kg⁄person × (6 + 2) people = 40kg…Answer
(original value − new value)
= × 100
original value 2004-7.
700 − 616 84 Let 𝑥 be the angle represented by the sector C.
= × 100 = × 100
700 700 70° + 140° + 𝑥 = 360°
= 12%…Answer 𝑥 = 150°
150°
the percentage = × 100
2008-14. 360°
Expenditure on food is 30 % so multiply 125
= % ≈ 41.7%(1d. p. ) … Answer
monthly salary by 30 %. 3
30
K38000 × 30 % = 38000 × = 380 × 30
100
= K11400…Answer
Section B
2005-18.
2007-11. a.
Let the total number of pupils at the school be 𝑥. Let the share of Chimwemwe = 𝑥
𝑥 × 60% = 48 then the share of Chikondi = 𝑥 + 80
60 𝑥 + (𝑥 + 80) = 540
𝑥× = 48
100 𝑥 = 230
100 The share of Chimwemwe is K230.
𝑥 = 48 ×
60 The share of Chikondi is K230 + K80 = K310.
𝑥 = 80 …Answer
∴ The total number of pupils are 80 …Answer

6 Proportions - Answers
 7 Probability

Section A
2012-8. 2009-7.
a. (total successful outcomes)
Favourable outcome P(success) =
(total possible outcomes)
P(6) =
Total possible outcome P(picking a white bean)
2 1
= = … Answer 20 20
6 3 = =
20 + 30 + 10 60
2011-14. 1
= … Answer
a. 3
P(picking a letter N)
number of letter N 2007-6.
=
total number of letters P(choosing the letter A)
2 1 number of the letter A
= = … Answer =
10 5 total number of letters
2
= … Answer
2010-6. 5
P(picking a blue counter)
number of blue counters 2003-5.
=
total number of counters b.
9 9 The probability of choosing tail
= =
6 + 9 15 1
= … Answer
3 2
= … Answer
5

7 Probability - Answers
 8 Statistics
2011-17.
Section A a.
2010-5. Football = 360° − 126° − 54° − 42° = 138°
sum of values ∴ Students who play football
Mean =
number of values 138°
= 60 × = 23 … nswer
7 + 11 + 6 + 7 + 9 40 360°
= =
5 5
= 8…Answer 2008-19.
a.
2009-2. 15 + 20 + 12 + 10 + 𝑘
= 14
mean = sum of values ÷ number of values 5
sum of values = mean × number of values 57 + 𝑘
= 14
sum of values = 400 × 3 = 1200 5
the mass of the third packet 57 + 𝑘 = 70
= 1200 − 900 = 300g…Answer 𝑘 = 70 − 57 = 13…Answer

2006-5. 2007-18.
a. a.
put the number in order 1,2,3,4,4,4,7
5
Median = the mean of the two middle values
3+4 4
= = 3 5 … nswer
Number of students

2
3
2005-4.
b. 2 + 1 + 6 + 10 + 6 = 25…Answer 2

1
Section B
0
2012-17. 6 7 8 9 10
Marks of students
12

10
Number of Pupils

0
May June July August
Month of Birth

8 Statistics - Answers
 9 Algebraic expressions
2010-7.
Section A a.
2012-3. 2𝑐 − 4 2(𝑐 − 2)
=
2𝑒 − 𝑓 + 𝑔 = 2(4) − (−2) + (−5) 𝑐 − 4 (𝑐 + 2)(𝑐 − 2)
2

= 8 + 2 − 5 = 10 − 5 2
= … Answer
= 5…Answer 𝑐+2

2012-5. 2009-1.
30ef = 2 × 3 × 5 × e × f
2𝑝𝑟 2 × (−1) × (−4) 8
10e2f = 2 × 5 × e × e × f = =
𝑡 −2 −2
5f 2 = 5 × f × f = −4…Answer
LCM = 2 × 3 × 5 × e × e × f × f
= 30e2f 2…Answer 2009-4.
b.
2012-14. 𝑕 = 𝑘 − (𝑔 + 𝑒)
a. 𝑕 = 𝑘−𝑔−𝑒
L = ra + n ∴ 𝑔 = 𝑘 − 𝑒 − 𝑕…Answer
L − n = ra
L − n ra 2009-14.
=
a a a.
L−n 𝑥 + 3𝑦 − (−𝑥) − 4𝑦
r= … Answer
a = 𝑥 + 3𝑦 + 𝑥 − 4𝑦 = 𝑥 + 𝑥 + 3𝑦 − 4𝑦
= 2𝑥 − 𝑦…Answer
2011-1.
2𝑘 − 7𝑓 − 4(2𝑘 − 3𝑓) = 2𝑘 − 7𝑓 − 8𝑘 + 12𝑓 2008-9.
= −6𝑘 + 5𝑓…Answer 𝑥𝑘 − 𝑥𝑚 − 𝑘𝑧 + 𝑚𝑧
= (𝑥𝑘 − 𝑥𝑚) − (𝑘𝑧 − 𝑚𝑧)
2011-5. = 𝑥(𝑘 − 𝑚) − 𝑧(𝑘 − 𝑚)
4𝑎𝑟 2 = 2 × 2 × 𝑎 × 𝑟 × 𝑟 = (𝑘 − 𝑚)(𝑥 − 𝑧) …Answer
6𝑎2 𝑏𝑟 = 2 × 3 × 𝑎 × 𝑎 × 𝑏 × 𝑟
8𝑎𝑐 2 𝑟 = 2 × 2 × 2 × 𝑎 × 𝑐 × 𝑐 × 𝑟 2008-11.
∴ HCF = 2 × 𝑎 × 𝑟 = 2𝑎𝑟…Answer 4𝑥 (4𝑥 − 4) 4𝑥 5
÷ = ×
5 5 5 (4𝑥 − 4)
2010-1. 4𝑥 4𝑥
𝑎2 − 𝑏 = (−2)2 − 5 = 4 − 5 = =
4𝑥 − 4 4(𝑥 − 1)
= −1…Answer 𝑥
=
𝑥−1

