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FIRST TERM E-LEARNING NOTE
SUBJECT: MATHEMATICS CLASS: JSS1
Weeks | Topic |
1 | Whole Numbers Counting and Writing (i) Millions (ii) Billions (iii) Trillions |
2 | Whole Numbers Continued: Problems solving in quantitative aptitude reasoning using large numbers |
3 | Lowest Common Multiple (L.C.M) and Highest Common Factor (H.C.F) of Whole Numbers. (a) Concepts of L.C.M and H.C.F (b) L.C.M and H.C.F of quantitative reasoning |
4 | Fractions: (a) Meaning of Fraction (b) Types of fractions (Proper & Improper) (c) Mixed numbers |
5 | Fractions continued: Equivalent Fractions ( Identify and apply equivalent fractions in showing commodities and problems solving in quantitative aptitude) |
6 | Fractions continued. (a) ordering of fractions (b) conversion of fractions to percentage and vice versa (c) conversion of fraction to decimal and vice versa |
7 | Review of the first half term’s work and periodic test |
8 | Fractions continued: Addition and subtraction of fractions |
9 | Fractions Continued: (a) Multiplication and Division of fractions (b) Prime numbers and factors |
10 | Estimation: (i) Concept of estimation and reasons (ii) Estimation of dimensions and reasons (iii) Estimation of capacity(volumes) and mass of objects (iv) Estimation of other things (v) Quantitative reasoning involving estimation |
11 | Revision of the 1^{st} term’s work and preparation for the first term examination |
12 | First term examination |
WEEK ONE
TOPIC: WHOLE NUMBERS
CONTENT
- Introduction
- System of Counting
- Counting in Millions
- Counting inBillions and Trillions
INTRODUCTION
- Counting
It is likely that mathematics began when people started to count and measure. Counting and measuring are part of everyday life.
Ancient people used fingers and toes to help them count or group numbers in different number bases. This led them to collect numbers in groups: sometimes 5s (fingers of one hand), sometimes 10s (both hands) and even in 20s (hands and feet). When people group numbers in 5s, we say they use a base five method. The most common bases used were five, ten and twenty. For example, a person with thirty two cows would say ‘I have six fives and two cows’ when counting in base ten. The most widely used base is base ten also called the denary system.
Other bases of counting: seven and sixty
7 days = 1 week
60 seconds = 1 minute
60 minutes = 1 hour
In English, ‘dozen’ means 12, ‘score’ means 20 and ‘gross’ means 144
System of Counting
- Tally System
Tally marks were probably the first numerals.
The ancient people employed tally marks to count large numbers. The tally marks were scratched on stones or sometimes cut on sticks but today we use tally marks to count or record large data, especially in statistics.
A tally mark of 5 is written by putting a line across a tally count of 4.
i.e = 4 and = 5
Example 1
Draw the tally marks for each of the following numbers:
- 34 (b) 15
Solution
- 34 =
- 15 =
EVALUATION
- During a dry season, it did not rain for 128 days. How many weeks and days is this?
- What is the number represented by
- Draw the tally marks for each of the following numbers: (a) 43 (b) 52
- Roman numerals
The Romans used capital letters of the alphabets to represent numbers. Many people believe that the Romans used the fingers to represent numbers as follows:
I for one finger, II for two fingers, III for three fingers, V for five fingers and X for the combination of two hands ( or two V’s) .
The Roman also used L for fifty, C for hundred, D for five hundred and M for one thousand as shown below.
Hindu-Arabic
Roman Numeral
Hindu-Arabic
Roman Numeral
1
I
20
XX
2
II
40
XL
3
III
50
L
4
IV
60
LX
5
V
90
XC
6
VI
100
C
7
VII
400
CD
8
VIII
500
D
9
IX
900
CM
10
X
1000
M
The Roman used the subtraction and addition method to obtain other numerals. For example
- IV means V- I i.e. 5- 4 = 4
- VI means V+ I, i.e. 5 + 1 = 6
- IX means X- I, i.e. 10 – 1 = 9
- XXIV means XX + IV = 20 + 4 = 24
- CD means D- C = 500 – 100 = 400
- MC means M + C = 1000 + 100 = 1100
Example 1
Change the following numbers to Roman numerals: (a) 2459 (b) 3282
Solution
- 2459— 2000 = MM
400 = CD
50 = L
9 = IX
2459 = MMCDLIX
- 3282 = 3000 + 200 + 80 + 2
= MMM CC LXXX II
i.e 3282 = MMMCCLXXXII
EVALUATION
- Write the following Roman figures in natural ( or counting) numbers:
- MMMCLIV (b) MMCDLXXI (c) MCMIX (d) DCCCIV
- Write the following natural numbers in Roman figures:
- 2659 (b) 1009 (c) 3498 (d) 1584
- The Counting board
A counting board is a block of stone or wood ruled in columns. Loose counters, pebbles, stones or seeds in the columns show the value of the numbers in the columns.
Counters in the right-hand column (U) represent units, counters in the next column (T) represent tens, and so on.
TH | H | T | U |
●●● | ● | ||
●● | ●●●● | ●●●● |
2 7 5
The diagram below is a counting board showing the number 275.
- The Abacus
An abacus is a frame consisting of beads or disks that can be moved up or down (i.e. slide) on a series of wires or strings. Each wire has its own value. Both abacus and counting board work in the same way when carrying out calculations.
Example 1
M HTH TH H T U
An Abacus showing 2703
- Place Value of Numbers
Numbers of units, tens, hundreds,…….., are each represented by a single numeral.
(a).For a whole number:
– the units place is at the right-hand end of the number.
– the tens place is next to the units place on the left, and so on
For example: 5834 means ↓
5 thousands, 8 hundreds, 3 tens, and 4 units.
See the illustration below:
5 8 3 4
(b) for decimal fraction, we count the places to the right from the decimal point as tenths, hundredths, thousandths, etc.
See the illustration below:
↓ ↓ ↓ ↓ ↓
6 . 7 9 8
6 → units
. → decimal
7 → tenths
9 → hundredths
8 → thousandths
Example 1:
What is the place value of each of the following?
- the 9 in 10269
- the 2 in 2984
Solution:
- the 9 in 10269 is = 9 units or nine units
- the 2 in 2984 is = 2 thousands or two thousands
Example 2
What is the value of each of the following?
- the 8 in 1.85
- the 0 in 16.08
Solution:
- the 8 in 1.85 is = 8 tenths or eight tenths
- the 0 in 16.08 is =0 in tenths or zero tenths
Example 3
What is the value of each digit in 3 865 742
Solution
3 | 8 | 6 | 5 | 7 | 4 | 2 | |||
M | H. Th | T.Th | Th | H | T | U | |||
Digit | Value | Word Form | |||||||
3 | 3 000 000 | Three million | |||||||
8 | 800 000 | Eight hundred thousand | |||||||
6 | 60 000 | Sixty thousand | |||||||
5 | 5 000 | Five thousand | |||||||
7 | 700 | Seven hundred | |||||||
4 | 40 | Forty | |||||||
2 | 2 | Two |
EVALUATION
1 (a) The place value of 5 in 5763 is ……………
(b)What is the place value 1 in 5.691?
2. Give the value of each digit in 489 734
3. Write down the number shown in the following figures:
(a)
READING ASSIGNMENT
- Essential Mathematics for JSS1 by AJS Oluwasanmi page 3-7
- New General Mathematic for Jss1 by M. F. Macrae et al page 17-18.
Counting and Writing in millions, billions and trillions
The figures 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are called digits or units.
The table below gives the names and values of some large numbers.
Name | Value |
One thousand | 1 000 |
Ten thousand | 10 000 |
One hundred thousand | 100 000 |
One million | 1 000 000 |
Ten million | 10 000 000 |
One hundred million | 100 000 000 |
One billion | 1 000 000 000 |
One trillion | 1 000 000 000 000 |
Large numbers can be read easily by grouping the digits in threes starting from the right hand side as shown below.
Billion Million TH H T U
25 800 074 4 3 0
The 1^{st} gap separates hundreds from thousands and the second gap separates thousands from millions and the third gap separates million from billion.
Thus 25 800 074 430 reads twenty five billion, eight hundred million, seventy four thousand, eight hundred and ninety.
Example
Write the following in figures:
- twelve billion, three hundred and nine million, ninety five thousand, six hundred and sixty three
- six trillion, four hundred and thirty billion, one hundred and five million, two hundred and one thousand and fifty four
- nine hundred and four billion, five hundred and forty million, three hundred and seventy thousand, seven hundred and fifty
Solution
- You can work it out as follows:
Nine hundred and four billion | = 904 000 000 000 |
Five hundred and forty million | = 540 000 000 |
Three hundred and seventy thousand | = 370 000 |
Seven hundred and fifty | = 750 |
Adding | = 904 540 370 750 |
EVALUATION
- Write the following in figures:
- Write in figures, the number referred to in the statement: Last year a bank made a profit of ‘two hundred and twenty billion, five hundred and one thousand, four hundred and ninety three Naira ( N)
WEEKEND ASSIGNMENT
- The value of 8 in 18214 is (a) 8 units (b) 8 tens ( c) 8 hundreds ( d) 8 thousands (e) 8 ten thousands
- The Roman numerals CXCIV represents the number (a) 194 (b) 186 (c ) 214 (d) 215 (e) 216.
- What is the number represented by ? (a) 32 (b) 40 (c) 28 (d) 39
- The value of 7 in 3.673 is (a) 7tenths (b) 7 hundredths ( c ) 7 units ( d) 7 hundredth.
- Three million and four in figures is (a) 300004 (b) 300040 (c) 30000004 (d) 3000004
THEORY
- Change this Roman figure to natural numbers
(i) MMCDLXXI (ii) MMMCLIV
- Write the following in figures:
(a) fifteen trillion, six hundred and seventy one billion, three hundred and ninety one million, eighty eight thousand, five hundred and fifty five.
(b) three hundred and twenty-nine billion, five hundred and sixty two million, eight hundred and one thousand, four hundred and thirty three.
WEEK TWO
TOPIC: WHOLE NUMBERS
CONTENT
- Ordering Large Numbers
- Using Mixture of Digits and Words with Large Numbers
- Problems Solving in Quantitative Aptitude Reasoning (QR) using large numbers
Ordering Large Numbers
Any 2-digit number is larger than every unit number, e.g 11 is larger than 9. Any 3-digit number is larger than every 2-digit numbere.g 132 is greater than 86, and so on.
When a set of numbers are given, it is useful to rearrange the numbers in such a way that those that start in such a way that those that start with the same digit can be compared.
Example 1
Find the smallest and the largest number from the following set of numbers:
2 675 571, 3 498 567, 2 670 781, 3 497 859
Solution
Comparing numbers that start with 2
2 6 7 5 5 7 1
2 6 7 0 7 8 1
Comparing numbers that start with 3
3 4 9 8 5 6 7
3 4 9 7 8 5 9
The smallest number is 2 670 781 and the largest number is 3 498 567.
