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) axΓ·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 β (i) listing or roster method (ii) rule method (iii) set builder notation. (c) Types of sets β (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:

π = πΌπ Γ 10π + πΌπβ1 Γ 10πβ1 + πΌπβ2 Γ 10πβ2 + β― + πΌπβπ Γ 10πβπ

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 Γ 104 + 5 Γ 103 + 0 Γ 102 + 7 Γ 101 + 8 Γ 100

= 4 Γ 10000 + 5 Γ 1000 + 0 Γ 100 + 7 Γ 10 + 8 Γ 1

(π) 0.0235 = 0 Γ 100 + 0 Γ 10β1 + 2 Γ 10β2 + 3 Γ 10β3 + 5 Γ 10β4

= 0 Γ 1 + 0 Γ

1 10

+ 2 Γ

1 102 + 3 Γ

1 103 + 5 Γ

1 104

(π) 930.133 = 9 Γ 102 + 3 Γ 101 + 0 Γ 100 + 1 Γ 10β1 + 3 Γ 10β2 + 3 Γ 10β3

= 9 Γ 102 + 3 Γ 101 + 0 Γ 1 + 1 Γ

1 101 + 3 Γ

1 102 + 3 Γ

1 103

EXAMPLE 2

Write the following in expanded notation form

(a) 32.516 (b) 0.10012

(a) 32.516 = 3 Γ 61 + 2 Γ 60 + 5 Γ 6β1 + 1 Γ 6β2

= 3 Γ 6 + 2 Γ 1 + 5 Γ

+ 1 Γ

1

6

1 62

(π) 0.10012 = 0 Γ 20 + 1 Γ 2β1 + 0 Γ 2β2 + 0 Γ 2β3 + 1 Γ 2β4

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= 0 Γ 1 + 1 Γ

1

21 + 0 Γ

1

22 + 0 Γ

1

23 + 1 Γ

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

READING ASSIGNMENT:

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 Γ 103 + 0 Γ 102 + 5 Γ 101 + 8 Γ 100 (b) 4 Γ 101 + 3 Γ 100 + 0 Γ 10β1 + 2 Γ 10β2 (c) 5 Γ 70 + 8 Γ 7β1 + 9 Γ 7β2

REFERENCE TEXTS:

ο· New General Mathematics for senior secondary schools 1 by M.F Macrae et al;

pearson education limited

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ο·

New school mathematics for senior secondary school et al; Africana publishers limited

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CONTENT:

WEEK 2:

NUMBER BASE SYSTEM

οΌ Concept of number base system οΌ Conversion from one base to base 10 οΌ 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 digitsβ0, 1βsuffice to represent a number in the binary system; 6 digitsβ0, 1, 2, 3, 4, 5βare needed to represent a number in the sexagesimal system; and 12 digitsβ0, 1, 2, 3, 4, 5, 6, 7, 8, 9, t (ten), e (eleven)βare needed to represent a number in the duodecimal system. The number 30155 in the sexagesimal system is the number (3 Γ 64) + (0 Γ 63) + (1 Γ 62) + (5 Γ 61) + (5 Γ 60) = 3959 in the decimal system; the number 2et in the duodecimal system is the number (2 Γ 122) + (11 Γ 121) + (10 Γ 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 Γ 53) + (2 Γ 52) + (4 Γ 51)(3 Γ 50) = 125 + 50 + 20 + 3 = 198ten

2. Convert 1111110two to a number in base ten.

Solution:

1111110two= (1 Γ 26) + (1 Γ 25) + (1 Γ 24)+(1 Γ 23) + (1 Γ 22)

(1 Γ 21) + (0 Γ 20) = 64 + 32 + 16 + 8 + 4 + 2 + 0

= 126 ten

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3. Convert 4ππ‘π‘π€πππ£π to base ten.

Solution

4ππ‘π‘π€πππ£π = (4 x 122) + (11 x 121) +( 10 x 120)

=( 4 x 144 )+ (11 x 12) + (10 x 1)

= 576 + 132 + 10

= 718π‘ππ

4. Convert π‘5ππ‘π€πππ£π to base ten.

Solution π‘5ππ‘π€πππ£π = 10 x 122 + 5 x121 + 11 x 120 = 10 x 144 + 5 x 12 + 11 x 1 = 1440 + 60 + 11 = 1511π‘ππ

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.

