﻿ Reciprocals | Mathematics Form 2

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Butterfly Chameleon #### Reciprocals - Mathematics Form 2

Background knowledge

1. Fractions

The four basic operations on fractions are:

b. Subtraction

c. Multiplication

d. Division

2. Decimals Decimals can be added and subtracted

Worked examples

a) 0.3457 + 0.06043

Solution 0.34570 0.06043 + 0.40613

b) 23.143 1.95

Solution 23.143 1.950 21.193

Decimals can be converted to fractions and vice versa

Worked examples

Standard form

Worked examples

Significant figures

Significant figures are used in approximations of numbers.

Worked example

(a) Write the following numbers to

3 significant figures

i. 3572 Solution 3572 = 3570 (3sf)

ii. 4.73478 Solution 4.73478 = 4.73 (3sf)

iii. 0.007235 Solution 0.007235 = 0.00724 (3sf) The following quadratic identities are useful in expansion of binomial expressions.

(a + b)

2

= (a + b) (a + b)

= a (a + b) + b(a + b)

= a2 + ab + ab + b2
= a2+ 2ab + b2

(a b) 2 mn <>o OpPT ABC(a b) (a b)

||||||||||||| ||||||||||||||||2

1, \ |||||x|||||,a

y||||||||2y+2|||b

= a (a b) b(a b)
= a2 ab ab + b2
= a2 2ab + b2

(a2 + b2) (a b)

= a (a b) + b (a b)    LESSON OBJECTIVES

By the end of the lesson the learner should be able to:-

find the reciprocals of numbers by division method

find the reciprocals of numbers by use of mathematical table

apply reciprocals of numbers.

LESSON OBJECTIVE
By the end of the lesson the you should be able to find the

volume of a frustrum.

RECIPROCALS

The reciprocal of a fraction is obtained by interchanging the numerator and the denominator, i.e. by inverting the fraction.

Reciprocals are very important in every day life.

Lesson Objectives

By the end of the topic you should be able to:

• solve quadratic equations by factorization
• form and solve quadratic equations

Objectives

By the end of the lesson you should be able to:

• Solve the linear inequalities in two unknown geographically
• Form simple linear inequalities from inequality graphs.

Background

Data refer to collection information obtained through experience or observation.

For example: the marks of a test of a class of 15 students are as follows :

25, 22, 25, 30, 28, 40, 30, 40, 34, 28, 26, 36, 42, 37, 28.

When data is not organized, it is referred to as raw data. The data above can be organized in order as follows.

22, 25, 25, 26, 28, 28, 28, 30, 30, 34, 36, 37, 40, 40, 42.

When the data is organized in such a way, it becomes easy to make comparisons and interpretation. It can be used to answer questions such as:

1. what is the highest and lowest marks?

2. what mark was scored by most students?

3. what is the difference between the highest and the lowest marks?

From the data above, some values appear more than once: the number of times a value appears is referred to as its frequency.

To show frequency (f), the data above can be organized further as follows. The number of times each value appears is indicated by tallying in the tally column. The values in the marks column are written in ascending order. Such a table that shows given values and their frequencies is called a frequency distribution table.

Example

The data below represents the marks obtained in a mathematics test.

61, 70, 40, 43, 50, 40, 30, 75, 80, 75, 61, 50, 30, 40, 80,

61, 50, 30, 40, 43,50,61, 30, 40, 30, 80, 30,50, 30, 50

Make a frequency distribution table and answer the following questions.

1. How many students sat for the test?
2. What were the lowest and highest marks?

What is the difference between the highest and lowest marks obtained?

Solution

The following is a frequency distribution table for the data The total number of students is found by adding the frequencies = 40.

The Highest mark =80 and

The Lowest mark = 30.

The difference between highest and lowest = 80-30 = 50.

Grouped data

Data can be arranged in groups or classes.The following table shows grouped data of marked obtained by 40 students in a mathematics rest. From the table 5 students scored between 10 and 19 marks,10 students scored 20 and 29 marks and so on. The table shows grouped data. In grouped data

1. class limits are the end values that specify a class interval.in the grouped data above ,10-19,20-29,50-59 are the classes definedd by their respective class limits.
2. A class boundary is obtained by calculating the average of the upper and the lower limits of corresponding classes. In the grouped data above ,class boundaries are obtained as follows  Objectives

By the end of the lesson you should be able to represent data in form of bar chart and histogram.

Background

Data refer to collection information obtained through experience or observation.
For example: the marks of a test of a class of 15 students are as follows
25 22 25 30 28
40 30 40 34 28
26 36 42 37 28
When data is not organized, it is referred to as raw data. The data above can be organized in order as follows.
22,25,25,26,28,28,2830,30,34,36,37,40,40,42.

