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CIE IGCSE MATHS0580

TOPICALSOLVEDQUESTIONSON THESYLLABUS

TABLE OF CONTENTS

2 CHAPTER 1 Numbers

5 CHAPTER 2 Algebra and Graphs

11 CHAPTER 3 Geometry

16 CHAPTER 4 Mensuration (Perimeters, Areas and Volumes)

18 CHAPTER 5 Trigonometry

20 CHAPTER 6 Matrices and Transformations

22 CHAPTER 7 Probability

23 CHAPTER 8 Statistics

CIE IGCSE MATHEMATICS//0580

1. NUMBERS

1.1 Integers, HCF/LCM, Prime numbers, Sig Figs, Dec Places

Question 1:

Find the lowest common multiple (LCM) of 36 and 48. [2]

Solution:

We can do this by writing out all of the multiples of the two numbers. The multiples of 36 are:

36, 72, 108, 144, 180, ... The multiples of 48 are

1.3 Square and Cube Numbers

3

Simplify (3210)5

Question 3:

[2]

Solution:

Apply the power to everything inside the brackets? and use the general rule that () =

3

325

?

(10?35)

Note

that

32

=

25

hence

1

325

=

2

=

13

(325)

?

6

= 236 =

48, 96, 144, 192, ... We can see that the lowest common multiple is:

144

1.4 Conversion - Percentages, Fractions & Decimals

Question 4:

1.2 Sets and Venn Diagram

Write the recurring decimal 0.32 as a fraction. [2]

Question 2:

[0.32 means 0.3222...]

a) = {x: 2 x 16, x is an integer}

Solution:

M = {even numbers}

P = {prime numbers}

i) Find n(M).

[1]

ii) Write down the set (P M). [1]

b) On the Venn diagram, shade A B.

[1]

Solution:

Part (a)(i)

() is the number of elements in set M. M is all the

even numbers between 2 and 16 inclusive which is

() =

Part (a)(ii)

( ) are the elements not in the union of sets P

and M.

( ) = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16} ( ) = {, }

Part (b)

We need to get rid of the recurring decimal by doing

the following

100 ? 0.32 = 32. 2

10 ? 0.32 = 3. 2

100 ? 0.32 - 10 ? 0.32 = 90 ? 0.32

= 32. 2 - 3. 2

90 ? 0.32 = 29

Now divide by 90

.

=

1.5 Order by Size

Question 5:

Write the following in order of size, smallest first. [2]

3.14

22 7

3.142 3

Solution:

The order of size can be found by writing all of these

numbers out to the same number of decimal places,

and then comparing. In order to do this, put each of

the values into the same format (decimals) using the

`SD' button (located above `DEL') on your calculator.

= 3.14159 (5. . )

3.14 = 3.14000 (5. . ) 22 7 = 3.14286 (5. . ) 3.142 = 3.14200 (5. . )

3 = 3.00000 (5. . )

Therefore, the order we get (smallest to largest) is:

3

<

3.14

<

<

3.142

<

PAGE 2 OF 26

CIE IGCSE MATHEMATICS//0580

1.6 Standard Form

Question 6:

Write 2.8 ? 102 as an ordinary number.

[1]

Solution:

We can write 2.8 x 102 as an ordinary number like this: 2.8 x 102 simply means 2.8 x 100

. ? =

Solution:

Write all numbers correct to one significant figure:

4 ? 30

9-3 Do the calculations.

2 ? 30 60 6 =6

We get the final answer:

1.7 Addition/Subtraction/Multiplication/

Division of Fractions & Decimals

Question 7:

Show that

13 1 2 ? 16 = 8 Do not use a calculator and show all the steps of your

working.

[2]

Solution:

This question is most simply done by converting

everything

to

proper

fractions.

We

want

to

change

1

1 2

into a proper fraction, which can be done as shown.

1

1

21 3

12 1+2 2+2 2

Our problem then becomes

33 2 ? 16 We can use `Keep-Change-Flip' to change this to a

multiplication problem. We keep 32, change ? into ?,

and

flip

3 16

to

136.

3 16 2? 3

We now can multiply the numerators and

denominators.

3 ? 16 48

2?3 = 6 =8

Hence

?

=

.

1.8 Estimation

Question 8:

By writing each number correct to 1 significant figure,

estimate the value of

3 9 ? 29 3

89 - 27

Show all your working.

[2]

1.9 Bounds

Question 9:

An equilateral triangle has sides of length 16.1 cm, correct to the nearest millimetre.

Find the lower and upper bounds of the perimeter of

the triangle.

[2]

Solution:

An equilateral

triangle has all sides and angles equal.

