In this chapter you will learn more about whole numbers, and you will strengthen your skills to do calculations and to solve problems. Show Properties of whole numbers
We say: addition and multiplication are commutative. The numbers can be swopped around and their order does not change the answer. This does not work for subtraction and division, however.
Lebogang and Nathi both have to calculate \(25 \times 24\). Lebogang calculates \(25 \times 4\) and then multiplies by 6. Nathi calculates \(25 \times 6\) and then multiplies by 4.
If three or more numbers have to be multiplied, it does not matter which two of the numbers are multiplied first. This is called the associative property of multiplication. We also say multipliaction is associative
The distributive property is a useful property because it allows us to do this: Both answers are 18. Notice that we have to use brackets in the first example to show that the addition operation must be done first. Otherwise, we would have done the multiplication first. For example, the expression \(3 \times 2 + 4\) means "multiply 3 by 2; then add 4". It does not mean "add 2 and 4; then multiply by 3". The expression \(4 + 3 \times 2\) also means "multiply 3 by 2; then add 4".
If you wish to specify that addition or subtraction should be done first, that part of the expression should be enclosed in brackets. The distributive property can be used to break up a difficult multiplication into smaller parts. For example, it can be used to make it easier to calculate \(6 \times 204\): \(6 \times 204\) can be rewritten as \(6 \times (200 + 4)\) (Remember the brackets!) \[ \begin{align} &= 6 \times 200 + 6 \times 4 \\ &= 1 200 + 24 \\ &= 1 224 \end{align} \] Multiplication can also be distributed over subtraction, for example to calculate \(7 \times 96\): \[ \begin{align} 7 \times 96 &= 7 \times (100 - 4) \\ &= 7 \times 100 - 7 \times 4 \\ &= 700 - 28 \\ &= 672 \end{align} \]
Two more properties of number are:
Calculations with whole numbers
What you have done when you tried to give answers to questions 1(a) to (d), is called estimation. To estimate is to try to get close to an answer without actually doing the required calculations with the given numbers.
An estimate may also be called an approximation.
Calculating with "easy" numbers that are close to given numbers is a good way to obtain approximate answers, for example:
The word compensate means to do things that will remove damage.
What you have done in question 1 to find the correct answer for \(386 + 3 435\) is called rounding off and compensating. By rounding the numbers off you introduced errors. You then compensated for the errors by making adjustments to your answer.
Subtraction can also be done in this way. For example, to calculate \(\text{R }5 362 - \text{R }2 687\), you may round R2 687 up to R3 000. The calculation can proceed as follows:
This means that \(\text{R }5 362 - \text{R }2 687 = \text{R }2 675\), because \(\text{R }5 362 - \text{R }2 687 = (\text{R }5 362 + \text{R }313) - (\text{R }2 687 + \text{R }313)\). Numbers can be added by thinking of their parts as we say the numbers. For example, we say 4 994 as four thousand nine hundred and ninety-four. This can be written in expanded notation as 4 000 + 900 + 90 + 4. Similarly, we can think of 31 837 as \(30 000 + 1 000 + 800 + 30 + 7\). 31 837 + 4 994 can be calculated by working with the various kinds of parts separately. To make this easy, the numbers can be written below each other so that the units are below the units, the tens below the tens and so on, as shown above.
The numbers in each column can be added to get a new set of numbers.
The work may start with the 10 000s or any other parts. Starting with the units as shown above makes it possible to do more of the work mentally, and write less, as shown below. To achieve this, only the units digit 1 of the 11 is written in the first step. The 10 of the 11 is remembered and added to the 30 and 90 of the tens column, to get 130. 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 130: only the digit "3" is written (in the tens column, so it means 30), and the 100 is carried to the next step.
There are many ways to find the difference between two numbers. For example, to find the difference between 267 and 859 one may think of the numbers as they may be written on a number line. We may think of the distance between 267 and 859 as three steps: from 267 to 300, from 300 to 800, and from 800 to 859. How big are each of these three steps? The above shows that \(859 - 267 \) is \(33 + 500 + 59\).
Like addition, subtraction can also be done by working with the different parts in which we say numbers. For example, \(8 764 - 2 352\) can be calculated as follows: \( \begin{align} 8 \text{ thousand} - 2 \text{ thousand} &= 6 \text{ thousand} \\ 7 \text{ hundred} - 3 \text{ hundred} &= 4 \text{ hundred} \\ 6 \text{ tens} - 5 \text{ tens} &= 1 \text{ ten} \\ 4 \text{ units} - 2 \text{ units} &= 2 \text{ units} \end{align} \) So, \(8 764 - 2 352 = 6 412\) Subtraction by parts is more difficult in some cases, for example \(6 213 - 2 758\): \( \begin{align}6 000 - 2 000 &= 4 000 \text{. This step is easy, but the following steps cause problems:} \\ 200 - 700 &= \ \ ? \\ 10 - 50 &= \ \ ? \\ 3 - 8 &= \ \ ? \end{align} \)
One way to overcome these problems is to work with negative numbers: \(200 - 700 = (-500) \\ 10 - 50 = (-40) \\ 3 - 8 = (-5) \\ 4000 - 500 \rightarrow 3500 - 45 =\) Fortunately, the parts and sequence of work may be rearranged to overcome these problems, as shown below:
This reasoning can also be set out in columns:
With some practice, you can learn to subtract using borrowing without writing all the steps. It is convenient to work in columns, as shown below for calculating \(6 213 - 2 758\).