The numbers in the example above can also be arranged in order of size starting with the smallest as follows:
2 670 781, 2 675 571, 3 497 859, 3 498 567
This arrangement is also called ascending order. The reverse is known as descending order.
Example 2
Arrange these numbers in order of size magnitude) starting with the smallest: 13456786, 24567432, 38479871, 24558011, 13498069, 38478817.
Solution
Always group large numbers in threes.
Arranging the numbers that start with 1 in order of size: 13 456 786, 13 498 069
Arranging the numbers that start with 2: 24 558 011, 24 567 432
Arranging the numbers that start with 3: 38 478 817, 38 479 871
Hence, arranging these numbers in order of magnitude gives: 13 456 786, 13 498 069, 24 558 011, 24 567 432, 24 558 011, 24 567 432
EVALUATION
- Arrange the following numbers in ascending order: 89728567, 89704567, 89693670, 89776909, 89735890.
- Arrange the following numbers in descending order: 217679057, 497378939, 234656452, 21023404895, 2100998969.
Using Mixture of Digits and Words with Large Numbers
People often get confused when reading and writing large numbers. To avoid the confusion, the editors of newspapers use a combination of digits and words to show large numbers. For example, 1 million people, $ 1.5 billion, N 3.6 trillion.
Some newspapers headlines are as follows:
- Unemployment soaring to 10.7 million
- HIV rose to an estimated 23 million in 2010
- Cost of ID rises to £10 billion in the UK.
Example 1
Write these numbers as a mixture of digits and words:
- £30 000 000 (b) N75 000 000 000 (c) $460 000 000 (d) £3 400 000 000 000
Solution
- £30 000 000 = £30 x 1 000 000
= £30 million
- N 75 000 000 000 = N 75 x 1 000 000 000
= N 75 billion
- $460 000 000 = $460 x 1 000 000
= $460 million
- £3 400 000 000 000 = £3.4 x 1 000 000 000 000
= £3.4 trillion
Example 2
Write the following numbers in digits
- 3.6 million (b) 2 billion
Solution
- 3.6 million = 3.6 x 1 000 000
= 3 600 000
- 2 billion = 2.75 x 1 000 000 000
= 2 750 000 000
Large Numbers (QR)
The S. I system of units is an internationally agreed method of measuring quantities such as length, mass, capacity and time.
Example 1.
Express the following in millimeter.
(a) 173 cm (b) 5.9km (c ) 200m
Solution
- 173 cm to mm
Since 1cm = 10mm
Then, 173cm = 173 x 10
= 1 730mm
- 5.9km to mm
1cm = 10mm, 100cm = 1m, 1000m = 1km. Therefore, 1 000 000mm = 1 km
- km = 5.9 x 1 000 000mm
= 5 900 000 or 5.9 million(mm)
- 200m to mm
1m = 1000mm
200m = 200 x 1 000mm
= 200 000mm or 200 thousand (mm)
EVALUATION
- billion (b) £0.85 trillion (c) million litres
- Write these numbers as a mixture of digits and words: (a) 780 000 barrels (b) 900 000 km (c) $ 900 000 000
- Change the following to the unit in brackets: (a) 5000kg (grams) (b) 1250 litres ( mL)
READING ASSIGNMENT
- Essential Mathematics for JSS1 by AJS Oluwasanmi page23-35.
- New General Mathematics for JSS 1 by M. F Macrae et al pg 15-20.
WEEKEND ASSIGNMENT
- What is the value of 1.2 km in metres? (a) 120m (b) 1 200m (c ) 12 000m (d ) 120 000m
- Which of the following numbers is the largest?(a) 727345565 (b) 727245565 (c)727445565 (d) 726778876.
- million in digits only is (a) $1 200 000 (b) $1 140 000 (c) $1 250 000 (d) $125 000
- Le 5 600 000 in digits and words is (a) Le 56 million (b) Le 5.6 billion (c) Le 0.56 billion (d) Le 5.6 million
- 13 500 000mm in km is (a) 13.5 km (b) 1.35 km (c) 1350 km (d) 13500 km
THEORY
- Write these numbers in digits only: (a) Le 0.5 billion (b) $ 9.1 million
- Write down the missing number (QR):
- 100 987 331, 101 987 331, 102 987 331, __________, __________, 105 987 331
- 980 231 680, 980 231 682, ________, 980 231 686, ____________, 980 231 690.
WEEK THREE
TOPIC: LEAST COMMON MULTIPLE AND HIGHEST COMMON FACTOR OF WHOLE NUMBERS
CONTENT
- Multiples
- Common Multiples
- Least Common Multiples (LCM)
- Highest Common Factors (HCF).
Multiples
Multiples mean finding the product of a positive integer with another positive integer. Simply put, when a whole number multiples another whole number, the result obtained is called the multiple of either of those numbers.
The first four multiples of 12 are
12 x 1 = 12
12 x 2 = 24
12 x 3 = 36
12 x 4 =48
Thus, we write 12, 24, 36 and 48 as the first for multiples of 12.
Note: Every whole number has an infinite number of multiples
Every whole number has a finite number of factors.
Example 1
Find the next five multiple of the following whole numbers.
(a) 4 (b) 8 (c ) 11.
Solution
In these questions, the numbers are not to be included because it reads next.
(a) 4, 4 x 1 = 4 (not included)
4 x 2 = 8
4 x 3 = 12
4 x 4 = 16
4 x 5 = 20
4 x 6 = 24
:. The next five multiples of 4 are 8, 12, 16, 20 and 24.
(b) 8, 8 x 1 =8 multiple of 11
8 x 2 =16 11 x 1 =11
8 x 3 =24 11 x 2 = 22
8 x 4 =32 11 x 3 = 33
8 x 5 =40 11 x 4 = 44
8 x 6 = 48 11 x 5 = 55
11 x 6 = 66
the next five multiples of the next five multiples of 11 are 22, 33,44,55 and 66
8 are 16, 24, 32, 40 and 48.
Example 2
Which of the following numbers 18, 20, 27 36 and 50 are
- multiples of 2
- multiples of 3
- multiples of 4.
Solution
When a number can be divided exactly by another number it means the quotient is a multiple of the divisor. 18, 20, 27 36, 50
- multiples of 2 are 18, 20, 27 36, 50
- multiples of 3 are 18, 27 and 36
- multiple of 4 are 20 and 36.
EVALUATION
- 1.Find the next 7 multiples of the following numbers.
(a) 15 (b) 25 (c ) 13.
- Which of the following whole numbers 37, 68, 51, 128, 85 and 187 are
(a) multiples of 2
(b) multiples of 3
(c) multiples of 5
(d) multiples of 17.
Common multiples
When two or more numbers have a multiple in common, then the numbers is known as a common multiple.
Example
Find the first two common multiples of 4.6 and 8.
Solution
Their multiple are as shown below;
4 = 4, 8, 12, 16, 20 (24) 28,32, 36, 40, 44, (48) 52,56,60,64……
6 = 6, 12, 18,(24), 30, 36, 42, (48), 54, 60, 66, 72.
8= 8, 16,(24), 32, 40, (48), 56, 64, 72,
Considering the three whole numbers, their first two common multiples are
24 and 48.
Examples
Write down three common multiples of the following sets of numbers
(a) 5 and 6
(b) 3, 10 and 15.
Solution
(a) First three common multiples of 5 and 6.
5 = 5, 10, 15, 20, 25, (30),35, 40 45, 50, 55 (60) 65, 70,75,80 85,(90) 95,100, 105,110,115,120.
6= 6, 12,18,24, (30),36, 42,48,54,(60),66,72,78,84, (90),96,102,106,114,120.
:.The first three common multiples of 5 and 6 are, 30 60 and 90.
(b) First three common multiples of 3, 10 and 15
3= 3,6,9,12,15,18,21, 24,27, (30),33,36,39,42,45,81,84,87,(90),93,96.
10= 10, 20, (30), 40, 50, (60) 70, 80 (90),100,110, etc
15 = 15, (30), 45,(60), 75, (90), 105, 120.
:. The first three common multiples of 3, 10, and 15 are 30, 60 and 90.
EVALUATION
Find the first four common multiples of the following sets of numbers
(a) 4 and 7 (b) 2,5 and 7 (c ) 3, 6, 9.
READING ASSIGNMENT
- Essential Mathematic for JSS1 by AJS Oluwasanmipg 32
- New General Mathematics for JSS1 by M.F Macraeetalpg 26-27.
Least Common Multiples (LCM)
You can find the least common multiples of two or more numbers by listing as many multiples as you need until you have one that is common to both or all the numbers.
For instance to find the LCM of 24 and 15
Multiples of 24 = 24, 48, 72,96, 120………
Multiples of 15 = 15, 30, 45, 60, 75,90, 105, 120.
Although the numbers will have many common multiples but, looking at what we are after, that is the least of the common multiples, the answer will be 120.
:. LCM of 24 and 15 = 120.
Rather than writing out a long list of multiples for each number, you can use the prime factors method to find the LCM. This is the method we are going to apply.
Example 1
Find the LCM of the following whole numbers:
(a) 24 and 15 (b) 8 and 45 ( c ) 16 and 18 (d) 90, 105 and 210.
Solution
(a) The LCM of 24 and 15
2 24 15 2 8 45
2 12 15 2 4 45
2 6 15 2 2 45
3 2 5 3 1 45
5 1 1 3 1 15
3 1 5
5 1 1
LCM = 2 x 2 x 2 x 3 x 5= 120 Lcm = 2 x 2 x 2 x 3 x 3 x 5 = 360
:.LCM of 24 and 15 = 120 ;. LCM of 8 and 45 = 360.
(c) LCM of 16 and 18 (d) LCM of 90, 105 and 210
2 16 18 2 90 105 210
2 8 9 3 45 105 105
2 4 9 3 15 35 35
2 2 9 5 5 35 35
2 1 9 7 1 7 7
3 1 3 1 1 1
3 1 1cc
LCM = 2 x 2 x 2 x x 2 x3 x3 = 144 LCM = 2 x 3 x 3 x 5 x 7 = 630
:. LCM of 16 and 18 = 144 :. LCM of 90, 105 and 210 is = 630.
Given that the numbers are expressed as a product of prime factors, the lcmis the product of the prime factors of the numbers without double counting.
Example 2
Find the LCM of the following .Leave your answers in prime factors.
(a) 2 x 2 x 3, (b )2 x 2 x 5
2 x 2 x 2 x 5 3 x 5 x 7
2 x 2 x5 2 x 3 x 3 x 3
2x 2 x 3 x 3 x 5 3 x 5 x 5 x 7.
Solution
(a) 2 x 2 x 3
2 x 2 x 2 x 5
2 x 2 x 5
2 x 2 x 3 x 3 x 5
LCM = 2 x 2 x2 x 3 x 3 x 5.
(b) 2 x 3 x 3
3x 5x 7
2 x 3 x 3 x 3
3 x 5 x 5 x 7
LCM = 2 x 3 x 3 x 3 x 5 x 5 x 7.