16.1

16.1

We know that each

side is 16.1 to the

nearest mm. This

means

16.1

that, each side could

be between 16.15 and16.05.

We can therefore calculate:

The minimum perimeter/lower bound is:

16.05 + 16.05 + 16.05 = .

And the maximum perimeter/upper bound is:

16.15 + 16.15 + 16.15 = .

1.10 Ratios

Question 10:

The scale on a map is 1: 20 000.

(a) Calculate the actual distance between two points

which are 2.7 cm apart on the map.

Give your answer in kilometres.

[2]

(b) A field has an area of 64 400 2 .

Calculate the area of the field on the map in 2

[2]

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CIE IGCSE MATHEMATICS//0580

Solution: Part (a)

Multiply the distance on the map by the scale factor to find the real distance in centimeters.

= 2.7 ? 20 000 = 54 000

Divide the distance by 100 to get the distance in meters. (1m = 100cm)

= 540 Divide the real distance in meters by 1000 to get the distance in kilometers (1km = 1000m)

=.

Part (b)

Multiply the area by 10 000 to get the area in square centimeters. (1m2 = 100cm x 100cm= 10 000 cm2)

= 644 000 000 2 Divide by the scale factor 20 0002to get the area on the map. (Note: Area scale factor is the square of the length scale factor)

644 000 000 2 = (20 000)2

= .

1.11 Percentages

Question 11:

In 1970 the population of China was 8.2 x 108. In 2007 the population of China was 1.322 x 109.

Calculate the population in 2007 as a percentage of

the population in 1970.

[2]

Solution:

The population in 2007 as a percentage of the

population in 1970 can be calculated by:

2007 1970

? 100,

Substituting in the values gives:

1.322 ? 109 8.2 ? 108

? 100

= 161. 2 1951

The answer after rounding is:

%

1.12 Using a calculator

Use your calculator to find the value of (cos30?)2 - (sin 30?)2 2(sin120?)(cos 120?)

Question 12:

[2]

Solution:

By inputting the values into your calculator, you get:

(cos(30))2 - (sin(30))2 2(sin(120)(cos(120))

= 2

3- 1

44

=

? 3 ? -1

0.5

-3

=

-3 3

2

2

2

So, the answer is:

=

-

1.13 Time

Question 13:

A train leaves Zurich at 22 40 and arrives in Vienna at 07 32 the next day. Work out the time taken. [1]

Solution:

We can count the time it takes to get us to the Vienna. Add 20 minutes to take it to the next hour:

22: 40 + 20 = 23: 00 Add 1 hour to take it to the next day (24:00 is equivalent to midnight, or 00:00)

23: 00 + 1 = 24: 00 (= 00: 00) Now add 7 hours and 32 minutes to get to the desired time

00: 00 + 7 32 = 07: 32 The time taken is all the hours and minutes added together like this

20 + 1 + 7 32 Total time = (8hr 52m)

1.14 Currency Conversions

Question 14:

(a) In 2007, a tourist changed 4000 Chinese Yuan into

pounds (?) when the exchange rate was ?1 =

15.2978 Chinese Yuan. Calculate the amount he

received, giving your answer correct to 2 decimal

places.

[2]

(b) In 2006, the exchange rate was ?1 = 15.9128

Chinese Yuan. Calculate the percentage decrease

in the number of Chinese Yuan for each ?1 from

2006 to 2007.

[2]

Solution: Part (a)

In order to change from Chinese Yuan into pounds, we can do this:

1 = ?0.06537 4000 = ?0.06537 ? 4000

4000 = ?261.4755 But we need this to the nearest penny

= ?.

PAGE 4 OF 26

CIE IGCSE MATHEMATICS//0580

Part (b)

To calculate the percentage decrease we need to do

the following: ?1

?1 2006

? 100

15.9128 - 15.2978

15.9128

? 100 = 3.8648

Hence the percentage decrease is 3.865%

1.15 Finance Problems

Question 15:

Emily invests $x at a rate of 3% per year simple

interest. After 5 years she has $20.10 interest. Find

the value of x.

[3]

Solution:

The equation for simple interest is

Where:

+ = (1 + 100 )

? = Principal investment value

? i = Interest gained

? R = Interest rate (%)

? t = Investment time 3?5

+ 20.1 = (1 + 100 ) 20.1

1 + = 1 + 0.06 20.1

= 0.06 =

1.16 Finance Problems

Question 16:

Zainab borrows $198 from a bank to pay for a new

bed. The bank charges compound interest at 1.9 % per

month. Calculate how much interest she owes at the

end of 3 months. Give your answer correct to 2

decimal places.