9 Algebraic expressions - Answers

2007-3. 2004-4.
(𝑥 + 𝑦) (2 + 4) 6 2 3 2 𝑥−2 3 𝑥
= = − = × − ×
(𝑥 − 𝑦) (2 − 4) −2 𝑥 𝑥−2 𝑥 𝑥−2 𝑥−2 𝑥
= −3…Answer 2(𝑥 − 2) − 3𝑥 −𝑥 − 4
= = … Answer
𝑥(𝑥 − 2) 𝑥(𝑥 − 2)
2007-4.
2𝑥 = 2 × 𝑥 2004-11.
12𝑥𝑦 = 2 × 2 × 3 × 𝑥 × 𝑦 substitute a = 4, b = −2 and c = −3 into
18𝑥 2 𝑦 = 2 × 3 × 3 × 𝑥 × 𝑥 × 𝑦 a(b + c):
∴ L. C. M. = 2 × 2 × 3 × 3 × 𝑥 × 𝑥 × 𝑦 4 × *(−2) + (−3)+ = 4 × (−5)
= 36𝑥 2 𝑦 … Answer = −20…Answer

2006-3. 2003-2.
2
𝑥 + 𝑥𝑦 = 𝑥(𝑥 + 𝑦) …Answer b.4 − 𝑥 2 = (2 − 𝑥)(2 + 𝑥) …Answer

2005-1. 2003-5.
2b − ac = 2 × 4 − 3 × (−2) a.
= 8 − (−6) = 8 + 6 4𝑚2 = 2 × 2 × 𝑚 × 𝑚
= 14…Answer 6𝑚𝑛2 = 2 × 3 × 𝑚 × 𝑛 × 𝑛
10𝑚𝑛 = 2 × 5 × 𝑚 × 𝑛
2005-15. The HCF is 2 × 𝑚 × 𝑛 = 2𝑚𝑛…Answer
a. −5(2𝑝 − 3𝑞) = −10𝑝 + 15𝑞…Answer
2003-6.
2004-2. Substitute − 5 for 𝑥, −2 for 𝑦, and 2 for 𝑚.
2 2
(3𝑑 − 2) = 9𝑑 − 12𝑑 + 4…Answer 3𝑥𝑦 2 3 × (−5) × (−2)2
=
2𝑚 2×2
2004-3. = −15…Answer
3 2
3 𝑥 𝑦𝑧 = 3 × 3 × 3 × 𝑥 × 𝑥 × 𝑦 × 𝑧
32 𝑥𝑧 3 = 3 × 3 × 𝑥 × 𝑧 × 𝑧 × 𝑧 2003-13.
∴ HCF = 3 × 3 × 𝑥 × 𝑧 b.
2 (2𝑏 + 4)(𝑑 − 3𝑏𝑑) = 2𝑏𝑑 − 6𝑏2 𝑑 + 4𝑑 − 12𝑏𝑑
= 3 𝑥𝑧…Answer
= 4𝑑 − 6𝑏2 𝑑 − 10𝑏𝑑…Answer

9 Algebraic expressions - Answers

 10 Linear equations
2008-12.
Section A 5𝑥 − 𝑦 = 8
2012-8. −𝑦 = 8 − 5𝑥
b. 𝑦 = 5𝑥 − 8
𝑃+1 3 Therefore,
+𝑃 =
5 5 (i) -8
𝑃 + 1 + 5𝑃 = 3 (ii) 5
6𝑃 = 3 − 1
6𝑃 = 2 2005-4.
1 a.
𝑃 = … Answer
3
5𝑚
=𝑚−2
3
2011-2. 5𝑚
2 3× = 3 × (𝑚 − 2)
=2 3
𝑟−1 5𝑚 = 3𝑚 − 6
2 𝑚 = −3…Answer
× (𝑟 − 1) = 2(𝑟 − 1)
(𝑟 − 1)
2 = 2𝑟 − 2
2𝑟 = 4 Section B
𝑟 = 2…Answer 2010-20.
a.
2011-8. 𝑟+2 1
50 − (𝑎 + 5) − (2𝑎 − 3) = 30 − =1
3 2
50 − 𝑎 − 5 − 2𝑎 + 3 = 30 𝑟+2 3
48 − 3𝑎 = 30 =
3 2
−3𝑎 = −18 𝑟+2 3
𝑎 = 6…Answer ×6= ×6
3 2
2𝑟 + 4 = 9
2009-15. 2𝑟 = 5
Let the width of the fence be W. 5 1
Let the length of the fence be L. 𝑟= = 2 … Answer
2 2
𝑊 =𝐿−4
𝐿 =𝑊+4
perimeter = 2(𝑊 + 𝐿)
∴ 36 = 2*𝑊 + (𝑊 + 4)+
(substitute perimeter = 36 and 𝐿 = 𝑊 + 4)
36 = 2(2𝑊 + 4)
36 = 4𝑊 + 8
−4𝑊 = −28
𝑊=7
the width of the fence=7metres …Answer
10 Linear equations - Answers
 11 Inequalities
2006-5.
Section A b.
2012-2. (8𝑦 + 40) + 2𝑦 < 80
𝑥 ≥ 3…Answer 10𝑦 + 40 < 80
10𝑦 < 40
2011-4. 𝑦 < 4 … Answer
𝑏 + 1 ≥ 13 − 3𝑏
4𝑏 ≥ 12 2005-5.
𝑏 ≥ 3…Answer 4𝑥 − 2 > 𝑥 + 4
3𝑥 > 6
2010-3. 𝑥 > 2…Answer
𝑛 + (𝑛 + 5) < 15
2𝑛 < 15 − 5 2004-6.
2𝑛 < 10 3−𝑥 <5
∴ 𝑛 < 5…Answer Take 3 from both sides:
−𝑥 < 2
2009-8. Multiply both sides by − 1:
4𝑦 − 1 > 𝑦 + 11 𝑥 > −2…Answer
4𝑦 − 𝑦 > 11 + 1
3𝑦 > 12 2003-7.
𝑦 > 4…Answer 𝑚 2+𝑚
− ≤0
2 3
2007-7. 𝑚 2+𝑚
6×( − )≤6×0
2𝑥 + 10 ≤ 4 2 3
2𝑥 ≤ −6 3𝑚 − 2(2 + 𝑚) ≤ 0
𝑥 ≤ −3…Answer 3𝑚 − 4 − 2𝑚 ≤ 0
𝑚 ≤ 4…Answer

