 # SS1-1ST-TERM-MATHS-E-NOTES

WEEK TOPIC
MATHEMATICS
FIRST TERM SCHEME OF WORK
THEME: NUMBER AND NUMERATION:
1. Revision of expanded notation 2. Number Base System: (a) Conversion from one base to base 10. (b) Conversion of
decimal fraction in one base to base 10.
3. Number Base System: (c) Conversion of number from one base to another base. (d) Addition, subtraction, multiplication and division of number bases. (e) Application to computer programming.
4. Modular Arithmetic: (a) Revision of addition, subtraction, multiplication and division
of integers. (b) Concept of module arithmetic.
5. Modular Arithmetic:(c) Addition, subtraction and multiplication operations in module
arithmetic. (d) Application to daily life.
6. Standard Form and Indices: (a) Revision of standard form (b) Introduce indices and
examples (c) Laws of indices; (i) ax x ay = ax+y (ii) ax÷ay = ax-y (iii) (ax)y = axy, etc. (d) Application of indices, simple indicial equation.
7. Logarithms: (a) Deducing logarithm from indices and standard form. (b) Definition of Logarithms (c) Graph of y = 10x (d) Reading of logarithm and the antilogarithm tables.
8. Logarithms:(e) Use of logarithm table and antilogarithm table in calculation involving (multiplication, division, powers and roots.(f) Application of logarithm in capital market and other real life problems.
9. Sets: (a) Definition of set (b) Set Notation – (i) listing or roster method (ii) rule method (iii) set builder notation. (c) Types of sets – (i) empty set (ii) Guide and infinite sets (iii) Universal sets.
10. Sets: (d) Set operations (i) Union (ii) Intersection (iii) Complement. (e) Venn diagram
and application up to 3 set problem.
11. Revision. 12. Examination.

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WEEK 1
REVISION OF EXPANDED NOTATION
CONTENT: 1. Concept of expanded notation
2. Selected difficult topics
1. Concept of expanded notation: Every decimal number X can be expressed uniquely in the form:
𝑋 = 𝐼𝑛 × 10𝑛 + 𝐼𝑛−1 × 10𝑛−1 + 𝐼𝑛−2 × 10𝑛−2 + ⋯ + 𝐼𝑛−𝑛 × 10𝑛−𝑛
This is known as the expanded notation
EXAMPLE 1
Express the following in expanded notation form
(a) 45078 (b) 0.0235 (c) 930.133 (a) 45078 = 4 × 104 + 5 × 103 + 0 × 102 + 7 × 101 + 8 × 100
= 4 × 10000 + 5 × 1000 + 0 × 100 + 7 × 10 + 8 × 1
(𝑏) 0.0235 = 0 × 100 + 0 × 10−1 + 2 × 10−2 + 3 × 10−3 + 5 × 10−4
= 0 × 1 + 0 ×
1 10
+ 2 ×
1 102 + 3 ×
1 103 + 5 ×
1 104
(𝑐) 930.133 = 9 × 102 + 3 × 101 + 0 × 100 + 1 × 10−1 + 3 × 10−2 + 3 × 10−3
= 9 × 102 + 3 × 101 + 0 × 1 + 1 ×
1 101 + 3 ×
1 102 + 3 ×
1 103
EXAMPLE 2
Write the following in expanded notation form
(a) 32.516 (b) 0.10012

(a) 32.516 = 3 × 61 + 2 × 60 + 5 × 6−1 + 1 × 6−2
= 3 × 6 + 2 × 1 + 5 ×
+ 1 ×
1
6
1 62
(𝑏) 0.10012 = 0 × 20 + 1 × 2−1 + 0 × 2−2 + 0 × 2−3 + 1 × 2−4
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= 0 × 1 + 1 ×
1
21 + 0 ×
1
22 + 0 ×
1
23 + 1 ×
1 24
= 0 +
1
2
+
0
4
+
0
8
+
1
16
EVALUATION:
Write the following decimal numbers in an expanded notation form
(a) 402 (b) 60.008 (c) 0.0153
GENERAL EVALUATION:
Write in the expanded notation form.
(a) 6.666 (b) 315.014 (c) 1
1 2
General revision
WEEKEND ASSIGNMENT:
(1) Write the following base ten numbers in expanded notation form
(a) 862051 (b) 27654 (c) 7569

(2) Express the following numbers as numbers in the denary scale using expanded form
(a) 111.112 (b) 13.628 (c) 312.334

(3) Write the following numbers in their ordinary form
(a) 6 × 103 + 0 × 102 + 5 × 101 + 8 × 100 (b) 4 × 101 + 3 × 100 + 0 × 10−1 + 2 × 10−2 (c) 5 × 70 + 8 × 7−1 + 9 × 7−2

REFERENCE TEXTS:
 New General Mathematics for senior secondary schools 1 by M.F Macrae et al;
pearson education limited
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New school mathematics for senior secondary school et al; Africana publishers limited

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CONTENT:
WEEK 2:
NUMBER BASE SYSTEM
 Concept of number base system  Conversion from one base to base 10  Conversion of decimal fraction in one base to base 10
SUB-TOPIC 1: CONCEPT OF NUMBER BASE SYSTEM
A number system is defined by the base it uses, the base being the number of different symbols required by the system to represent any of the infinite series of numbers. A base is also a number that, when raised to a particular power (that is, when multiplied by itself a particular number of times, as in 102 = 10 x 10 = 100), has a logarithm equal to the power. For example, the logarithm of 100 to the base 10 is 2.
SUB-TOPIC 2: CONVERSION FROM ONE BASE TO BASE 10
Two digits—0, 1—suffice to represent a number in the binary system; 6 digits—0, 1, 2, 3, 4, 5—are needed to represent a number in the sexagesimal system; and 12 digits—0, 1, 2, 3, 4, 5, 6, 7, 8, 9, t (ten), e (eleven)—are needed to represent a number in the duodecimal system. The number 30155 in the sexagesimal system is the number (3 × 64) + (0 × 63) + (1 × 62) + (5 × 61) + (5 × 60) = 3959 in the decimal system; the number 2et in the duodecimal system is the number (2 × 122) + (11 × 121) + (10 × 120) = 430 in the decimal system.
To convert from any base to base ten, expand the given number(s) in the powers of their bases and simplify.
Examples:
1. Convert 1243five to base ten Solution:
1243five= (1 × 53) + (2 × 52) + (4 × 51)(3 × 50) = 125 + 50 + 20 + 3 = 198ten
2. Convert 1111110two to a number in base ten.
Solution:
1111110two= (1 × 26) + (1 × 25) + (1 × 24)+(1 × 23) + (1 × 22)
(1 × 21) + (0 × 20) = 64 + 32 + 16 + 8 + 4 + 2 + 0

= 126 ten

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3. Convert 4𝑒𝑡𝑡𝑤𝑒𝑙𝑣𝑒 to base ten.
Solution
4𝑒𝑡𝑡𝑤𝑒𝑙𝑣𝑒 = (4 x 122) + (11 x 121) +( 10 x 120)
=( 4 x 144 )+ (11 x 12) + (10 x 1)
= 576 + 132 + 10
= 718𝑡𝑒𝑛
4. Convert 𝑡5𝑒𝑡𝑤𝑒𝑙𝑣𝑒 to base ten.
Solution 𝑡5𝑒𝑡𝑤𝑒𝑙𝑣𝑒 = 10 x 122 + 5 x121 + 11 x 120 = 10 x 144 + 5 x 12 + 11 x 1 = 1440 + 60 + 11 = 1511𝑡𝑒𝑛
Thus, the decimal system in universal use today (except for computer application) requires ten different symbols, or digits, to represent numbers and is therefore a base-10 system.

 Conversion from other base less than ten to base ten  Conversion from other base greater than ten to base ten  Conversion of decimal fraction in one base to base ten  Conversion of fractions in base ten to any base
SUB-TOPIC 1: CONVERSION FROM OTHER BASE LESS THAN TEN TO BASE TEN
Firstly, we shall consider conversion of numbers in a base less than ten to a base ten number. A number in base ten is known as a decimal or denary number. All numbers in a given n can be written using only the following digits 0, 1, 2 , ….., n -1. For instance in base two, the only digits that can be used are only 0 and 1. In base three, you can only use digits 0, 1 or 2.
Generally our normal counting is done in base ten when doing this, the base is normally indicated. E.g in the denary number 546 is 546ten.
The first digit from the right towards left is 6 and is called the unit digit. The next digit is 4 and is called the tens digit and has value 4×10.
The next digit 5 is called the hundred digit i.e 500. Hence 52 41 60 5×102+4×101+6×100
ten= 5 4 6 =
= 500 + 400+ 6
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= 546
Notice that the digits listed on top of the denary numbers 546 are the powers of the base.
Example
Convert 1203 in base five to a denary number.
Solution
Circle the digits of the number 1203 from the last to the first, beginning at zero
i.e 1203five
Therefore expand number raising the base five to the grades listed on the top of the number as shown below
i.e1 2 0 3 five= 1×53+2×52+0x51+3×50
= 1×125+2×25+0x5+3×1
= 125+50+0+3
= 178
The new number is in base ten i.e 1203five= 178ten