In fact, by doing more work mentally, you may learn to save more paper by writing even less as shown below. Do not use a calculator when you do question 5, because the purpose of this work is for you to come to understand methods of subtraction. What you will learn here will later help you to understand algebra better.
\(7 \times 4 598\) can be calculated in parts, as shown here: \[ \begin{align} 7 \times 4 000 &= 28 000 \\ 7 \times 500 &= 3 500 \\ 7 \times 90 &= 630 \\ 7 \times 8 &= 56 \end{align} \] The four partial products can now be added to get the answer, which is 32 186. It is convenient to write the work in vertical columns for units, tens, hundreds and so on, as shown below.
The answer can be produced with less writing, by "carrying" parts of the partial answers to the next column, when working from right to left in the columns. Only the 6 of the product \(7 \times 8\) is written down instead of 56. The 50 is kept in mind, and added to the 630 obtained when \(7 \times 90\) is calculated in the next step.
The municipal gardener wants to work out exactly how many trees, at R27 each, he can buy with the budgeted amount of R9 400. His thinking and writing are described below. Step 1 What he writes: What he thinks: I want to find out how many chunks of 27 there are in 9400 Step 2 What he writes: What he thinks: I think there are at least 300 chunks of 27 in 9 400. \(300 \times 27 = 8 100\). I need to know how much is left over. Step 3 (He has to rub out the one "0" of the 300 on top, to make space.) What he writes:
What he thinks: I think there are at least 40 chunks of 27 in 1 300. \(40 \times 27 = 1 080\). I need to know how much is left over. I want to find out how many chunks of 27 there are in 220.Perhaps I can buy some extra trees. Step 4 (He rubs out another "0".) What he writes:
What he thinks: I think there are at least 8 chunks of 27 in 220. \(8\times 27 = 216\). So, I can buy 348 young trees and will have R4 left. Do not use a calculator to do questions 3 and 4. 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.
Multiples, factors and prime factors
If \(n\) is a natural number,6n represents the multiples of 6.
Of which numbers is 20 a multiple? \(20 = 1 \times 20 = 2 \times 10 = 4 \times 5 = 5\times 4 = 10 \times 2 = 20 \times 1\)
20 is a multiple of 1; 2; 4; 5; 10 and 20 and all of these numbers are factors of 20. Factors come in pairs. The following pairs are factors of 20:
The number 36 can be formed as \(2 \times 2 \times 3 \times 3\). Because 2 and 3 are used twice, they are called repeated factors of 36.
A number that cannot be expressed as a product of two whole numbers, except as the product \(1\times\) the number itself, is called a prime number.
Composite numbers are natural numbers with more than two different factors. The sequence of composite numbers is 4; 6; 8; 9; 10; 12; ...
To find all the factors of a number you can write the number as the product of prime factors, first by writing it as the product of two convenient (composite) factors and then by splitting these factors into smaller factors until all factors are prime. Then you take all the possible combinations of the products of the prime factors.
Every composite number can be expressed as the product of prime factors and this can happen in only one way. Example: Find the factors of 84. Write 84 as the product of prime factors by starting with different known factors: \[ \begin{align} 84 &= 4 \times 21 \\ &= 2 \times 2 \times 3 \times 7 \end{align} \] or \[ \begin{align} 84 &= 7 \times 12 \\ &= 7 \times 3 \times 4 \\ &=7 \times 3 \times 2 \times 2 \end{align} \] or \[ \begin{align} 84 &= 2 \times 42 \\ &= 2 \times 6 \times 7 \\ &=2 \times 2 \times 3 \times 7 \end{align} \] A more systematic way of finding the prime factors of a number would be to start with the prime numbers and try the consecutive prime numbers 2; 3; 5; 7; ... as possible factors. The work may be set out as shown below.
\[ 1 430 = 2 \times 5 \times 11 \times 13 \]
\[2 457 = 3 \times 3 \times 3 \times 7 \times 13 \] We can use exponents to write the products of prime factors more compactly as products of powers of prime factors. \[ \begin{align} 2 457 &= 3 \times 3 \times 3 \times 7 \times 13 &&= 3^3 \times 7 \times 13 \\ 72 &= 2 \times 2 \times 2 \times 3 \times 3 &&= 2^3 \times 3^2 \\ 1 500 &= 2 \times 2 \times 3 \times 5 \times 5 \times 5 &&= 2^2 \times 3 \times 5^3 \end{align} \]
We use common multiples when fractions with different denominators are added. To add \( \frac{2}{3} + \frac{3}{4} \) the common denominator is \(3 \times 4\), so the sum becomes \( \frac{8}{12} + \frac{9}{12} \) In the same way, we could use \(6 \times 8 = 48\) as a common denominator to add \( \frac{1}{6} + \frac{3}{8}\), but 24 is the lowest common multiple (LCM) of 6 and 8. Prime factorisation makes it easy to find the lowest common multiple or highest common factor.When we simplify a fraction, we divide the same number into the numerator and the denominator. For the simplest fraction, use the highest common factor (HCF) to divide into both numerator and denominator.
