Place value, ordering and rounding – Year 7 maths lesson to download as PDF

Example 1

Write down the value of the 5 in 35 072.48 and in 3.052.

In 35 072.48 the 5 is in the thousands column, so it is worth 5 000. In 3.052 the 5 is in the hundredths column, so it is worth 0.05.

Example 2

Put 0.45, 0.405, 0.5 and 0.045 in order, smallest first.

With three decimal places: 0.450, 0.405, 0.500, 0.045. The order is 0.045, 0.405, 0.45, 0.5.

Example 3

Put \(-2.5\), \(-2.05\), 0 and \(-2.55\) in order, smallest first.

For negative numbers, the larger the digits, the smaller the number. The order is \(-2.55\), \(-2.5\), \(-2.05\), 0.

Example 4

Work out \(3.47\times100\) and \(52\div1000\).

\(3.47\times100=347\). \(52\div1000=0.052\).

Example 5

Round 48 726 to the nearest 100 and to the nearest 1 000.

Nearest 100: the tens digit is 2, so round down to 48 700. Nearest 1 000: the hundreds digit is 7, so round up to 49 000.

Example 6

Round 6.7849 to 1, 2 and 3 decimal places.

1 d.p.: the next digit is 8, so 6.8. 2 d.p.: the next digit is 4, so 6.78. 3 d.p.: the next digit is 9, so 6.785.

Example 7

Round 0.003 846 to 2 significant figures and 48 739 to 3 significant figures.

0.003 846: the first two significant figures are 3 and 8, the next digit is 4, so 0.0038. For 48 739: the first three are 4, 8, 7, the next digit is 3, so 48 700.

Example 8

Estimate \(\dfrac{4.8\times 31}{0.52}\).

Round: \(\dfrac{5\times30}{0.5}=\dfrac{150}{0.5}=300\).

Worked example in context

A café sells 148 cups of tea a day for 5 days. Estimate the total number of cups, then find the exact total.

Estimate: \(150\times5=750\). Exact: \(148\times5=740\). The estimate is close, and it tells you that a calculator answer of 7400 or 74 must be wrong.

Example 1

Round 0.006 483 to 2 significant figures.

The zeros at the start are not significant; they only show the size of the number. The first significant figure is 6, and the second is 4. The next digit is 8, which is 5 or more, so the 4 rounds up to 5. The answer is 0.0065. Notice that we keep the zeros after the decimal point, because removing them would change the size of the number completely.

Example 2

Estimate the cost of 28 school jumpers at £19.75 each.

Round each number to one significant figure: 28 becomes 30 and £19.75 becomes £20. The estimate is \(30\times20=£600\). Both numbers were rounded up, so the real cost must be a little less than £600. The exact cost is \(28\times19.75=£553\), which agrees with our expectation. Knowing whether an estimate is too big or too small is just as useful as the estimate itself.

Example 3

Order these numbers from smallest to largest: 0.6, 0.06, 0.66, 0.606.

The easiest way is to give every number the same number of decimal places by adding zeros at the end, which does not change their value: 0.600, 0.060, 0.660 and 0.606. Now we can compare them like whole numbers of thousandths: 60, 600, 606, 660. So the order is 0.06, 0.6, 0.606, 0.66. Comparing the number of digits is a trap: 0.606 has more digits than 0.66 but it is smaller.