This expansion method can be used in converting from any base to base ten.
Evaluation
Convert the following to denary numbers
(i) 10.2three (ii) 214seven (iii) 780nine (iv) Find if 200x + 144nine = 14Btwelve. (v) Solve for x and y if 32x + 53y + 61nine (vi) 24x + 35y = 45ten
SUB-TOPIC 2: CONVERSION FROM OTHER BASE GREATER THAN TEN TO BASE TEN
Expansion method can be used to convert numbers in base say base thirteen to base ten, Remember in base thirteen the digits we have are 0, 1, 2, 3,4,5, 6,7,8,9, A, B, C. where A represents ten B represents eleven and C represents twelve. Letters are used for two- digits numbers less than the base thirteen.
Example
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Convert 1B9thirteen to denary number
Solution
1B9thirteen= 1×132+Bx131+9×130
= 1×169+11×13+9×1
= 169+143+9
= 321ten
Example: Convert 206fifteen to a denary number
Solution
20Cfifteen= 2×152+0x151+12×150
= 2 x 225 + 0 x 15 + 12 x 1
= 450 + 0 x 15 + 12 x 1
= 462ten
Evaluation
Convert the following to denary numbers
1. 1024eleven 2. 2059twelve 3. 51Cfourteen
ASSIGNMENT
A. Convert the following numbers to denary numbers
(i) 10011two (ii) 768nine (iii) 10Aeleven (iv) B12twelve (v) 7B3Atwelve (vi) 6D4Fsixteen
B. Read example 17, 2nd method on page 52 of New General Mathematics SS1
SUB-TOPIC 3: CONVERSION OF DECIMAL FRACTIONS IN ONE BASE TO BASE TEN
Sometimes we are faced with numbers which are not whole numbers. Hence it is very necessary to study also the conversion of fractional parts of numbers. The following
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examples can be used in our study of the conversion of fractional parts of other bases to decimal system.
Example 1:
Convert 6.47 to denary number
Solution
6.47 = 6 × 70 + 4 × 7−1
= 6 × 1 + 4 ×
1 7

= 6 +
4 7
(6
4 7
)10
Example 2:
Convert 101.011𝑡𝑤𝑜to base ten.
Solution
101.011𝑡𝑤𝑜𝑡 = 1 × 22 + 0 × 21 × 1 × 20 + 0 × 2−1 + 1 × 2−2 + 1 × 2−3
= 1 × 4 + 0 × 2 + 1 × 1 + 0 ×
= 4 + 0 + 1 + 0 +
1 2
1 4
1 ×
1 22 + 1 ×
1 23
+
1 8
= 5 +
1 4
+
1 8

= 5
3 8𝑡𝑒𝑛
𝑜𝑟 5.375𝑡𝑒𝑛
Example 3:
Convert 32.516 to base ten.
Solution
32.516 = 3 × 61 + 2 × 60 + 5 × 6−1 + 1 × 6−2
= 3 × 6 + 2 × 1 + 5 ×
1 6
+ 1 ×
1 62
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= 18 + 2 +
5 6
+
1 36

= 20 +
= (20
30 + 1
36

31 36
)10
32.516 = (20
31 36
)10
= 20.86 (2 𝑑. 𝑝)
EVALUATION
1. Convert 11.011𝑡𝑤𝑜 to a given number in base ten. 2. Convert the binary number11011.11 to base 10.
SUB-TOPIC 4:
CONVERSION OF FRACTIONS IN BASE TEN TO ANY OTHER BASE
Fraction in base ten can be converted to other bases using various methods.
Example 1:
Express
7 to bicimals. 8
Solution
Change
7 to decimal fraction i.e 8
7
8
= 0.875 and multiply by 2.
0.875
X 2
1750
X 2
1500
X 2
1000
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As we multiply 2 × 0.875, we get 1.750. Keep the 1 and multiply 750 by 2, and get 1.500. = Keep the 1 and multiply 500 by 2 and get 1.000. Stop when all is zero. The value of 0.111𝑡𝑤𝑜or convert 7 and 8 to base two and then divide.
7
8
2
2 2
2
2
2
2

7
3 R 1 1 R 1 0 R 1 7𝑡𝑒𝑛 = 111𝑡𝑤𝑜

8
4 R 0
2 R 0
1 R 0

0 R 1
8𝑡𝑒𝑛 = 1000𝑡𝑤𝑜
∴ [
111 1000
] = 0.111𝑡𝑤𝑜
Example 2:
Express (19
Solution
13
25
)𝑡𝑒𝑛 to base five.
5
5

5
5

19
3 R 4
0 R 3

13
2 R 3
0 R 2
= 345
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5
5
5

= 23𝑓𝑖𝑣𝑒

25
5 R 0
1 R 0
0 R 1 = 100𝑓𝑖𝑣𝑒

= [34 +
23
] 100 𝑓𝑖𝑣𝑒

= [34 + 0.23]𝑓𝑖𝑣𝑒
= 34.23𝑓𝑖𝑣𝑒
EVALUATION
1. Express 2. Express (11011.10012) to decimal base.
to base two
5
13
WEEKEND ACTIVITY
Convert the following numbers in denary to the base indicated.
(𝑎)37.31𝑡𝑒𝑛 to base 6
(b) 10.8𝑡𝑒𝑛to base 3.
Find the value of 𝑥 in each of the following equations.
(a) 23𝑥 + 14𝑥 = 42𝑥 (b) 53𝑥 − 24𝑥 = 25𝑥 (c) 113𝑥 + 121𝑥 = 300𝑥 (d) 562𝑥 − 153𝑥 = 407𝑥
Find the value of 𝑥 and 𝑦 in the following pairs of equation.
(a) 32𝑥 + 53𝑦 = 55

24𝑥 + 35𝑦 = 45
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(b) 64𝑥 − 53𝑦 = 25

(c) 54𝑥 − 11𝑦 = 36

(d) 34𝑥 + 21𝑦 = 26
WEEKEND ASSIGNMENT
47𝑥 − 34𝑦 = 21
25𝑥 + 10𝑦 = 21
42𝑥 − 12𝑦 = 17
New General Mathematics for Senior Secondary Schools 1, page 56, question 2 (a-c)
REFERENCE TEXTS
1. M.F Macrae etal (2011),New General Mathematics for Senior Secondary Schools 1. 2. MAN Mathematics for senior Secondary Schools 1. 3. New school mathematics for senior secondary school et al; Africana publishers
limited
4. Fundamental General Mathematics For Senior Secondary School by Idode G. O

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WEEK 3
CONVERSION OF NUMBER FROM ONE BASE TO
ANOTHER
Content
1. Conversion from base ten to other bases 2. Conversion from one base to another base via base ten.
We convert from base ten to other bases by repeated division
Example. Convert 137ten to a base five number
5
5
5
5
137
27 r 2

5 r 2
1 r 0
0 r 1
: . 137ten = 1022five
Example converts the denary number 102 to a base twelve number

12
12

107
8 r 11

0 r 8
107ten = 8Btwelve
To convert from a base to another you may have to pass through base ten
Example, Convert 301four to a base six number.
Solution
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First 301fourwill be converted to a base ten number
301four = 3×42 + 0x41 + 1×40
= 48 + 0 + 1
= 49ten
49tenwill now be converted to a base six number by repeated division
6
6
6

49
8 r 1
1 r 2

0 r 1
301four = 121six
Evaluation
1. Convert 2210three to a base five number 2. Convert 5201seven to a binary (base two) number
Addition, Subtraction and Multiplication of number
Operation in other bases other than base ten are carried out in a manner similar to what is obtained in base ten. We can illustrate the procedure as shown in the example below.
Example. 167eight + 125eight
Solution
167eight
+ 145eight

7+5 = 12. This exceeds the value of the base. 12 contain a bundle of 8 and 4 units. That one bundle of 8 is carried to the next column as 1
1 + 6 + 4 = 11
11 is another single bundle of 8 and three, Hence we write 3 and carry the bundle to the next column as 1
167eight
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+ 145eight
334eight
Example 501twelve – 3Btwelve
501eight
+ 3Beight

Recall B in base twelve is eleven.
If 1 is ‘borrowed from 5 in the third column, getting to the next column on the right becomes a twelve. From it we can take one to the next column to the right again. To get 12+1 = 13 from which we finally subtract B (i.e eleven)
501twelve
+ 3Btwelve
482 twelve
Notice that after borrowing 1 from the middle column, eleven was left. If is out of this eleven that 3 is subtracted to get 8 in the second column of the answer.
Example. Simply 134six x 5six
154six
+ 5six

5 x 4 = 20 i.e 3 bundles of 6 plus 2 units. Write 2 add 3 to the product of 5 x 5 of second column to get 28. 28 = 4(sixes) plus 4. Take the 4 bundles to next column. 4 + 5x 1 = 9 which is 13six. So 154six x 5six = 1342 six
Example. Simplify 134five x 24five
134five
x 24five

4 x 4 = 16 i.e 3(fives) and 1 unit.
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These 3 bundle of 5 is added to the product of (3×4) of the second column. 3×4+3 = 15.
15 = 3(fives) and zero
This new 3 bundles of 3 is to be added to the product 1x 4 of the third column
1 x 4 + 3 = 7 which will written as 12five similar thing is done with 134five times the distance 2, thus
134five
x 24five
1201
323
4431five
Evaluation
1. 1205six x 3six 2. 143five + 24five 3. 211four + 32four 4. 103four x 32four
Division of numbers bases
Since in binary (base two) system, the digits we have are 0 and 1. Each digit of the quotient 110111 ÷ 101 must be either 1 or 0.