ο Conversion from other base less than ten to base ten ο Conversion from other base greater than ten to base ten ο Conversion of decimal fraction in one base to base ten ο 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 , β¦.., 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 Γ 70 + 4 Γ 7β1

= 6 Γ 1 + 4 Γ

1 7

= 6 +

4 7

(6

4 7

)10

Example 2:

Convert 101.011π‘π€πto base ten.

Solution

101.011π‘π€ππ‘ = 1 Γ 22 + 0 Γ 21 Γ 1 Γ 20 + 0 Γ 2β1 + 1 Γ 2β2 + 1 Γ 2β3

= 1 Γ 4 + 0 Γ 2 + 1 Γ 1 + 0 Γ

= 4 + 0 + 1 + 0 +

1 2

1 4

1 Γ

1 22 + 1 Γ

1 23

+

1 8

= 5 +

1 4

+

1 8

= 5

3 8π‘ππ

ππ 5.375π‘ππ

Example 3:

Convert 32.516 to base ten.

Solution

32.516 = 3 Γ 61 + 2 Γ 60 + 5 Γ 6β1 + 1 Γ 6β2

= 3 Γ 6 + 2 Γ 1 + 5 Γ

1 6

+ 1 Γ

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 π. π)

EVALUATION

1. Convert 11.011π‘π€π 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 Γ 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 0.111π‘π€πor 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 7π‘ππ = 111π‘π€π

8

4 R 0

2 R 0

1 R 0

0 R 1

8π‘ππ = 1000π‘π€π

β΄ [

111 1000

] = 0.111π‘π€π

Example 2:

Express (19

Solution

13

25

)π‘ππ 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

= 23πππ£π

25

5 R 0

1 R 0

0 R 1 = 100πππ£π

= [34 +

23

] 100 πππ£π

= [34 + 0.23]πππ£π

= 34.23πππ£π

EVALUATION

1. Express 2. Express (11011.10012) to decimal base.

to base two

5

13

READING ASSINGMENT

Read about the operations with bases.

WEEKEND ACTIVITY

Convert the following numbers in denary to the base indicated.

(π)37.31π‘ππ to base 6

(b) 10.8π‘ππto base 3.

Find the value of π₯ in each of the following equations.

(a) 23π₯ + 14π₯ = 42π₯ (b) 53π₯ β 24π₯ = 25π₯ (c) 113π₯ + 121π₯ = 300π₯ (d) 562π₯ β 153π₯ = 407π₯

Find the value of π₯ and π¦ in the following pairs of equation.

(a) 32π₯ + 53π¦ = 55

24π₯ + 35π¦ = 45

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(b) 64π₯ β 53π¦ = 25

(c) 54π₯ β 11π¦ = 36

(d) 34π₯ + 21π¦ = 26

WEEKEND ASSIGNMENT

47π₯ β 34π¦ = 21

25π₯ + 10π¦ = 21

42π₯ β 12π¦ = 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 β 3Btwelve

501eight

+ 3Beight

Recall B in base twelve is eleven.

If 1 is βborrowed 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 Γ· 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 240sixΓ· 20six

12

20

240

20

40

40

So 240sixΓ· 20six= 12π ππ₯

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. 387nineΓ· 25nine

ASSIGNMENT

New General Mathematics SS1. Ex 3i Nos 1 β 5 page 54.

