In this chapter you will learn more about whole numbers. You will learn about different ways to express whole numbers as sums and products. You will learn about different ways of doing calculations, and different ways of recording your work when doing calculations. You will strengthen your skills to do calculations and to solve problems. Show RevisionDo not use a calculator at all in section 1.1.
Suppose you want to know how many black dots there are in the arrangement on page 6. One way is to count in groups of three. When you do this, you may have to point with your finger or pencil to keep track. The counting will go like this: three, six, nine, twelve, fifteen, eighteen . . . Another way to find out how many black dots there are is to analyse the arrangement and do some calculations. In the arrangement, there are ten rows of threes from the top to the bottom, and three columns of threes from left to right, just as in the table alongside.
One way to calculate the total number of black dots is to do \(3 \times 10 = 30\) for the dots in each column, and then \(30 + 30 + 30 = 90\). Another way is to add up in each row \((3 + 3 + 3 = 9)\) and then multiply by 10: \(10 \times 9 = 90\). A third way is to notice that there are \(3 \times 10 = 30\) groups of three, so the total is \(3 \times 30 = 90\).
10 tens are a hundred: \(10 \times 10 = 100\) 10 hundreds are a thousand:\( 10 \times 100 = 1 000\) 10 thousands are a ten thousand: \(10 \times 1 000 = 10 000\) 10 ten thousands are a hundred thousand: \(10 \times 10 000 = 100 000\) 10 hundred thousands are a million: \(10 \times 100 000 = 1 000 000\)
Here is some advice that can make it easier to count in certain counting units, for example in seventies. It feels easier to count in fifties than in seventies because you get to multiples of 100 at every second step: fifty, hundred, one hundred and fifty, two hundred, two hundred and fifty, 300, 350, 400, 450, 500 ... and so on. When you count in seventies, this does not happen: seventy, one hundred and forty, two hundred and ten, two hundred and eighty ... It may help you to cross over the multiples of 100 in two steps each time, like this: In this way, you make the multiples of 100 act as "stepping stones" for your counting.
Doubling can be used to do multiplication. For example, \(29 \times 8\) can be calculated as follows: 8 doubled is 16, so \(16 = 2 \times 8\) (step 1) 16 doubled is 32, so \(32 = 4 \times 8\) (step 2) 32 doubled is 64, so \(64 = 8 \times 8\) (step 3) 64 doubled is 128, so \(128 = 16 \times 8\) (step 4). Doubling again will go past \(29 \times 8\). \(16 \times 8 + 8 \times 8 + 4 \times 8 = (16 + 8 + 4) \times 8 = 28 \times 8\). So \(28 \times 8 = 128 + 64 + 32\) which is 224. So \(29 \times 8 = 224 + 8 = 232\).
Halving can also be used to do multiplication. For example, \( 37 \times 28\) can be calculated as follows: \(100 \times 28 = 2 800\). Half of that is \(50 \times 28\) which is half of 2 800, that is 1 400. Half of \(50 \times 28\) is half of 1 400, so \(25 \times 28\) is 700. \(10 \times 28 = 280\), so \(25 \times 28 + 10 \times 28 = 980\), so \(35 \times 28 = 980\). \(2 \times 28 = 2 \times 25 + 2 \times 3 = 56\), so \(37 \times 28\) is \(980 + 56 = 1 036\).
If chickens cost R27 each, how many chickens can you buy with R2 400? A way to use halving to work this out is shown on the next page. 100 chickens cost (100 \(\times\) 27 = R2 700. That is more than R2 400. 50 chickens cost half as much, that is R1 350. So I can buy 50 chickens and even more. Half of 50 is 25 and half of R1 350 is R675. So 75 chickens cost R1 350 + R675, which is R2 025. So there is R375 left. 10 chickens cost R270, so 85 chickens cost R2 025 + R270 = R2 295. There is R105 left. 3 chickens cost 3 \( \times\) R25 + 3 \(\times\) R2 = R81. I can buy 88 chickens and that will cost R2 376.
What you actually did in questions 1, 2 and 3 was to calculate \(7 500 \div 27\), \(4 580 \div 17\) and \(1 800 \div 26\). You solved division problems. Yet most of the work was to do multiplication, and a little bit of subtraction. When you had to calculate \(1 800 \div 26\) in question 3, you may have asked yourself: With what must I multiply 26, to get as close to 1 800 as possible?