1011
101 110111
101
11
0

111
101
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101
101
Once you start the division, the digits are brought down one after the other.
Example 240six÷ 20six
12
20
240

20
40
40
So 240six÷ 20six= 12𝑠𝑖𝑥
Evaluation simplify the following
1. 4 7 7eight

2. BBtwelve 3. 1011two x 111two
+36 7eight + A1twelve

4. Which is bigger E5Asixteen or 1271fifteen
5. 387nine÷ 25nine
ASSIGNMENT
New General Mathematics SS1. Ex 3i Nos 1 – 5 page 54.
Functional Mathematics Graduated Exercise page 17 – 18, No 1- 30

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CONTENT:
WEEK 4
MODULAR ARITHMETIC
 Revision of addition and subtraction of integers  Revision of multiplication and division of integers  Concept of modular arithmetic/Cyclic events

SUB-TOPIC 1 & 2: Revision of addition, subtraction, multiplication and division of integers
Recall: Integer is a counting whole numbers, either positive or negative. Examples 1,5,20, −1, −5 𝑒𝑡𝑐 These numbers can be added, subtracted, divided or multiplied. (i)
Addition of integers; (a) 486 + 289 = 775 (b) −25 + (−78) = −103 Subtraction of integers; (a) 582 − 328 = 254 (b) 902 − 437 = 465
(iii) Multiplication of integers;
(ii)
(iv)
(a) 181 × 42 = 7602 (b) 208 × 5 = 1040 Division of integers; (a) 972 ÷ 27 =
972
= 36
27
1008
= 84
12
(b) 1008 ÷ 12 =
EVALUATION:
Solve the following;
(i) (ii) (iii) (iv)
3092 + 216 + 1801 = ̇ 2968 − 989 = ̇ 318 × 322 = ̇ 420 ÷ 35 = ̇
SUB-TOPIC 3: Concept of Modular Arithmetic
The word Modular implies consisting of separate parts or units which can be put together to form something, often in different combinations.
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Arithmetic− the science of numbers involving adding, subtracting, multiplying and dividing of numbers
Modular Arithmetic is simply the arithmetic of remainders when an integer is divided by a fixed non-zero integer.
Examples;
Reduce 65 to its simplest form in:
(a) modulo 3 (b) modulo 4 (c) modulo 5 (d) modulo 6

(a) 65 ÷ 3 = 21, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 2
65 = 2(𝑚𝑜𝑑 3)
(b) 65 ÷ 4 = 16, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1
65 = 1(𝑚𝑜𝑑 4)
(c) 65 ÷ 5 = 13, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0
65 = 0(𝑚𝑜𝑑 5)
(d) 65 ÷ 6 = 10, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 5
65 = 5(𝑚𝑜𝑑 6)
EVALUATION:
Reduce 72 to its simplest form
(a) Modulo 3 (b) Modulo 4 (c) Modulo 5 (d) Modulo 6 (e) Modulo 7
Cyclic Events: Cyclic means happening in cycles.
Just as you ride your bicycle, the wheel rotates from a point to another. There are events that have constant ice day’s interval of three days, four, five or a week.
Examples: If ice cream is served every three days. If you are served on Thursday, the next serving will be 𝑡ℎ𝑢𝑟𝑠𝑑𝑎𝑦 + 3𝑑𝑎𝑦𝑠 = 𝑠𝑢𝑛𝑑𝑎𝑦
Find the number which results from the following additions on the number cycle below of ice cream
(a) 2 + 9 = 11
11 ÷ 3 = 3, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 2 ∴ 2 + 9 ≡ 2
(b) 𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦 3 + 14 𝑖𝑛 4 𝑐𝑦𝑐𝑙𝑖𝑐 𝑒𝑣𝑒𝑛𝑡𝑠,
17 ÷ 4 = 4, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1
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∴ 3 + 14 ≡ 1
EVALUATION:
Use the number cycle 5 to simplify (a) 1 + 6 (b) 2 + 32
New General Mathematics for SSS 1, pages 227; exercises 20a, 20b
Mathematical Association of Nigeria (MAN) pages 14-24
WEEKEND ASSIGNMENT
New General Mathematics for SSS 1, pages 227; exercises 20a, 20b, 20c
Mathematical Association of Nigeria (MAN) pages 14-24

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CONTENT:
WEEK 5
MODULAR ARITHMETIC
 Addition, Subtraction, Multiplication and Division operations in module arithmetic  Application to daily life
SUB-TOPIC 1 Addition and Subtraction: This can be either addition or subtraction tables where the number of digits given represents the modulo. Examples; (a)

0
1
2
0
0
1
2
1
2
1
2
0
2
0
1
The table above shows addition (mod 3)
Θ 0
1
2
0
0
1
2
3

1

0

3
1

(b) The table above shows subtraction (mod 4)
(c) (i) Find 39 29(mod 6)
Solution: 39 29= 68
= (6×11+2)
= 2(mod 6)
N.B 68 ÷ 6 = 11, 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 2
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= 2(𝑚𝑜𝑑 6)
(ii) Calculate the following in the given moduli (a) 12Θ 5(mod 4) (b) 38 Θ 42(mod 7)
Solution: (a) 12Θ 5 = 7
7 = 4 + 3
= 3(mod 4)
(b) 38 Θ 42 = −4
−4 = −7 + 3
= 3(mod 7)
EVALUATION:
(1) Find the following additions modulo 5
(a) 3 9 (b) 65 32 (c) 41 52 (d) 8 17
(2) Find the simplest positive form of each of the following numbers modulo 5
(a) −9 (b) −32 (c) −75 (d) −256
SUB-TOPIC 2
Multiplication of modulo
Examples: Evaluate the following modulo 4
(a) 2 2 (b) 5 7 (c) 6 73
Solution: (a) 2 2 = 4
= 4 + 0(mod 4)
= 0(mod 4)
(b) 5 7 = 35
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= 4 x 8 + 3
= 3(mod 4)
(c) 6 73 = 438
= 4 x 109 + 2 = 2(mod 4)
EVALUATION:
Find the values in the moduli written beside them
(a) 16 7(mod 5) (b) 21 18(mod 10) (c) 8 25(mod 3) (d) 27 4(mod 7) (e) 80 29(mod 7)
SUB-TOPIC 3
Division of modulo
Examples: Find the values of the following;
(a) 2 (÷) 3(mod 4) (b) 7 (÷) 2(mod 5) (c) 2 (÷) 2(mod 4)
Solution: (a) If 2 (÷) 3 = 𝑥
1
𝑥

=
2 ⇒ 3 Cross-multiply , 3𝑥 = 2 Add 4 to RHS 3𝑥 = 2 + 4(mod 4) 3𝑥 = 6(mod 4) Divide both sides by 3 𝑥 = 2(mod 4)
(b) 7 (÷) 2 = 𝑥

1
𝑥
=
7 2 2𝑥 = 7(mod 5) 2𝑥 = (5 x 1) + 2(mod 5) 2𝑥 = 2 𝑥 = 1
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(c) 2 (÷) 2 = 𝑥
𝑥

2
2
1
=
2𝑥 = 2(mod 4) Divide both sides by 2 𝑥 = 1 Or 2𝑥 = 2 + 4(mod 4) 2𝑥 = 6(mod 4) 𝑥 = 3(mod 4)
N.B If 3(÷)2 = 𝑥, then 2𝑥 = 3
No multiple of 4 can be added to 3 to make it exactly divisible by 2. There are no values of 3(÷) 2 in modulo 4.
EVALUATION:
Calculate the following division in modulo 5
(a) 28(÷)7 (b) 29(÷)2 (c) 58(÷)4 (d) 74(÷)7
N.B Educators should also solve various examples.
GENERAL EVALUATION:
(1) Copy and complete the table for addition (mod 5)

0
1
2
3
4
0

4
1

3

2
3

4 4

(2) Copy and complete the table for subtraction modulo 6
Θ 0
1
2
3
4
5
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0
1
2
3
4

(3) Complete the multiplication modulo 5 in the table below

0
1
2
3
4
5
0
0
0
0
0

0
1
0

2
0

1

3
0

4
5

1
0

New General Mathematics for SSS 1, pages 227; exercises 20a, 20b
Mathematical Association of Nigeria (MAN) pages 14-24
WEEKEND ASSIGNMENT:
New General Mathematics for SSS 1, pages 227; exercises 20a, 20b, 20c
Mathematical Association of Nigeria (MAN) pages 14-24

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Content:
WEEK 6
STANDARD FORM AND INDICES
 Revision of standard form  Introduce indices and examples  Laws of indices  Application of indices, simple indicial equation
Sub-topic 1: Revision of standard form
Example 1; Express the following in standard form
(a) 5.37 (b) 53.7 (c) 537 (d) 35.65 (e) 7500 (f) 1403420
Solution: (a) 5.37 = 5.37 x 1
= 5.37 x 100 (b) 53.7 = 5.37 x 101 (c) 537 = 5.37 x 100
= 5.37 x 10 x 10 = 5.37 x 102 (d) 35.65 = 3.565 x 10
= 3.565 x 101
(e) 7500 = 7.5 x 1000 = 7.5 x 103
(f) 1403420 = 1.403420 x 1000000
= 1.403420 x 106
Example 2; Express the following in standard form
(a) 0.037 (b) 0.00065 (c) 0.0058 (d) 0.61
Solution:
Method 1:
(a) 0.037 = 3.7 x 0.01
= 3.7 x 10−2
(b) 0.00065 = 6.5 x 0.0001
= 6.5 x 10−4 (c) 0.0058 = 5.8 x 0.001 = 5.8 x 10−3
(d) 0.61 = 6.1 x 0.1
= 6.1 x 10−1

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Method 2:
(a) 0.037 =
=
=
0.037
1 3.7
100 3.7 102
= 3.7 x 10−2

(b) 0.00065 =
=
0.00065
1 6.5

10000 6.5 104
0.0058
1 5.8
1000 5.8 103
= = 6.5 × 10−4
(c) 0.0058 =
=
= = 5.8 × 10−3
(d) 0.61 =
0.61
1 6.1
= = 6.1 × 10−1
10
EVALUATION:
Express the following in standard form
1. (a) 86000 (b) 4730 (c) 307 (d) 1903000 2. (a) 0.075 (b) 0.00059 (c) 0.22 (d) 0.0000036

Sub-topic 2: Introduction of indices and examples
Indices is the plural of the word index. An index is the power of a given number. Numbers are sometimes expressed in index form e.g. 8 can be expressed as 23 in index form, 81 can be expressed as 34 in index form, 1/125 can be expressed as 1/5
in index form etc.
3 or 5-3
A number when expressed in index form must have a base and a power, e.g. when 9 is expressed in index form, we have 32 . In this case, 3 is the base and 2 is the index or power .When 625 is expressed in the index form, we have 54 . Here, 5 is the base and 4 is the index or power.
In general, index numbers are written in the form xm where x is the base and m is the index. It should be noted that mathematical operations (ie +, _ ,× and ÷) involving these types of
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numbers does not follow the conventional way. There are laws that govern its mathematical operations.
Hence, this topic indices deals with mathematical operations involving index numbers.
Sub-topic 3: Laws Of Indices
The following are the laws governing the mathematical operations involving index numbers. These laws are true for all values of m, n and x ≠ 0.
(1) x m
× x n
(2) x m
÷ x n
(3) x –n
(4) x o

(5) (x m)n
(6) x1/n
(7) x m/n

=
=
=
=
=
=
=
=
x m+ n
x m / x n = x m – n
1/ x n

1
x m.n
n x
(x m) 1/n
n x m
Sub-topic 4:Applications Of The Laws
The following are some examples of how to apply the laws of indices stated above.

