International Baccalaureate · MYP 2

Whole Numbers, Rounding and Estimation

⌚ ~146 min ▢ 16 blocks ✓ 69 exercises ➣ Prerequisites : Whole Numbers and Operations, Number Properties, Decimals

I Properties for Mental Calculation

At the shop, Léa buys $4$ posters at $\dollar 19.99$ each. Before she can take out her phone, the shopkeeper announces: “$\dollar 79.96$!” — in two seconds, without any calculator. How does he do it?

Answer

The shopkeeper does not compute $4 \times 19.99$ digit by digit: he replaces $19.99$ by $20 - 0.01$, $$4 \times 19.99 = 4 \times 20 - 4 \times 0.01 = 80 - 0.04 = 79.96.$$ He is allowed to spread the $4$ over the difference — that right and the right to reorder sums and products first are two properties of the operations, stated below.

Proposition — Commutativity and Associativity

In a sum, the terms can be swapped and grouped in any order. In a product, the factors can be swapped and grouped in any order. Choosing pairs that make simple numbers — a whole number, $1$, $10$, $100$ — speeds up mental calculation. $$\begin{aligned} 1.2+27.9+0.8&=(1.2+0.8)+27.9 &&\text{(swap and group the terms that make a whole number)}\\ &=2+27.9\\ &=29.9 \end{aligned}$$ $$\begin{aligned} 2.5\times 6.7\times 4&=(2.5\times 4)\times 6.7 &&\text{(swap and group the factors that make a simple number)}\\ &=10\times 6.7\\ &=67 \end{aligned}$$ These properties hold for addition and multiplication; they do not hold for subtraction or division.

Proposition — Distributivity

Multiplication distributes over addition and subtraction — here the letters $k$, $a$ and $b$ simply stand for any numbers: $$k\times(a+b)=k\times a+k\times b \qquad\qquad k\times(a-b)=k\times a-k\times b$$ A factor close to a simple number can be replaced by a sum or a difference: $$\begin{aligned} 25\times 0.99&=25\times(1-0.01) &&\text{(since } 0.99=1-0.01\text{)}\\ &=25\times 1-25\times 0.01 &&\text{(distributivity)}\\ &=25-0.25\\ &=24.75 \end{aligned}$$

Example

Compute mentally, stating the property used.

$4.8 + 13.5 + 5.2$
$2 \times 8.3 \times 0.5$
$12 \times 1.01$

Answer

Group the terms that make a whole number: $$\begin{aligned} 4.8 + 13.5 + 5.2&=(4.8+5.2)+13.5 &&\text{(commutativity and associativity)}\\ &=10+13.5\\ &=23.5 \end{aligned}$$
Group the factors that make $1$: $$\begin{aligned} 2 \times 8.3 \times 0.5&=(2\times 0.5)\times 8.3 &&\text{(commutativity and associativity)}\\ &=1\times 8.3\\ &=8.3 \end{aligned}$$
Replace $1.01$ with the sum $1+0.01$: $$\begin{aligned} 12 \times 1.01&=12\times(1+0.01) &&\text{(since } 1.01=1+0.01\text{)}\\ &=12+0.12 &&\text{(distributivity)}\\ &=12.12 \end{aligned}$$

Example

Compute cleverly: $7 \times 2.5 + 3 \times 2.5$.

Answer

Both products share the factor $2.5$: the distributivity identity can also be read from right to left, to collect instead of spreading: $$7 \times 2.5 + 3 \times 2.5 = (7 + 3) \times 2.5 = 10 \times 2.5 = 25.$$

Example

Calculate $6 \times (10 + 3)$ in two ways, then $6 \times 10 + 3$. Why do the brackets matter?

Answer

Following the order of operations, the parentheses first: $$\begin{aligned} 6 \times (10 + 3) &= 6 \times 13 &&\text{(parentheses first)}\\ &= 78 \end{aligned}$$ With distributivity: $$\begin{aligned} 6 \times (10 + 3) &= 6 \times 10 + 6 \times 3 &&\text{(distributivity)}\\ &= 60 + 18 &&\text{(the multiplications first)}\\ &= 78 \end{aligned}$$ The same result — and the second expression needs no parentheses: its multiplications come first anyway. But $6 \times 10 + 3 = 60 + 3 = 63$: without parentheses, the $6$ multiplies only the $10$. Distributivity turns a product whose factor is in parentheses into a sum of products, which needs none.