EVALUATION
1.Find the LCM of the following
(a) 4,6, and 9 (b) 6, 8, 10 and 12 (c ) 9, 10,12 and 15 (d) 108 and 360.
2. Find the Lcm of the following leaving your answers in index form.
(a) 2 x 2 x 2 x 3 x 3 (b) 3 x 3 x 5
2 x 3 x 5 x5 2 x 3 x 7
2 x 2 x 3 x 3 x 5 2 x 5 x 5 x 7
(c) 2^{3} x 3^{2} x 5
3 x 5^{3} x 7^{2}
2^{4} x 3 x 7^{2},
3^{2} x 5^{2} x 7^{3}
Highest Common Factor
Highest common factor (HCF) of two or more numbers is the largest number that divides exactly into all the numbers.
Example 1
Find the HCF of 21 and 84.
Solution
3 21 2 84
7 7 2 42
1 3 21
7 7
1
21 = 3 x 7
84 = 2 x 2 x 3 x 7
HCF = 3 x 7 = 21
Example 2
Find the HCF of 195 and 330.
Solution
3 195
5 65
- 13
11
HCF of 195 and 330
195 = 3 x 5 x 13
330 = 2 x 3x5x11
HCF = 3 x 5 = 15
- 330
- 165
5 55
11 11
1
Example 3
Find the HCF of 288, 180 and 106. leave your answer in index form.
Solution
2 288 2 180 2 108
2 144 2 90 2 54
2 72 3 45 3 27
2 36 3 15 3 9
2 18 5 5 3 3
3 9 1 1
3 3
1
288 = 2 x 2 x 2 x2x2 x 3 x 3
180 = 2×2 x 3 x 3
108 = 2×2 x3 x 3×3
HCF = 2x 2x 3 x 3 = 2^{2} x 3^{2} …….index form
= 36 (ordinary form).
Example 4
Find the HCF of the following . Leave the answers in prime factors and use index notation.
(a) 2^{3} x 3^{2} x 7
2^{2} x 3 x 5^{2}
2^{2} x 3^{3} x 5
(b) 2^{3} x 5^{2} x 7
2^{2} x 3^{2} x 5
3^{3} x 5^{3} x 7^{2}
Solution
(a)2^{3} x 3^{2} x 7 = 2 x 2^{2} x 3 x 3 x7
2^{2} x 3 x 5^{2 }= 2^{2} x 3 x 5^{2}
2^{2} x 3^{3} x 5 = 2^{2}x 3 x 32 x 5
HCF = 2^{2} x 3 index form
4 x 3 = 12.
(b) 2^{3} x 5^{2} x 7 = 2^{3}x 5^{2} x 7
2^{2} x 3^{2} x 5 = 2^{2} x 3^{2}x 5
3^{3} x 5^{3} x 7^{2}= 3^{3} x 5^{3} x 7^{2}
The factor that is common is in 5
:. HCF = 5
EVALUATION
1.Findthe HCF of the following;
i. leaving your answers in index form
ii. leaving your answers in whole number
(a) 160, 96 and 224
(b) 189, 279and 108
(c) 126, 234 and 90.
2. Find the HCF of the following .Leave your answers in prime factors and use index notation.
22x 33 x 5
21 x 34 x 5
2 x 35 x 72
(b) 23 x 33 x 53
24 x 3 x 52 x 7
25 x 32 x 5 x 72
READING ASSIGNMENT
- New General Mathematics for jss 1 by m.Fmacraeetalpg 25-26
- Essential Mathematics for jss1 by AjSOluwasanmi, pg 31.
WEEKEND ASSIGNMENT
1. The value of 23 x 32 is (a) 1291 (b) 658 (c) 729 (d)7 36 (e) 54
2. The LCM of 12 and 15 is (a) 90 (b) 60 (c) 30 (d) 120 (e) 180
3. The HCF of 63and 90 is (a) 7 (b) 3 (c) 12 (d) 6 (e) 9
4. The first three common multiples of 3 and 11 are (a) 3, 33, 66, (b) 11, 33, 66 ( c ) 33, 66, 99 (d) 33, 44, 55 (e) 33, 22, 11.
5. Which of the following whole numbers 22, 11, 54, 35, 40, 75, and 105 is /are multiples of 5? (a) 11, 22, 35 (b) 35, 40, 75, 105 ( c ) 54, 35, 40, 75, 105 (d) 35, 54, 40, 75 (e) 105,75,40,35,54.
THEORY
1. Give the first five multiples of the following
I. 5 II 7 III. 11
B Write down four common multiples of the following sets of numbers
I. 3, 4 and 5 II. 3, 10 and 15.
2a. Find the LCM of
I. 9, 24, 32, and 90 II. 2^{3} x 3^{2} x 5 x 7
3 x 5^{3} x 7^{2}
2^{4} x 3 x 7^{2}
3^{2} x 5^{2} x 7^{3}
b. Find the HCF of
i. 126, 234 and 90 ii. 2^{3} x 3^{3} x 5^{3}
b. 63, 42, and 21 2^{4} x 3 x 5^{2} x 7
2^{5} x 3^{2} x 5 x 7^{2}
WEEK FOUR
TOPIC: FRACTIONS
CONTENT
A: Common Fractions
B. Types
(i) Proper Fractions
(ii) Improper Fractions
(iii) Simple Conversion
A: COMMON FRACTIONS
A fraction is a number which is represented by one integer – the numerator – divided by another integer – the denominator ( or the divisor).
Simply put, a fraction is a part of a whole number. It is not always possible to use whole numbers to describe parts of quantities. It is therefore, important to know that to describe parts of quantities, fraction is used for example.
GENERAL FORM OF A FRACTION
From the explanation given above, we can write fraction in the form
a/b where
a = the numerator
b= the denominator
Fraction is divided into
1. Common (Vulgar) Fractions
Here, the fraction is written as one number over another .
Numerator is the term given to the number on the top part of a fractions.
Denominator is the term given to the number at the bottom part of a fraction .For example
9 Numerator
11 Denominator
II. Decimal Fractions
Decimal fractions are simply called decimal numbers. It has numbers to the left and right of a decimal point. See week 5 for detail.
B.Types of Fractions (Common)
Common fractions are grouped under three headings. becausefractions are written as one integer divided by another – a ratio – they are called rational numbers.
Fractions are either proper,improper or mixed.
1.Proper Fractions : This is a common fraction having its numerator less than its denominator. Example
(a) ^{4}/_{7} (b) ^{3}/_{5} (c) ^{2}/_{5} etc
2 Improper Fraction: This is a common fraction having its numerator greater than its denominator .examples.
^{11}/_{5} (b) ^{4}/_{3} ^{39}/_{11} (d) etc
3. Mixed Numbers: This type of fraction is in the form of an integer and a fraction. That is it has two parts.
– a whole number, and
– a fraction (usually a proper fraction)
Example is shown in the figure below
From the diagram, we can describe the fraction as 1 ½ oranges
Where,
1=whole number part and
½ = fractional part
EVALUATION
1.Give five examples each of the types of fraction
2. Using the shapes/figures below, write out the fraction stating whether it is a proper, improper or a mixed fraction.
I II
READING ASSIGNMENT
1. Essential Mathematics for JSS1 by AJS Oluwasanmipg 35 – 36
2. New General Mathematics for JSS1 by M.F Macraeetalpg 29 – 30.
3. Simple Conversions: Conversions can be made from improper fractions to a mixed fraction and vice versa .Lets see some examples.
Example 1
Express the following improper fractions as mixed fractions
(a) ^{4}/_{3 } (b) 5^{7}/_{10} (c ) ^{93}/_{20} (d) 1^{13}/ _{3}
Solution
(a) = = = 1 + =
Alternatively, divide the numerator by the denominator and express the remainder as the numerator of the fractional part of the mixed fractions. The number of times the numerator can be divided before the remainder is the whole number part.
Hence, 4 = 4 ÷ 3 = 1 remainder 1
3
= = 1 ^{1}/_{3}
(b) 57/10 = = 50/10 + 7/10 = 5 + 7/10 = 5^{7}/_{10}
(c )^{93}/_{20} (d) ^{113}/_{3}
= 80 + 13 = 111 + 2
20 3
80 +13 = 111 + 2
20 20 3 3
= 4 + ^{13}/_{20} = 37+ ^{2}/_{3}
= 4 ^{13}/_{20} = 37 ^{2}/_{3}
Example 2
Conversion from Mixed numbers to Improper Fraction
Let A be the general form of a mixed number, where
A = whole number part
x/y = fractional part.
To convert to improper fraction, the following steps are followed.
1.Multiply the denominator of the fractional part by the whole number
2. Add the numerator of the fractional part to the result in (1) above.
3. Express the result in (2) above as the numerator of the improper fraction with the original denominator of the fractional part as the same denominator.
A = A y + x
y
Example 2
Express the following mixed fractions as improper fractions
(a) 5 ½ (b) 3 ^{2}/_{5}( c) 7 ^{1}/_{8} (d) 10 ^{1}/_{3}
Solution
(a) 5 ½ (b) 3 ^{2}/_{5}
= 5 x 2 + 1 3 x 5 + 2
2 5
= 10 + 1 = 15 + 2
2 5
=^{11}/_{2} = ^{17}/_{5}
( c) 7 ^{1}/8 (d ) 10 ^{1}/3
= 7 x 8 + 1 = 10 x 3 + 1
8 3
= 56 + 1 =
8
= =
EVALUATION
1. Express the following as mixed fractions
(a) 503/10 (b) 113/2
2. Express the following as improper fractions.
(a) 7 ½ (b) 33
READING ASSIGNMENT .
- Essential Mathematics for JSS 1 by AJS Oluwasanmipg
- New General Mathematics for JSS1 by M. F Macrae et al pg
WEEKEND ASSIGNMENT
1. Which of the following is not a proper fraction?(a) ¼ (b) ¾ (c )^{3}/_{2} (d) ^{5}/_{8}
2. Express 3 ^{1}/_{7} as an improper fraction is (a) ^{11}/7 (b)( c) ^{7}/_{22 }(d) ^{22}/ 7
3.What fraction of the figure shown is shaded?(a) ^{2}/_{11} (b) ^{3}/_{9} ( c)^{ 8}/_{3} (d) ^{4}/_{11}.
4. Express ^{99}/_{5} as a mixed fraction (a) 19 ^{4}/_{5} (b) 18 ^{4}/_{5} ( c) 19 ^{5}/_{4} (d) 18 ^{5}/_{4}
5.The figures above can best be described as
(a) 2 ½ – mixed numbers
(b) 2 ¾ – proper fraction
(c) 2 ¾ -improper fraction
(d) 2 ¾ -decimal
THEORY
1. Distinguish clearly the various types of fraction known to you, giving 2 examples each.
2. Express (a) ^{103}/_{5} as mixed fraction (b) 115 ^{2}/_{5} as improper fraction .
WEEK FIVE
TOPIC: FRACTIONS
CONTENT
I. Equivalent fractions
II. Concept of equivalent fractions in sharing commodities.
III. Problem solving in quantitative aptitude (QR).
Equivalent Fraction: Two or more fractions are said to be equivalent if they have the same values. Equivalent fractions can be obtained by multiplying or dividing the numerator and the denominator by the same number.