The HCF is divided into the numerator and the denominator to write the fraction in its simplest form. So \( \frac{36}{144} = \frac{2 \times 2 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 3 \times 3} = \frac{1}{4} \) Use prime factorisation to determine the LCM and HCF of 32, 48 and 84 in a systematic way: \[ \begin{align} 32 &= 2 \times 2 \times 2 \times 2 \times 2 &&= 2^5 \\ 48 &= 2 \times 2 \times 2 \times 2 \times 3 &&= 2^4 \times 3 \\ 84 &= 2 \times 2 \times 3 \times 7 &&= 2^2 \times 3 \times 7 \end{align} \] The LCM is a multiple, so all of the factors of all the numbers must divide into it. All of the factors that are present in the three numbers must also be factors of the LCM, even if it is a factor of only one of the numbers. But because it has to be the lowest common multiple, no unnecessary factors are in the LCM. The highest power of each factor is in the LCM, because then all of the other factors can divide into it. In 32, 48 and 84, the highest power of 2 is \(2^5\), the highest power of 3 is 3 and the highest power of 7 is 7. LCM = \(2^5 \times 3 \times 7 = 672\) The HCF is a common factor. Therefore, for a factor to be in the HCF, it must be a factor of all of the numbers. 2 is the only number that appears as a factor of all three numbers. The lowest power of 2 is \(2^2\), so the HCF is \(2^2\).
You may use a calculator for this investigation.
Solving problemsYou may use a calculator for the work in this section.
Instead of saying " . . . per day", people often say "at a rate of " . . . per day". Speed is a way to describe the rate of movement.
The word per is often used to describe a rate and can mean for every, for, in each, in, out of, or every.
The patterns in question 4 can be described like this: In pattern A, the ratio of yellow beads to red beads is 4 to 5. This is written as 4 : 5. In pattern B, the ratio between yellow beads and red beads is 3 : 6. In pattern C the ratio is 2 : 7. In question 5, machine A produces 2 screws for every 3 screws that machine B produces. This can be described by saying that the ratio between the production speeds of machines A and B is 2 : 3.
We use ratios to show how many times more, or less, one quantity is than another. A ratio is a comparison of two (or more) quantities. The number of hours that Nathi, Paul and Tim worked are in the ratio 5 : 4 : 3. To be fair, the money should also be shared in that ratio. That means that Nathi should receive 5 parts, Paul 4 parts and Tim 3 parts of the money. There were 12 parts, which means Nathi should receive \( \frac{5}{12} \) of the total amount, Paul should get \( \frac{4}{12}\) and Tim should get \( \frac{3}{12} \).
Remember that if you multiply by 1, the number does not change. If you multiply by a number greater than 1, the number increases. If you multiply by a number smaller than 1, the number decreases.
To increase 40 in the ratio 2 : 3 means that the 40 represents two parts and must be increased so that the new number represents 3 parts. If 40 represents two parts, 20 represents 1 part. The increased number will therefore be \(20 \times 3 = 60\).
Rashid is a furniture dealer. He buys a couch for R2 420. He displays the couch in his showroom with the price marked as R3 200. Rashid offers a discount of R320 to customers who pay cash.
The amount for which a dealer buys an article from a producer or manufacturer is called the cost price. The price marked on the article is called the marked price and the price of the article after discount is the selling price.
The discount on an article is always less than the marked price of the article. In fact, it is only a fraction of the marked price. The discount of R320 that Rashid offers on the couch is 10 hundredths of the marked price. Another word for hundredths is percentage, and the symbol for percentage is %. So we can say that Rashid offers a discount of 10%. A percentage is a number of hundredths. 18% is 18 hundredths, and 25% is 25 hundredths.
%is a symbol for hundredths. 8% means 8 hundredths and 15% means 15 hundredths. The symbol % is just a variation of the \( \frac{}{100} \) that is used in the common fraction notation for hundredths. 8% is \( \frac{8}{100} \). A discount of 6% on an article can be calculated in two steps: Step 1: Calculate 1 hundredth of the marked price (divide by 100). Step 2: Calculate 6 hundredths of the marked price (multiply by 6).
You may use a calculator to do questions 6, 7 and 8.
When a person borrows money from a bank or some other institution, he or she normally has to pay for the use of the money. This is called interest.
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