11 Inequalities - Answers
 12 Linear simultaneous equations
3𝑥 − (𝑥 − 1) = 5
Section A 3𝑥 − 𝑥 + 1 = 5
2012-15. 𝑥=2
a. b i e he 𝑥 a e i e ai (1)
𝑥 + 3𝑦 = 11 (× 4) 𝑦 − 2 = −1
5𝑥 + 4𝑦 = 22 (× 3) 𝑦=1
∴ 𝑥 = 2, 𝑦 = 1 …Answer
4𝑥 + 12𝑦 = 44
−)15𝑥 + 12𝑦 = 66 2003-14.
−11𝑥 = −22 7𝑥 + 𝑦 = 3 (1)
𝑥=2 𝑥 − 2𝑦 = 9 (2)
𝑥 + 3𝑦 = 11 ea a ge e ai (1) ef 𝑦
2 + 3𝑦 = 11 𝑦 = 3 − 7𝑥
3𝑦 = 11 − 2 b i e he 𝑦 a e i e ai (2)
3𝑦 = 9 𝑥 − 2(3 − 7𝑥) = 9
𝑦=3 𝑥 − 6 + 14𝑥 = 9
∴ 𝑥 = 2, 𝑦 = 3 …Answer 𝑥=1
b i e he 𝑥 a e i e ai (1)
2008-15. 7+𝑦=3
2𝑥 + 𝑦 = 5 ⋯ (a) 𝑦 = −4
𝑥 + 3𝑦 = 5 ⋯ (b) ∴ 𝑥 = 1, 𝑦 = −4 …Answer
Multiply equation (b) by 2
2𝑥 + 6𝑦 = 10 ⋯ (c)
Now subtract (a) from (c)
Section B
2𝑥 + 6𝑦 = 10 ⋯ (c) 2010-17.
− ) 2𝑥 + 𝑦 = 5 ⋯ (a) 2𝑚 − 𝑛 = 4 ⋯ (i)
5𝑦 = 5 2𝑚 + 3𝑛 = −4 ⋯ (ii)
𝑦=1 Rearrange (i)
substitute 𝑦 = 1 in (b) 2𝑚 = 4 + 𝑛 ⋯ (iii)
𝑥 + 3𝑦 = 5 Substitute (iii) for (ii)
𝑥+3×1 = 5 (4 + 𝑛) + 3𝑛 = −4
𝑥+3 = 5 4𝑛 = −8
𝑥=2 𝑛 = −2 ⋯ (i )
∴ 𝑥 = 2, 𝑦 = 1 …Answer Substitute (iv) for (i)
2𝑚 − (−2) = 4
2004-9. 2𝑚 = 2
𝑦 − 𝑥 = −1 (1) 𝑚=1
3𝑥 − 𝑦 = 5 (2) ∴ 𝑚 = 1 a d 𝑛 = −2…Answer
ea a ge e ai (1) ef 𝑦
𝑦 =𝑥−1
b i e he 𝑦 a e i e ai (2)
12 Linear simultaneous equations - Answers
2009-17. 2006-18.
b. a. 𝑥 + 𝑦 = 29
Let the cost of each banana be B. b. 5𝑥 + 2𝑦 = 100
Let the cost of each mango be M. c.
B + B + M + M + M = 28 𝑥 + 𝑦 = 29 (1)
(two bananas and three mangoes is K28) 5𝑥 + 2𝑦 = 100 (2)
∴ 2B + 3M = 28 ⋯ (1) m(1),
B + B + B + B + B + M = 31 𝑦 = 29 − 𝑥 (3)
(five bananas and one mango is M31) b i e(29 − 𝑥)f 𝑦 i (2)
∴ 5B + M = 31 ⋯ (2) 5𝑥 + 2(29 − 𝑥) = 100
M = 31 − 5B ⋯ (3) 5𝑥 + 58 − 2𝑥 = 100
substitute (3) to (1) 3𝑥 = 42
2B + 3 × (31 − 5B) = 28 𝑥 = 14
2B + 93 − 15B = 28 b i e 14 f 𝑥 i (3)
2B − 15 = 28 − 93 𝑦 = 29 − 14 = 15
−13B = −65 h 𝑥 = 14 a d 𝑦 = 15 …Answer
B=5
M = 31 − 5B
M = 31 − 5 × 5
M = 31 − 25
M=6
∴ ba a a K5, ma g K6 …Answer

12 Linear simultaneous equations - Answers

 13 Quadratic equations

Section A Section B
2012-10. 2009-18.
𝑥(𝑥 − 2) = 35 b.
2
𝑥 − 2𝑥 = 35 𝑦 2 + 4𝑦 − 32 = 0
𝑥 2 − 2𝑥 − 35 = 0 𝑦 2 + (8 − 4)𝑦 + 8 × (−4) = 0
(𝑥 − 7)(𝑥 + 5) = 0 (𝑦 + 8)(𝑦 − 4) = 0
∴ 𝑥 − 7 = 0 or 𝑥 + 5 = 0 𝑦 = −8 or 𝑦 = 4…Answer
∴ 𝑥 = 7 or 𝑥 = −5
∴ 𝑥 = 7 (∵ 𝑥 ≥ 0) …Answer 2007-20.
a.
2011-10. 4
− 4𝑡 2 = 0
b. 9
𝑘 2 = 10𝑘 − 25 4
−4𝑡 2 = −
𝑘 3 − 10k + 25 = 0 9
(𝑘 − 5)2 = 0 1
𝑡2 =
𝑘 = 5…Answer 9
1
𝑡 = ± … Answer
2010-14. 3
𝑡 2 + 7𝑡 + 10 = 0
𝑡(𝑡 + 2) + 5(𝑡 + 2) = 0 2006-16.
(𝑡 + 5)(𝑡 + 2) = 0 2 1 1
− =
∴ 𝑡 = −5 or 𝑡 = −2…Answer 𝑚 𝑚+2 3
2(𝑚 + 2) − 𝑚 𝑚(𝑚 + 2)
=
2009-13. 𝑚(𝑚 + 2) 3𝑚(𝑚 + 2)
2𝑓 2 − 2 3*2(𝑚 + 2) − 𝑚+ 𝑚(𝑚 + 2)
2 2 2+
=
= 2(𝑓 − 1) = 2*𝑓 + (1 − 1)𝑓 − 1 3𝑚(𝑚 + 2) 3𝑚(𝑚 + 2)
= 2(𝑓 + 1)(𝑓 − 1)…Answer 6𝑚 + 12 − 3𝑚 = 𝑚2 + 2𝑚
3𝑚 + 12 = 𝑚2 + 2𝑚
2006-14. 𝑚2 − m − 12 = 0
𝑥 2 + 𝑥 = 42 (𝑚 + 3)(𝑚 − 4) = 0
𝑥 2 + 𝑥 − 42 = 0 𝑚 = −3, 4…Answer
(𝑥 + 7)(𝑥 − 6) = 0
𝑥 = −7 or 𝑥 = 6 2005-18.
Since an age cannot be negative value, 𝑥 = 6 . b.
∴ The woman is 62 = 36 years old…Answer 𝐴𝐵 = 𝐴𝐶
𝑥 + 20 = 𝑥 2
𝑥 2 − 𝑥 − 20 = 0
(𝑥 − 5)(𝑥 + 4) = 0
𝑥 = −4, 5…Answer