When the operation performed on the original fraction to get the new fraction is division, it is referred to as simplification .Here their common factor is used in dividing the numerator and the denominator.
i.Multiplication
:. The fraction 3/5, 6/10 and 12/20 are said to be equivalent fractions.
ii. Division
= 150 = 75 = 15
200 100 20
:. The fractions ^{150}/_{200}, ^{75}/_{100} and ^{15}/_{ 20} are said to be equivalent fractions.
iii. Simplification: by dividing both numerator and denominator by a common factor.
44 = 22
70 35
Example 1.
Find the missing numbers
(a) ^{1}/_{3}= ^{3}/_{9} = ^{6}/_{A} = ^{B}/_{24} = ^{50}/_{C} = ^{D}/_{900} = ^{100}/_{E}
(b) ^{1}/_{5} = ^{10}/_{50} = = =
solution
(a) = ^{1}/_{3} =^{3}/_{9} = ^{6}/_{A} =^{B}/_{24} = ^{50}/_{C}= ^{D}/_{900} = ^{100}/_{E}
= ^{1}/_{3} =^{6}/_{A}, ^{1}/_{3} = 1 x 6 =^{ 6}/_{18} A = 18.
= ^{1}/_{3} =^{B}/_{24}, ^{1}/_{3} = 1 x 8 = ^{8}/_{24} B = 8
3 x 8
= ^{1}/_{3} = ^{50}/_{C}, ^{1}/_{3} = 1 x 50 = ^{50}/_{150} , C = 150
= ^{1}/_{3} =^{D}/_{900} , ^{1}/_{3} = 1 x 300 = ^{300}/_{900} , D = 300
3 x 50
= ^{1}/_{3} = ^{100}/_{E}, 1/_{3} = 1 x 100 = ^{100}/_{300} , E = 300
3 x 100.
Thus, the missing numbers calculated will make the fractions equivalent.
= ^{1}/_{3} =^{3}/_{9} = ^{6}/_{18} = ^{8}/_{24} = ^{50}/_{150} = ^{300}/_{900} = ^{100}/_{300}.
(b)
= ^{1}/_{5} = , ^{1}/_{5} x ^{2}/_{2}, missing number = 2
= ^{1}/_{5}= ^{1}/_{5} x ^{4}/_{4} =^{ 4}/_{20} , missing number = 20
= ^{1}/_{5}= ^{1}/_{5} x ^{20}/_{20} = ^{20}/_{100}, missing number = 20
= ^{1}/_{5}= , ^{1}/_{5} x ^{24}/_{24} = ^{24}/_{120},missing number = 120
Example
Find the missing numbers.
(a) =
(b) =
( c ) =
(d) 7 = 14
9
Solution
(a) 2 = ……. think of what will multiply 3 to give 18..
3 18
= 2/3 x 6/6 =2 x 6 = 12/18
3 x 6
:. The missing number is 12.
(b) 5 = 20 think of what will multiply 5 to give 20
6
= 5 x 4 = 20
6 4 24
the missing number is 24
(c ) 3 = ……
5 15 .think of what will multiply 5 to give 15.
= 3 x 3 = 3 x 3 = 9
5 3 5 x 3 15
:.the missing number is 9.
(d) 7 x 2 = 7 x 2 = 14
9 2 9 x 2 18 think of what will multiply 9 to give 18
:. The missing number is 18
EVALUATION
1.Find the missing numbers
¾ =^{6}/_{8} = ^{15}/_{A} = ^{24}/_{B} = ^{C}/_{28} =^{ D}/_{100} = ^{E}/_{24}
2. Find the missing numbers
i. 3 =
8 48
ii 5 =
9 36
iii. 5 = 20
6
READING ASSIGNMENT
1. Essential Mathematics for JSS1 by AJS Oluwasanmipg 38
2.New General Mathematics for JSS1 by M.F Macrae eta l pg 30-31.
II.Equivalent Fractions in Sharing Commodities
Problems involving sharing of commodities can be resolved with the knowledge of fractions. Some examples below will help us to understand this aspect of fraction.
Example1
Some notebooks where shared into 18 equally. If 5 exercise books were given to Ojo, what fraction is left?
Solution
Number of notebooks =18
Ojo’s share = 6
Number left = 18 – 6 = 12
Fraction left =^{12}/_{18 } = ^{12}/_{18} ÷^{ 6}/_{6}
= ^{2}/_{3 }
Example 2
Amarket woman had 90 yams. She sold^{2}/_{3} of it. How many yam did she sell?
Solution
No of yams = 90
No sold = ^{2}/_{3} of 90
= ^{2}/_{3} x 90 = 2 x 30
= 60 yams
:. 60 yams where sold.
Examples
Some oranges were shared out. Olu got ^{3}/_{8}of them. He gave 5 to his brother and 4 to her sister and had 6 left. How many oranges were there altogether?
Solution
Fraction received by Olu = ^{3}/_{8}
No of oranges he gave out = 5 + 4 = 9
No of oranges left with him = 6 = 9 + 6 = 15.
As equivalent fraction, ^{3}/_{8 } =^{15}/? = 3 x 5 =^{15}_{/40}.
8 x 5
:.40oranges were there altogether.
Example 4
In a shop, the price of a radio is reduced by one third .If the original price of the radio is N24, 000 what is the reduced price?
Solution
Original price = N2400
^{1}/_{3} of this price = ^{1}/_{3} x 2400 =N800
reduced price =2400- 800 = N1600.
Alternatively,
Consider the original price as unit
:. Reduced price = 1 – ^{1}/_{3} = ^{2}/_{3}
^{2}/_{3} of the unit price =^{2}/_{3} x 2400 = N1600
EVALUATION
1. In a class of 40 students, ¼ are doing science subjects, ^{1}/_{5}are doing arts and the remaining are doing commercial subjects. How many students are :(a) doing science (b) commercial?
2. In a service. ¼ of the people are men , 1/3 are women and the rest are children. If there are 50 children, how many (a) people are there altogether (b) more women than men are there ?
READING ASSIGNMENT
1. Essential Mathematics for JSS1 by AJS Oluwasanmipg 49-51
2. New General Mathematics for JSS1 by MF Macraeetalpg 33 -36.
III. Problem Solving in Quantitative Aptitude
Some of the examples under quantitative aptitude(reasoning ) have been seriously dealt with at the early part of this topic. Let us take some more examples .
Example 1
Find the missing numbers
( a) 16 = (B ) 10 =
48 3 168
( c ) 3 =
749
Solution
(a) 16 = think of a number that will divide 48 to give 3
48 3
= 16 ÷ 16 = 1
48 ÷ 16 3
:. The missing number is 1.
(b) 10 =
16 8 ‘ think of a number which will divide 16 to give 8′
= 10 ÷ 2 = 5
16 ÷ 2 8.
The missing number is 5.
( c) 3 = think of a number that will multiply 7 to give 49
7 49
= 3 x 7 = 3 x 7 = 21
7 7 7 x 7 49
:. The missing number is 21.
Example 2
Reduce the following fractions to their lowest terms/
(a) 5 (b) 24 ( c ) 14
100 54 21
Solution
The concept of equivalent fraction using division as the operation can be very helpful.
(a) 5 = 5 ÷ 5 = 1
100 100 ÷5 20
(b) 24 = 24 ÷2 = 12 = 12 ÷ 13 = 4
54 54 ÷2 27 27 ÷ 3 9
(c) 14 = 14 ÷7 = 12
21 21 ÷ 7 3
Example 3
What fraction of
- 6 weeks is 6 days?
- 650m is 1km?
- 4mm is 10cm?
- 500g is 2kg?
Solution
Before reducing fractions, the quantities must be in the same units.
(a) 6 weeks ………..6 days.
= 6 days
6 weeks
= 6 days= 6
6 x 7 days 6 x 7
= 1 the fraction = 1
7 7
(b) 650m ——– 1km
= 650 m = 650m = 65 ÷ 5 = 13
1km 1000m 1000 ÷ 5 20
(c) 4mm ……….10cm
= 4mm = 4mm = 4
10cm 10 x10 100
= 4 ÷ 4 = 1
100 ÷ 4 25
The fraction = 1/25.
(d) 500g ……….2kg
= 500g = 500g = 500 = 5
2kg 2000g 2000 20
= 5÷ 5 = 1
20 ÷ 5 4
the fraction is ¼
EVALUATION
1. Find the missing numbers
(a) 55 = 11
305 ?
(b) 9 = ?
11 99
2. Express 13 weeks as a fraction of 1 year.
READING ASSIGNMENT
1. Essential Mathematics for JSS 1 by AJS Oluwasanmi ,pg
2. New General Mathematics for JSS1 by Mf Macraeetalpg 37.
WEEKEND ASSIGNMENT
1. Which of the following is not equivalent to ½ ?
a. ^{9}/_{18} b. ^{11}/_{22 }c. ^{15}/_{30} d. ^{16}/_{32} e. ^{24}/_{42}
2. To express the fraction 30/48 in its lowest term, divide the numerator and demominator by
A. 2 B. 3 C.5 D. 6 E. 8
3.45 minutes , expressed as a fraction of one hour is
a. 1/60 b. 1/45 c. ¾ d. 4/5 e. 4/3
4. The missing number in the fraction 3 = ?
4 20 is
a. 6 b. 9 c. 12 d. 5 e. 15.
5. A woman bought 2 crates of eggs. ¼ of them are bad. How many of the eggs are good?
a. 3/6 b. 24 c. 48 d.12 e. 32.
THEORY
1.Find the missing numbers
1 = ? = ? = ? = 5
4 8 12 16 ?
(b) 2 = ? = ? = 8 = 10
5 10 15 ? ?
2. A drum holds 2 ½ litres of water when itsis ¾ full. How many litres of water can it hold when it is
(a) full, b, two-third (c)empty.
WEEK SIX
TOPIC: FRACTIONS
CONTENT
Ordering of Fractions
Percentages – Conversion
Conversion of Fractions to Decimals and Vice–versa.
Ordering of Fractions
It is much easier to compare the size of fractions, when they have the same denominator.
Example 1
Which is the larger fraction: ^{5}/_{7} or ^{6}/_{8}?