Functional Mathematics Graduated Exercise page 17 β 18, No 1- 30

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CONTENT:

WEEK 4

MODULAR ARITHMETIC

ο Revision of addition and subtraction of integers ο Revision of multiplication and division of integers ο 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, β1, β5 ππ‘π These numbers can be added, subtracted, divided or multiplied. (i)

Addition of integers; (a) 486 + 289 = 775 (b) β25 + (β78) = β103 Subtraction of integers; (a) 582 β 328 = 254 (b) 902 β 437 = 465

(iii) Multiplication of integers;

(ii)

(iv)

(a) 181 Γ 42 = 7602 (b) 208 Γ 5 = 1040 Division of integers; (a) 972 Γ· 27 =

972

= 36

27

1008

= 84

12

(b) 1008 Γ· 12 =

EVALUATION:

Solve the following;

(i) (ii) (iii) (iv)

3092 + 216 + 1801 = Μ 2968 β 989 = Μ 318 Γ 322 = Μ 420 Γ· 35 = Μ

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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Arithmeticβ 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 Γ· 3 = 21, πππππππππ 2

65 = 2(πππ 3)

(b) 65 Γ· 4 = 16, πππππππππ 1

65 = 1(πππ 4)

(c) 65 Γ· 5 = 13, πππππππππ 0

65 = 0(πππ 5)

(d) 65 Γ· 6 = 10, πππππππππ 5

65 = 5(πππ 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 dayβs 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 π‘βπ’ππ πππ¦ + 3πππ¦π = π π’ππππ¦

Find the number which results from the following additions on the number cycle below of ice cream

(a) 2 + 9 = 11

11 Γ· 3 = 3, πππππππππ 2 β΄ 2 + 9 β‘ 2

(b) ππππππππ¦ 3 + 14 ππ 4 ππ¦ππππ ππ£πππ‘π ,

17 Γ· 4 = 4, πππππππππ 1

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β΄ 3 + 14 β‘ 1

EVALUATION:

Use the number cycle 5 to simplify (a) 1 + 6 (b) 2 + 32

READING ASSIGNMENT

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

ο Addition, Subtraction, Multiplication and Division operations in module arithmetic ο 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)

Ξ 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 Γ· 6 = 11, πππππππππ 2

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= 2(πππ 6)

(ii) Calculate the following in the given moduli (a) 12Ξ 5(mod 4) (b) 38 Ξ 42(mod 7)

Solution: (a) 12Ξ 5 = 7

7 = 4 + 3

= 3(mod 4)

(b) 38 Ξ 42 = β4

β4 = β7 + 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) β9 (b) β32 (c) β75 (d) β256

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 (Γ·) 3(mod 4) (b) 7 (Γ·) 2(mod 5) (c) 2 (Γ·) 2(mod 4)

Solution: (a) If 2 (Γ·) 3 = π₯

1

π₯

=

2 β 3 Cross-multiply , 3π₯ = 2 Add 4 to RHS 3π₯ = 2 + 4(mod 4) 3π₯ = 6(mod 4) Divide both sides by 3 π₯ = 2(mod 4)

(b) 7 (Γ·) 2 = π₯

1

π₯

=

7 2 2π₯ = 7(mod 5) 2π₯ = (5 x 1) + 2(mod 5) 2π₯ = 2 π₯ = 1

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(c) 2 (Γ·) 2 = π₯

π₯

2

2

1

=

2π₯ = 2(mod 4) Divide both sides by 2 π₯ = 1 Or 2π₯ = 2 + 4(mod 4) 2π₯ = 6(mod 4) π₯ = 3(mod 4)

N.B If 3(Γ·)2 = π₯, then 2π₯ = 3

No multiple of 4 can be added to 3 to make it exactly divisible by 2. There are no values of 3(Γ·) 2 in modulo 4.

EVALUATION:

Calculate the following division in modulo 5

(a) 28(Γ·)7 (b) 29(Γ·)2 (c) 58(Γ·)4 (d) 74(Γ·)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

Ξ 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

READING ASSIGNMENT:

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

ο Revision of standard form ο Introduce indices and examples ο Laws of indices ο 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 10β2

(b) 0.00065 = 6.5 x 0.0001

= 6.5 x 10β4 (c) 0.0058 = 5.8 x 0.001 = 5.8 x 10β3

(d) 0.61 = 6.1 x 0.1

= 6.1 x 10β1

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Method 2:

(a) 0.037 =

=

=

0.037

1 3.7

100 3.7 102

= 3.7 x 10β2

(b) 0.00065 =

=

0.00065

1 6.5

10000 6.5 104

0.0058

1 5.8

1000 5.8 103

= = 6.5 Γ 10β4

(c) 0.0058 =

=

= = 5.8 Γ 10β3

(d) 0.61 =

0.61

1 6.1

= = 6.1 Γ 10β1

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 +, _ ,Γ and Γ·) 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 β 0.

