1.4 Rounding off | Algebraic expressions

1.4 Rounding off (EMA8)

Rounding off a decimal number to a given number of decimal places is the quickest way to approximate a number. For example, if you wanted to round off \(\text{2,6525272}\) to three decimal places, you would:

count three places after the decimal and place a \(|\) between the third and fourth numbers;
round up the third digit if the fourth digit is greater than or equal to \(\text{5}\);
leave the third digit unchanged if the fourth digit is less than \(\text{5}\);
if the third digit is \(\text{9}\) and needs to be rounded up, then the \(\text{9}\) becomes a \(\text{0}\) and the second digit is rounded up.

So, since the first digit after the \(|\) is a \(\text{5,}\) we must round up the digit in the third decimal place to a \(\text{3}\) and the final answer of \(\text{2,6525272}\) rounded to three decimal places is \(\text{2,653}\).

The following video explains how to round off.

Video: 2DD8

Worked example 4: Rounding off

Round off the following numbers to the indicated number of decimal places:

\(\dfrac{120}{99}=\text{1,}\dot{2}\dot{1}\) to \(\text{3}\) decimal places.
\(\pi =\text{3,141592653...}\) to \(\text{4}\) decimal places.
\(\sqrt{3}=\text{1,7320508...}\) to \(\text{4}\) decimal places.
\(\text{2,78974526}\) to \(\text{3}\) decimal places.

Mark off the required number of decimal places

If the number is not a decimal you first need to write the number as a decimal.

\(\dfrac{120}{99} = \text{1,212}|121212\ldots\)
\(\pi =\text{3,1415}|92653\ldots\)
\(\sqrt{3}=\text{1,7320}|508\ldots\)
\(\text{2,789}|74526\)

Check the next digit to see if you must round up or round down

The last digit of \(\frac{120}{99}=\text{1,212}|121212\dot{1}\dot{2}\) must be rounded down.
The last digit of \(\pi =\text{3,1415}|92653\ldots\) must be rounded up.
The last digit of \(\sqrt{3}=\text{1,7320}|508\ldots\) must be rounded up.
The last digit of \(\text{2,789}|74526\) must be rounded up.

Since this is a \(\text{9}\) we replace it with a \(\text{0}\) and round up the second last digit.

Write the final answer

\(\dfrac{120}{99}=\text{1,212}\) rounded to \(\text{3}\) decimal places.
\(\pi =\text{3,1416}\) rounded to \(\text{4}\) decimal places.
\(\sqrt{3}=\text{1,7321}\) rounded to \(\text{4}\) decimal places.
\(\text{2,790}\)

temp text

Textbook Exercise 1.2

\(\text{12,56637061...}\)

Mark off the required number of decimal places: \(\text{12,566}|37061\ldots\). The next digit is a \(\text{3}\) and so we round down: \(\text{12,566}\).

\(\text{3,31662479...}\)

Mark off the required number of decimal places: \(\text{3,316}|62479\ldots\). The next digit is a \(\text{6}\) and so we round up: \(\text{3,317}\).

\(\text{0,2666666...}\)

Mark off the required number of decimal places: \(\text{0,266}|6666\ldots\). The next digit is a \(\text{6}\) and so we round up: \(\text{0,267}\).

\(\text{1,912931183...}\)

Mark off the required number of decimal places: \(\text{1,912}|931183\ldots\). The next digit is a \(\text{9}\) and so we round up: \(\text{1,913}\).

\(\text{6,32455532...}\)

Mark off the required number of decimal places: \(\text{6,324}|55532\ldots\). The next digit is a \(\text{5}\) and so we round up: \(\text{6,325}\).

\(\text{0,05555555...}\)

Mark off the required number of decimal places: \(\text{0,055}|55555\ldots\). The next digit is a \(\text{5}\) and so we round up: \(\text{0,056}\).

\(\text{345,04399906}\) to \(\text{4}\) decimal places.

\[\text{345,04399906} \approx \text{345,0440}\]

\(\text{1 361,72980445}\) to \(\text{2}\) decimal places.

\[\text{1 361,72980445} \approx \text{1 361,73}\]

\(\text{728,00905239}\) to \(\text{6}\) decimal places.

\[\text{728,00905239} \approx \text{728,009052}\]

\(\dfrac{1}{27}\) to \(\text{4}\) decimal places.

