Calculations using whole numbers

1.4 Calculations using whole numbers

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. The difference between an estimate and the actual answer is called the error.

Calculating with “easy” numbers that are close to given numbers is a good way to obtain approximate answers (or estimates), for example:

• To approximate \(764 + 829\), you could calculate \(800 + 800\) to get the approximate answer \(1\ 600\), with an error of \(7\).
• To approximate \(84 \times 178\), you could calculate \(80 \times 200\) to get the approximate answer \(16\ 000\), with an error of \(1\ 048\).

Exercise 1.1: Approximate the answers to large products

Calculate with easy numbers close to the given numbers to produce approximate answers for each product below. Do not use a calculator. When you have made your approximations, use a calculator to find the precise answers.

1. \[88 \times 56\]
2. \[77 \times 98\]
3. \[94 \times 275\]
4. \[71 \times 198\]
5. \[1\ 250 \times 79\]

1. \[88 \times 56\]

   Approximation: \(90 \times 50 = 4\ 500\)

   Actual answer: \(4\ 928\)

2. \[77 \times 98\]

   Approximation: \(80 \times 100 = 8\ 000\)

   Actual answer: \(7\ 546\)

3. \[94 \times 275\]

   Approximation: \(90 \times 300 = 27\ 000\)

   Actual answer: \(25\ 850\)

4. \[71 \times 198\]

   Approximation: \(70 \times 200 = 14\ 000\)

   Actual answer: \(14\ 058\)

5. \[1\ 250 \times 79\]

   Approximation: \(1\ 000 \times 80 = 80\ 000\)

   Actual answer: \(98\ 750\)

Methods for adding, subtracting and multiplying

Let’s revise some basic addition, subtraction and multiplication facts.

• Remember that the sum is what you get after you have done your addition. We can add in any order.

  \[44 + 41 = 85\]

  The sum of \(44\) and \(41\) is \(85\).

• The difference is what is left after you have done your subtraction. We must subtract in the given order.

  \[55 - 15 = 40\]

  The difference between \(55\) and \(15\) is \(40\).

• There are lots of ways to subtract numbers from each other. This regrouping method is just one example.

  \[(55 - 5) - 10 = 50 - 10 = 40\]

• The product is what you get after you have done your multiplication. We can multiply in any order.

  \[5 \times 10 = 50\]

  The product of \(5\) and \(10\) is \(50\).

• The quotient is what is left after you have done your division. We must divide in the given order.

  To determine the quotient when \(30\) is divided by \(5\), we must divide \(30\) by \(5\).

  \[30 \div 5 = 6\]

  The quotient is \(6\).

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Methods for addition

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 \(30\ 000 + 10\ 000 + 800 + 30 + 7\).

\(31\ 837 + 4\ 994\) 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 here.

\[\begin{array}{ccccc} 3 & 1 & 8 & 3 & 7 \\ & 4 & 9 & 9 & 4 \end{array}\]

Always align the numbers starting from ones and moving to the bigger units, in other words, from right to left.

We can now expand each number:

\[\begin{array}{rccccc} \mathbf{31\ 837} & 30\ 000 & 1\ 000 & 800 & 30 & 7 \\ \mathbf{4\ 994} & & 4\ 000 & 900 & 90 & 4 \end{array}\]

Add the numbers in each column separately, to get a new set of numbers that you can write at the far left.

\[\begin{array}{rccccc} \mathbf{31\ 837} & 30\ 000 & 1\ 000 & 800 & 30 & 7 \\ \mathbf{4\ 994} & & 4\ 000 & 900 & 90 & 4 \\ \hline 11 & & & & & 11 \\ 120 & & & & 120 & \\ 1\ 700 & & & 1\ 700 & & \\ 5\ 000 & & 5\ 000 & & & \\ 30\ 000 & 30\ 000 & & & & \\ \hline 36\ 831 & & & & & \end{array}\]

It is easy to add the new set of numbers to get the answer:

\[30\ 000 + 5\ 000 + 1\ 700 + 120 + 11 = 36\ 831\]

Starting with the ones as shown above makes it possible to do more of the work mentally, and write less, as shown below.

\[\begin{array}{ccccc} & 3 & ^{1}1 & ^{1}8 & ^{1}3 & 7 \\ & & 4 & 9 & 9 & 4 \\ \hline & 3 & 6 & 8 & 3 & 1 \end{array}\]

To achieve this, the first step is to add the digits in the ones column: \(7 + 4 = 11\). Only write the digit “\(1\)” of the \(11\). The \(10\) of the \(11\) is remembered and added to the \(3\) in the tens column.