(1) x m

Γ x n

(2) x m

Γ· x n

(3) x βn

(4) x o

(5) (x m)n

(6) x1/n

(7) x m/n

=

=

=

=

=

=

=

=

x m+ n

x m / x n = x m β 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 Γ· x n = x m-n

Example 2:

simplify the following

(i) 37 Γ· 34

(ii) 21a4 b3

7ab2

(iii) 9a5 b3

Γ· 3a2

b-2

Solution:

(i) 37 Γ· 34

= 37 – 4

= 33

= 27

(ii) 21a4b3

7ab2

= 3a4 β 1b3 β 2

= 3a3b

(iii) 9a5b3 Γ· 3a2b-2 = 3a5 β 2 b3 β (-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 οΈ 57

(vi) 15a3b5

5ab2

(vii) 8a6b2 οΈ 4ab-2

LAW 3: x βn = 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 . xβ 9. y6 . yβ 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 Γ· 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 Γ· 9x-3

=

=

=

=

=

33x2x3 Γ· 9x-3

27×6 Γ· 9x-3

3×6 β (-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 Β½

52

= 4 2 x Β½

5 2 x Β½

= 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

ο 27

(ii)

51/16 -3

4

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ο

(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 β΄ x = (ii)

3x β 1 = 81

3x β 1 = 34

x β 1 = 4

x = 4 + 1

β΄ x = 5

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=

=

=

=

2

(iii)

8x

23x

3x

x

β΄ 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 = 2Β½ 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

ο Deducing logarithm from indices and standard form ο Definition of Logarithms ο Antilogarithms ο 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 β€ 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. βSomethingβ

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

π

π

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

Add

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

Add

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 Γ· 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 Γ· 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 Γ· 2.581 = 29.16

Evaluation

Use table to evaluate the following

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(1) 53.81 Γ· 16.25 (2)632.4 Γ· 34.25 (3) 63.75 Γ· 8.946 (4) 875.2 Γ· 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

π₯ππππ

Log M =

Log M = Log βππ π π₯ (c) Log Mx/n = π Example 1. Evaluate the following (53.75)3 Solution

π

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. β35.81 3. 19.183 4. (67.9/5.23)3 5. β679.5 x 92.6

SUB-TOPIC 4 Calculation of roots using Logarithms and Anti-Logarithms Use tables of Logarithms and antilog to calculate β27.41 Solution

5

Number

Standard Form

Log

Operation

5 β27.41

(2.741 x 101)1/5 1.4380 Γ· 5

Divide Log by 5

Log

Anti-Log

0.2876

1939

0.2876

Standard form

Number

1.939 x 100

1.939

Therefore β27.41

= 1.939

5

Example 2: use table to find β218 3.12

3

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Solution

Hint: Workout 218 Γ· 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 Γ· 3

Division

0.6148

Standard form

Number

4.119 x 100

4.119

218

3.12

Log

0.6148

4119

Therefore β218 3.12

3

= 4.119

Example 3

Evaluate 63.752 β 21.392

We can use difference of two squares

i.e A2 β B2 = (A+B)(A-B)

63.752 β 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

Addition

…. 3606

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EVALUATION

Evaluate the following

1. (39.652 β 7.432)1/2 2. 84.352 β 36.952 3. 64.742 β 55.262 4. 94.682 β 43.252 5. 25.142 β 7.522

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Content:

ο Definition of sets ο Set notations ο 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) SetβBuilders Notations

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A set can also be specified using the set β builder notation. Set β builder notation is an algebraic way of representing sets using a mixture of word, letters , numbers and inequality symbols e.g. B = {x : 6 β€ x < 11, x Ρ ΖΆ} or B = {x/6 β€ x < 11, x Ρ ΖΆ}. The expression above is interpreted as βB 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)β

–

The stroke (/) or colon (:) can be used

interchangeably to mean βsuch thatβ

– The letter Z or I if used represents integer or whole numbers.