LAW 1: x m X x n = x m + n

Example 1: Simplify the following
{i} 52 x 57

{ii} 32 x 34 x 33 x 3
{iii} 23a-2 b x 2a5b3
{iv} x3/4 X x5/8

Solution:
(i) 52 x 57 = 52+7
= 59
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(ii) 32 x 34 x 33 x 3 = 32+4+3+1
= 310 (Since 3 = 31)
(iii) 23a-2b x 2a5b3 = 23 x 21 x a-2 x a5 x b1 x b3

= 23+1 a-2+5 b1+3
= 24a3b4
= 16a3b4
(iv) x3/4 X x5/8 = x(6+5)/8
= x11/8

LAW 2: x m ÷ x n = x m-n

Example 2:
simplify the following
(i) 37 ÷ 34
(ii) 21a4 b3
7ab2
(iii) 9a5 b3
÷ 3a2
b-2
Solution:
(i) 37 ÷ 34
= 37 – 4

= 33
= 27
(ii) 21a4b3
7ab2
= 3a4 – 1b3 – 2

= 3a3b
(iii) 9a5b3 ÷ 3a2b-2 = 3a5 – 2 b3 – (-2)
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= 3a3 b3 + 2
= 3a3b5

EVALUATION
Simplify the following
(i) 73 x 76 (ii) 25 x 23 x 22
(iii) 34a-2b x 3a4b2 (iv) a1/2 x a1/4

(v) 54  57
(vi) 15a3b5
5ab2
(vii) 8a6b2  4ab-2

LAW 3: x –n = 1/xn

Example 3:
Simplify the following
(i) 2-4

(ii)
6a-2b3

3a3 b-5
(iii) (27/8 )-2/3

(iv) 24x4y6
(v) _1_
8x9y3
3-2
Solution:
(i)
2-4 = _1_

24
= _1_
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16

(ii)
6a-2b3 = 6b3b5

3a3b-5
3a3a2
= 2b3+5

a3+2
= 2b8
a5
(iii)
27 -2/3 8 2/3

8 = 27

= 23 2/3

33

= 2 3×2/3

3 3×2/3
= 22
32
= 4

9
(iv)
24x4y6 = 3×4 . x– 9. y6 . y– 3

8x9y3

= 3×4-9 y6-3
= 3x-5y3
= 3y3
x5
(v)
_1_

3-2
= 32
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= 9

LAW 4:
x0 = 1

Example 4:
Simplify the following
(i)
(ii)
(20ab7 x 15a6bc5)0
(17x4y2)0 + 1
(iii)
32 x 3-3 x 3
Solution:
(i)
(ii)

(20ab7 x 15a6bc5)0 = 1
(17x4y2)0 + 1
= 1 + 1

= 2
(iii)
32 x 3-3 x 3
= 32 x 3-3 x 31

= 32-3+1
= 30
= 1
LAW 5:
(xm)n =
xm.n
Example 5:
Simplify the following
(i)
(ii)
(ab2)3 x (2a4b)2
5a3b2 x (2ab)-2
(iii)
(3×2)3 ÷ 9x-3

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Solution:
(i)
(ab2)3 x (2a4b)2

= a3 b2x3 X 22 X a4x2 b2
= a3b6 X 4a8b2
= 4a3+8 b6+2
= 4a11b8
(ii)
5a3b2 x (2ab)-2

=
=
=

=
5a3b2 x 2-2a-2b-2
5 x 2-2a3 x a-2 x b2 x b-2
5 x 1 a3-2 b2-2
22
5 ab0
4
= 5a
4

(iii)
(3×2)3 ÷ 9x-3

=
=
=
=
=
33x2x3 ÷ 9x-3
27×6 ÷ 9x-3
3×6 – (-3)
3×6 + 3
3×9

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EVALUATION
Simplify the following (i) 3-5 (ii) 2a-3b2
3a2b-4
(iii) 15a3b5 (iv) 30x3y5
5ab2 6x7y2
(v) 3 -4
2
(vi) (300ab2)0
(vii) (27xy4)0 + 1
(viii) 23 x 2-4 x 2
(ix)
(x3y)2 x (2xy2)5
(x)
2(ab2)3 x a2b
(xi)
3a2b x (2ab)-3

LAW 6: x1/n =
n x

Example 6
Simplify the following.
(i.)
3 27
(ii.)
16

25
(iii.)
(0.027)1/3
Solution:
(i.)
3 27 = (33)1/3

= 33 x 1/3
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= 3
(ii).
16
16 1/3

25 = 25
= 42 ½

52
= 4 2 x ½

5 2 x ½
= 4

5

(iii)
(0.027)1/3 =
0.027 1/3

1
=
_27_ 1/3
1000

=
33 1/3
103
=
33 x 1/3

103 x 1/3
=
_3_

10

LAW 7: xm/n = (xm)1/n
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= n xm
Simplify the following.

(i)
3 a8 x 2a6 x 4a -2
(ii) 3 8y-6

Solution:
(i)
3 a8 x 2a6 x 4a -2 = 2x4x a8 +6-2 1/3

= 8a12 1/3
= 23a12 1/3
= 23×1/3a12x1/3 = 2a4
(ii) 3 8y-6 = 8y-6 1/3

= 23y-6 1/3 = 23×1/3xy-6×1/3
= 2y-2
= 2/ y2
EVALUATION
Simplify the following (i) 3 8 2
 27
(ii)
51/16 -3
4
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(iii) 33/8 -2/3

(vi)
2a5b-6 -1/2

50a3 b-2
(vii)
(1/81)-1/4
(viii) (ix) 0.091/2

Sub-topic 5: simple indicial equation
We shall consider the application of the laws of indices in solving index equations
Example
(i)
(ii)
2×2
= 50
3x-1 = 81
(iii) 8x = 64
Solution:
50 50 2 25 52
= = = = 5
2×2 (i) x2 x2 x2 ∴ x = (ii)
3x – 1 = 81

3x – 1 = 34
x – 1 = 4
x = 4 + 1

∴ x = 5
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=
=
=
=
2
(iii)

8x
23x
3x
x
∴ x =

64
26
6
6/3
EVALUATION
Solve the following index equation
1.
2.
3.
4.
5.
6.
7.
32x
9x
5a3
a-1/2
4x
x -2

4x + 1 = 64
= 27
= 81
= 40
= 5
= 32
= 4
8. 4x x 8 = 64
9. 32x-1 = 27
10. 27 x 3x = 81
12. 2×1/3 = 16
13. 3×2 = 27

GENERAL EVALUATION
(1) Given that3x91+x=27-x, find x SSCE, June
1994.
(2) If 4x = 2½ x 8, find x (3) If 8x/2 = 23/8 x 43/4, find x [WAEC]
[WAEC]
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(4) Given that 4x = 32, find the value of x. 1985 [WAEC] (5) Solve the equation 125x+3 = 5, find x. 1981 [WAEC]
(6) Given that 8 x 4x-2 = 64, find x. (7) Given that 1252x+1 = 625 x 25-x, find x.
(8) If 3m x 27(2m-1) = 81, find m.
[WAEC]
(9) Find the value of x in 8-1 x 2(2x+1) = 64. [WAEC].
(10) 7 x 49(x+2) =
Find the value of x, given that
1
343

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CONTENT
WEEK 7
LOGARITHMS
 Deducing logarithm from indices and standard form  Definition of Logarithms  Antilogarithms  The graph of y = 10x
There is a close link between indices and logarithms
100 = 102. This can be written in logarithmic notation as log10100 = 2.
Similarly 8 = 23 and it can be written as log28 = 3.
In general, N = bx in logarithmic notation is LogbN = x.
We say the logarithms of N in base b is x. When the base is ten, the logarithms is known as common logarithms.
The logarithms of a number N in base b is the power to which b must be raised to get N.
Evaluation.
Re-write using logarithmic notation (i) 1000 = 103 (ii) 0.01 = 10-2 (iii) 24 = 16 (iv)
1 = 2-3 8
Change the following to index form
(i)
Log416 = 2 (ii) log3 (
) = -3
1
27
The logarithm of a number has two parts and integer (whole number) then the decimal
point. The integral part is called the characteristics and the decimal part is called mantissa.
To find the logarithms of 27.5 form the table, express the number in the standard form as 27.5 = 2.75 x 101. The power of ten in this standard form is the characteristics of Log 27.5. The decimal part is called mantissa.
Remember a number is in the standard form if written as A x 10n where A is a number
such that 1 ≤ A < 10 and n is an integer.
27.5 = 2.75 x 101, 27.5 when written in the standard form, the power of ten is 1. Hence the characteristic of Log 27.5 is 1. The mantissa can be read from 4-figure table. This 4-figure table is at the back of your New General Mathematics textbook.
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Below is a row from the 4-figure table Differences
X 0 431 2 7 4
1 433 0
2 434 6
3 436 2
4 437 8
5 439 3
6 440 9
7 442 5
8 444 0
9 445 6
1 2 3 4 5 6 7 8 9 1 1 3 4 6 7 9 1 1 4
1 2