When the data is organized in such a way, it becomes easy to make comparisons and interpretation. It can be used to answer questions such as:

1. what is the highest and lowest marks?
2. what mark was scored by most students?
3. what is the difference between the highest and the lowest marks?

from the data above, some values appear more than once: the number of times a value appears is referred to as its frequency.
To show frequency (f) the data above can be organized further as follows.

 MARKS TALLY FREQUENCY 22 / 1 25 // 2 26 / 1 28 /// 3 30 // 2 34 / 1 36 / 1 37 / 1 40 // 2 42 / 1

The number of times each value appears is indicated by tarrying in the tarry column. The values in the marks column are written in ascending order. Such a table that shows given values and their frequencies is called a frequency distribution table.

Example

The data below represents the marks obtained in a mathematics test.
61 70 40 43 50
40 30 75 80 75
61 50 30 40 80
61 50 30 40 43
50 61 30 40 30
80 30 50 30 50
Make a frequency distribution table and answer the following questions .
1. How many students sat for the test?
2. What were the lowest and highest marks?

What is the difference between the highest and lowest marks obtained?

Solution

The following is a frequency distribution table for the data
 Marks Tally F 30 ///// // 7 40 ///// 5 43 // 2 50 ///// / 6 61 //// 4 70 / 1 75 // 2 80 /// 3

The total number of students is found by adding the frequencies =40
Highest mark =80
Lowest mark =30
Difference between highest and lowest =80-30 =50
Grouped data
Data can be arranged in groups or classes. The following table shows grouped data of marked obtained by 40 students in a mathematics rest.

 Marks F 10-19 5 20-29 10 30-39 12 40-49 7 50-59 6

From the table 5students scored between 10 and 19 marks,10 students scored 20 and 29 marks and so on. The table shows grouped data.
In grouped data

1. class limits are the end values that specify a class interval.in the grouped data above ,10-19,20-29,50-59 are the classes defund by their respective class limits.
2. A class boundary is obtained by calculating the average of the upper and the lower limits of corresponding classes. In the grouped data above ,class boundaries are obtained as follows
 class limits boundaries Lower Upper 10-19 ? 20-29  30-39  40-49  50-59 ?

(highlight the diagonals whose average is worked out to get the class boundaries)
Therefore, class boundaries
? - 19.5
19.5 - 29.5
29.5 - 39.5
39.5 - 49.5
49.5 - ?
To get the missing class boundaries of the first and last classes you assume there us a class above the first class and a class below the last class and then calculate the class boundaries as follows: The class boundary is 10 - 19
Similarly for the Last work out the class boundary as follows:

60 -69

10 59 =59.5

Or these two values can be obtained by observation i.e.

Example

Using the data below;
1. find the class boundaries of each class
2. Calculate the class intervals.

Class

9 - 19
20 - 25
26 - 29
30 - 49
Solution
The class boundaries are worked out as follows:
Therefore class boundaries are
8.5 - 19.5
19.5 - 25.5
25.5 - 29.5
29.5 - 49.5
The clas interval is obtained by getting the difference beteen the upprt class boundary and lower class boundary.
 Class boundary Class interval 8.5 - 19.5 19.5 8.5 = 11 19.5 - 25.5 25.5 19.5 = 6 25 - 29.5 29.5 25.5= 4 29.5 - 49.5 49.5 29.5 = 20

Introduction

Reciprocal of a number is 1 divided by the number itself. In general, the reciprocal of x is 1 divided by x which is 1/x.

Examples

<!--[if !supportLists]--> (a)    <!--[endif]--> The reciprocal of 3 is 1/3.

<!--[if !supportLists]--> (b)   <!--[endif]--> The reciprocal of 100 is 1/100

<!--[if !supportLists]--> (c)    <!--[endif]--> The reciprocal of 3/2 is 1/3/2 = 2/3

<!--[if !supportLists]--> (a)    <!--[endif]--> Reciprocal of 0.1 is  1 but 0.1 = 1

10

Therefore of 0.1 is 1/1/10 = 1 x 10/1 = 10

<!--[if !supportLists]--> (b)   <!--[endif]--> Reciprocal of 0.05 is 1 but 0.05 = 5

0.05      100

Therefore reciprocal of 0.05 is 1    = 1 x 100

5        5

100

= 100

5

= 20

NB

Reciprocals of 3 is 1/3. Use long division to evaluate the reciprocal as follows:

<!--[if !vml]--> <!--[endif]-->     0.333..

3  10

9

10

9

10

9

1

The reciprocal of 3 is 0.333. which cab stated to the required approximation (decimal places or significant figures).

The reciprocal of 2/3 is  13

2  2

3

1.5

<!--[if !vml]--> <!--[endif]--> <!--[if !vml]--> <!--[endif]-->          = 2  3

2

10

10

. .