Hence, the elements of the set A = {x : 6 β€ x < 11, x Ρ ΖΆ} are A = { 6, 7, 8, 9,10}.

NB:

–

–

The values of x starts at 6 because 6 β€ x

The values ends at 10 because x < 11

and 10 is the first integer less than 11.

The set builderβs 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 β 4 = 1, x Ρ ΖΆ}

(b) P = {x : x is the prime factor of the LCM of 15 and 24}

Β© Q = {The set of alphabets}

(d) R = {x : x β₯ 5, x is an odd number}

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

Examples 1:

List the elements of the following sets

(i)

(ii)

A = {x : 2 4, x Ρ ΖΆ}

(iii)

C = {x : -3 β€ x β€ 18, x Ρ ΖΆ}.

(iv)

D = {x : 5x -3 = 2x + 12, x Ρ Z}.

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(v)

E = {x : 3x -2 = x + 3, x Ρ I}

(vi)

F = {x : 6x -5 β₯ 8x + 7, x Ρ ΖΆ}

(vii)

P = {x : 15 β€ 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 β€ 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 β€ x, and ends at 18 because x β€

To be able to list the elements of this set, the equation defined has to be

solved

(iv) i.e.

5x β 3 = 2x + 12

5x β 2x = 12 + 3

3x = 15

x

= 15/3

β΄ x = 5

β΄ D = {5}

(v) We also need to solve the equation to get the set values

3x – 2 = x + 3

3x β x = 3 + 2

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2x

= 5

β΄ x = 5/2

Since 5/2 is not an integer (whole number) therefore the set will contain no element.

β΄Ρ

= { } or Γ

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

6x β 5 β₯ 8x + 7

6x β 8x β₯ 7 + 5

-2x β₯ 12

x β€ 12/-2

β΄ x β€ -6

β΄ F = {β¦, -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 Ρ ΖΆ} OR

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A = {x : 8 β€ x < 15, x Ρ ΖΆ} OR

A = {x : 7 2, x Ρ ΖΆ} OR

B = {x : x β₯ 3, x Ρ ΖΆ}

(iii)

C = {x : x 8 or x β₯ 7, x is odd, x Ρ ΖΆ}

P={1,2} suggests the solutions of a quadratic equation. Therefore , the equation or

(v) set-builders notation can be obtained from :

x2 β (sum of roots)x + product of roots = 0

x2 β(-1)x + (1 x -2) = 0

x2 + x – 2 = 0

P = {x : x2 + x – 2 = 0, x Ρ ΖΆ}

(vi) Q = {x : x is a vowel}

EVALUATION

1. List the elements in the following Sets

(a)

A = {x : -2 β€ x < 4, x Ρ ΖΆ}

(b)

B = {x : 9 < x < 24, x Ρ N}

(c)

C = {x : 7 < x β€ 20, x is a prime number, x Ρ I}

(d)

D = {x / 2x β 1 = 10, x Ρ Z}

(e)

P = {x : x are the prime factor of the LCM of 60 and 42}

2. Rewrite the following using Set β

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 = {β¦,-4,-3,-2,-1,0,1,2,3,β¦}

(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 ΓΈ 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 Ρ ΖΆ }

Solution:

3x β 5 < x + 1 and x + 1 < 2x + 3

3x β x < 1 + 5 and x β 2x < 3 β 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}

β΄ 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 Γ since 8, 9, 10 are no prime numbers. β΄ n(A) = 0

Example 6:

Find the number of elements in the following sets:

(i)

B = {x : x β€ 7, x Ρ ΖΆ}

(ii)

C = {x : 3 < x β€ 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) = β

(ii)

C = {4, 6, 8}. β΄ 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 ΞΎor ΞΌ.