We first write the fraction as a decimal and then we can round off.

\begin{align*} \frac{1}{27} &= \text{0,037037...} \\ & \approx \text{0,0370} \end{align*}

\(\dfrac{45}{99}\) to \(\text{5}\) decimal places.

We first write the fraction as a decimal and then we can round off.

\begin{align*} \frac{45}{99} &= \text{0,45454545...} \\ & \approx \text{0,45455} \end{align*}

\(\dfrac{1}{12}\) to \(\text{2}\) decimal places.

We first write the fraction as a decimal and then we can round off.

\begin{align*} \frac{1}{12} &= \text{0,08333...} \\ & \approx \text{0,08} \end{align*}

Calculate the area of \(ABDE\) to \(\text{2}\) decimal places.

\(ABDE\) is a square and so the area is just the length squared.

\begin{align*} A &= l^{2} \\ &= \pi^2 \\ & = \text{9,86904...} \\ & \approx \text{9,87} \end{align*}

Calculate the area of \(BCD\) to \(\text{2}\) decimal places.

\(BCD\) is a right-angled triangle and so we have the perpendicular height. The area is:

\begin{align*} A & = \frac{1}{2} b h\\ & = \frac{1}{2} \pi^2 \\ & = \text{4,934802...} \\ & \approx \text{4,93} \end{align*}

Using you answers in (a) and (b) calculate the area of \(ABCDE\).

The area of \(ABCDE\) is the sum of the areas of \(ABDE\) and \(BCD\).

\begin{align*} A & = \text{9,87} + \text{4,93} \\ & \approx \text{14,80} \end{align*}

Without rounding off, what is the area of \(ABCDE\)?

\begin{align*} A_{ABCDE} & = A_{ABDE} + A_{BCD} \\ & = l^{2} + \frac{1}{2}bh \\ &= \pi ^2 + \frac{1}{2} \pi^2 \\ &= \text{14,8044...} \end{align*}

Calculate \(i\) correct to \(\text{2}\) decimal places.

\begin{align*} i & = \frac{r}{600} \\ & = \frac{\text{7,4}}{600} \\ & = \text{0,01233} \\ & \approx \text{0,01} \end{align*}

Using you answer from (a), calculate \(A\) in \(A = P(1+ i)^n\).

\begin{align*} A &= P(1+ i)^n \\ &= \text{200 000}\left(1+ \text{0,01}\right)^{96} \\ &= \text{519 854,59} \end{align*}

Calculate \(A\) without rounding off your answer in (a), compare this answer with your answer in (b).

\begin{align*} A &= P(1+ i)^n \\ A &= \text{200 000}\left(1+ \frac{\text{7,4}}{600}\right)^{96} \\ &= \text{648 768,22} \end{align*}

There is a \(\text{128 913,63}\) difference between the answer in (b) and the one calculated without rounding until the final step.

If it takes \(\text{1}\) person to carry \(\text{3}\) boxes, how many people are needed to carry \(\text{31}\) boxes?

Each person can carry 3 boxes. So we need to divide 31 by 3 to find out how many people are needed to carry 31 boxes.

\[\frac{31}{3} = \text{10,3333...}\]

Therefore \(\text{11}\) people are needed to carry \(\text{31}\) boxes. We cannot have \(\text{0,333}\) of a person so we round up to the nearest whole number.

If \(\text{7}\) tickets cost \(\text{R}\,\text{35,20}\), how much does one ticket cost?

Since 7 tickets cost \(\text{R}\,\text{35,20}\), 1 ticket must cost \(\text{R}\,\text{35,20}\) divided by 7.

\[\frac{\text{35,20}}{7} =\text{5,028571429}\\\]

Therefore one ticket costs \(\text{R}\,\text{5,03}\). Money should be rounded off to \(\text{2}\) decimal places.

1.3 Rational and irrational numbers

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1.5 Estimating surds