13 Quadratic equations - Answers

2004-18. 2003-16.
2
The area of the rectangle = 63cm Let 𝑥 = Mavuto′s age,
𝑥(𝑥 + 2) = 63 then sister′ s age = 𝑥 + 12.
𝑥 2 + 2𝑥 − 63 = 0 𝑥(𝑥 + 12) = 108
(𝑥 + 9)(𝑥 − 7) = 0 𝑥 2 + 12𝑥 − 108 = 0
𝑥 = −9,7 (𝑥 + 18)(𝑥 − 6) = 0
The width of the rectangle is 7cm …Answer 𝑥 = 6 (∵ 𝑥 ≥ 0)
∴ Mavuto is 6 years old …Answer

13 Quadratic equations - Answers

 14 Coordinates

Section A 2004-14.
2006-7. 𝑄(0,2) 𝑅(−3, −4)
(5, 1)

14 Coordinates - Answers
 15 Linear graph

Section A Section B
2012-12. 2011-18.
a. 𝑦 = −3 a.
b. 𝑦 = 5 − 3 × 2 = 5 − 6 = −1…Answer
A(−1, −3), B(2,3) b.
𝑦2 − 𝑦1
Gradient =
𝑥2 − 𝑥1
3−(−3) 6
= 2−(−1) = 3 = 2…Answer

2010-10.
𝑦 = 3𝑥 + 𝑐
Substitute the point (1, 4)
4=3×1+𝑐
4−3=𝑐
∴ 𝑐 = 1…Answer

2006-12.
a. 𝑦 = 3 (−2 ≤ 𝑥 ≤ 3)

2005-9.
𝑥 -3 0 3
𝑦 0 3 6 c. 𝑥 = 2, 𝑦 = −1

15 Linear graph - Answers

2009-16. 2005-20.
a. a.
𝑥 0 1 2 𝑦 = 5 − 2𝑥

𝑦 -3 -1 -1 𝑥 -1 2 5
𝑦 7 1 -5
b.
𝑦 =𝑥−4
y
𝑥 -1 2 5
𝑦 -5 -2 1
3
b.
2

1
x
-3 -2 -1 0 (0, 0) 1 2 3
-1

-2

-3

c.
On the graph, the line 𝑦 = 5 − 2𝑥 and
𝑦 = 𝑥 − 4 meet at the point (3,-1).
∴ 𝑥 = 3 and 𝑦 = −1…Answer

15 Linear graph - Answers

 16 Travel graph
Section A
2012-4.
10 litres = 70 km
7
7 litres = 70 km × = 49 km … Answer
10

Section B
2010-16.
a.

20

16
Distance (km)

12

0
6:30 7:00 7:30 8:00 8:30 9:00
Time(hours)

b. 2.5hours

16 Travel graph - Answers

2008-20.
a.
200

175

150

125
Distance (km)

100

75

50

25

0
12:00 13:00 14:00 15:00 16:00 17:00 18:00
Time(hours)
b.
First find the speed after breakdown to arriving.
distance 200 − 100 100
speed = = = = 40𝑘𝑚/ℎ
time 17: 30 − 15: 00 2.5
Subtract breakdown time from arriving.
16:54 − 15:00 = 1:54 (1 hour and 54 minutes)
54 114
So distance is 40 𝑘𝑚/ℎ × (1 + 60) ℎ = 40 × 60
= 76𝑘𝑚

Therefore, 100 km + 76 km = 176km…Answer

c.
distance 200 − 100 100 100
speed = = = = = 40𝑘𝑚⁄ℎ … Answer
time 17: 30 − 15: 00 2: 30 2.5
16 Travel graph - Answers
2007-17.
a.
(i) 60km
(ii)
Distance from C to D 60km
= = 20km⁄h … Answer
Time 11.30am to 14.30pm 3 hours

2006-20.
a.

250
Station Q
Train B

200
Distance (km)

150

100

Train A
50

Station P
0
6.00am 8.00 10.00 12.00pm 14.00 16.00 18.00 20.00
Time(hours)

b. 13.30pm

2004-19.
a.
(i) 300km
(ii) 30minutes

2003-20.
60km
The speed of the car = = 40 km⁄h
1.5hour
60km
The speed of the bus = = 30 km⁄h
2hour
The difference = 40 − 30 = 10 km⁄h…Answer