Solution
= ^{5}/_{7} or^{6}/_{8}
to have a common denominator
= ^{5}/_{7} x ^{8}/_{8} or^{6}/_{8} x ^{7}/_{7}
= ^{40}/_{56} or ^{42}/_{56}
hence 6/_{8} is larger than ^{5}/_{7},
Examples
Which has the greater mass: 3054g or 3.56kg
Solution
= 3054g or 3.56kg
= 3054kg or 3.56kg
1000
= 3.054kg or 3.56kg
therefore, 3.56kg is greater than 3054kg
Examples
Which is the larger fraction in this pairs?
a. 3 ^{21}/_{50} or 3 ^{31}/_{60} b. ^{37}/_{45} or^{19}/_{24}
Solution
a. 3 ^{21}/_{50} or 3 ^{31}/_{60}
The whole number “3” can be ignored in the working . Consider the fractional part of the mixed fraction.
= ^{21}/_{50} or ^{31}/_{60}
= ^{21}/_{50} x^{ 6}/_{6} or ^{31}/_{60} x^{ 5}/_{5}
= ^{126}/ _{300} or ^{155}/_{300}.
Considering the values of the numerator 155 > 126
Therefore, 3 ^{31}/_{60} is larger than 3 ^{21}/_{50}.
(b) ^{37}/_{45} or^{19}/_{24}
= ^{37}/_{45} x^{8}/_{8} or ^{19}/_{24} x ^{15}/_{15}
= ^{296}/_{360} or^{285}/_{360}
Considering the values of their numerators,
296 > 285.
:. The fraction ^{37}/_{45} is larger than ^{19}/_{24}.
Example
Arrange the following fractions in ascending order
- ^{1}/_{3} , ^{1}/_{9}, ^{5}/1_{8}
- ^{2}/_{3},^{ 5}/_{6}, ^{7}/_{12}, ¾
Solution
a. ^{1}/_{3}, ^{ 1}/_{9}, ^{5}/_{18}
= ^{1}/_{3} x ^{6}/_{6} =^{6}/ _{18}
= 1/_{9} x^{2}/_{2} = ^{2}/ _{18}
= ^{15}/_{18 }x ^{1}/_{1} =^{5}/ _{18}.
Comparing their numerator, 2,5,6,
:. The fractions are
^{1}/_{9}, ^{5}/_{18}, ^{1}/_{3}.
(b) ^{2}/_{3}, ^{5}/_{6}, ^{7}/_{12}, ^{3}/_{4}/
= ^{2}/_{3} =^{2}/_{3} x ^{4}/_{4} =^{8}/ _{12}
= ^{5}/_{6} =5/_{6} x^{2}/_{2} = ^{10}/_{12}
= ^{7}/_{12} =^{7}/_{12} x ^{1}/1 =^{7}/ _{12}
¾ =^{3}/_{4}x ^{3}/_{3 } =^{9}/_{12}.
Comparing their numerators, 7,8,9 10.
The fractions are
^{7}/_{12}, ^{2}/_{3}, ^{3}/_{6}, ^{5}/_{6}.
READING ASSIGNMENT
Essential Mathematics for JSS 1 by AJS Oluwasanmipg 51
New General Mathematics for JSS 1 by MF.Macraepg 31-32.
EVALUATION
1. Which of the following fractions is larger?
a.^{ 2}/_{5} or ^{5}/_{7} b.^{ 5}/_{6} or^{4}/_{9}
2. Arrange the following fractions in ascending order
^{3}/_{5}, ^{8}/_{15}, ^{17}/_{30 } (b) ^{3}/_{5}, ^{5}/_{8}, ^{7}/_{10}, ^{13}/_{20}.
PERCENTAGES
“Per cent’ means per hundred or ‘out of ‘hundred’ or ‘in every hundred’. For example, when we say a student obtained 63 percent in a test, what we mean is that he or she had 63 marks out of 100 marks this is usually written as 63%. Where the symbol % means per cent.
a. Converting From percentage to fraction.
Here, the given value in percentage is divided by 100.
A% = in fraction or A ÷ 100, A x 1/_{100.}
Express the following as a fraction in its simplest form
i. 30% ii. 75% iii.7 ½ % iv. 13 ¾ %
Solution.
i. 30% = 30 = 3
100 10
ii. 75% = = ¾
iii. 7 ½ % = 15 = 3
100 x 2 40
iv. 13 ¾ % = 55 x 1 = 11
4 x 100 80
b. Converting a percentage into a decimal
To convert a percentage to a fraction divide the percentage by 100.
Examples
Change these to decimals
I 45% ii. 34 ¾ % iii. 5.8%
Solution
i.45% = 45/100 = 0.45
ii.34 ¾ %= 34.75/100 = 0.3475
iii.5.8% = 5.8/100 = 0.0058.
c. Converting a fraction into percentage
To convert a fraction into a percentage, multiply it by 100.
Examples
Express these fractions as percentages
i. ¼ ii.25/400 iii. 5/8
Solution
¼ = ¼ x 100% = 25%
ii. 125/400 = 125 x 100
400 = 125/4 = 31.25%
iii. 5/8 = 5/8 x 100 % =500/6 = 62.5%.
d. Converting a decimal into a percentage.
To change a decimal to a percentage multiply it by 100
Example.
Express the following as a percentage
a.0.75 b.0.045
Solution
a.0.75 =0.75 x 100 = 75
b. 0.048 =0.048 x 100 = 4.8%.
e.Finding the percentage of a quantity
To find the percentage of a quantity, express the percentage as a fraction, then multiply by the quantity.
Examples
i. 4.5% of N 248 ii. 205 of N250
Solution
i. 4.5% of N 248
= 4.5 x 248
100
=N1116/100 =N11.16.
ii. 20% of N250
= 20/100 x 250
=N50.
f.Expressing one quantity as a percentage of another .
To express one quantity as percentage of another write, the first quantity as fraction of the second and then multiply by 100.
Examples
i.8 students did not do their assignments in a class of 40.
a. What is this as a percentage?
b. What percentage of the class did their assignment?
Solution
a. Writing the first quantity as a fraction of the second gives 8/40.
Multiply the fraction by 100
Therefore, 8/40 x 100= 2 x 10 = 20%
20% of the student did not do their assignment .
c.Those who did their assignment were:
40 -8 = 32 students 32/40 x 100 = 32/ 4 x10 =80
80% did their assignments.
2.What percentage of N5 is 150 kobo?
Solution
Convert N5 to kobo first.
N5 = 5 x 100 = 500kobo
Expressing as a fraction , we have 150/500
Therefore, 150 x 100
501
the percentage is 30%
3.What percentage of 15km is 20,000cm?
Solution
Convert both quanities to same unit first
1 km = 100,000cm
15km = 100000 x 15 =1500 000cm
Expressing as a fraction 20000/1500 000
Then multiply by 100
20 000/1500 000 x 100 = 20/15 = 1.33%
EVALUATION
1. Calculate the following :
(a) 5% of N500 (b) 18% of 144km.
2. Convert the following fraction into decimal:
(a) 4/5 (b) 1 2/5
READING ASSIGNMENT
- Essential Mathematics for JSS1 by AJS Oluwasanmipg 53-56
- New General Mathematics for JSS I by MF. Macraepg 36-38.
III.Converting Fractions to Decimal
To convert a fraction into decimal first re-write the number as a decimal then divide it by the denominator
Terminating decimal
When the denominator divides exactly into numerator a terminating decimal is obtained.
Example
Change ¾ into a terminating decimal number
Solution
0.75
4 30
25
20
20
¾ = 0.75.
Recurring or Repeating Decimals
Sometimes when changing fractions to decimal gives the same figure or group figures repeating themselves on and on. These types of fraction are called non-terminating decimals or recurring decimals.
Examples
Change the following into decimals :
(a) 4/9 (b) 6/11
Solution
a. 4/9 = 0.444
9 40
36
40
36
4
Therefore 4/9 = 0.444
= 0.4.
b. 6/11 0.545454
11 60
55
50
44
60
55
50
44
60
55
50
44
6
Therefore, 6/11 = 0.545454…..= 0.54.
Converting the following into fractions
i. 0.4 ii.0.067
Solution
i. 0.4 = 4/10 = 2/5
ii. 0.067 = 67/1000
(d) Addition and subtraction in decimal
Simplify the following :
i. 0.6 + 1. 7 ii. 0.59 – 0.55 iii. 7.5 + 1.8 iv.9.3 – 6.2
Solution
i.0.6 + 1.7
0.6
+1.7
2.3
ii. 0.59
– 0.55
0.44
iii. 7.5
+ 1.8
9.3
iv.9.3
– 6.2
3.1
e. Multiplication and Division of Decimals
examples
Simplify the following :
i. 0.08 x 0.7 ii. 0.5 x 7 iii. 0.18 ÷ 1.2
Solution
i. 0.08
x0.7
0.056
i. 0. 5
x 7
3.5
iii. 0.18 ÷ 6 = 0.03
iv. 1.56 ÷ 1.2
1.3
12 15.6
12
36
therefore, 1.56 ÷ 1.2 = 1.3
EVALUATION
Simplify the following :
i. 14.5 – 2.5 x 3.14
ii. 0.6 x 0.08
0.8
READING ASSIGNMENT
1. Essential Mathematics for JSS1 by AJS Oluwasanmi.
2. New General Mathematics for JSS1 by M.F Macrae et al.
WEEK SEVEN
TOPIC : ADDITION AND SUBTRACTION OF FRACTIONS
CONTENT
i. Introduction
ii. Addition of Fractions
iii. Subtraction of Fractions
iv. Further Examples.
I. Introduction
Two or more fractions can be added or subtracted immediately if they both possess the same denominator, in which case we add or subtract the numerators and divide by the common denominator . For example
2/5 + 1/5 = 2 + 1 = 3/5
5
If they do not have the same denominator they must be rewritten in equivalent form so that they do have the same denominator – called the common denominator e.g
2/7 + 1/5 = 10/35 + 7/35 = 10 + 7 = 17
35 35.
The common denominator of the equivalent fraction is the LCM of the two original denominator that is,
2/7 + 1/5 = 5 x 2 + 7 x 1 = 10 + 7 = 10 + 7 = 17
5 x 7 7 x 5 35 35 35 35
From the explanation, the above example has its LCM = 35.
Can you try this,
5/8 + 1/6 ?
the correct answer is 19/24
Summary
If fractions have different denominators:
- Find a common denominator by expressing each fractions as an equivalent fraction
- Add or subtract their numerators.
II. Addition of Fractions
Example: Simplify the following fractions
(a) ¼ + ½ (b) 2/3 + 5/6 (c) 2/5 + ½ + ¼
Solution
a. ¼ + ½ = ¼ + 2 x 1 = ¼ + 2 = 1+ 2 = 3
2 x 2 4 4 4
(b) 2 + 5 = 2 x 2 + 5 = 4 + 5 = 4 + 5 = 9 = 1 ^{3}/6
3 6 3 x 2 6 6 6 6 6
= 1 ½ mixed fraction
(c ) 2 + ½ + ¼ = 2 x 4 + 1 x 10 + 1 x 5
5 5 x 4 2 x 10 4 x 5
= 8/20 |10/20 + 5/20 = 8 + 10 + 5 = 23/24 = 1 3/20
20.
Example 2:Simplify the following fractions.