Division is called the inverse of multiplication. Multiplication is called the inverse of division. Multiplication and division are inverse operations. Ordering and comparing whole numbers
The first row in the table shows the average distances of the planets from the Sun. These distances are given in millions of kilometres.
The distances from the Sun are called average distances, because the planets are not always the same distance from the Sun. Their orbits are not circles. One million kilometres is 1 000 000 km.
The information in the table is also given in the drawings on page 2. Study the top drawing to find out what equatorial diameter means.
Sometimes we do not need to know the exact number or exact amount. We say a loaf of bread costs about R10, or a bag of mealie meal costs about R20. The loaf of bread may cost R8 or R12 but it is close to R10. The mealie meal may cost R18 or R21 but it is close to R20. When you read in a newspaper that there were 15 000 spectators at a soccer game, you know that that is not the actual number. In the language of mathematics we call this process rounding off or rounding.
To round off to the nearest 5, we round numbers that end in 1 or 2, or 6 or 7 down to the closest multiple of 5. We round numbers that end in 3 or 4, or 8 or 9 up to the closest multiple of 5. For example, 233 is rounded down to 230, 234 is rounded up to 235, 237 is rounded down to 235 and 238 is rounded up to 240.
To round off to the nearest 10, we round numbers that end in 1, 2, 3 or 4 down to the closest multiple of 10 (or decade). We round numbers that end in 5, 6, 7, 8 or 9 up to the closest multiple of 10. For example, if you want to round off 534 to the nearest 10, you have to look at the units digit. The units digit is 4 and it is closer to 0 than to 10. The rounded off number will be 530.
When rounding to the nearest 100, we look at the last two digits of the number. If the number is less than 50 we round down to the lower 100. If the number is 50 or more we round up to the higher 100.
When rounding to the nearest 1 000, we look at the hundreds. Is the hundreds value less than, equal to or greater than 500? If less than 500, round down (the thousands value stays the same), if equal to 500 round up, and if greater than 500 round up too. When rounding to the nearest 10 000, we look at the thousands. Is the thousands value less than, equal to or greater than 5 000 ?If less than 5 00,round down (the ten thousands value stays the same ),if equal to 5 00 or greater than 5 000 round up
Factors, prime numbers and common multiplesThe number 80 can be produced by multiplying 4 and 20: \(4 \times 20 = 80\). The number 80 can also be produced by multiplying 5 and 16.
The number 80 can also be produced by multiplying 2, 10 and 4: \(2 \times 10 = 20\) and \(20 \times 4 = 80\) or \(10 \times 4 = 40\) and \(40 \times 2 = 80\). We can use brackets to describe what calculation is done first. So instead of writing "\(2 \times 10 = 20\) and \(20 \times 4 = 80\)" we may write \((2 \times 10) \times 4\). Instead of writing "\(10 \times 4 = 40 \)and \(40 \times 2\)" we may write \(2 \times (10 \times 4)\).
The number 105 can be produced by multiplying 3, 5 and 7, hence we can write \(105 = 3 \times 5 \times 7\). Mathematicians often describe this by saying "105 is the product of 3, 5 and 7" or "105 can be expressed as the product \(3 \times 5 \times 7\)".
The whole numbers that are multiplied to form a number are called factors of the number. For example, 6 and 8 are factors of 48 because \(6 \times 8 = 48\). But 6 and 8 are not the only numbers that are factors of 48. 2 is also a factor of 48 because \(48 = 2 \times 24\). And 24 is a factor of 48. The numbers 3 and 16 are also factors of 48 because \(48 = 3 \times 16\).
The number 36 can be formed by \(2 \times 2 \times 3 \times 3\). Because 2 is used twice, it is called a repeated factor of 36. The number 3 is also a repeated factor of 36.
Eratosthenes, a Greek mathematician who lived a long time ago, designed a method to find the prime numbers. The process is called "the sieve of Eratosthenes".
We can say that Manare found the prime factors of 840, or Manare factorised 840 completely. We write: \(2\times 2 \rightarrow 4\times 2\rightarrow 24\times 5 \rightarrow 120 \times 7 = 840. \)
Here is a method to find the prime factors of a number: If the number is even, divide it by 2. If the answer is even, divide by 2 again. Continue like this as long as it is possible. If the answer is odd, divide by 3, if it is possible. Continue to divide by 3 as long as it is possible. Then switch to 5. Continue like this by each time trying to divide by the next prime number.