16 Travel graph - Answers

 17 Sets
2007-5.
Section A A ∩ B = *2,4,6+
2011-11.
a. 2006-10.
ξ *3+, *4+, *3,4+, Φ…Answer
(write only three subsets)
A B
coke
fanta cocopina 2005-8.
sprite cherry 𝑋∪𝑌
plum = *𝑛𝑗𝑖𝑤𝑎, 𝑝𝑢𝑚𝑏𝑤𝑎, 𝑘𝑎𝑘𝑜𝑤𝑎, 𝑘ℎ𝑤𝑎𝑛𝑔𝑤𝑎𝑙𝑎,
𝑛𝑘ℎ𝑢𝑛𝑑𝑎, 𝑡𝑖𝑚𝑏𝑎, 𝑘𝑎𝑏𝑎𝑤𝑖, 𝑘𝑎𝑑𝑧𝑖𝑑𝑧𝑖+
The number of elements is 8 …Answer
b. A ∩ B = Φ
2004-10.
2010-9.
ξ

x z A B
4 3
5
6 2 5 9
4 3
8 7 6
8 9 7

2009-4.
a. 2003-13.
(i) D = {Maize, Cassava, Soya, Rice, Sweet potato, a.
Sorghum, Millet} (i) 𝑃 = *𝑎, 𝑏, 𝑑, 𝑗, 𝑚+
the number of crops grown on farm D (ii) 𝑃 ∩ 𝑄 = *𝑏, 𝑚+
= 7…Answer
(ii) R∩D = *Maize, Cassava, Soya+
crops grown on both farm
Section B
= Maize, Cassava, Soya…Answer 2012-18.
b.
2008-3. (i) P ∩ Q = *2, 4, 6+

(ii) P ∪ Q = *1, 2, 3, 4, 5, 6, 7, 8+

17 Sets - Answers
 18 Vectors
2003-9.
Section A
1 1 2 2
⃗⃗⃗⃗⃗ + 𝐷𝐶
(𝐴𝐵 ⃗⃗⃗⃗⃗ ) = [( ) + ( )]
2009-11. 2 2 3 5
−4 1 4 2
⃗⃗⃗⃗⃗ = −3𝐴𝐵
𝐶𝐷 ⃗⃗⃗⃗⃗ = −3 × ( ) = ( ) = ( ) … Answer
2 2 8 4
−3 × (−4) 12
=( ) = ( ) … Answer
−3 × 2 −6
Section B
2008-8. 2012-18.
⃗⃗⃗⃗⃗ = 𝑃𝑅
𝑃𝑄 ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ + 𝑅𝑄 a.
6 𝑥 1 2 −1
( )= ( )+( ) 𝑎 − 3𝑏 = ( ) − 3 ( )
5 𝑦 3 −3 2
6=𝑥+1 𝑥+1 = 6 𝑥 =6−1=5 2 −3 2+3
( )=( )=( ) =( )−( )=( )
5 = 𝑦+3 𝑦+3=5 𝑦 = 5−3 = 2 −3 6 −3 − 6
5 5
⃗⃗⃗⃗⃗
𝑃𝑅 = ( ) … Answer = ( ) … Answer
2 −9

2006-9. 2011-19.
⃗⃗⃗⃗⃗
(i) 𝐴𝐵 b.
⃗⃗⃗⃗⃗ = 𝐵 − 𝐴
𝐴𝐵 𝑎 −4 −8
( ) = 2( ) = ( )
= (5, −2) − (2, 3) = (3, −5)…Answer 10 5 10
∴ 𝑎 = −8…Answer
⃗⃗⃗⃗⃗
(ii) 𝐵𝐴
⃗⃗⃗⃗⃗ = A − B = (2, 3) − (5, −2)
BA 2010-18.
= (−3,5)…Answer a.
2 2
⃗⃗⃗⃗⃗ + 𝐿𝑀
𝐾𝐿 ⃗⃗⃗⃗⃗⃗ = ( ) + ( )
∴ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = (3, −5) − (−3,5)
AB − BA 6 0
= (6, −10) = 2AB ⃗⃗⃗⃗⃗ 2+2 4
=( ) = ( ) … Answer
6+0 6
2005-7.
1 3
a. 𝐻 ( ) , 𝐾 ( )
−2 3

⃗⃗⃗⃗⃗⃗ = (3) − ( 1 ) = (2)

b. 𝐻𝐾
3 −2 5

18 Vectors - Answers
2007-19.
b.

𝑦
5

3 𝐵 𝐶

𝐴
0 1 2 3 4 5 6 𝑥

2004-19.
b.
−3 2 −3 (−3) 5
𝑎 − ( ) = ( ) − ( ) = (2 − )=( )
1 4 1 4−1 3
3 2 3 2+3 5
𝑏⃗ + ( ) = ( ) + ( ) = ( )=( )
5 −2 5 −2 + 5 3
−3 3
∴ 𝑎 − ( ) = 𝑏⃗ + ( )
1 5

18 Vectors - Answers
 19 Line and angles
2006-8.
Section A ∠DEF = 180° − 110° = 70° (straight line)
2012-7. ∴ 𝑢 = 70°
∠ABH + ∠GHB = 180° (CD ∥ EF, alternate ∠s are equal)
(allied angles, AB ∥ GH) ∠BID = 180° − 140°
∠ABH + 125° = 180° = 40° (straight line)
∠ABH = 180° − 125° = 55° ∴ ∠IDC = 40° (AB ∥ CD, alternate ∠s are equal)
∠ABH + ∠DBE = ∠BEF ∴ 𝑤 = 180° − 𝑢 − ∠IDC = 180° − 70° − 40°
(alternate angles, AC ∥ DF) = 70° (straight line)
∠DBE + 55° = 105° ∴ 𝑢 = 70°, 𝑤 = 70°…Answer
∠DBE = 105° − 55°
∠DBE = 50°…Answer 2003-8.
∠𝐴𝐵𝐹 = ∠𝐴𝐶𝐸(corresponding angles)
2010-2. 78° = 2𝑥
5𝑦° = 130° 𝑥 = 39°
∴ 𝑦° = 26°…Answer ∠𝐴𝐶𝐷 = 2𝑥 + 𝑥 = 3𝑥 = 3 × 39°
= 117°…Answer
2008-5.
n + 62° = 147° (alternate angle)
n = 147° − 62° = 85°…Answer
Section B
m = 180° − 62° = 118°…Answer 2005-19.
(alternate angle or corresponding angle) a.
(i)
2007-1. ∠ABH + ∠HBK + ∠KBC = 180°(straight line)
𝑥 = 180° − 80° 54° + 𝑥° + 2𝑥° = 180°
= 100° (straight line) …Answer 𝑥° = 42°
𝑥 = 42…Answer

(ii)
∠KBC = 2𝑥 = 2 × 42° = 84°
∠KBC = ∠PCB
Since alternate angles are equal, BK ∥ PC.