We say the \(10\) is carried from the ones column to the tens column. The same is done when the tens parts are added to get \(13\): only the digit “\(3\)” is written (in the tens column, so it means \(30\)), and the \(1\) is carried to the next column. This \(1\) is actually \(100\), so it needs to be carried to the hundreds column.

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Methods of subtraction

There are many ways to find the difference between two numbers. For example, to find the difference between \(267\) and \(859\), you can 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?

This number line shows that \(859 - 267\) is \(33 + 500 + 59\). This means that \(859 - 267 = 592\).

Like addition, subtraction can also be done by working with the different parts in which we say (or expand) the numbers. For example, \(8\ 764 − 2\ 352\) can be calculated as follows:

• \(8\) thousand − \(2\) thousand = \(6\) thousand
• \(7\) hundred − \(3\) hundred = \(4\) hundred
• \(6\) tens − \(5\) tens = \(1\) ten
• \(4\) units − \(2\) units = \(2\) units

So, \(8\ 764 − 2\ 352 = 6\ 412\).

Subtraction by parts is more difficult in some cases, for example \(6\ 213 − 2\ 758\):

\(6\ 000 - 2\ 000 = 4\ 000\) This step is easy, but the following steps cause problems:

• \(200 - 700 = \ ?\)
• \(10 - 50 = \ ?\)
• \(3 - 8 = \ ?\)

One way to overcome these problems is to work with negative numbers:

• \(200 - 700 = - 500\)
• \(10 - 50 = - 40\)
• \(3 - 8 = - 5\)

Now, we add positive and negative answers to get the result:

\[\begin{align} 6\ 213 - 2\ 758 &= 4\ 000 + ( - 500) + ( - 40) + ( - 5) \\ &= 3\ 500 + ( - 45) \\ &= 3\ 455 \end{align}\]

In this method, it is easy to make mistakes if you are not careful with the minus sign. Fortunately, the parts and sequence of work may be rearranged to overcome these problems, as shown below. We regroup both numbers to make subtraction in columns easier:

\[\begin{align} 6\ 213 &= 6\ 000 + 200 + 10 + 3 \\ &= 5\ 000 + 1\ 100 + 100 + 13 \end{align}\]

	\(\mathbf{+ 1\ 000}\) (borrow \(1\ 000\))	\(\mathbf{+ 100}\) (borrow \(100\))	\(\mathbf{+ 10}\) (borrow \(10\))
\(1^{\text{st}}\) number: \(6\ 213\)	\(6\ 000 = 5\ 000 + 1\ 000\)	\(200 = 100 + 100\)	\(10\)	\(3\)
	\(5\ 000\)	\(1\ 100\)	\(100\)	\(13\)
\(2^{\text{nd}}\) number: \(2\ 758\)	\(2\ 000\)	\(700\)	\(50\)	\(8\)
	\(5\ 000 − 2\ 000 = 3\ 000\)	\(1\ 100 − 700 = 400\)	\(100 − 50 = 50\)	\(13 − 8 = 5\)

Now it is easier to find the answer: \(3\ 000 + 400 + 50 + 5 = 3\ 455\).

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.

Exercise 1.2: Subtraction with large numbers

Calculate each of the following

1. \[4\ 340 - 3\ 870\]
2. \[1\ 653 - 1\ 156\]
3. \[3\ 671 - 2\ 528\]

1. \[4\ 340 - 3\ 870 = 470\]
2. \[1\ 653 - 1\ 156 = 497\]
3. \[3\ 671 - 2\ 528 = 1\ 143\]

Estimate the difference between the two car prices in each case to the nearest \(\text{R } 1\ 000\) or closer. Then calculate the difference.