Skills to practice

Reordering Sums and Products
Using Distributivity with Whole Numbers
Using Distributivity
Solving Real-World Problems

II Significant Figures

Discover

The final of a football cup was watched by $48\,736$ spectators in the stadium. The next day, a radio station announced “about $49\,000$ spectators”, a newspaper wrote “nearly $50\,000$ spectators”, and the website of the club “$48\,700$ fans”.
How many digits of $48\,736$ did each of them keep? Why did none of them simply write $49$?

Answer

Each of them rounded $48\,736$, keeping its first digits and replacing the others by zeros.

The radio kept two digits: $48\,736$ is closer to $49$ thousands than to $48$ thousands, since the next digit is $7$.
The newspaper kept one digit: $4$ ten-thousands became $5$ ten-thousands, since the next digit is $8$.
The website kept three digits: $487$ hundreds, since the next digit is $3$.

The zeros written after these digits are not part of the count: they keep the number at the right size. The number $49$ alone would mean forty-nine spectators, a thousand times fewer. The digits kept, counted from the first non-zero digit, are called the significant figures of the rounded value.

Definition — Significant Figures

The significant figures of a number are the digits that give its precision, counted from the first non-zero digit, reading from left to right, by the rules below; the other zeros only put the digits in their places and keep the size of the number.

Every non-zero digit is significant.
A zero between two non-zero digits is significant: $307$ and $4.05$ each have $3$ significant figures.
The zeros before the first non-zero digit are place holders, never significant: $0.0038$ has $2$ significant figures, $3$ and $8$.
A zero at the end of the decimal part is significant: $2.50$ has $3$ significant figures — its last zero shows that the value is given to the hundredth.
The zeros at the end of a whole number, as in $5\,300$, are read as place holders: $5\,300$ has $2$ significant figures, unless it is stated otherwise — “$5\,300$, to $3$ significant figures” makes its first zero significant.

“Significant figures” is often shortened to “s.f.”. Rounding to $2$ decimal places keeps $2$ digits after the decimal point, whatever the size of the number; rounding to $2$ significant figures keeps $2$ digits counted from the first non-zero digit, wherever the decimal point is.

Example

Give the number of significant figures of each number.

$4\,072$
$0.0306$
$5.20$
$250\,000$

Answer

The significant figures are counted from the first non-zero digit; they are coloured below.

$\textcolor{colordef}{4\,072}$ has $4$ significant figures: the zero lies between the non-zero digits $4$ and $7$.
$0.0\textcolor{colordef}{306}$ has $3$ significant figures: the first non-zero digit is $3$; the two zeros before it are place holders, and the zero between $3$ and $6$ is significant.
$\textcolor{colordef}{5.20}$ has $3$ significant figures: the final zero of the decimal part is significant.
$\textcolor{colordef}{25}0\,000$ has $2$ significant figures: nothing states otherwise, so the zeros at the end of the whole number are read as place holders.

Skills to practice

Deciding Whether a Zero Is Significant
Counting Significant Figures

III Rounding to Significant Figures

Method — Rounding to a Number of Significant Figures

To round a number to $n$ significant figures:

Find the first significant figure: the first non-zero digit from the left.
Count $n$ figures from it and underline the $n$-th one: it is the last figure kept.
Look at the next digit. If it is $0$, $1$, $2$, $3$ or $4$, the underlined figure stays the same; if it is $5$, $6$, $7$, $8$ or $9$, the underlined figure increases by $1$ — a $9$ then becomes $0$ and carries $1$ to the digit on its left.
Keep the size of the number: before the decimal point, replace each dropped digit by a zero; after the decimal point, drop the digits. A zero that is one of the $n$ figures is written, as in $3.997 \approx 4.00$ to $3$ s.f.

The rounded value follows the sign $\approx$, with the accuracy stated: $48\,736 \approx 49\,000$ (to $2$ s.f.).

Example

Round $48\,736$ to $1$, to $2$ and to $3$ significant figures.

Answer

The first significant figure is the $4$ of the ten-thousands.