19 Line and angles - Answers

 20 Reflection and rotations

Section A
2010-15.
a. omitted (same as the figure in the question)
b.
CA = CA′ and CB = CB ′
∴ Centre of rotation is (4, 0)…Answer

2006-13.
a. b.

2005-11.

2004-15.
(1,0)

20 Reflection and rotations - Answers

 Section B
2012-19.

20 Reflection and rotations - Answers

2009-18.
a.
(i) omitted (same as the figure in the question)
(ii)

A(－6, 1), B(－2, 3), C(－2, 1)

A’(1, 6), B’(3, 2), C’(1, 2)

A’
6

2
C’ B’

x
－6 －4 －2 0 (0 , 0) 2 4 6 8

－2

20 Reflection and rotations - Answers

2008-16.
a. b.

10 Y

A 6 A’
‘

4
B C C’ B’

0
-20 -15 -10 -5 5 10 15 20X
-2

-4

-6
c. (5,4)

20 Reflection and rotations - Answers

 21 Triangles
2007-14.
Section A b. In triangle PSR and triangle QRT,
2011-12. SR = RT (given)
SN = RN (given) PS = QR (PQRS is a parallelogram)
PS = QR (PQRS is a rectangle) ∠PSR = ∠QRT
∠PSN = ∠QRN = 90°(PQRS is a rectangle) (PS ∥ QR, corresponding angles are equal)
∴ Triangle PSN ≡ Triangle QRN(S. A. S) ∴ triangle PSR ≡ triangle QRT(S. A. S)

2011-13. 2004-1.
∠PQU = ∠QRS = 68° ∠𝑅 + ∠𝑃 + ∠𝑄 = 180°
(UQ ∥ SR, corresponding angles are equal) ∠𝑅 + 65° + 50° = 180°
∠QRT = 68° − 18° = 50° ∠𝑅 = 65°…Answer
∠QRT = ∠QTR = 50° (the base angles of
isosceles triangle QTR are equal)
∠QTV + ∠QTR = 180° (angle on a straight line)
Section B
∴ ∠QTV = 180° − ∠QTR = 180° − 50° 2006-17.
= 130°…Answer a. ∠KLS = ∠LSM = 30°
(LK ∥ MS, altermate ∠s are equal)
2009-12. ∠LSM = ∠SLM = 30°
∠𝑥 + ∠CBF = 180° (supplementary angles) (base angles of isosceles triangles)
∴ ∠𝑥 + 92° = 180° ∴ ∠LMS = 180° − 30° − 30° = 120°
∠𝑥 = 180° − 92° = 88° ∴ ∠KMS = 120° − 70°
∠BEF = ∠ABE (alternate angles) = 50°…Answer
∴ ∠BEF = 35°
∠𝑦 + ∠𝑥 + ∠𝐵𝐸𝐹 = 180° b. Acute angled triangle
(the angle sum of a triangle = 180°)
∠𝑦 + 88° + 35° = 180° 2005-16.
(substituted ∠𝑥 = 88° and ∠BEF = 35°) Let angle NDC = 𝑥,
∠𝑦 = 180° − 88° − 35° = 57° then angle NCD = 𝑥(triangle NDC is isosceles)
∴ ∠𝑥 = 88°, ∠𝑦 = 57°…Answer 70° + 𝑥 + 𝑥 = 180°
𝑥 = 55°
2009-14. ∠ADN = ∠ADC − ∠NDC
b. = 90° − 55° = 35°…Answer
In △ WME and △ XLW,
∠WME = ∠XLW = 90° 2003-18.
(EM and XL are perpendicular to WN) a. In triangles LMQ and MNP,
∠EWM = ∠WXL (given angle) QL = PM (given side)
WE = XW (sides of a square are equal) LM = MN (given side)
∴△ WME ≡ △ XLW(A.A.S) ∠QLM = ∠PMN = 90° (given angle)
∴ triangle LMQ ≡ triangle MNP(S.A.S)

21 Triangles - Answers
 22 Convex polygons
2006-15.
Section A Let the number of sides of the polygon be n.
2011-6. Sum of the interior angles of a n-sided
n polygon × 𝑛
n × 𝑛
𝑛 9
n 𝑛 …Answer
n
There are 10 sides in this polygon. 2005-14.
Th xt io angl i
is one interior angle.
The sum of the exterior angle of a polygon is
always .
The size of each exterior angle of the polygon is ÷
n Th numb of id …Answer

2010-7. 2004-13.
b. The sum of the interior angles in an n-sided
Sum of the interior angles of an n-sided polygon polygon × 𝑛
× 𝑛 th sum of th int io angl
× 𝑛 9 × …Answer
× 𝑛 9
2003-4.
𝑛 ∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐷 + ∠𝐵𝐴𝐷 + ∠𝐴𝐷𝐶
𝑛 …Answer 9 + 5 +7 + ∠𝐴𝐷𝐶
∠𝐴𝐷𝐶 …Answer
2007-12.
Let the number of sides of the polygon be n.
Sum of the interior angles of a n-sided
Section B
polygon × 𝑛 2012-16.
× 𝑛 y + + +y +
𝑛 ＝5 angl um of a p ntagon
𝑛 5 y +y + + + ＝5
Th numb of id of th polygon i 5. y + ＝5
y ＝5
Th int io angl of th polygon
5 y＝
5 y ＝5 …Answer
Th xt io angl of th polygon
5 …Answer

22 Convex polygons - Answers

2008-17. 2003-18.
a. b.
CDEF is a quadrilateral. The sum of interior L t 𝑥 b th numb of id .
angles of a quadrilateral is 360°.
Th n th xt io angl i
∠C + ∠D + ∠E + ∠F 𝑥
∠C 5 , ∠D 5 , ∠E
Th int io angl i
5 + 5 + + ∠F 𝑥
+ ∠F
×
∠F …Answer 𝑥 𝑥
𝑥
b. Th numb of id Answer
ABCDE is a pentagon. The sum of interior angles
of a pentagon is 540°.
∠C + ∠D + ∠E + ∠E B + ∠ BC 5
∠ BC
5 + 5 + + ∠E B + ∠ BC 5
∠E B + 5
∠E B 5
∠E B ∠ BC
∠F B ∠FB 7
Triangle AFB has 2 angles of the same size so it
is an isosceles triangle.