To check for Log27.5, look for the first two digits i.e 27 in the first column.
Now look across that row of 27 and stop at the column with 5 at the top. This gives the figure 4393.
Hence Log27.5 = 1.4393
To find Log275.2, 2.752 x 102
The power of 10 is the standard form of the number is 2. Thus, the characteristic is 2. Log275.2 = 2. ‘Something’
For the mantissa, find the figure along the row of 27 with middle column under 5 as before (4393). Now find the number in the differences column headed. This number is 3. Add 3 to 4393 to get 4396. Thus Log275.2 = 2.4396.
Evaluation
Use table to find
i. ii. iii.
Log 37.1 Log 64.71 Log 7.238
LESSON 2
If the logarithm of a number is given, one can determine the number for the antilogarithm table.
Example: Find the antilogarithms of the following (a) 0.5670 (b)2.9504
Solution
(a) The first two digits after the decimal point i.e .56 is sought for in the extreme left column of the antilogarithm table then look across towards right till you, get to the column with heading 9. (Read 56 under 9) there you will see 3707. Since the integral part of 0.5690 is 0, it means if the antilog (3707) is written in the standard form the power of 10 is zero. i.e antilog of 0.5690 = 3.707 x 100 = 3.707
(b) The antilog of 2.9504 is found by checking the decimal part .9504 in the table.
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(c) Along the row beginning with 0.95 look forward right and pick the number in the column with 0 at the top. This gives the figures 8913 proceed further to the difference column with heading 4 to get 8. This 8 is added to 8913 to get 8921. The integral part of the initial number (2.9504) is 2. This shows that there are three digits before the decimal point in the antilog of 2.9504 so antilog 2.9504 = 892.1. LESSON 3 The graph of y = 10x can be used to find antilogarithm (and logarithm). Below is the table of values. X Y = 10x
0.8 6.3
0.6 4.0
0 1
0.1 1.3
0.9 7.9
1 10
0.7 5.0
0.2 1.6
0.3 2.0
0.4 2.5
0.5 3.2
For example the broken line shows that the antilog 0.5 is approximately 3.2 or inversely that Log 3.2 ~0.5 Evaluation Find the logarithms of (i) (ii) (iii)
32.7 61.02 3.247
2. Use antilog tables to find the numbers whose logarithms are
(1) (1.82) (ii)2.0813 (iii) 0.2108 ASSIGNMENT New General Mathematics SS1, Ex 1h No 8, Ex 1i, No 4 and No 5.

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WEEK 8
LOGARITHMS
CONTENT: Use Of Logarithm Table And Antilogarithm Table In Calculation Involving
i. Multiplication ii. Division iii. Powers iv. Roots v. Application of logarithm in capital market and other real life problems
PERIOD 1
Logarithm and antilogarithm tables are used to perform some arithmetic basic operations namely: multiplication and division. Also, we use logarithm in calculations involving powers and roots.
The basic principles of calculation using logarithm depends strictly on the laws on indices. Recall from week 7 that.
(a) Log MN = Log M + Log N (b) Log
= Log M – Log N
𝑀
𝑁
Hence, we conclude that in logarithm;
1. When numbers are multiplied, we add their logarithms 2. When two numbers are dividing, we subtract their logarithms.
SUB- TOPIC 1
Multiplication of numbers using logarithm tables
Example 1
Evaluate 92.63 x 2.914
Solution

Number
Standard Form

Log

Operation
9.263 x 101
2.914 x 100

1.9667
0.4645

2.4312

92.63
2.914

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Standard form
Number
2.699 x 102
269.9

Log

Anti-Log
2.4312
2699

Therefore 92.63 x 2.914 = 269.9
Example 2
Evaluate 34.83 x 5.427
Solution

Number
Standard Form

Log

Operation
1.5420
0.7346

2.2766
Standard form
Number
1.891 x 102
189.1
34.83
5.427

Log

3.483 x 101
5.427 x 100

Anti-Log
2.2766
1891

Therefore 34.83 x 5.427= 189.1
EVALUATION
Evaluate the following
1. 6.26 x 23.83 2. 409.1 x 3.932 3. 8.31 x 22.45 x 19.64 4. 431.2 x 21.35

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SUB-TOPIC 2 – DIVISION OF NUMBERS USING LOGARITHM
Example 1
Evaluate 357.2 ÷ 87.23
Solution1
Number
Standard Form

Log

Operation
357.2
87.23

Log

3.572 x 102
8.723 x 101

Anti-Log
0.6123
4096

Therefore 357.2 x 87.23 = 4.096
Example 2
2.5529
Subtraction
1.9406

0.6123
Standard form
Number
4.096 x 100
4.096
Use a logarithm table to evaluate 75.26 ÷ 2.581
Solution

Number
Standard Form

Log

Operation
1.8765
Subtraction
0.4118

1.4647
Standard form
Number
2.916 x 101
29.16
75.26
2.581

Log

7.526 x 101
2.581 x 100

Anti-Log
1.4647
2916

Therefore 75.26 ÷ 2.581 = 29.16
Evaluation
Use table to evaluate the following
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(1) 53.81 ÷ 16.25 (2)632.4 ÷ 34.25 (3) 63.75 ÷ 8.946 (4) 875.2 ÷ 35.81

SUB-TOPIC 3 Calculation Of Powers Using Logarithm Study these examples. At times, calculations can involve powers and roots. From the laws of logarithm, we have (a) Log Mn = nLogM (b) Log M1/n =

1

𝑥𝑙𝑜𝑔𝑀
Log M =
Log M = Log √𝑀𝑛 𝑛 𝑥 (c) Log Mx/n = 𝑛 Example 1. Evaluate the following (53.75)3 Solution
𝑛
Number
Standard Form
Log

Operation
(53.75)3
(5.375 x 101)3

1.7304
Multiply Log by 3

Log

x 3

5.1912
Anti-Log
Standard form
Number
5.1912
1553

1.553 x 105
155300
Therefore 53.753 = 155300
Example 2: 64.592
Solution
Number
Standard Form
Log

Operation
(64.59)2
(6.459 x 101)2
1.8102
Multiply Log by 2

Log

Anti-Log
x 2

3.6204
Standard form
Number
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3.6204
4173

4.173 x 103
4173
Therefore 64.592= 4173
Evaluation
Evaluate the following
1. 5.6324 2. √35.81 3. 19.183 4. (67.9/5.23)3 5. √679.5 x 92.6

SUB-TOPIC 4 Calculation of roots using Logarithms and Anti-Logarithms Use tables of Logarithms and antilog to calculate √27.41 Solution

5
Number
Standard Form
Log

Operation
5 √27.41

(2.741 x 101)1/5 1.4380 ÷ 5
Divide Log by 5

Log

Anti-Log
0.2876
1939

0.2876
Standard form
Number
1.939 x 100
1.939
Therefore √27.41
= 1.939
5
Example 2: use table to find √218 3.12
3

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Solution
Hint: Workout 218 ÷ 3.12 before taking the cube root.
Number
Standard Form

Log

Operation
(2.18 x 102)
(3.12 x 100)

2.3385
Subtraction
0.4942

Anti-Log
1.8443 ÷ 3
Division

0.6148
Standard form
Number
4.119 x 100
4.119
218
3.12

Log

0.6148
4119

Therefore √218 3.12
3
= 4.119
Example 3
Evaluate 63.752 – 21.392
We can use difference of two squares
i.e A2 – B2 = (A+B)(A-B)
63.752 – 21.392 = (63.75+21.39)(63.75-21.39)
= (85.14)(42.36)
= 85.14 X 42.36
Number
Log

85.14

(1.9301)

42.36 (1.6260) 3606 3.5561

operation

…. 3606

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EVALUATION
Evaluate the following
1. (39.652 – 7.432)1/2 2. 84.352 – 36.952 3. 64.742 – 55.262 4. 94.682 – 43.252 5. 25.142 – 7.522

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Content:
 Definition of sets  Set notations  Types of sets
SUB-TOPIC 1: Definition
WEEK 9
SETS
A
set is a well-defined list or collection of objects with some characteristics which are unique to its members. Examples:
(i)
a set of mathematics text books
(ii)
a set of cutleries
(iii)
a set of drawing materials etc.
Sometimes there may be no obvious connection between the members of a set.
Example: {chair, 3, car, orange, book, boy, stone}.
Each item in a given set are normally referred to as member or element of the set.
SUB-TOPIC 2: SET NOTATION
This is a way of representing a set using any of the following.
Listing method
(i) (ii) Rule method or word description (iii) Set builders notation. (i)
Listing Method
A set is usually denoted by capital letters and the elements in it can be defined either by making a list of its members. Eg A = {2, 3, 5, 7}, B = {a, b, c, d, e, f, g, h, i} etc.
Note that the elements of a set are normally separated by commas and enclosed in curly brackets or braces
(ii)
Rule Method. The elements in a set can be defined also by describing the rule or property that connects its members. Eg C = {even number between 7 and 15. D= {set of numbers divisible by 5 between 1 and 52.}, B = {x : x is the factors of 24}etc
(iii) Set–Builders Notations
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A set can also be specified using the set – builder notation. Set – builder notation is an algebraic way of representing sets using a mixture of word, letters , numbers and inequality symbols e.g. B = {x : 6 ≤ x < 11, x є ƶ} or B = {x/6 ≤ x < 11, x є ƶ}. The expression above is interpreted as “B is a set of values x such that 6 is less than or equal to x and x is less than 11, where x is an integer (z)”

The stroke (/) or colon (:) can be used
interchangeably to mean “such that”
– The letter Z or I if used represents integer or whole numbers.
Hence, the elements of the set A = {x : 6 ≤ x < 11, x є ƶ} are A = { 6, 7, 8, 9,10}.
NB:

The values of x starts at 6 because 6 ≤ x
The values ends at 10 because x < 11
and 10 is the first integer less than 11.
The set builder’s notation could be an equation, which has to be solved to obtain the elements of the set. It could also be an inequality, which also has to be solved to get the range of values that forms the set.
EVALUATION
(a) Define Set (a) C = {x : 3x – 4 = 1, x є ƶ}
(b) P = {x : x is the prime factor of the LCM of 15 and 24}
© Q = {The set of alphabets}
(d) R = {x : x ≥ 5, x is an odd number}

SUB-TOPIC 3: Set – Builders Notations (conts.)