- 1 ¾ + 2 ^{2}/3 + ½
- 3 ¾ + ^{5}/8 1 ^{7}/12
- 5 ^{4}/9 +7 ^{1}/3 + ^{1}/12
Solution.
1 ¾ + 2 ^{2}/3 + ½
convert to improper fractions
^{7}/4 +^{8}/3 + ½
7 x 3 + 8 x 4 + 1 x 6
4 x 3 3x 4 2 x 6
^{21}/12 +^{ 32}/12 +^{ 6}/12
= 21+ 32 + 6
12
= ^{59}/ 12
4 ^{11}/12
b. 3 ¾ +^{ 5}/8 + 1 ^{7}/12
convert to improper fractions
^{15}/4 + ^{5}/ 8 +^{19}/ 12
15 x 6 + 5 x 3 + 19 x 2
4 x 6 8 x 3 12 x 12
= ^{90}/24 +^{15}/ 24 + ^{38}/24
= 90 + 15 + 38
24
143
24
5 ^{23}/24
c. 5^{ 4}/9 +7^{ 1}/3 + ^{1}/12
convert to improper fractions
^{49}/9 + ^{22}/2 + ^{1}/12
= 49 x 4 + 22 x 12 + 1 x 3
9 x 4 3 x 12 12 x 3
^{196}/36 +^{264}/36 +^{ 3}/36
196 + 264 + 3
36
463
36
= 12
EVALUATION
Simplify the following:
a. 3 ^{7}/8 + 2 ^{3}/4
b. 1 ½ + 2 ^{1}/3 + 3 ¼
c. 5 + 1 ¾ + 2 ^{2}/3
READING ASSIGNMENT
1. Essential Mathematics for JSS 1 by AJS Oluwasanmipg 32 – 45
2. New General Mathematics for JSS1 by M.F. Macraepg 32 – 33.
III. Subtraction of Fractions
Example 1: simplify the following:
a. ^{2}/3 – ¼ b. ¾ – ^{5}/8 c. 5 ¾ – 2 ^{4}/5
Solution
^{2}/3 – ¼
= 2 x 4 – 1 x 3
3 x 4 4 x 3
= 8 – 3
12 12
8 – 3
12
5
12.
b. ¾ – ^{5}/8
3 x 2 – 5
4 x 2 8
6 – 5
8 8
6 – 5
8
1
8
c. 5 ¾ – 2 ^{4}/5
convert to improper fraction,
23/4 –^{14}/5 = 23 x 5 – 14 x 4
4 x 5 5 x 4
= 115 –56
20 20
= 115 – 56
20
59
20 = 2 ^{19}/20.
Example 2: simplify the following :
a. 5 ^{1}/_{6} – 3^{ 2}/_{3} + 6 ^{7}/_{12}
b. 2 ½ + 3 + ^{7}/_{10} – ^{2}/_{5} – 2
c. 2 ½ + ¾ – ^{11}/_{6} + 4 – 1^{ 2}/_{3}
Solution
a. 5^{1}/_{6} – 3 ^{2}/_{3} + ^{6 7}/_{12}
= ^{31}/_{6} –^{11}/_{3} + ^{79}/_{12}
= 31 x 2 – 11 x 14 + 79
6 x 2 3 x 4 12
^{62}/_{12} –^{44}/_{12} +^{ 79}/_{12}
= 62 – 44 + 79
12
97
12 = 8 ^{1}/_{12}
b. 2 ½ + 3 + ^{7}/_{10} – ^{2}/_{5} – 2
= ^{5}/_{2 }– ^{3}/_{1} +^{7}/_{10} – ^{2}/_{5} –^{ 2}/_{1}
5 x 5 – 3 x 10 + 7 – 2 x 2 – 2 x 10
2 x 5 1 x10 10 5 x 2 1x 10
25- 30+ 7 – 4 – 20
10 10 10 10 10
= 25 – 30 + 7 – 4 – 20
10
= 25 + 7 – 30 – 4 -20
10
= 32 – 30 – 4 – 20
10
= –22
10
2 ^{2}/_{10} = – 2 ^{1}/_{5}
c. 2 ½ + ¾ – 1 ^{1}/_{6} + 4 – 1 ^{2}/_{3}
^{5}/_{2 }+ ¾ –^{7}/_{6} +^{4}/1 – ^{5}/_{3}
5 x 6 +3x 3 – 7 x 2 + 4 x 12 – 5 x 4
2 x 6 4 x 3 6 x 2 1×12 3 x 4
30 + 9 – 14 + 48 – 20
12 12 12 12 12
30+99 – 14 + 48 – 20
12
30 + 9- 14 + 48 – 20
12
30 +9 + 48 – 14 – 20
12
87 – 34
12
^{53}/12
4 ^{5}/_{12}
EVALUATION
Simplify the following :
1. 2 ½ – 1 ^{4}/_{5} + 2 ^{3}/2 – 1
2.7 ½ + 3 ^{1}/_{6} – 3 ¼
3.14 ^{4}/_{15} – 4 ^{2}/_{3} + 7 ^{1}/_{5}
III. Further examples
Example 1
What is the sum of 2 ¾ and 2^{ 4}/_{5}?
Solution
Sum = addition 9 + 0
Hence, sum of 2 ¾ and 2 ^{4}/_{5} is
= 2 ¾ + 2 ^{4}_{5}
11+ 14
4 5
11 x 5 + 14 x 4
4 x 5 5 x 4
= 55 + 56
20 20
55 + 56
20
111
20 = 5 11/20
Example 2
A 2 ¼ kg piece of meat is cut from a meat that weighs 3 2/5kg. What is the weight of the meat left?
Solution
Original weights of meat = 2 2/5kg
Weight of meat cut = 2 ¼ kg
Final weight of meat = 3 2/5 – 2 ¼
= 17/5 – 9/4
= 17 x 4 – 9 x 5
5 x 4 4 x 5
68 – 45
20 20.
68 – 45
20
23
- = 2 3/20
The weight of the meat left = 2 3/20 kg.
Example 3
A fruit grower uses 1/3 of his land for bananas, 3/8 for pineapples, 1/6 for mangoes and the remainder for oranges. What fraction of his land is used for oranges.
Solution.
The entire land is a unit = 1
Every other fractions add up to give 1
.oranges + bananas + pineapple + mango = 1
:. Orange = 1 – ( 1/3 + 3/8 + 1/6)
= 1 – ( 1 x 8 + 3 x 3 + 1 x 4 )
3 x 8 8 x 3 6 x 4
= 1 – ( 8/4 + 9/24 + 4/24 )
= 1 – 8 (8 + 9 + 4 )
24
1/1 – 21/24
= 24 – 21
24
= 3/24 = 1/8.
:. The fruit grower used 1/8 for oranges.
EVALUATION
1. By how much is the sum of 2 4/5 and 4 ½ less than 8 1/10?
2. A boy plays football for 13/4 hours, listens to radio for ¾ hours and then spends 1 ¼ hours doing his homework. How much time does he spend altogether doing these things?
READING ASSIGNMENT
1. Essential Mathematics for JSS 1 by AJS Oluwasanmipg 45
2. New General Mathematics for JSS1 by M.F. Macraepg 33
WEEKEND ASSIGNMENT
1. Simplify 2 ½ + ¼
(a) 3 ¾ (b).2 1/8 (c) 1 ¾ (d) 2 ¾.
2. Simplify 4 2/5 – 3 ¼
(a) 1 3/20 (b) 3 2/5 (c) 1 7/20 (d) 1 5/8
3. The common denominator of the fractions is
(a) 8 (b) 12 (c ) 6 (d) 15
4. Simplify
(a) 1 43/45 (b) 43/45 (c) 2 37/45 1 41/45
5. What is the sum of 1 ¾, 2 3/5 and 5 ¾
(a0 3 1/30 (b) 5 1/60 (c) 7 1/60 (d) 8 1/50.
THEORY
1. Simplify the following;
( a) 3^{7}/8 + 2 ¾ (b) 2 ^{5}6 + 5 ^{7}/8
(c)
2. Mr. Hope spends 1/3 of his earnings on food and ¼ on clothes. He then saves the rest. What fraction does he
(a) spend altogether
(b) save?
WEEK NINE
TOPIC : MULTIPLICATION AND DIVISION
- Simple Examples on Multiplication and Division of Fractions
- Harder Examples
- Word Problems.
- Prime Numbers and Fractions
Introduction
Multiplication of fractions is simply a direct method compared to division of fractions. In multiplication, there is direct multiplication of the numerator of one fraction with the other and the denominator with the other.
Division is usually technical as there is reversal of the sign ( ÷) to multiplication sign 9x ) thereby leading to the reciprocal of the right-hand value.
A x x = A x α
B y B x y
But A ÷ x = A x y = Ay
B y B x Bx.
BODMAS
When signs are combined as a result of combination of fractons, it is therefore important to apply some rules that will enable us know where to start from. Such guide is BODMAS. It states that when there is combination of signs, they should be taken in order of their arrangement
B= bracket
O= of
D= division
M= multiplication
A= addition
S= subtraction.
Simple example
Example 1
Simplify the following;
(a) 1 ^{3}/_{5} x 6 (b) ^{4}/11 of 3 ^{2}/_{3}
(c) 3 ¾ x^{4}/_{9} x 1 ^{1}/_{5} (d) ^{12}/_{25} of (1 ¼ )^{2}
Solution
(a) 1 ^{3}/_{5} x 6 (b) ^{4}/_{11} of 3 ^{2}/_{3}
= ^{8}/5 x 6 = ^{4}/11 x ^{11}/_{3}
= 8 x 6 = 4/_{3}
5 x 1 = 1 1/3
= 48
5
= 9 ^{3}/_{5}
c. 3 ¾ x ^{4}/_{9} x 1^{ 1}/_{5} (d) ^{12}/_{25} of ( 1 ¼ ) ^{2}
^{15}/_{4} x ^{4}/_{9} x ^{6}/_{5} ^{12}/_{25} x 5/4
15 x 6 =^{ 12}/_{25} x ^{5}/_{4} x ^{5}/_{4} =^{12}/_{16} = 2
9 x 5
Example 2
Simplify (a) 7 ^{1}/_{5} ÷ 25 (b) ^{12}/_{25} ÷9/ _{10}
(c) 7 ^{7}_{/8} ÷ 6 5/_{12}
Solution.
(a) 7 ^{1}/_{5} ÷ 25 (c ) 7 ^{7}/_{8} ÷ 6 ^{5}/_{12}
= ^{36}/_{5 }÷ ^{25}/_{1} = ^{63}/_{8} ÷ ^{77}/_{12}
= ^{36}/_{5} x ^{1}/_{25} = ^{63}/_{8} x^{12}/_{77}
= ^{36}/_{125}
b. ^{12}/_{2} ÷ ^{9}/_{10} = 9 x 3
^{12}/_{25} x ^{10}/_{9} 2 x 11
4 x 2 = ^{27}/_{22}
5 x 3 = 1 ^{5}/_{22}
= ^{8}/_{15}.