22 Convex polygons - Answers

 23 Similarity
2004-12.
Section A triangle ~ triangle
2012-13.
=
b.
In triangle BEA and triangle DFA, 3
=
∠AFD = ∠BEA (given) 4 6
∠ADF = ∠ABE = 90° (ABCD is a square) 18 9
= = = 4 5cm … Answer
∠DAF = ∠BAE 4 2
(remaining angles of △ BEA and △ BCD)
∴ triangle BEA is similar to triangle DFA
Section B
2011-9. 2008-18.
△ CED:△ AEB = ED: EB = EC: EA b. Triangle ABC is similar to triangle BDC.
ED: EB = 9: 9 + 6 = 9: 15 = 3: 5 Therefore
EC: EA = 4: 4 + 𝑥 AB BC
=
∴ 3: 5 = 4: 4 + 𝑥 BD DC
12 + 3𝑥 = 20 AB 10cm
=
3𝑥 = 8 9 6cm
8 10
𝑥= (cm) … Answer AB = × 9 = 15cm … Answer
3 6

2010-8. 2007-19.
In triangle AFE and triangle BCD, ZY 15
a. scale factor = AB = 3
=5
∠AFE = ∠BCD (given)
∠AED = ∠BDE = 90° (ABDE is a rectangle) ZX = scale factor × AC
∴ ∠AEF = ∠BDC = 90° = 5 × 7 = 35cm…Answer
(sum of adjacent angles on a straight line)
∴ ∠FAE = ∠CBE 2003-19.
(remaining angles of ΔAFE and ΔBCD) a. (i)
∴Triangle AFE and triangle BCD are similar. In triangles CAB and CBD,
(corresponding angles are equal) ∠CAB = ∠CBD(given angle)
∠ACB = ∠BCD(common angle)
2005-15. wo angles are respectively equal
b. ∴ riangles CAB and CBD are similar
△ 𝐷𝐶𝐸 ~ △ 𝑁𝐶𝑀 (ii) Calculate, the length of CD. (6 marks)
𝐶𝐷 𝐶𝑁 𝐴𝐵 𝐵𝐷
= =
𝐶𝐸 𝐶𝑀 𝐶𝐵 𝐶𝐷
10 𝐶𝑁 8 6
= =
12 30 7 𝐶𝐷
𝐶𝑁 = 25cm…Answer 21
𝐶𝐷 = = 5 25cm … Answer
4
23 Similarity - Answers
 24 Pythagoras theorem

Section A Section B
2011-15. 2009-17.
2 2
(d + 4) = d + 8 2 a.
2 2 FC = DB
d + 8d + 16 = d + 64
8d = 48 (opposite sides of a parallelogram are equal)
d = 6…Answer ∴ FC = 5cm
AF = AC − FC
2003-11. ∴ AF = 10 − 5 = 5cm
2 2 2 △ AEF is a right angled triangle, so
3 + 𝐵𝐶 = 7
2
𝐵𝐶 = 40 𝐴𝐸 2 + 𝐸𝐹 2 = 𝐴𝐹 2 (pythagoras ′ theorem)
𝐵𝐶 = √40 = 2√10 = 2 × 3.162 𝐴𝐸 2 + 32 = 52
= 6.324 ≈ 6.3 (2 s. f. )…Answer (substituted 𝐸𝐹 = 3 and 𝐴𝐹 = 5)
AE 2 + 9 = 25
AE 2 = 25 − 9
AE 2 = 16
AE = 4cm…Answer

2007-18.
b.
XY 2 + YZ 2 = 32 + 42
= 9 + 16 = 25 = 52
= XZ 2
∴ The triangle XYZ is right angled.

24 Pythagoras’s theorem - Answers

 25 Quadrilaterals

Section A Area of rhombus =△ ABD +△ CBD

2012-9. 1 1
= 12 × 8 × + 12 × 8 ×
(2𝑥 + 30)° = (7𝑥 − 25)° 2 2
2𝑥 − 7𝑥 = −25 − 30 = 48 + 48 = 96cm2…Answer
−5𝑥 = −55
𝑥 = 11 2008-19.
∴ ∠RUT = (2𝑥 + 30)° b.
= (2 × 11 + 30)° = (22 + 30)° WXYZ is a parallelogram so opposite angles are
= 52°…Answer equal.
(𝑎2 − 2𝑎) = 63
2005-12. 𝑎2 − 2𝑎 = 63
∠HRJ = ∠HJR (Isosceles triangle) 𝑎2 − 2𝑎 − 63 = 0
∠HJR = ∠GKR (given angle) (𝑎 − 9)(𝑎 + 7) = 0
∴ ∠HRJ = ∠GKR 𝑎 = 9, −7
Since the corresponding angles are equal,
GK ∥ HR. 2004-20.
Since GH ∥ KJ, GH ∥ KJ. b.
∴ GHRK is a parallelogram. In the rhombus ABCD,
let AC = 10cm and BD = 24cm.
Let E be the point where the diagonals meet.
Section B Then AE = CE = 5cm, BE = DE = 12cm
2010-20. In triangle AEB, use the Pythagorean theorem:
b. AB 2 = AE2 + BE 2
AB 2 = 52 + 122 = 169
AB = 13
The length of a side is 13cm …Answer