To $1$ s.f.: in $\underline{4}8\,736$ the next digit is $8$, so the $4$ becomes $5$, and the four dropped digits become zeros: $48\,736 \approx 50\,000$.
To $2$ s.f.: in $4\underline{8}\,736$ the next digit is $7$, so the $8$ becomes $9$: $48\,736 \approx 49\,000$.
To $3$ s.f.: in $48\,\underline{7}36$ the next digit is $3$, so the $7$ stays: $48\,736 \approx 48\,700$.

The three rounded values are close to $48\,736$ and have its size: five digits before the decimal point.

Example

Round each number as indicated.

$0.004567$ to $2$ significant figures
$2.3961$ to $3$ significant figures
$0.0997$ to $2$ significant figures

Answer

The first significant figure is $4$, in the thousandths. In $0.004\underline{5}67$ the next digit is $6$, so the $5$ becomes $6$: $0.004567 \approx 0.0046$. The zeros after the decimal point stay: they place the $4$ and the $6$.
The figures kept are $2$, $3$ and $9$. In $2.3\underline{9}61$ the next digit is $6$, so the $9$ becomes $0$ and carries $1$ to the $3$, which becomes $4$: $2.3961 \approx 2.40$. The final $0$ is the third significant figure: it is written.
The first significant figure is the $9$ of the hundredths. In $0.09\underline{9}7$ the next digit is $7$: the underlined $9$ becomes $0$ and carries $1$, the other $9$ becomes $0$ and carries $1$ to the tenths: $0.0997 \approx 0.10$. The carry has moved the first significant figure: the two figures are now the $1$ of the tenths and the $0$ of the hundredths.

Example

Round $3\,846.2$ and $0.00483$ to $2$ decimal places, then to $2$ significant figures. Which of the two kinds of rounding keeps useful information about $0.00483$?

Answer

To $2$ decimal places, two digits are kept after the decimal point: $3\,846.2$ has only one decimal place, so it is written $3\,846.20$ with no rounding; $0.00483 \approx 0.00$, since the digit after the hundredths is $4$.
To $2$ significant figures, two digits are kept from the first non-zero digit: $3\,846.2 \approx 3\,800$ (next digit $4$) and $0.00483 \approx 0.0048$ (next digit $3$).

Rounded to $2$ decimal places, $0.00483$ becomes $0.00$: all its digits are lost. Rounded to $2$ significant figures, it keeps its size and its first two figures: significant figures suit numbers of every size, while decimal places suit values measured in the same unit, such as amounts of money to the cent.

Rounding Keeps the Size of the Number

The most frequent error is to forget the place holders. Rounded to $2$ significant figures, $48\,736$ is $49\,000$, not $49$, which is a thousand times smaller; and $0.004567$ is $0.0046$, not $46$. A rounded value is always close to the number it replaces: comparing their sizes, as with an order of magnitude, catches the error at once.

Report an error

Skills to practice

Checking a Rounding
Rounding Whole Numbers
Rounding Decimals and Carrying

IV Orders of Magnitude

Definition — Order of Magnitude

An order of magnitude of a number is a simple number close to it: a whole number, or a multiple of $10$, of $100$, of $1\,000$, …

$20$ is an order of magnitude of $19.8$.
$500$ is an order of magnitude of $487.25$.
$3$ is an order of magnitude of $2.95$.

An order of magnitude of a result is obtained by computing with orders of magnitude of the numbers. It does not give the exact result: it gives its size.

Method — Estimating with Orders of Magnitude

Step 1: Round each number to a simple nearby number — often the nearest whole number; for a number smaller than $1$, keep its first non-zero digit: $0.29 \approx 0.3$ and $0.048 \approx 0.05$.
Step 2: Compute mentally with the rounded numbers.
Step 3: Conclude. The result obtained is an order of magnitude: the exact result is close to it.

Example

Give an order of magnitude of each result.

$503.7 + 98.2$
$41.6 \times 1.9$

Answer

For the sum: $$\begin{aligned} 503.7 + 98.2 &\approx 500 + 100 &&\text{(orders of magnitude of each number)}\\ &= 600 &&\text{(mental addition)} \end{aligned}$$ The exact sum, $601.9$, is indeed close to $600$.
For the product: $$\begin{aligned} 41.6 \times 1.9 &\approx 40 \times 2 &&\text{(orders of magnitude of each number)}\\ &= 80 &&\text{(mental multiplication)} \end{aligned}$$ The exact product, $79.04$, is indeed close to $80$.