[3]

Solution:

To calculate how much interest she owes on $198 at

the end of the 3 months we first must calculate the

total amount after interest at the end of the 3 months.

This is done as follows:

198

?

(1

+

100)

3

where is the interest rate. As we know that the

interest rate is 1.9%,

r

=(1

+

1.9 )

100

3

=

1.019

and hence the total amount after interest is:

198 ? 1.0193 = $209.50.

Hence Zainab owes the bank

$209.5 - $198 = $11.5

So, the answer is:

$.

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2. ALGEBRA AND GRAPHS

2.1 Using Algebra to Solve Problems

Simplify 16 - 4(3 - 2)2.

Question 17:

[3]

Solution:

Simplifying the equation gives: 16 - 4(3 - 2)2 = 16 - 4(92 - 12 + 4) = 16 - 362 + 48 - 16 = -362 + 48

= 12(4 - 3)

So, the answer is:

( - )

PAGE 5 OF 26

CIE IGCSE MATHEMATICS//0580

2.2 Factorisation (Linear)

2.5 Linear equations

Factorise completely.

Question 18:

a) 2 + 4 + + 2

[2]

b) 162 ? 82

[2]

Solution:

Part (a)

( + )( + )

We can check this by expanding it back out:

+ 2 + 2 + 4

Part (b)

We can start off by factorising out the common factor

of 2 2(81 - 42)

Then we can see that this is the difference of two

squares

= 2(92 - (2)2)

= ( + )( - )

2.3 Algebraic fractions

Question 19:

Write as a single fraction in its simplest form.

[3]

+ 2 3 - - 1

Solution:

Multiply

3

by

-1 -1

to

create

a

common

denominator:

3( - 1) + 2

= - 1 - - 1

Combine the fractions:

3 - 3 - ( + 2)

=

- 1

3 - 3 - - 2

= - 1

-

-

Solve the equation. 5 ? 2 = 3 ? 19

5 - 2 = 3 - 19 Add 2x to both sides of the equality:

5 = 5 - 19 Add 19 to both sides:

5 = 24 Divide both sides by 5:

24 = 5 = .

Question 21:

[2]

Solution:

2.6 Simultaneous Linear Equations

Solve the simultaneous equations.

Question 22:

[3]

0.4 ? 5 = 27 2 + 0.2 = 9

Solution:

Rearrange one of the equations to get just x or just y on one side:

0.4 = 27 + 5 Substitute this into the second equation:

5(27 + 5) + 0.2 = 9 Simplify:

135 + 25.2 = 9 Solve:

25.2 = -126 = - Substitute your answer into one of the equations:

2 - 1 = 9 Solve for x:

2 = 10 = So the answer is:

= , = -

2.4 Indices

Simplify

5

3

2

?

1

-52

8

2

Question 20:

[2]

Solution:

To simplify the equation, we use the fact that

? = -

Hence:

53 8 2 ?

1 -5 2 2

=

5 (8

?

1 2)

32--25

=

5 4

4

So, the answer is:

.

=

PAGE 6 OF 26

CIE IGCSE MATHEMATICS//0580

2.7 Linear inequalities

2.8 Graphical inequalities

Solve the inequality. 3x - 1 11x + 2

Question 23:

[2]

Question 25:

Solution:

To solve the inequality 3 - 1 11 + 2 we must

rearrange for .

3 - 1 11 + 2

-3 from both sides

-1 8 + 2

-2 from both sides

-3 8

Divide both sides by 8

Hence we get

3 - 8

-

2.7 Quadratic Equations

Question 24:

= 2 + 7 ? 5 can be written in the form

= ( + )2 + .

Find the value of a and the value of b.

[3]

Solution:

If we expand ( + )2 and collect terms we get

= ( + )2 + = 2 + 2 + 2 +

If we now compare coefficients of the powers of x we

have

1: 2 = 7

=

or

3.5

0 (): 2 + = -5

49

4 + = -5

20 49

= - 4 - 4

=

-

or

-17.25

Find four inequalities that define the region, R, on the

grid.

[4]

Solution:

The lines on the grid that border R are

= 4

= 3

= 2

=

R is above 3 and below 4, to the right of 2 and to the

left of the diagonal line. Note that solid lines mean we

include them in the inequality. This is written as

<

>

2.9 Sequences and nth term

Question 26:

Find the nth term of each of these sequences.

(a) 16, 19, 22, 25, 28, ...

[2]

(b) 1, 3, 9, 27, 81, ...

[2]

Solution: Part (a)

The difference between the terms is 3. The sequence can then be written as

3 + Where a is some real number and n is the term. The first term is then

3 + = 16 = 13 Hence = +

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