1. \(\text{R } 102\ 365\) and \(\text{R } 98\ 128\)
2. \(\text{R } 63\ 378\) and \(\text{R } 96\ 889\)

1. \(\text{R } 102\ 365\) and \(\text{R } 98\ 128\)

   Estimate: \(\text{R } 102\ 000 − \text{R } 98\ 000 = \text{R } 4\ 000\)

   Actual: \(\text{R } 4\ 237\)

2. \(\text{R } 63\ 378\) and \(\text{R } 96\ 889\)

   Estimate: \(\text{R } 97\ 000 − \text{R } 63\ 000 = \text{R } 34\ 000\)

   Actual: \(\text{R } 33\ 511\)

First estimate the answers to the nearest \(100\ 000\) or \(10\ 000\) or \(1\ 000\). Then calculate.

1. \[238\ 769 - 141\ 453\]
2. \[856\ 333 - 739\ 878\]
3. \[65\ 244 - 39\ 427\]

1. \[238\ 769 - 141\ 453\]

   Estimate: \(240\ 000 − 140\ 000 = 100\ 000\)

   Actual: \(97\ 316\)

2. \[856\ 333 - 739\ 878\]

   Estimate: \(860\ 000 − 740\ 000 = 120\ 000\)

   Actual: \(116\ 455\)

3. \[65\ 244 - 39\ 427\]

   Estimate: \(65\ 000 − 40\ 000 = 25\ 000\)

   Actual: \(25\ 817\)

A method of multiplication

Just as with addition and subtraction, we can use the method of breaking up numbers into parts. For example, to find the product \(7 \times 4\ 598\), we can expand \(4\ 596\) like this:

• \(7 \times 4\ 000 = 28\ 000\)
• \(7 \times 500 = 3\ 500\)
• \(7 \times 90 = 630\)
• \(7 \times 8 = 56\)

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.

\[\begin{array}{rccccc} & & 4 & 5 & 9 & 8 \\ \times & & & & & 7 \\ \hline & & & & 5 & 6 \\ & & & 6 & 3 & 0 \\ & & ^{1}3 & 5 & 0 & 0 \\ + & ^{1}2 & 8 & 0 & 0 & 0 \\ \hline & 3 & 2 & 1 & 8 & 6 \end{array}\]

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.

\[\begin{array}{rccccc} & & ^{4}4 & ^{6}5 & ^{5}9 & 8 \\ \times & & & & & 7 \\ \hline & 3 & 2 & 1 & 8 & 6 \end{array}\]

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.

Exercise 1.3

Calculate each of the following. Do not use a calculator now.

1. \[29 \times 549\]
2. \[78 \times 1\ 750\]
3. \[641 \times 99\]

Use your calculator to check your answers for Question 1. Redo the questions for which you had the wrong answers.

1. \[29 \times 549 = 15\ 921\]
2. \[78 \times 1\ 750 = 136\ 500\]
3. \[641 \times 99 = 63\ 459\]

Calculate each of the following. Do not use a calculator now.

1. \[97 \times 176\]
2. \[74 \times 197\]

1. \[97 \times 176 = 17\ 072\]
2. \[74 \times 197 = 14\ 578\]

Long division method

Long division is a useful method for dividing large numbers without a calculator. Essentially, you break the division problem into lots of similar steps, and follow a sequence.

Exercise 1.4: Calculations with division

Brenda bought \(64\) goats, all at the same price. She paid \(\text{R} 5\ 440\) in total. What was the price for each goat?

Your first step can be to work out how much Brenda would have paid if she paid \(\text{R} 10\) per goat, but you can start with a bigger step if you wish.

\[5\ 440 \div 64 = \text{R}85\]

Mosibudi has \(\text{R} 2\ 850\) and he wants to buy candles for his sister’s wedding reception. The candles cost \(\text{R} 48\) each. How many candles can he buy?

\(2\ 850 \div 48 = 59 \text{ rem } 18\), so he can buy \(59\) candles.

Calculate each of the following, without using a calculator.

1. \[7\ 234 \div 48\]
2. \[3\ 267 \div 24\]
3. \[9\ 500 \div 364\]
4. \[8\ 347 \div 24\]

1. \[7\ 234 \div 48 = 150 \text{ rem } 34\]
2. \[3\ 267 \div 24 = 136 \text{ rem } 3\]
3. \[9\ 500 \div 364 = 26 \text{ rem } 36\]
4. \[8\ 347 \div 24 = 347 \text{ rem } 19\]