Examples 1:
List the elements of the following sets
(i)
(ii)
A = {x : 2 4, x є ƶ}
(iii)
C = {x : -3 ≤ x ≤ 18, x є ƶ}.
(iv)
D = {x : 5x -3 = 2x + 12, x є Z}.
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(v)
E = {x : 3x -2 = x + 3, x є I}
(vi)
F = {x : 6x -5 ≥ 8x + 7, x є ƶ}
(vii)
P = {x : 15 ≤ x < 25, x are numbers divisible by 3}
(viii) Q = {x : x is a factor of 18, }
Solution:
(i)
A = {3, 4, 5, 6, 7}

Note that:
– the values of x start at 3, because 2 < x
-The values of x ends at 7 because x ≤ 7 i.e. because of the equality sign.
(ii)
B = {5, 6, 7, 8, 9, .. .}

Note that:
the values of x start from 5 because 5 is the first number told that x is greater than 4)
greater than 4 (i.e. we are
(iii)
C = {-3, -2, -1, 0, 1, . . , 15, 16, 17, 18}

Note that:

18 (there is equality sign at both
The values of x starts from -3
ends).
because -3 ≤ x, and ends at 18 because x ≤
To be able to list the elements of this set, the equation defined has to be
solved
(iv) i.e.
5x – 3 = 2x + 12

5x – 2x = 12 + 3
3x = 15

x
= 15/3
∴ x = 5
∴ D = {5}
(v) We also need to solve the equation to get the set values

3x – 2 = x + 3
3x – x = 3 + 2
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2x
= 5
∴ x = 5/2
Since 5/2 is not an integer (whole number) therefore the set will contain no element.
∴є
= { } or Ø

(vi) Solving the inequality to get the range of values for the set, we have

6x – 5 ≥ 8x + 7
6x – 8x ≥ 7 + 5
-2x ≥ 12
x ≤ 12/-2
∴ x ≤ -6
∴ F = {…, -8, -7, -6}
(vii)
P = {15, 18, 21, 24}

Note that:
The values of x start at 15 because it is the first number divisible by 3 and
within the range defined.
falls
(viii) Q = {1, 2, 3, 6, 9, 18}
Example 3:
Rewrite the following using set builder notation
(i)
A = {8, 9, 10, 11, 12, 13, 14}
(ii)
B = {3, 4, 5, 6 . . . }
(iii)
C = {. . . 21, 22, 23, 24}
(iv)
D = {7, 9, 11, 13, 15, 17 . . .}
(v)
P = {1, -2}
(vi) Q = {a, e, i, o, u}
Solution:
(i)
A = {x : 7 < x < 15, x є ƶ} OR
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A = {x : 8 ≤ x < 15, x є ƶ} OR
A = {x : 7 2, x є ƶ} OR

B = {x : x ≥ 3, x є ƶ}
(iii)
C = {x : x 8 or x ≥ 7, x is odd, x є ƶ}
P={1,2} suggests the solutions of a quadratic equation. Therefore , the equation or
(v) set-builders notation can be obtained from :
x2 – (sum of roots)x + product of roots = 0
x2 –(-1)x + (1 x -2) = 0
x2 + x – 2 = 0
P = {x : x2 + x – 2 = 0, x є ƶ}
(vi) Q = {x : x is a vowel}

EVALUATION
1. List the elements in the following Sets
(a)
A = {x : -2 ≤ x < 4, x є ƶ}

(b)
B = {x : 9 < x < 24, x є N}
(c)
C = {x : 7 < x ≤ 20, x is a prime number, x є I}
(d)
D = {x / 2x – 1 = 10, x є Z}
(e)
P = {x : x are the prime factor of the LCM of 60 and 42}
2. Rewrite the following using Set –
builder notations.

(a)
Q = {. . . 2, 3, 4, 5}
(b)
A = {2, 5}
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(c)
B = {2, 4, 6, 8, 10, 12 . . .}
(d)
A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
(e)
C = {1, 3, -2}

SUB-TOPIC 4: TYPES OF SETS
Finite Sets
Refers to any set, in which it is possible to count all the elements that make up the set.
These types of sets have end. E.g.
A = {1, 2, 3, . . , 8, 9, 10}
B = {18, 19, 20, 21, 22}
C = {Prime number between 1 and 15} etc.
Infinite Sets
Refers to any set, in which it is impossible to count all the elements that make up the set. In other words, members or elements of these types of set have no end. These types of set, when listed are usually terminated with three dots or three dots before the starting values showing that the values continue in the order listed. E.g.
(i)
A = {1, 2, 3, 4, . . }
(ii)
B = {…,-4,-3,-2,-1,0,1,2,3,…}
(iii)
C = {Real numbers} etc.
Empty or null Set
A set is said to be empty if it contains no element. Eg {the set of whole number that lies between 1 and 2}, {the set of goats that can read and write}, etc Empty sets are usually represented using ø or { }.
It should be noted that {0} is NOT an empty set because it contains the element 0, Another name for empty set is null set.
Number of Elements in a Set
Given a set A = {-2, -1, 0, 1, 2, 3, 4, 6} the number of elements in the set A denoted by n(A)
is 9; i.e. n (A) = 7
If B = {2,3, 5} then n (B) = 3
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If Q = {0} then n (Q) = 1
Other examples are as follows:
Example 4:
Find the number of elements in the set:
P = {x : 3x -5 < x + 1 < 2x + 3, x є ƶ }
Solution:
3x – 5 < x + 1 and x + 1 < 2x + 3
3x – x < 1 + 5 and x – 2x < 3 – 1
2x < 6
– x < 2
x < 6/2
x -2
-2 < x

-2 < x < 3
The integers that form the solution set are
P = {-1, 0, 1, 2}
∴ n {P} = 4
Example 5:
Find the number of elements in the set
A = {x : 7 < x < 11, x is a prime number}
Solution:
The set A = { } or Ø since 8, 9, 10 are no prime numbers. ∴ n(A) = 0
Example 6:
Find the number of elements in the following sets:
(i)
B = {x : x ≤ 7, x є ƶ}
(ii)
C = {x : 3 < x ≤ 8, x is a number divisible by 2}.

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Solution:
(i)
B = {. . . 3, 4, 5, 6, 7}.

The values of the set B has no end hence it is an infinite set i.e. n(B) = ∞
(ii)
C = {4, 6, 8}. ∴ n(C) = 3
The Universal Set
This is the Set that contains all the elements that are used in a given problem. Universal
Sets vary from problem to problem. It is usually denoted using the symbols ξor μ.
Note that when the Universal Set of a given problem is defined, all values outside the
universal set cannot be considered i.e. they are invalid.
EVALUATION
(1) List the elements in the following Sets
(a)
A = {x : -2 ≤ x < 4, x є ƶ}

(b)
B = {x : 9 < x < 24, x є N}
(c)
C = {x : 7 < x ≤ 20, x is a prime number, x є I}
(d)
D = {x / 2x – 1 = 10, x є Z}
(e)
P = {x : x are the prime factor of the LCM of 60 and 42}
(2). Find the number of elements in the sets in question (1) above

(3). If.
(a) A= {3,5,7,8,9,10,}, Then n(A) =
(b) B= {1, 3, 1, 2, 1, 7}, Then n(B) =
© Q= {a, d, g, a, c, f, h, c,} , Then n(Q) =
(d) P= {4,5,6,7,…,12,13}, Then n(P) =
(e) D ={ days of the week} , then n(D)=
(4)
State if the following are finite, infiniteor null set
(i)
Q = {x : x ≥ 7, x Є Z}
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(ii)
P = {x : -4 ≤ x < 16, x Є I}
(iii)
A = {x : 2x – 7 = 2, x Є Z}
(iv) B= { sets of goats that can fly}
(v) D={sets of students with four legs}
5. list the following sets in relation to the universal set. Given that the universal set ξ = {x: 1< x< 15, x ε z}
(i) A= {x; -3≤ x≤7, x ε Z}
(ii) B= {x: 5<x<6, xεz},
(iii) C={x: X≥ 5, xε z }

(6) list the elements of the following universal sets.
(i) The set of all positive integers
(ii) The set of all integers
(iii) (iv) (v)
ξ ={ x: 1 < x < 30, x are multiples of 3} ξ = { x: 7≤ x<25, x are odd numbers} ξ = { x: x≥10, xεz}
WEEKEND ASSIGNMENT New General Mathematics for SSS, Book 1 Pages 97- 100 Exercise 8c Question no. 1,4 and 5
1. New General Mathematics for SSS, Book 1 Pages 97 – 100. 2. Man Mathematics for SSS, Book 1, Pages 45 – 60

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Content:
WEEK 10
SETS
 Set operations  Venn diagram and application up to 3 set problem.
SUB-TOPIC 5: Operations on Sets

The Union of Sets:
The Union of Sets A and B is the Set that is formed from the elements of the two Sets A and B. This is usually denoted by “A ⋃B” meaning A Union B. Thus A ⋃B is the Set which consists of elements of A or of B or of both A and B.
Using Set notations, the Union of two Sets A and B is represented as follows
Example 10:
Given that A = {3, 7, 8, 10}
and B = {3, 5, 6, 8, 9} then
(A ⋃ B) = {3, 5, 6,7, 8, 9,10}
Example 11:
If A = {a, b, c, d}, B = {1, 2, 3, 4} and C ={a, 3, θ} Then A ⋃B ⋃C = {a, b, c, d,1, 2, 3, 4, θ}
Intersection of Sets
The intersection of Sets A and B is the set of elements that are common to both A and B. This is usually denoted by “A ∩ B” meaning A intersection B. When represented using Venn diagram we have

ξ

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If A ∩ B = Ø, then the Sets A and B are said to be disjoint. Disjoint Sets are Sets that have no element in common.