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 β€ x < 4, x Ρ ΖΆ}

(b)

B = {x : 9 < x < 24, x Ρ N}

(c)

C = {x : 7 < x β€ 20, x is a prime number, x Ρ I}

(d)

D = {x / 2x β 1 = 10, x Ρ 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) =

Β© Q= {a, d, g, a, c, f, h, c,} , Then n(Q) =

(d) P= {4,5,6,7,β¦,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 β₯ 7, x Π Z}

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(ii)

P = {x : -4 β€ x < 16, x Π I}

(iii)

A = {x : 2x β 7 = 2, x Π 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 ΞΎ = {x: 1< x< 15, x Ξ΅ z}

(i) A= {x; -3β€ xβ€7, x Ξ΅ Z}

(ii) B= {x: 5<x<6, xΞ΅z},

(iii) C={x: Xβ₯ 5, xΞ΅ 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)

ΞΎ ={ x: 1 < x < 30, x are multiples of 3} ΞΎ = { x: 7β€ x<25, x are odd numbers} ΞΎ = { x: xβ₯10, xΞ΅z}

WEEKEND ASSIGNMENT New General Mathematics for SSS, Book 1 Pages 97- 100 Exercise 8c Question no. 1,4 and 5

WEEKEND READING

1. New General Mathematics for SSS, Book 1 Pages 97 β 100. 2. Man Mathematics for SSS, Book 1, Pages 45 – 60

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Content:

WEEK 10

SETS

ο Set operations ο 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 βA βBβ meaning A Union B. Thus A βB 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 β 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, ΞΈ} Then A βB βC = {a, b, c, d,1, 2, 3, 4, ΞΈ}

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 βA β© Bβ meaning A intersection B. When represented using Venn diagram we have

ΞΎ

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

ΞΎ

B

A

Example 12:

Given that A = {5, 7, 8, 10} and B = {3, 5, 6, 8, 9}, then

A β© 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 β© Q β© R = {c, g}

Example 14:

If A = {1, 2, 3} and B = {6, 8, 10}, then

A β© B = { } or Γ. The Set A and B are disjoint.

EVALUATION

1. Given that ΞΎ= {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 ΞΎ.. Find:

(i) P βQ

(ii) P β©Q

(iii) Q β©R

(iv)

(P β©Q) βR

(v) P β©Q β©R

(vi) (P βQ) β©(Q βR)

(vii) (P β©Q) β(Q β©R)

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If A = {1, 2, 3, 4} and

B = {3, 5, 6}, find:

(i) A β©B

(ii) A β B

(iii) (A β© B) β B

(iv) (A β B) β© A

(WAEC)

2.

3.

If ΞΎ = {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 Γ

(f) ΞΎ

4.

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

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

following:

(a) A βB

(b) B β C

(c) A βB β C (d) A β©B

(e) A β©C

(f) B β©C

(g) (A β B) β©(A β C)

(h) A β©( B β C)

5.

Let ΞΎ = {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 β B) β©(B β C)

A β©( B β C)

B β©(A β C)

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(d)

Show that:

(A β©B) β (A β©C) = A β©(B β C)

SUB-TOPIC 6: The Complement of Set

If A is a Subset of the Universal Set ΞΎ, then, the complement of the Set A are made up of elements that are not in A, but are found in the Universal Set ΞΎ. This is usually denoted by Ac or Aβ². for example

If ΞΎ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {3, 5, 6, 9} then Ac or Aβ² = {1, 2, 4, 7, 8, 10}

Using Venn diagram, this is represented by the shaded portion below:

1

2

3 5 6

ΞΎ

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

ΞΎ = {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) Aβ²

(ii) Bβ² (iii) (A βB)β²

(iv) (A β©B)β² (v) Aβ² β©Bβ² (vi) Aβ² βBβ²

(vii) (Aβ²)β²

Solution:

(i)

Aβ² = {12, 14, 15, 17, 19}

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(ii)

Bβ² = {11, 13, 15, 17}

(iii)

A βB = {11, 12, 13, 14, 16, 18, 19, 20}

(A βB)β² = {15, 17}

(iv)

A β©B = {16, 18, 20}

(A β©B)β² = {11, 12, 13, 14, 15, 17, 19}

(v)

(Aβ² β©Bβ²) = {15, 17}

(vi)

Aβ² β Bβ² = {11, 12, 13, 14, 15, 17, 19}

(vii) Aβ² = {12, 14, 15, 17, 19}

(Aβ²)β² = {11, 13, 16, 18, 20} = A

NB:

From the example above, observe that from (iii) and (v) (A β B)β² = Aβ² β© Bβ²

Also, from (iv) and (vi)

(A β© B)β² = Aβ² β Bβ² and from (vii) (Aβ²)β² = A

Example 16:

Given that ΞΎ = {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) Aβ²

(ii) Bβ²

(iii) Aβ² β Bβ²

(iv) Aβ²β© Bβ²

(v) (A β© B)β²

(vi) (A β B)β²

(vii) (Bβ²)β²

Solution:

(i)

(ii)

Aβ² = {b, d, f, h, j}

Bβ² = {a, e, g, h}

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(iii) Aβ² β Bβ² = {a, b, d, e, f, g, h, j}

(iv)

Aβ² β© Bβ² = { h }

(v)

A β© B ={c, i }

(A β© B)β² ={a, b, d, e, f, g, h, j}

(vi)

A β B = {a, b, c, d, e, f, g, i, j}

(A βB)β² =

{ h}

(vii) Bβ² =

{a, e, g, h}

(Bβ²)β² = {b, c, d, f, i, j} = B

From the example above, we also observe that Aβ²β Bβ² = (A β© B)β² — From (iii) and (v)

Aβ² β© Bβ² = (A β B)β² — From (iv) and (vi)

And (Bβ²)β² = B — From (vii) From the last two examples we can clearly see that (AβB)β² = Aβ²β© Bβ²,

(A β© B)β² = Aβ² β Bβ²

and (Aβ²)β² = A

Generally, for any two Subsets A and B of a Universal Set ΞΎ, the following are true:

(i)

(A β B)β² = Aβ² β© Bβ²

(ii)

(A β© B)β² = Aβ² β Bβ²

(iii)

(Aβ²)β² = A or (Bβ²)β² = B

These are known as De Morganβs 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 β 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 β‘ B (A is equivalents to B) since n(A) = n(B).

EVALUATION

1.

If ΞΎ = {x : 0 < x β€ 15 x Ξ΅ Ζ΅}

A = {1, 5, 7, 11} and

B = {5, 7, 14}

Find the following:

(a) A β© B

(b) Aβ² (c) Bβ²

(d) (A β B)β²

(e) Aβ² β© B (f)A β© Bβ²

(g) Show that: (Aβ©B)β² = Aβ² β Bβ²

2.

Given that ΞΎ = {x : 0 β€ x < 10, x Ξ΅ Ζ΅}

P = {1, 2, 4, 5}

Q = {2, 4, 6, 8} and

R = {1, 3, 4, 8, 9}

Find:

(a) Pβ² (b) Qβ² (c) Rβ² (d) Q β© Rβ²

(e) Pβ² β© Qβ²

(f) Pβ² β Qβ² (g)(Q β© Rβ²)β²

(h) (Pβ² β© Rβ²)β² (i) (P β© ΞΎ) β²

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(j) Pβ² β© Qβ² β© Rβ²

3.

P, Q and R are Subsets of the

Universal Set Β΅ such that Β΅ = {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 β© Q (ii) P β Qβ² β© R

(iii) Q β R

(iv) Pβ² (v) Pβ² β© 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.

οΈ

A

B

I

II

III

IV

Compartment I: represent the set of elements in A only. i.e. A ο B/ using set notations. Compartment II: represents the set of elements common to both A and B i.e. AοB

Compartment III: represents the set of elements in B only i.e. A/ ο B

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Compartment IV:

(AοB) / or A/ ο 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(οΈ) = 40

n(L) = 19

n(B) = 17

n(L ο B)/ = 13

Let x represents those that have visited both Cities

i.e. n(LοB) = x

οΈ

L

B

19 β x x

17 β x

13

19 β x + x +17 β x + 13 = 40

49 β x = 40

49 β 40 = x

9 = x

ο9 Students have visited both Cities

(b)

Those that have visited Benin City only are = 17 β x

= 17 β 9

= 8

ο 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(οΈ) = 45

n(A) = 21

n(B) = 25

n(A οB) = 6

Let n(AοB) / = x

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

οΈ

A

B

21 β 6

6 25 β 6

x

21 – 6 + 6 +25 β 6 + x = 45

15 + 6 + 19 + x = 45

40 + x = 45

x = 45 β 40

ο x = 5

οThose that offer neither,

i.e. n(AοB) / = 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 (2xβ5) ate none of the two. How many ate

both fried rice and salad?

