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Decimals, Rounding & Approximating

Real answers rarely come out as neat whole numbers. Rounding sensibly, and estimating before you calculate, keeps your work readable and catches mistakes before they cost you.

MYP 4Standard MathsNumberCriteria A · D~9 min read

Written by Fayek Jawad Haider · MYP student

Rounding to decimal places

The number of decimal places (d.p.) is simply how many digits you keep after the decimal point. To round, look at the very next digit along, the first one you are about to drop:

  • if it is 5 or more, round the last kept digit up;
  • if it is 4 or less, leave the last kept digit as it is.

That single "look at the next digit" rule is all rounding ever is.

Worked example

Round \(3.14159\) to 2 decimal places.

1
Keep two digits after the point: \(3.14\,|\,159\). The bar marks where you are cutting.
2
The next digit is 1, which is 4 or less, so the kept digits do not change.
\(3.14159 \approx 3.14\) (2 d.p.)

Watch the carry

When rounding up pushes a 9 over, it carries. Rounding \(4.98\) to 1 d.p. gives \(5.0\), not \(4.10\). Keep the trailing zero so the reader can see you rounded to one place.

Significant figures

Significant figures (s.f.) count the digits that carry real information, starting from the first non-zero digit. Leading zeros, like the ones in \(0.00408\), just show size and do not count.

The rounding rule is exactly the same as before: find your cut-off, then look at the next digit. The only new skill is spotting where the significant figures begin.

Worked example

Round \(0.04628\) to 2 significant figures.

1
The first significant figure is the 4 (leading zeros do not count). The second is the 6.
2
The next digit is 2, which is 4 or less, so we round down and keep the 6.
\(0.04628 \approx 0.046\) (2 s.f.)

For a large number, hold the place value with zeros. Rounding \(45872\) to 2 s.f.: the first two significant figures are 4 and 5, the next digit is 8 (round up), and zeros fill the gap to keep the size, giving \(46000\).

Estimating by rounding

An estimate is a quick, rough answer you get by rounding each number, usually to 1 significant figure, before calculating. It will not be exact, but it tells you the size of answer to expect, which is the best defence against a slipped decimal point or a mis-keyed calculator.

Worked example

Estimate the value of \(4.87 \times 19.6\).

1
Round each number to 1 s.f.: \(4.87 \approx 5\) and \(19.6 \approx 20\).
2
Multiply the rounded values: \(5 \times 20 = 100\).
\(4.87 \times 19.6 \approx 100\) (the exact value is \(95.452\), so the estimate is a good check)

Where this is assessed

Rounding sits in Criterion A, but estimating shows off Criterion D: using maths to judge whether an answer is reasonable. In multi-step problems, round only at the very end, an early round-off error grows as it is carried through.

Check yourself

1. Round \(7.8351\) to 2 decimal places. +

Keep two digits after the point: \(7.83\,|\,51\). The next digit is 5, which is 5 or more, so round the last kept digit up: \(7.8351 \approx \mathbf{7.84}\) (2 d.p.).

2. Round \(0.0708\) to 2 significant figures. +

The first significant figure is 7, the second is 0. The next digit is 8, so round up: the 0 becomes 1, giving \(0.0708 \approx \mathbf{0.071}\) (2 s.f.).

3. Estimate \(612 \div 2.9\) by rounding to 1 significant figure. +

Round each: \(612 \approx 600\) and \(2.9 \approx 3\). Then \(600 \div 3 = \mathbf{200}\). (The exact value is about \(211\), so the estimate lands close.)

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