Skills to practice

Placing the Decimal Point

V One-Figure Estimates

Method — Making a One-Figure Estimate

A one-figure estimate of a calculation gives its size quickly, without a calculator.

Round every number to $1$ significant figure: $7\,235 \approx 7\,000$, $0.48 \approx 0.5$, $0.032 \approx 0.03$.
Compute mentally with the rounded numbers.
Write the result after the sign $\approx$: it is an estimate, and in most calculations the exact result has the same size.
Check a calculator result against the estimate: the two must have the same size. A result about $10$ or $100$ times too large or too small reveals a keying error — often a misplaced decimal point: redo the calculation.

The estimate can mislead when two close numbers are subtracted: $704 - 696 \approx 700 - 700 = 0$, while the exact difference is $8$. Such a difference is computed exactly rather than estimated.

Example

Make a one-figure estimate of each calculation.

$7\,235 \times 591$
$38.7 \times 0.52$
$\dfrac{812}{0.39}$

Answer

$$\begin{aligned}[t] 7\,235 \times 591 &\approx 7\,000 \times 600 && \text{(each number to 1 s.f.)}\\ &= 4\,200\,000. \end{aligned}$$ The exact product, $4\,275\,885$, is indeed about four million.
$$\begin{aligned}[t] 38.7 \times 0.52 &\approx 40 \times 0.5 && \text{(each number to 1 s.f.)}\\ &= 20. \end{aligned}$$ The exact product is $20.124$.
$$\begin{aligned}[t] \frac{812}{0.39} &\approx \frac{800}{0.4} && \text{(each number to 1 s.f.)}\\ &= \frac{8\,000}{4} && \text{(both multiplied by 10: the quotient does not change)}\\ &= 2\,000. \end{aligned}$$ The exact quotient is $2\,082.05\ldots$ Dividing by a number smaller than $1$ gives a result larger than the number divided.

Example

Tom computed $4.85 \times 61.3$ with a calculator and read $2\,973.05$.
Check his result with a one-figure estimate.

Answer

$$\begin{aligned}[t] 4.85 \times 61.3 &\approx 5 \times 60 && \text{(each number to 1 s.f.)}\\ &= 300. \end{aligned}$$ The result $2\,973.05$ is about ten times too large: it cannot be correct. Tom probably typed $48.5$ instead of $4.85$, since $48.5 \times 61.3 = 2\,973.05$. Done again, the calculation gives $4.85 \times 61.3 = 297.305$, which has the size of the estimate $300$.

Skills to practice

Choosing the Correct Size
Making One-Figure Estimates

VI Giving an Answer to Three Significant Figures

When to Give an Answer to Three Significant Figures

When a calculation gives a short exact value, such as $15$ or $2.75$, the answer is written in full. When the result goes on — a division that does not end, such as $20 \div 7 = 2.857142\ldots$ — and the question does not say how to round, a common convention is to give the answer to $3$ significant figures: $20 \div 7 \approx 2.86$. Three figures are enough for most uses; for a result computed from measurements, more figures would even claim a precision that the measurements do not have. Two rules go with this convention:

keep the full calculator value during the calculation, and round only the final answer;
give an amount of money to the cent, that is to $2$ decimal places, rather than to $3$ significant figures.

Report an error

Example

Compute $10 \div 3 \times 300$ in two ways.

Round $10 \div 3$ to $3$ significant figures first, then multiply by $300$.
Keep the exact value of $10 \div 3$ until the end.

Answer

$10 \div 3 = 3.333\ldots \approx 3.33$, then $3.33 \times 300 = 999$.
$$\begin{aligned}[t] 10 \div 3 \times 300 &= \frac{10}{3} \times 300\\ &= \frac{3\,000}{3} && \text{(multiply the numerator by 300)}\\ &= 1\,000. \end{aligned}$$

The exact result is $1\,000$. Rounding in the middle of the calculation gave $999$: the small error of the rounding, $0.00333\ldots$, was multiplied by $300$. The rounding is done once, on the final answer.

Skills to practice

Deciding How to Round an Answer
Rounding a Final Answer

VII Going further

Skills to practice

Solving Problems with Rounding and Estimation
Investigating Rounding
Quiz