(i) (ii) salad but not fried rice?

(ii)

neither fried rice nor salad?

(3) The Venn diagram below represents a universal set οΈ of integers and its subsets

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

(a) P ο Q

(b) P ο Q

οΈ

P 1 9

(c ) οΈ ο Q

2

3

8, 4

(d) P ο οΈ

10 5

6

7 Q

SSCE, NOV. 1995 Nα» 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: οΈ

B

A

V II VI

III I

IV

VII

C VIII

Compartment I represents AοBοC

(elements common to the three sets A, B and C).

Compartment II represents AοBοC/

(elements common to both A and B only).

Compartment III represents AοB/οC

(elements common to both A and C only).

Compartment IV represents A/οBοC

(elements common to both B and C only).

Compartment V represents A οB/οC/

(elements of A only).

Compartment VI represents A/ ο B ο C /

(elements of B only).

Compartment VII represents A/ οB/ οC

(elements of C only).

Compartment VIII represents (AοBοC)/ or A/οB/οC/ 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(οΈ) = 80

Let V be Volleyball F be Football H be Handball

n(VοF/οH/) i.e. Volleyball only = 15 n(V/οFοH/ ) i.e. Football only = 18 n(V/οF/οH) i.e. Handball only = 21 n(VοFοH/) i.e. Volleyball and Football only = 5 n(VοF/οH) i.e. Volleyball and Handball only = 8 n(V/οFοH) i.e. Football and Handball only =10 Let n(VοFοH) = x . i.e. Those that play the three games = x

(a)

οΈ 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 ο x = 3

x = 80 β 77

The number that plays football

n(F) = 18 + 5 + 10 + x = 33 + 3 = 36

ο 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(οΈ) = 80 n(V) = 31 n(F) = 36 n(H) = 42 n(V ο F ο H) = 80 (Since each play at least one of the games). n(V ο F) = 8 n(V ο H) = 11 n(F ο H) = 13. n(V ο F ο H) = x.

Let

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To fill the Venn diagram we start with the centre Compartment

where n(V ο F ο H) = x. οΈ V F 12 + x 8 β x 15 + x

x

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

13 – x

11 β x

18 + x H

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

= 42 β [ (11 β x ) + x + (13 β x ) ] = 42 β [24 β x ] = 42 β 24 + x

11β x 13 – x

H

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

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

x

13 β x

i.e.n (V ο F/ ο H/)

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

= 31 β 19 + x

= 12 + x

8 – x

x

11 β 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 β x + 15 + x + x + 11 β x + 13 β x + 18 + x = 80 ο

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

οΈ

15

5

18

3 8 10

21

H

b (ii) Exactly two of the three games

= n(V ο F ο H/) + n(V ο F/ ο H) + n(V/ ο F ο H) = 8 β x + 11 β x + 13 β x

= 32 β 3x = 32 β 3(3) ο 23 of them play exactly two of the three games. b (iii) Exactly one of the three games

= 32 β 9 = 23

= 45 + 9 = 54

= n(V ο F/ ο H/) + n(V/ ο F ο H/) + n(V/ ο F/ ο H) = 12 + x + 15 + x + 18 + x = 45 + 3x

= 45 + 3(3) ο 54 of them play exactly one of the three games. b (iv) For Handball only ο 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/οF/ ο 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, β 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 Ξ΅ = {1,2,3,β¦,14,15} (a) draw a Venn diagram to illustrate the given information. (b) use your Venn diagram to find (i) C ο A/ (ii) A/ ο (B ο C) WASSCE, June 2002, No. 2

(WAEC).

WEEKEND ASSIGNMENT New General Mathematics for SSS, Book 1 Pages 101 β 104 Exercise 8d Question no. 7,9,11,12,13, 14 and 15

WEEKEND READING

3. New General Mathematics for SSS, Book 1 Pages 97 β 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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