ξ

B
A

Example 12:
Given that A = {5, 7, 8, 10} and B = {3, 5, 6, 8, 9}, then
A ∩ B = {5, 8}.
Example 13:
If P = {a, b, c, d, e, f, g}, Q = {b, c, e, g} and R = {a, c, d, f, g}
Then, P ∩ Q ∩ R = {c, g}

Example 14:
If A = {1, 2, 3} and B = {6, 8, 10}, then
A ∩ B = { } or Ø. The Set A and B are disjoint.

EVALUATION
1. Given that ξ= {21, 22, 23, 24, . . ., 29, 30},
P = {21, 23, 25, 26, 28},
Q = {22, 24, 26, 27, 28} and
R = {21, 25, 26, 27, 30} are Subsets of ξ.. Find:
(i) P ⋃Q
(ii) P ∩Q
(iii) Q ∩R
(iv)
(P ∩Q) ⋃R

(v) P ∩Q ∩R
(vi) (P ⋃Q) ∩(Q ⋃R)
(vii) (P ∩Q) ⋃(Q ∩R)

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If A = {1, 2, 3, 4} and
B = {3, 5, 6}, find:

(i) A ∩B
(ii) A ⋃ B
(iii) (A ∩ B) ⋃ B
(iv) (A ⋃ B) ∩ A

(WAEC)
2.

3.
If ξ = {a, e, i, o, u, m, n} find the
complement of the following Sets
(a) A = {e, o, u}

(b) B = {i, o, u}
(c) C = {a, u, m, n}
(d) P = {u}
(e) Q = { } or Ø
(f) ξ

4.
If A = {7, 8, 9, 10}, B = {8, 10, 12, 14}

and C = {7, 9, 10, 14. 15} find the
following:

(a) A ⋃B
(b) B ⋃ C
(c) A ⋃B ⋃ C (d) A ∩B
(e) A ∩C
(f) B ∩C
(g) (A ⋃ B) ∩(A ⋃ C)
(h) A ∩( B ⋃ C)
5.
Let ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},

A = {2, 3, 4, 5, 6,}
B = {3, 4, 6, 8, 10} and
C = {2, 4, 6, 7, 10}
Find:

(a)
(b)
(c)
(A ⋃ B) ∩(B ⋃ C)
A ∩( B ⋃ C)
B ∩(A ⋃ C)
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(d)
Show that:
(A ∩B) ⋃ (A ∩C) = A ∩(B ⋃ C)
SUB-TOPIC 6: The Complement of Set
If A is a Subset of the Universal Set ξ, then, the complement of the Set A are made up of elements that are not in A, but are found in the Universal Set ξ. This is usually denoted by Ac or A′. for example
If ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {3, 5, 6, 9} then Ac or A′ = {1, 2, 4, 7, 8, 10}
Using Venn diagram, this is represented by the shaded portion below:

1

2

3 5 6
ξ

9
9
8
Note that to find the complement of a Set, the Universal Set must be properly defined.

4 10
7
Example 15:
Given that
ξ = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {11, 13, 16, 18, 20} and
B = {12, 14, 16, 18, 19, 20}.
Find the following:
(i) A′
(ii) B′ (iii) (A ⋃B)′

(iv) (A ∩B)′ (v) A′ ∩B′ (vi) A′ ⋃B′
(vii) (A′)′

Solution:
(i)
A′ = {12, 14, 15, 17, 19}
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(ii)
B′ = {11, 13, 15, 17}
(iii)
A ⋃B = {11, 12, 13, 14, 16, 18, 19, 20}

(A ⋃B)′ = {15, 17}
(iv)
A ∩B = {16, 18, 20}

(A ∩B)′ = {11, 12, 13, 14, 15, 17, 19}
(v)
(A′ ∩B′) = {15, 17}
(vi)
A′ ⋃ B′ = {11, 12, 13, 14, 15, 17, 19}
(vii) A′ = {12, 14, 15, 17, 19}

(A′)′ = {11, 13, 16, 18, 20} = A
NB:
From the example above, observe that from (iii) and (v) (A ⋃ B)′ = A′ ∩ B′
Also, from (iv) and (vi)
(A ∩ B)′ = A′ ⋃ B′ and from (vii) (A′)′ = A
Example 16:
Given that ξ = {a, b, c, d, e, f, g, h, i, j}
A = {a, c, e, g, i} and
B = {b, c, d, f, i, j}.
Find the following:
(i) A′

(ii) B′
(iii) A′ ⋃ B′
(iv) A′∩ B′
(v) (A ∩ B)′

(vi) (A ⋃ B)′
(vii) (B′)′
Solution:
(i)
(ii)
A′ = {b, d, f, h, j}
B′ = {a, e, g, h}
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(iii) A′ ⋃ B′ = {a, b, d, e, f, g, h, j}
(iv)
A′ ∩ B′ = { h }
(v)
A ∩ B ={c, i }

(A ∩ B)′ ={a, b, d, e, f, g, h, j}
(vi)
A ⋃ B = {a, b, c, d, e, f, g, i, j}

(A ⋃B)′ =
{ h}
(vii) B′ =
{a, e, g, h}
(B′)′ = {b, c, d, f, i, j} = B

From the example above, we also observe that A′⋃ B′ = (A ∩ B)′ — From (iii) and (v)
A′ ∩ B′ = (A ⋃ B)′ — From (iv) and (vi)
And (B′)′ = B — From (vii) From the last two examples we can clearly see that (A⋃B)′ = A′∩ B′,

(A ∩ B)′ = A′ ⋃ B′
and (A′)′ = A
Generally, for any two Subsets A and B of a Universal Set ξ, the following are true:
(i)
(A ⋃ B)′ = A′ ∩ B′
(ii)
(A ∩ B)′ = A′ ⋃ B′
(iii)
(A′)′ = A or (B′)′ = B
These are known as De Morgan’s Laws of Complementation.

Equality of Set
Two Sets A and B are said to be equal if they have exactly the same elements. This means that every element of the first Set also belongs to the second Set and vice versa. E.g.
If A = {2, 3, 4, 5} and B = {2, 3, 4, 5} then the Set A = the Set B because n(A) = n (B) = 4 and
all members of the Set A are also members of the Set B.
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If P = {8, 2, 4} and Q = {2, 4, 8} the Set P = Q (order of arrangement does not matter, the elements in both Sets are the same).
If A = {1, 2, 3} and B = {3, 5, 6} then A ≠ B.

Equivalent Sets
Two Sets are said to be equivalent if the Sets have equal number of elements. E.g.
If A = {2, 3, 4, 5} and B = {a, b, c, d} then the Sets A ≡ B (A is equivalents to B) since n(A) = n(B).

EVALUATION
1.
If ξ = {x : 0 < x ≤ 15 x ε Ƶ}

A = {1, 5, 7, 11} and
B = {5, 7, 14}
Find the following:

(a) A ∩ B
(b) A′ (c) B′
(d) (A ⋃ B)′
(e) A′ ∩ B (f)A ∩ B′
(g) Show that: (A∩B)′ = A′ ⋃ B′

2.
Given that ξ = {x : 0 ≤ x < 10, x ε Ƶ}

P = {1, 2, 4, 5}
Q = {2, 4, 6, 8} and
R = {1, 3, 4, 8, 9}

Find:
(a) P′ (b) Q′ (c) R′ (d) Q ∩ R′
(e) P′ ∩ Q′
(f) P′ ⋃ Q′ (g)(Q ∩ R′)′
(h) (P′ ∩ R′)′ (i) (P ∩ ξ) ′
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(j) P′ ∩ Q′ ∩ R′

3.
P, Q and R are Subsets of the
Universal Set µ such that µ = {3, 4, 5, 6, . . . , 16, 17, 18}

P = {x : x is a number divisible by 3}
Q = {x : x is odd}
R = {x : x is a factor of 35}
Find: (i)
P ∩ Q (ii) P ⋃ Q′ ∩ R
(iii) Q ⋃ R
(iv) P′ (v) P′ ∩ R

SUB-TOPIC 7: THE USE OF VENN DIAGRAMS IN PROBLEM SOLVING between sets with diagrams. Venn diagrams.
A Mathematician by name John Venn was the man to first represent the relationship Ever since sets may be represented by diagrams called
The rectangle is used to represent the Universal set, and Circles for other sets, as we
shall see later.
PROBLEMS INVOLVING TWO SETS.
For two intersecting sets, the diagram is given below with the labels of what each
compartment represents.