Example 3
Simplify
(a) ^{3}/_{10} x ^{35}/_{36} (b) 5 ¼ ÷ 2 /_{5}
14/_{15} 3 ¾
= 7 x 1 ^{21}/_{4} ÷ /_{5}
2 x 12 ^{15}/_{4}
^{7}/_{24} ÷^{14}/_{15} =^{21}/_{4} x 5/_{14}
^{15}/_{14}
^{7}/_{24} x ^{15}/_{14} = ^{15}/_{8}
^{15}/_{4}
= 15 = ^{15}/_{8} ÷ ^{15}/_{4}
24 x 2
= 5 = 15 x 4
8 x 2 8 15
= 5 = 4
16 8 = ½
^{5}/_{16.}
EVALUATION
Simplify the following
1. 8 ^{1}/_{6} X 3 ^{3}/_{7}
11 ^{2}/_{3}
2. 7^{ 3}/_{7}÷ ^{5}/_{21}
9 ¾ x 2/3
READING ASSIGNMENT
1.Essential mathematics for jSSI by AJS Oluwasanmipg 52
2. New General Mathematics for JSSI by MG macrae et al pg 36.
Harder examples
Example I
Simplify the following fractions
a. ^{5}/_{8} x 1^{ 3}/_{5} b. ¾ of 3 ^{3}/_{7} c. ^{9}/16 ÷3 ^{3}/_{8}
Solution
^{5}/_{8} x 1 ^{3}/_{5} b ¾ of 3 ^{3}/_{7}
^{5}/_{8 }x ^{8}/_{5} =^{3}/_{4} x ^{24}/_{7}
5 x 8 = 3 x 24
8 x 5 4 x 7
= 1 3 x 6
7 = ^{18}/_{7} = 2 ^{4}/_{7}
c.^{9}/_{16} ÷ 3^{ 3}/_{8}
^{9}/_{16} ÷ ^{27}/_{8}
^{9}/_{16} x^{18}/_{27} = ^{6}/_{16} = ^{3}/_{8}.
Example 2.
Simplify 2 ^{4}/_{9} x 1 ^{7}/_{8} ÷ 2 ^{1}/_{5}
Solution
2 ^{4}/_{9} x 1 ^{7}/_{8} ÷ 2 ^{1}/_{5}
= ^{22}/_{9} x ^{15}/_{8} ÷ ^{11}/_{5}
BODMAS: application
= 22/9 x 15/8 x 15/11
5 x 5
3 x 4
= ^{25}/_{12}
= 2 ^{1}/^{12}.
Example 3
Simplify 3 ¾ ÷ ( 2 ^{1}/_{7} of 11 ^{2}/_{3} – 5)
Solution
3 ¾ ÷ ( 2^{ 1}/_{7} of 11 ^{2}/_{3} – 5)
BODMAS – application ( the bracket first0
= 3 ¾ ÷ ( ^{15}/_{7} of ^{35}/_{5} – 5)
3 ¾ ÷ (^{15}/_{7 }x ^{35}/_{3} – ^{5}/_{1})
see BODMAS also multiplication first
3 3/_{3} ÷ ( 5 x 5 – 5)
^{15}/_{4} ÷ ( 25 – 5)
^{15}/_{4} ÷ ^{20}/_{1}
= 3
4 x 4
3
1 6
EVALUATION
Simplify the following
- 9 ¾ – ^{1}/_{3} ) x 4 ^{1}/_{3} ÷ 3 ¼
- ( 2 ¾) 2 ÷( 3 ^{1}/_{3} of 2 ¾ )
III. Word problems
Example I
What is the area of a rectangle of length 12 2/3m and breadth 7 ¼ m?
Solution
7 ¼ m
12 ^{2}/_{3}m
Area of rectangle, A = L x B
L = 12 ^{2}/_{3}m, B= 7 ¼
Area = 12 ^{2}/_{3} x 7 ¼
Area = ^{28}/_{3} x ^{29}/_{4}
Area = 19 x 29
6
Area = 551
6
Area = 91 ^{5}/6m^{2}
Example 2
Divide the difference between 4 ^{1}/_{5} and 2 ^{2}/_{3} by 1 ^{2}/_{5}.
Solution
Interpreting the question
= 4 ^{1}/_{5} – 2 ^{2}/_{3}
1 ^{2}/_{5}
= ( ^{21}/_{5} –^{ 8}/_{3} ) ÷ 1 ^{2}/_{5}
( ^{21}/_{5} x ^{3}/_{3} – ^{8}/_{3} x ^{5}/_{5} ) ÷ 1 ^{2}/_{5}
( ^{63}/_{15} – ^{40}/_{15}) ÷ ^{7}/_{5}
( 63 – 40 ) ÷^{7}/_{5}
15
^{23}/_{15} ÷ ^{7}/_{5}
^{23}/_{15} x ^{5}/_{7}
^{23}/_{21}
= 1 ^{2}/_{21}
Example 3
What is three-quarters of 3 ^{3}/_{7} ?
Solution
= three-quarter = ¾
= ¾ of 3 ^{3}/_{7}
= ¾ x ^{24}/_{7}
= ^{3}/_{1} x ^{24}/_{7}
^{3}/_{1} x ^{6}/_{7}.
Example 4
In a school, 9/10 of the students play sports. 2/3of these play football. What fraction of the students play football.
Solution
Fraction who play sports = ^{9}/_{10}
Fraction that play football = ^{2}/_{3} of ^{9}/_{10}.
^{2}/_{3} x^{ 9}/_{10}
1 x3
5
=3/_{5}.
:.3/_{5} of the students play football.
Example 5
Three sisters share some money. The oldest gets 5/11 of the money. The next girl gets 7/_{12} of the remainder. What fraction of the money does the youngest girl get?
Solution
Let the total money; be a unit = 1
1^{st} girl gets = ^{5}/_{11} of 1 = ^{5}/_{11}
remainder = 1 –^{5}/_{11} = 11 – 5 = ^{6}/_{11}
11
2^{nd} girl gets = 7/_{12} of the remainder
=^{7}/_{12} of ^{6}/_{11}
= ^{7}/_{12}x 6/_{11} = ^{7}/_{22}
3^{rd} girl will get ^{6}/_{11} – ^{7}/_{22}
= ^{6}/_{11} x ^{2}/_{2} – ^{7}/_{22}
= ^{12}/_{22} – ^{7}/_{22}
= 12 – 7
22
=
:.the fraction of the money that the youngest girl will get =
EVALUATION
- A boy eats ¼ of a loaf at breakfast and 5/8 of it for lunch. What fraction of the loaf is left?
- In a class of 4/5 of the students have mathematics instrument . ¼ of these students have lost their protractors. What fraction of students in the class has protractors?
IV. PRIME NUMBERS
A prime number is a number that has only two factors, itself and 1. Some examples are 2,3,5,7,11,13,…
1 is not a prime number because it has only one factor, that is, itself unlike 2 which has itself and 1 as its factors. All prime numbers are odd numbers except 2 which is an even number.
Example 1
Write out all prime numbers between I and 30
Solution
Between 1 and 30.
Prime numbers = 2,3,5,7,119………..
1 is not a prime number because it has only one factor, that is, itself unlike 2 which has itself and 1 as its factors. All prime numbers are odd numbers except 2 which is an even number.
Example 1
Write out all prime numbers between I and 30
Solution
Between 1 and 30.
Prime numbers = 2,3,5,7,11, 13, 17,19,23,29.
II. Factors
A factor of a given number is a number which divides the given number without leaving any remainder. For instance, 10÷ 2 = 5 without a remainder, therefore, we say 5 is a factor of 10. Thus, we say that 3 is not a factor of 10.
Example 1
Find the factors of 32
Solution
32 = 1 x 32 =2 x 16 =4x 8
= 8 x 4 = 16 x 2 = 32 x 1
:. Factors of 32 = 1, 2,4,8, 16, and 32 or
Using table
32 1 x 32
2 x 16
4 x 8
Factors of 32 = 1, 2,4,8, 16 and 32
Note; A case were you have a particular number occurring two times, duplication is not allowed. Pick it once. See the example below;
Example 2
Find the factors of 144
Solution
144 1 x 144
144 2 x 72
144 3 x 48
144 4 x 36
144 6 x 24
144 8 x 18
144 9x 16
144 2 x 12
Factors of 144 = 1,2,3,4,6, 8,9,12,16, 18, 24, 36, 48, 72 and 144
from the example, 12 occurred twice, but only one was picked.
Example 3
Find the factors of 120
Solution
120 1 x 120
- 2 x 60
120 3 x 40
120 4 x 30
120 5 x 24
120 6x 20
120 8 x15
120 10 x 12
Factors of 120 = 1,2,3,4,5,6,8, 10,12,15,20,24,30,40,60 and 120.
Prime Factors:
From our definition of prime numbers, it will be easy joining factor to it and getting the meaning of prime factors.
Prime factors of a number are the factors of the number that are prime.
To find the prime factors of a number.
1.Start by dividing the number with the lowest number that is its factor and progress in that order.
2. If you divide with a particular number, check if it can divide the new number again before moving to the next prime number.
Example 1
Express the following whole numbers as product of prime factors.
(a) 12 (b) 18 (c) 880 (d) 875.
Solution
- 12 2 18
- 6 3 9
2 2 3 3
1 1
12 is expressed as a product of 18 = 2 x 3 x 3
primes.
12 = 2 x 2 x 3
( c ) 2 880 5 875
2 440 5 175
2 220 5 35
2 110 7 7
5 55
11 11 875 =5 x 5x 5 x 7
1
= 880 = 2 x 2 x 2 x 2 x 5 x 11
example 2
Express 1512 as a product of prime factors.
Solution
Following the example above
2 1512
- 756
- 378
- 189
- 63
- 21
7 27
1512 = 2 x 2 x 2 x3 x 3 x 3 x 7
EVALUATION
Express the following as product of prime numbers
- 108
- 216
- 800
- 900
- 17325
Index Form
If we have to write the following 4, 18, 16 as a product of prime factors, it will pose no challenge
4 = 2 x 2
8 = 2 x 2 x 2
16 = 2 x 2 x 2 x 2
As their products increase, the challenge of how to write 2 or whichever number is multiplying itself will arise.
A way of writing this in a shorter form is called index form.
The general form is x^{n}
Where x = the base, that is the multiplicative value and
n=index or power or the number of times a particular number multiplies itself.
Example 1
Express the following index
- 3 x 3 x 3 x 3
- 5 x 5 x 5
- 2 x 2 x 2 x 2 x 2 x 2 x 2
Solution
(a) 3 x 3 x 3 x 3, this shows that four 3’s are to be multiplied together.