The reciprocal of 2/3 is 1.5

The reciprocal of 0.05 is   1     but 0.05  = 5

0.05     100

Therefore  1     is   1  = 100

0.05      5

100

20

= 5    100

10

0

The reciprocal of 0.05 is 20

Examples

1. Find the reciprocal of 8 by division method

Solution

1. Find the reciprocal of 8 by division method
Step (1) The reciprocal of 8 is 1 (leave box for filling)

||||||||||||||||||||||||||||||||||||||||||_

||||||||||||||||||||||||||||||||||||||||||8

|||||||||||||||||||0.125

||||||||||||||||||__________
Step (2) = 8 |||1

Step (iii) Therefore the reciprocal of 8 =

2. Find the reciprocal of 1.6 by division method
Step (1) 1.6 =1

Step (ii) =
=
Step (iii) The reciprocal of =
Step (iv)5 = 8

Step (v) The reciprocal of 1.6 = 0.625

3. Use reciprocal tables to find the reciprocal f 0.924
Step 1: Write the number in standard form as
9.25 x 10-1

Step 2: The reciprocal of 9.24 = 0.1082

Step 3: Reciprocal of 10-1 =

= = 10-1
Reciprocal of 10-1 = 10
Reciprocal of 9.24 x 10-1 = 0.1082 x 10

Reciprocal of 0.924 = 1.082
4. Use reciprocal tables to find the reciprocal of 157.5
Step 1: Write 157.5 in standard form as
1.575 x 102
Step 2: The reciprocal of 1.575 = 0.6369
21
0.6348

Step 3: The reciprocal of 102 =

Step 4: The reciprocal of 1.575 x 102 = 0.6348 x
The reciprocal of 157.5 = 0.006348

Application of reciprocals

Worked examples
Use reciprocal tables to evaluate each of the following
a)

can be written as 5 x
The reciprocal of 5.723 is 0.1748
1 -
0.1747

5 x = 5 x 0.1747
= 0.8735

= 0.8735

b)

Solution

First consider
=

0.0612 in standard form is 6.12 x 10-2
The reciprocal of 6.12 x 10-2 is 0.1634 x 102 = 16.34

= 7 x 16.34
= 114.38

Now consider

=

Write 20.5 in standard form i.e. 2.05 x 101
Reciprocal of 20.5 x 101 = 0.4878 x 101
= 0.04878
= 3 x 0.04878
= 0.14634

= 114.234

Exercise
MA1-100101e1
1. Use reciprocal tables to evaluate 8
0.735
Step I: 8 can be written as 8 x 1
0.735 0.735
Step II: Write 0.735 in standard form as 7.35 x 10-1

Step III: The reciprocal of 7.35 x 10-1 = 0.1361 x 101
= 1.361
Step IV: = 8 x 1.361
Step V: = 10.888

MA1-100101e2
2.Use reciprocal tables to evaluate

Step 1: 2 can be written as 2 x 1
0.0457 0.0457

Step II: Write 0.0457 in standard form as 4.57 x 10-2

Step III: The reciprocal of 4.57 x 10-2 = 0.2188 x 102
= 21.88
Step IV: 2 x 1 = 2 x 21.88
0.0457

2 = 43.76
0.0457
Step V: 9 can be written as 9 x 1
253 253

Step VI: Write 253 in standard form as 2.53 x 102

Step VII: The reciprocal of 2.53 x 102 = 0.3953 x 10-2
= 0.003953

Step VIII: 9 x 1 = 9 x 0.003953
253

9 = 43.76 + 0.035577
253
= 43.795577

Step (1) The reciprocal of 8 is (leave box for filling)
ma2-reci002

0.125
Step (2) = 8 1

ma2-reci003

Step (iii) Therefore the reciprocal of 8 =

2. Find the reciprocal of 1.6 by division method
ma2-reci003A
Step (1) 1.6 =1
ma2-reci200

Step (ii) =
ma2-reci003B
=
ma2-reci003C
Step (iii) The reciprocal of =
ma2-reci003D
Step (iv)5 = 8
ma2-reci003E

Step (v) The reciprocal of 1.6 = 0.625

ma2-reci300

3. Use reciprocal tables to find the reciprocal f 0.924
Step 1: Write the number in standard form as
9.25 x 10-1

Step 2: The reciprocal of 9.24 = 0.1082

Step 3: Reciprocal of 10-1 =

= = 10-1
Reciprocal of 10-1 = 10
ma2-100400
Reciprocal of 9.24 x 10-1 = 0.1082 x 10

Reciprocal of 0.924 = 1.082
ma2-100500
4. Use reciprocal tables to find the reciprocal of 157.5
Step 1: Write 157.5 in standard form as
1.575 x 102
Step 2: The reciprocal of 1.575 = 0.6369
21
0.6348