When a number is a factor of two or more other numbers, it is called a common factor of the other numbers. For example, the number 5 is a common factor of 195 and 385. The factors of a certain number are 2; 2; 5; 7; 7; 11 and 17. The factors of another number are 2; 3; 3; 7; 7; 11; 13 and 23. The common prime factors of these two numbers are 2; 7; 7 and 11.
The biggest number that is a factor of two or more numbers is called the highest common factor (HCF) of the numbers.
The numbers in pattern A are called the multiples of 12. The numbers in pattern B are called the multiples of 15. The numbers, for example 60 and 120, that occur in both patterns, are called the common multiples of 12 and 15. The smallest of these numbers, namely 60, is called the lowest common multiple (LCM) of 12 and 15.
Properties of operationsSuppose you want to tell another person to do some calculations. You may do this by writing instructions. For example, you may write the instruction 200 - 130 - 30. This may be called a numerical expression. Suppose you have given the instruction \(200 - 130 - 30\) to two people, whom we will call Ben and Sara. This is what Ben does: \(200 - 130 = 70\) and \(70 - 30 = 40\). This is what Sara does: \(130 - 30 = 100\) and \(200 - 100 = 100\). To prevent such different interpretations or understandings of the same numerical expression, mathematicians have made the following agreement, and this is followed all over the world:
An agreement like this is called a mathematical convention.
In a numerical expression that involves addition and subtraction only, the operations should be performed from left to right, unless otherwise indicated in some way.
In question 3, all your answers should be the same. When three or more numbers are added, the order in which you perform the calculations makes no difference. This is called the associative property of addition. We also say: addition is associative.
The above convention creates a problem. How can one describe the calculations in question 1(b) with a numerical expression, without using words? To solve this problem, mathematicians have agreed to use brackets in numerical expressions. Brackets are used to specify that the operations within the brackets should be done first. Hence the numerical expression for 1(b) above is \((20 + 5) \times 5\), and the answer is 125.
Examples The expression \(12 + 3 \times 5 \) means "multiply 3 by 5, then add 12". It does not mean "add 12 and 3, then multiply by 5". If you wish to say "add 5 and 12, then multiply by 3", the numerical expression should be \( 3 \times (5 + 12)\) or \( (5 + 12) \times 3\). They mean the same.
If there are no brackets in a numerical expression, it means that multiplication and division should be done first, and addition and subtraction only later. If you wish to specify that addition or subtraction should be done first,that part of the expression should be enclosed in brackets.
Your answers for 4(a) and 4(b) should be the same. Your answers for 4(c) and 4(d) should also be the same.
The fact that your answers for calculations like those in 7 b and 7 c are equal, for any numbers that you may choose, is called the distributive property of multiplication over addition.
It may be described as follows: \(\text{first number} \times \text{second number} + \text{first number} \times \text{third number}\) \(= \text{first number } \times (\text{second number } + \text{third number })\). This can be described by saying that multiplication distributes over addition.
It is quite fortunate that multiplication distributes over addition, because it makes it easier to multiply. For example, \( 8 \times 238\) can be calculated by calculating \(8 \times 200\),\( 8 \times 30 \)and \(8 \times 8\), and adding the answers:\( 8 \times 238 = 8 \times 200 + 8 \times 30 + 8 \times 8 = 1 600 + 240 + 64 = 1 904\).
Basic operationsTo add two numbers, the one may be written below the other. For example, to calculate \(378 539 + 46 285 \) the one number may be written below the other so that the units are below the units, the tens below the tens, and so on. Writing the numbers like this has the advantage that
This makes it possible to work with each kind of part separately.
The numbers in each column can be added to get a new set of numbers:
Note that you can do the above steps in any order. Instead of starting with the units parts as shown above, you can start with the hundred thousands, or any other parts. Starting with the units parts has an advantage though: it makes it possible to do more of the work mentally and to write less, as shown below: To achieve this, only the units digit 4 of the 14 is written in the first step. The 10 of the 14 is remembered and added to the 30 and 80 of the tens column, to get 120. We say the 10 is carried from the units column to the tens column. The same is done when the tens parts are added to get 120: only the digit "2" is written (in the tens column, so it means 20), and the 100 is carried to the next step.