Writing index form
= X ^{n}
x = 3, n = 4
:. 3 ^{4}
(b) 5 x 5 x 5 = three 5’s in general form X^{n}
= x =5, n = 3
= 5 ^{3}
(c) 2 x 2 x2 x 2 x 2 x 2 x 2
= seven 2’s
= X ^{n}.
x = 2, n =7
= 2 ^{7}
As product of primes
800 …… 2 x 2 x 2 x 2 x 2 x 5 x 5
800 = 2^{5} . 5^{2}
example 3
Express the following as a product of primes in index form.
(a) 720 (b) 1404
(a) 720 (d) 1404
2 720 2 1404
2 360 2 702
2 180 3 351
2 90 3 117
3 45 3 39
3 15 3 13
5 5 13 1
720 = 2 x 2 x2 x 2 x3 x 2 x 5 1404 = 2 x 2 x 3 x 3 x3 x 13
= 2^{4}x 3^{2} x 5 = 2^{2} x3^{3} x 13
READING ASSIGNMENT
1.Essential mathematics for JSSI by AJS Oluwasanmipg29, 46-51
2. New General Mathematics for JSSI by MG macrae et al pg25
WEEKEND ASSIGNMENT
1.The fractions ^{C}/_{D} ÷ a/b is same as
(a) ^{Ca}/_{Db} (b) ^{Cb}/_{Da} (c ) C x a (d) a x C (e) a x b
D x b b x D D x b
2. Simplify ^{11}/_{25 }x 1 ^{4}/_{11}
(a) ^{2}/_{3} (b) ^{3}/_{5} (c) ^{2}/_{5} (d)^{ 4}/_{5} (e) 1 ¼
3.Find the length of a rectangle whose breadth and area are ^{7}/_{20}m and 8 ^{1}/_{5}m^{2}
(a) 23 ^{3}/_{7} (b) no answer ( c) 21 ^{2}/_{7} (d) 1 ^{7}/_{20} (e) 8 ^{11}/_{20}.
4. Simplify 5 ¼ + 1 ^{1}/_{6} – 3 ^{2}/_{3}
(a) 5 ^{11}/_{4} (b) 2 ¾ ( c) 3 ^{1}/_{12} (d) 1 ¾ ( e) 3 – ^{3}/_{2}
5. The product of prime factor of 28 is
(a) 2 x 3 x 7 (b) 2 x 4 x 7 (c ) 4 x 7 (d) 2 x 2 x 7 (e) 2 x 2 x2 x 7.
THEORY
1. Simplify 2 ^{2}/_{5} – 1 ¾
^{4}/5 + ½
2. Express the following as a product of prime factors; i. 105 ii. 75 iii. 60.
WEEK TEN
TOPIC: ESTIMATION
CONTENT
Rounding Off Numbers
Dimensions ,Distances, Capacity and Mass
Costing
Rounding off Number
Estimation is making guess of the nearly correct calculation in distance, weight, price or capacity of things without the actual measurement or calculation. Even though it is not accurately done, it gives a good idea of the correct answer.
Estimation help us to have a rough idea of the answer when we add, subtract, multiply or divides given quantity. Sometimes rounding off numbers and approximation are used in making estimation.
B. Rounding Off Number
Example;
1, Round 1234 to the nearest 10
II. Round 1834 to the nearest hundred
III. Round these numbers to the nearest thousands
- 3512 (b) 4265
Solution to the examples:
i. 1234 = 1230 to the nearest ten
ii. 1834 = 1800 to the nearest hundred
iii. 3512 = 4000 to the nearest thousand
iv. 4265 = 4000 to the nearest thousand.
2. Rounding Off Numbers to a Specific number of Decimal Places
Examples
Give 474.4547 correct to the nearest hundredth and thousandth
Solution to the example.
474.447 = 474.45 to the nearest tenth
=474.45 to the nearest hundredth
=474.455 to the nearest thousandth.
3. Rounding Decimal number to the Nearest Whole Number.
Examples
Round ( I ) 13.73 (ii)34.245 to the nearest whole number
Solution.
i. 13.73 = 14 to the nearest whole number
ii. 34.245 = 34 to the nearest whole number.
3 Significant Figure:The word significant means important and it is another way of approximating numbers. A figure position in a numbers. A figure’s position in a number show what the figure worth.
Note; That the first significant is always the first non-zero figure as you read a number from left. Again notice that zeros in the middle of a number are significant . Zero before or at the end of another are significant. Measurements and currencies are usually given to a specified number of significant figures.
Examples.
i. Give 5754 correct to
(a) 1. s.f (ii) 2. s.f (iii) 3. s.f.
ii. Give 147 .006 to
(a)1 s.f (ii) 2 s.f. (iii) 3. s.f. (iv) 4 s.f (e) 5 s.f
iii. Give 0.007025 to
(i) 1 s.f (ii)2 s.f (iii) 3.s.f (d) 4 s.f
iv. Give 0.0007004
(a) 2.s.f (b) 2.s.f.
Solution to the examples
i. 5754 = 60000 to 1 s.f
= 5800 to 2 sf
=5750 to 3s.f
ii. 147.006
= 100 correct to 1 s.f
= 150 correct to 2 s.f
=147 correct to 3s.f
= 147.0 correct 4 s.f
= 147.01 correct to 5 s.f
ii. 0.007025
= 0.007 correct to 1 s.f
=0.0070 correct to 2 s.f
= 0.00702 correct to 3 s.f
0.007025 correct to 4 s.f
iv.0.0007004
=0.00070 correct to 2 s.f
=0.000700 correct to 3 s.f
remember that the zeros at the end are necessary to show the number of significant figures.
EVALUATION
1. Read off to nearest ten (i) 95 (ii) 127
2. Give 3.9998 to (i) 1 s.f (ii) 2 s.f
Solution to evaluationquestions
1.i. 95 = 100 correct to nearest ten
ii. 127 = 13 correct to nearest ten
2. i. 4 = ( s.f)
ii. 4.0 ( s.f)
READING ASSIGNMENT
1. New General Mathematics JSS1 by J.B. Channonpg 152 Ex 24 a no 1a-e
2. Essential Mathematics by Oluwasanmipg 168 Ex 16.5 No.1 a,b, 2a,b,c.
II. Dimension and Distances, Capacity and Mass
The common units of length(i.e km, m,cm, mm) mass (i.etonne, kg. g ) capacity (i.e cl.ml) and time (hour, min. seconds ) are widely used. The most common unit for length are millimeter(mm) centimeter (cm). Meter(m) and centimeter for short length and the higher units (meter and kilometer) for larger distances.
The common units of mass are the gramme(g), kilogramme kg and tone (t). The common units of capacity and the milliliter (ml) centiliter (cl) and litre (l) as unit length, we use the lower units for smaller quantities.
It is important to be able to choose the most appropriate meter units of measurement to use. For example to measure distance less than a metre,smaller such as millimeter (mm) and centimeters are used. To measure a large distance metres (m) a kilometers (km) are used.
For example;
i. to measure the distance between Lagos and Benin City, we use km.
ii. to measure the height of a man, we use meters and centimeter.
iii. to measure the time it will take to run 200m, we use seconds etc.
Examples
1.State the metric units at length you would use to measure the following;
(a) length of your classroom (b) length of your fingernail
2. State the appropriate metric units of mass (weight) you would use to measure the following;
(a) your weight (b) the weight of a diary.
3. State the appropriate metric unit capacity you would use to measure the following;
a. the amount of water in a glass cup b. the amount of medicine in a tea spoon
4. State the appropriate metric unit of time you would use to estimate the following :
a . the time it takes a sportman to run 100m b. the time it takes to walk or travel to your school.
Solution to the examples.
1. (a) m (b)mm
2. (a) kg (b) g
3. (a) ml (b) ml
4. (a) second (b) min.
EVALUATION
1. State the units of length you would use to measure the following:
(a) height of a desk (b) height of yourself
2. State the units of mass you would use to measure the mass of the following
(a) a parcel (b) a large land
3. State the units of capacity you would use to measure the capacity of the following
(a) cup (b) car petrol tank (c)a tin of peak milk (d) the amount of water in a reservoir
4. State the appropriate metric units of time you would use to measure the following:
a time it takes to fill an empty tank b. the time it takes to travel from Lagos to Ado Ekiti.
Solution
1. a. cm b. cm
2. a. g or kg b. t
3. a. ml b. liter c. ml d. ml
4. a. min b. hour.
READING ASSIGNMENT
New General Mathematics by J.B Channonpg 24 Ex 24C nos 2 c-d
Essential Mathematics pg. 170 Ex.16. 7 nos 1 c-d.
Costing:
It is important to know the prices of items or goods in your area. This will enable you know the best place to buy a particular item at a reasonable price. In general, the more you buy the more you must pay.
Examples.
1.James bought 5 exercise books at a bookshop at N50.50 each> How much did he spend.
Solution
1. 1 exercise book cost = N50.50
therefore 5 exercise books will cost N50.50 x 5 = N225.50.
That means he spent N252.25
2. one candle cost 19C. Calculate the cost of five candles.
Solution
1 candle cost 19C
therefore 5 candles will cost 19C x 5 =95C
3. Find the cost of three tins of margarine at N48.00per tin
Solution
1 tin costs N48.00
Therefore 3 tins will cost N48.000 x 3 = N144.00
EVALUATION
1. Eggs cost N6.00 each. How much will one dozen of eggs cost?
2. 30 mangoes cost N150. What would be the cost of N140 similar margarine/
Solution
1 egg cost 12 eggs will cos N6.00
1 dozen (12) eggs will cost N6.00 x 12 eggs = N72.00
one dozen = N72.00
2. 30 mangoes cost N150.00
therefore 1 mango will cost 150.00/30 mangoes =N5
therefore 140 similar mangoes will cost N5 x 140 = N700.00
140 mangoes will cost N700.00
READING ASSIGNMENT
1. New General Mathematics JSS 1 by J.bChannonpg 159 Ex 23C No 4.
2. Essential Maths by AJS Oluwasanmipg 172 Ex 16 No 3
WEEKEND ASSIGNMENT
1. Round 567 to the nearest hundred
(a)500 (b)520 (c )540 (d) 580 (e) 600
2. What is 1.99961 correct to 2 d.p
(a) 1.99 (b) 2.00 (c ) 3.00 (d) 4.00 (e) 5.00
3. Write 7.0149 correct to the nearest thousandth
(a) 7.000 (b)7.014 (c ) 7.015 (d) 7.0145 (e) 7.0146
4. Give 0.000057891 to 4 s.f.
(a)0.00005789 (b)0.00005790 (c) 0.00005781 (d)0.000057892 (e)0.00005793.
5. Give 45698 correct to 3 s.f.
(a) 45600 (b)45700 (c )45800 (d)45690 (e) 45000.
THEORY
1. A sack of rice holds 64 basins of rice. How much will a woman get if she sells each basin of rice for N48.50.
2.Calculate the total cost of the following :
i. 3 textbooks at N400.00 each
ii.9 mathematical set at N2500.50
iii. 3 pens at 120.25 each
iv. 5 pencils at N20.00 each.