A
B

I
II
III

IV
Compartment I: represent the set of elements in A only. i.e. A  B/ using set notations. Compartment II: represents the set of elements common to both A and B i.e. AB
Compartment III: represents the set of elements in B only i.e. A/  B
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Compartment IV:
(AB) / or A/  B/
Example 15:

represents the set of elements that are neither in A nor B i.e.
In a Survey of 40 Students in a class, 19 have visited Lagos and 17 have visited Benin City.
If 13 have visited neither. How many Students have visited:
(i) Both Cities; (ii) Benin City but not Lagos (i.e. Benin City only)
Solution:

n() = 40
n(L) = 19
n(B) = 17
n(L  B)/ = 13
Let x represents those that have visited both Cities
i.e. n(LB) = x

L
B
19 – x x
17 – x

13

19 – x + x +17 – x + 13 = 40

49 – x = 40
49 – 40 = x
9 = x
9 Students have visited both Cities
(b)
Those that have visited Benin City only are = 17 – x

= 17 – 9
= 8
 8 have visited Benin City only.
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Example 16:

In a Class of 45 Students, if 21 offer Agricultural Science, 25 offer Biology and 6 offer both
subjects. Find
(i) those that offer neither.
(ii)the number that offers Biology but not
Agricultural Science (i.e. Biology only)
Solution:
(i)
n() = 45

n(A) = 21
n(B) = 25
n(A B) = 6
Let n(AB) / = x

i.e. Let those that offer neither be x.

A
B
21 – 6

6 25 – 6
x

21 – 6 + 6 +25 – 6 + x = 45

15 + 6 + 19 + x = 45

40 + x = 45

x = 45 – 40

 x = 5
Those that offer neither,
i.e. n(AB) / = 5
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EVALUATION
(1). In a gathering of 30 people, x speak Hausa and 15 speak Yoruba. If 5 people
speak both languages, find how many people that speak (i) Hausa.
(ii)Yoruba only

(iii)Hausa only.
(2) In a Birthday party attended by 22
people, 10 ate fried rice and 13 ate salad. If x ate both fried rice and salad and (2x–5) ate none of the two. How many ate
(i) (ii) salad but not fried rice?
(ii)
(3) The Venn diagram below represents a universal set  of integers and its subsets

P and Q. List the elements of the following sets;

(a) P  Q

(b) P  Q

P 1 9

(c )   Q

2
3
8, 4
(d) P  
10 5

6
7 Q

SSCE, NOV. 1995 Nọ 1. (WAEC)

SUB-TOPIC 8: PROBLEMS INVOLVING THREE SETS.
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The Venn diagram is made up of eight compartments as shown below: 

B
A

V II VI
III I

IV

VII
C VIII

Compartment I represents ABC
(elements common to the three sets A, B and C).
Compartment II represents ABC/
(elements common to both A and B only).
Compartment III represents AB/C

(elements common to both A and C only).
Compartment IV represents A/BC

(elements common to both B and C only).
Compartment V represents A B/C/

(elements of A only).
Compartment VI represents A/  B  C /
(elements of B only).
Compartment VII represents A/ B/ C
(elements of C only).
Compartment VIII represents (ABC)/ or A/B/C/ elements that are not in any of the three sets but are in the Universal set.
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Example 18:
The are 80 people in a sports camp. Each play at least one of the following games, volleyball, football and handball. 15 play volleyball only, 18 play football only, and 21 play handball only .If 5 play volleyball and foot ball only, 8 play volleyball and handball only, and 10 play football and Handball only.
(a) Represent the above information in a Venn diagram

(b) How many people play the three games?
(c) How many people play football ?
Solution:
List of information given in the question is as follows n() = 80
Let V be Volleyball F be Football H be Handball

n(VF/H/) i.e. Volleyball only = 15 n(V/FH/ ) i.e. Football only = 18 n(V/F/H) i.e. Handball only = 21 n(VFH/) i.e. Volleyball and Football only = 5 n(VF/H) i.e. Volleyball and Handball only = 8 n(V/FH) i.e. Football and Handball only =10 Let n(VFH) = x . i.e. Those that play the three games = x
(a)

 V
F

15 5 18

8 X 10
21 H
(b) 15 + 5 + 18 + 8 + x + 10 + 21 = 80
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77 + x = 80  x = 3
x = 80 – 77
The number that plays football
n(F) = 18 + 5 + 10 + x = 33 + 3 = 36
 3 people play the three games.
(C) N.B Since the word ONLY was used all through, the values are written directly in to each compartment without any manipulation as shown in the figure above. Suppose the question 18 above is framed as shown in the example 19 below, then the Approach would be different. Example 19: There are 80 people in a sports camp and each plays at least one of the following games: volleyball, football and handball. 31 play volleyball, 36 play football and 42 play handball. If 8 play volleyball and football, 11 play volleyball and handball and 13 play football and handball. (a) that play the three games. (b)
How many of them play: (i) All the three games, (ii) Exactly two of the three games, (iii) Exactly one of the three games (iv) handball only?
Draw a Venn diagram to illustrate this information, Using x to represent the number

Solution: Step 1: list out all information given in the question. Let (a)
V be Volley ball F be Football H be Hand ball n() = 80 n(V) = 31 n(F) = 36 n(H) = 42 n(V  F  H) = 80 (Since each play at least one of the games). n(V  F) = 8 n(V  H) = 11 n(F  H) = 13. n(V  F  H) = x.
Let
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where n(V  F  H) = x.  V F 12 + x 8 – x 15 + x
x

How we obtained the value for each of the other compartments is shown below.

13 – x
11 – x
18 + x H
For Volleyball and football only i.e. n(V  F  H/) (Since x is already in the circle of V  F) = n(V  F) – x = 8 – x For Volleyball and Handball only i.e. n(V  F/  H) (Since x is already in the circle of V  H) = n(V  H) – x = 11 – x For Football and Handball only i.e. n(V/  F  H) (Since x is already in the circle of F  H) = n(F  H) – x = 13 – x For Handball only i.e. n(V/  F/  H) n(H) – (All values already written in the circle of Handball) = 18 + x x
= 42 – [ (11 – x ) + x + (13 – x ) ] = 42 – [24 – x ] = 42 – 24 + x

11– x 13 – x
H

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n(F) – ( All values already written in the circle of Football) = 36 – [ (8 – x) + x + (13 – x) ]

For Football only i.e. n(V/  F  H/) = 36 – [21 – x ] = 36 – 21 + x = 15 + x 8 – x F For Volleyball only
x
13 – x

i.e.n (V  F/  H/)

n(V) – (All values already written in the circle of Volley ball). = 31 – [ (8 – x ) + x + ( 11 – x ) ] = 31 – [19 – x] V

= 31 – 19 + x
= 12 + x

8 – x
x

11 – x
To get the value of x, which
(b) represent those that play all three games, we add all the Compartments of the Venn diagram together and equate it to the total value in the Universal set and solve for x. i.e. 12 + x + 8 – x + 15 + x + x + 11 – x + 13 – x + 18 + x = 80 
77 + x = 80 x = 80 – 77 x = 3
3 people play all three games

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NOTE THAT If this value, x =3, is substituted into the Venn diagram, the answer obtained in the previous example would be got. i.e. V
F

15
5
18
3 8 10

21

H
b (ii) Exactly two of the three games

= n(V  F  H/) + n(V  F/  H) + n(V/  F  H) = 8 – x + 11 – x + 13 – x
= 32 – 3x = 32 – 3(3)  23 of them play exactly two of the three games. b (iii) Exactly one of the three games
= 32 – 9 = 23
= 45 + 9 = 54
= n(V  F/  H/) + n(V/  F  H/) + n(V/  F/  H) = 12 + x + 15 + x + 18 + x = 45 + 3x
= 45 + 3(3)  54 of them play exactly one of the three games. b (iv) For Handball only  21 of them play Handball only. EVALUATION (1) In a Class of 80 undergraduate Students, 21 took elective Courses from Botany only, 16 took from Zoology only, 13 took from Chemistry only. If each of the Students took elective from at least one of the above-mentioned Courses, 7 took Botany and Zoology only, 3 took Zoology and Chemistry only and 8 took Botany and Chemistry only.
n(V/F/  H) = 18 + x = 18 + 3 = 21

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(i) Value of x
(a) Draw a Venn diagram to illustrate the information above using x to represent those that took the three. (b) Find the: (ii) Number that took Botany (iii) Number that took Zoology and Chemistry. (2). In a Group of 120 Students, 72 of them play Football, 65 play Table Tennis and 53 Play Hockey. If 35 of the Students play both Football and Table Tennis, 30 play both Football and Hockey, 21 play both Table Tennis and Hockey and each of the Students play at least one of the three games, (a) Draw a Venn diagram to illustrate this information. (b) How many of them play: (i) All the three games; (ii) Exactly two of the three games; (iii) Exactly one of the three games (iv) Football alone? SSCE, NOV. 1996, № 6 (3) The set A = {1,3,5,7,9,11}, B = {2,3,5,7,11,15} and C={3,6,9,12,15} are subsets of ε = {1,2,3,…,14,15} (a) draw a Venn diagram to illustrate the given information. (b) use your Venn diagram to find (i) C  A/ (ii) A/  (B  C) WASSCE, June 2002, No. 2
(WAEC).
WEEKEND ASSIGNMENT New General Mathematics for SSS, Book 1 Pages 101 – 104 Exercise 8d Question no. 7,9,11,12,13, 14 and 15

3. New General Mathematics for SSS, Book 1 Pages 97 – 104. 4. Man Mathematics for SSS, Book 1, Pages 39- 60

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REFERENCES
1. M.F Macrae etal (2011), New General Mathematics for Senior Secondary Schools 1. 2. MAN Mathematics for senior Secondary Schools 1. 3. New school mathematics for senior secondary school et al; Africana publishers
limited
4. Fundamental General Mathematics For Senior Secondary School by Idode G. O

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WEEK 11

REVISION

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WEEK 12

Examination

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