A municipal manager is working on the municipal budget for a year. He has to try to keep the total expenditure on new office equipment below R800 000. He still has to budget for new computers that are badly needed, but this is what he has written so far:
There are many ways to subtract one number from another. For example, R835 234 - R687 885 can be calculated by "filling up" from R687 885 to R835 234: \(687 885 + 15 → 687 900 + 100→688 000 + 12 000 →700 000 + 135 234→ 835 234\)
The difference between R687 885 and R835 234 can now be calculated by adding up the numbers that had to be added to 687 885 to get 835 234. So \(\text{R}835 234 - \text{R}687 885 = \text{R}147 349\). Another easy way to subtract is to round off and compensate. For example, to calculate \(\text{R }3 224 - \text{R }1 885\), the R1 885 may be rounded up to R2 000. The calculation can proceed as follows:
Instead of calculating \( \text{R }3 224 - \text{R }1 885 \), which is a bit difficult, \( \text{R }3 339 - \text{R } 2 000 \) may be calculated. This is easy: \( \text{R }3 339 - \text{R }2 000 = \text{R }1 339 \). This means that \(\text{R }3 224 - \text{R }1 885 = \text{R }1 339\), because \(\text{R }3 224 - \text{R }1 885 = (\text{R }3 224 + \text{R }115) - (\text{R }1 885 + \text{R }115)\). To do question 1, you may use any one of the above two methods, or any other method you may know and prefer. Do not use a calculator, because the purpose of this work is for you to come to understand how subtraction may be done. What you will learn here, will later help you to understand algebra.
Another method of subtraction is to think of the numbers in expanded notation. For example, to calculate \(\text{R}835 234 - \text{R}687 885\), which was already done in a different way on the previous page, we could work like this:
Unfortunately, it is not possible to subtract in the columns now. However, the parts of the bigger number can be rearranged to make the subtraction in each column possible:
The rearrangement, also called "borrowing", was done like this: 10 was taken from the 30 in the tens column, and added to the 4 in the units column. 100 was taken from the 200 in the hundreds column, and added to the 20 that remained in the tens column. 1 000 was taken from the 5 000 in the thousands column, and added to the 100 that remained in the hundreds column.
It is not practical to write the expanded notation and the rearrangements each time you do a subtraction. However, with some practice you can learn to do it all in your mind without writing it down. Some people make small marks above the digits of the bigger number, or even change the digits, to keep track of the rearrangements they make in their minds.
\(6 \times \text{R}3 258 \) can be calculated in parts, as shown below. \(6 \times \text{R}3 000 = \text{R}18 000\) \(6 \times \text{R}200 = \text{R}1 200\) \(6 \times \text{R}50 = \text{R}300\) \(6 \times \text{R}8 = \text{R}48\)
The four partial products can now be added to get the answer, which is R19 548. It is convenient to write the work in vertical columns for units, tens, hundreds and so on, as shown on the right above. In fact, if you are willing to do some hard thinking you can produce the answer with even less writing. You can achieve this by working from right to left to calculate the partial products, and by "carrying" parts of the partial answers to the next column, as you can do when working from right to left in columns. It works like this:
When \( 6 \times 8 = 48\) is calculated, only the "8" is written down, in the units column. The "4" that represents 40 is not written. It is kept "on hold" in your mind. When \(6 \times 50 = 300 \) is calculated, the 40 from the previous step is added to 300 to get 340. Again, only the "4" that represents 40 is written. The 300 is kept on hold or "carried" to add to the answer of the next step. The work continues like this.
To calculate \(36 \times 378\), the work can be broken up in two parts, namely \(30 \times 378 \)and \(6 \times 378\).
A complete write-up of calculating \(76 \times 2 348 \)in columns is shown below.
The principal wants to work out exactly how many T-shirts, at R67 each, she can buy with R8 500. Her thinking and writing are described below. Step 1
Step 2
Step 3 (She has to rub out the one "0" of the 100 on top, to make space.)
Step 4 (She rubs out another "0".)
Step 5 (She rubs out the "5".)
Do not use a calculator in the questions that follow. The purpose of this work is for you to develop a good understanding of how division can be done. Check all your answers by doing multiplication.
Problem solvingYou may use a calculator for doing the work in this section.
A man borrows R12 000 from a bank for one year. He has to pay 15% interest to the bank. This means that, apart from paying the R12 000 back to the bank after a year, he has to pay 15 hundredths of R12 000 for the privilege of using the money that actually belongs to the bank.
15% is read as 15 per cent, and it is just a different way to say 15 hundredths. One hundredth of R12 000 can be calculated by dividing R12 000 by 100. This amount can then be multiplied by 15 to get 15 hundredths of R12 000.
The money paid for using another person's house is called rent. The money paid for using another person's money is called interest. Do not use a calculator when you do the following questions.
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