Step 3: The reciprocal of 102 =

Step 4: The reciprocal of 1.575 x 102 = 0.6348 x
The reciprocal of 157.5 = 0.006348

Application of reciprocals

Worked examples
Use reciprocal tables to evaluate each of the following
a)

can be written as 5 x
The reciprocal of 5.723 is 0.1748
1 -
0.1747

5 x = 5 x 0.1747
= 0.8735

= 0.8735

b)

Solution

First consider
=

0.0612 in standard form is 6.12 x 10-2
The reciprocal of 6.12 x 10-2 is 0.1634 x 102 = 16.34

= 7 x 16.34
= 114.38

Now consider

=

Write 20.5 in standard form i.e. 2.05 x 101
Reciprocal of 20.5 x 101 = 0.4878 x 101
= 0.04878
= 3 x 0.04878
= 0.14634

= 114.234

Exercise
MA1-100101e1
1. Use reciprocal tables to evaluate 8
0.735
Step I: 8 can be written as 8 x 1
0.735 0.735
Step II: Write 0.735 in standard form as 7.35 x 10-1

Step III: The reciprocal of 7.35 x 10-1 = 0.1361 x 101
= 1.361
Step IV: = 8 x 1.361
Step V: = 10.888

MA1-100101e2
2.Use reciprocal tables to evaluate

Step 1: 2 can be written as 2 x 1
0.0457 0.0457

Step II: Write 0.0457 in standard form as 4.57 x 10-2

Step III: The reciprocal of 4.57 x 10-2 = 0.2188 x 102
= 21.88
Step IV: 2 x 1 = 2 x 21.88
0.0457

2 = 43.76
0.0457
Step V: 9 can be written as 9 x 1
253 253

Step VI: Write 253 in standard form as 2.53 x 102

Step VII: The reciprocal of 2.53 x 102 = 0.3953 x 10-2
= 0.003953

Step VIII: 9 x 1 = 9 x 0.003953
253

9 = 43.76 + 0.035577
253
= 43.795577

Exercises:
Evaluate the following using logarithm tables:

(i) 0.09137 x 0.00378
0.0537

Solution
Number log
0.09137
0.00378

0.0537

6.4313x10-3

(ii)
NO Log
0.60322
x 2

Step II
NO LOG

1.022x10-1

(iii)
Solution
Step 1
NO LOG
0.3789

0.03487

0.3878

Step 2
=
Find the antilog of
=4.296 x 10-1 = 0.4296

Volume of a Cone

To find the volume of the above cone you proceed as follows Example 1 Find the volume of a cone whose slant height is 6cm and radius is 4cm. We use Pythagoras theorem to find height of the cone i.e. = 36-16= 20 h = 4.4721cm h = 4.472cm Therefore the area of cone = base Area x height =74.94cm3

Horizontal Bar Chart

The following is horizontal bar chart representing the given data.

Note that the marks are indicated on the x-axis while the subjects appear on the y-axis.

Vertical Bar Charts

In this case the marks are plotted on the y-axis. The marks represents frequency

Note that:

1. The bars are of equal width
2. The distance between the bars is the same.
3. the length of the bars represents the frequency

Example 2 The table below shows marks obtained in an English ndraw a bar chart to represent the data.How many students scored 50 marks and above? If the pass mark was 40%, how many students scored 50 marks and above?

Solution.

The vertical bar chart is obtained by plotting the frequency on the vertical axis and the classes on the horizontal axis as follows. 1. The number of students who scored 50marks and above is worked out as follows

50 - 59 10 students

60 - 69 8

70 - 79 8

Therefore total of students =10 + 18 + 8 = 36

2. the students who failed the examinations scored 39% and below The number of students is worked out as follows:

10 - 19 5 students

20 - 29 11

30 - 39 15

Therefore total number of students =5 + 11 + 5 = 31  The histogram is drawn by plotting the class boundaries on the horizontal axis and the frequency density on the vertical axis. Both axis must have a scale

Note:The width of each bar is proportional to the size of the class it represents.The area of the bar represents the frequency of a particular class  The following is the required histogram

The distance from the centre of the circle to any point on the circumference.It is also half the diameter. An Arc

It is a part of the circumference.

The chord.

It is a straight line joining two points on the circumference. The Diameter

It is also a chord which divides the circle into two equal parts. The Sector

It is an area bounded by two radii and an arc. Objectives

By the end of the lesson you should be able to:

relate and compute angle subtended by an arc at the circumference correctly.

relate and compute angle subtended by an arc at the centre and at the circumference correctly.

Angles Subtended by Same Arc at Cicumference

Example 1

Given the diagram below

What do you notice about arc XY?

It subtends angle X at the circumference.

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