25 Quadrilaterals - Answers
 26 Mensuration

Section A Section B
2007-14. 2009-19.
a. (the total surface area of a cylinder)
22 = (curved surface area) + (area of two circles)
Volume = πr2 ℎ = × 72 × 10
7 ∴ (2𝜋𝑟 × ℎ) + (2 × 𝜋𝑟 2 )
= 1540cm3…Answer 22 7
= (2 × × × 20)
7 2
2003-10. 22 7 2
+ {2 × ×( ) }
22 7 2
The base area = 7 × 7 × ( ) = 154cm2
7 = 440 + 77
The volume = the base area × the height = 517 cm2…Answer
= 154 × 10 = 1540cm3…Answer
2004-16.
Area of base = 8 × 6 = 48cm2
Area of triangle VAB = 8 × 14 ÷ 2 = 56cm2
Area of triangle VBC = 6 × 12 ÷ 2 = 36cm2
Area of lateral face = 56 × 2 + 36 × 2
= 184cm2
The total surface area = 48 + 184
= 232cm2 … Answer

26 Mensuration - Answers
 27 Geometrical constructions
Section A
2007-15.
a.

Section B
2012-20.
a.

b. 2.3cm

27 Geometrical constructions - Answers

2011-20.
(i)(ii)

(iii) ∠BAC = 50°

2010-19.
a. (i)
Step1. Using a ruler, draw a line of 5cm. Label the ends, P and Q.
Step2. Set the compasses to the second length, 5cm. Place the point of the compasses at P and draw
an arc.
Step3. Set the compasses to the third length, 5cm. Place the point of the compasses at Q and draw
another arc. Label the intersection, R.
Step4. Join P and Q to the intersection R.

27 Geometrical constructions - Answers

2010-19.
a. (ii)
Step1. Set the compasses to slightly more than half the length of the side PQ. Place the point at P and
draw an arc.
Step2. Without changing the setting of the compasses, place the point at Q and draw another arc.
Step3. Join the intersections of the arcs to find the perpendicular bisector.

Step4. Do “Step1 to Step3” also about sides PR and QR.The intersection of the three perpendicular
bisectors is the centre (O) of the circumscribed circle.
Step5. Set the compasses to the length of OP. Place the point of the compasses at O and draw the
circle.

27 Geometrical constructions - Answers

2009-20.
a.

UV=5cm
U V

UW = 6cm

b.

UV=5cm
U V X

UW = 6cm

c. UZ=13.3cm

27 Geometrical constructions - Answers

2008-18.
a. (i)

To draw 60° only compass and ruler.

1. Draw a straight line (8 cm long) and mark on point B.
2. With centre A, and draw an arc to cut AB at D.
3. With centre D and the same radius, draw an arc to cut the previous one at E. Join AE.

(ii)

1. With centre A, and draw an arc to cut AB at D and AC at E.

2. With centre D, and draw an arc.
3. With centre E, and draw an arc same radius as Step 2.
4. Label the point where the arcs cut F. Join AF. AF is the required line.

27 Geometrical constructions - Answers

2006-19.
a. b.

c. ∠APB = 78°

2004-20.
a.
Step1. Draw a line of 9cm. Label the ends A and B. Extend the line past A.

Step2. Open a pair of compasses to about 2cm wide. Place the point of the compasses at A and draw
an arc to cut the extended line. Label the intersection D.
Step3. Without changing the radius, place the point at D and draw an arc. Label the intersection with
the arc drawn in Step2, E.

Step4. Without changing the radius, place the point at E and draw an arc. Label the intersection with
the arc drawn in Step3, F.
Step5. Draw a line through A and F.
Step6. Open the compasses to 4.5cm wide. Place the point of the compasses at A and draw an arc to
cut the line AF. Label the intersection C.

Step7. Join B and C.

∠ABC ≈ 10°

27 Geometrical constructions - Answers

2003-17.
Step1. Draw PQ which is 8cm long.
Step2. Open a pair of compasses to 7cm wide. Place the point of the compasses at Q and draw an arc.
Step3. Open the compasses to 6cm wide. Place the point at P and draw an arc.
Step4. Label the intersection of the two arcs R. Join PR and QR.

b.
Step1. Place the point of the compasses at P and draw an arc to cut the lines PQ and PR.
Step2. Place the point at one intersection and draw an arc.
Step3. Without changing the radius, place the point at the other intersection and draw an arc.

Step4. Draw a line through P and the intersection of the arcs drawn in Step2 and 3.
Step5. Open the compasses to 10cm wide. Place the point at P and draw an arc. Label the intersection
with the arc, S.

𝑄𝑆 ≈ 5cm

27 Geometrical constructions - Answers

MATHEMATICS

*ONLY QUESTIONS
 Acknowledgements  
 
It  was  a  long  journey.  This  project  was  completed  thanks  to  the  united  efforts  
of  a  large  number  of  people.  Our  great  thanks  to  MANEB  for  permitting  our  using  
past-­‐papers,  many  schools  for  lending  us  past-­‐papers,  JICA  Malawi  Office  for  their  
great  support,  members  of  Japan  Overseas  Cooperation  Volunteers  and  Secondary  
School  teachers  who  worked  together  to  achieve  this  common  goal.  
 
Finally  we  thank  to  you,  students  and  teachers,  who  are  using  this  book  and  
who  will  make  future  of  Malawi  more  fruitful  and  more  colourful!  
 
TEACHERS  WHO  COOPERATED  
Charles  Zinam’dala   Lyson  Kamwana  

Franchise  Mambo   Martin  Njolomole  

Langson  Chigalu   Yohane  Tembo  

 
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Akihito  Miyamoto  (Chembera  CDSS)     Rie  Nomura  (Monkey  Bay  CDSS)  
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Erika  Atarashi  (Nsaru  SS)   Seiichi  Kittaka  (Balaka  SS)  
Fumie  Chikaoka  (Ching’ombe  CDSS)   Takeshi  Kiriyama  (Mbenjele  SS)  
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Junya  Nakakita  (Mangochi  SS)   Yohei  Matsuda  (Chembera  CDSS)  
Kanae  Higuchi  (Magomero  CDSS)   Yu  Yamazaki  (Bilila  CDSS)  
Kiyoshi  “uncle  k.”  Furusawa  (Mwatibu  CDSS)   Yutaka  Murahashi  (Lilongwe  TTC)  
Kotaro  Kijima  (Kabudula  CDSS)   Yutaka  Nishiyama  (Nsanama  CDSS)
Mitsuhiro  Uchida  (Ching’ombe  CDSS)  

 
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