Estimation

Rounding to decimal places

Rounding replaces a number with a simpler one that is close to it. The question tells you how simple: to 2 decimal places (2 d.p.) means keep two digits after the decimal point.

The method:

1. Count the digits you are keeping and draw a line after the last one.
2. Look at the next digit only, the first one after the line.
3. If it is 5 or more, round up: add 1 to the last digit kept. If it is 4 or less, leave the last digit kept as it is.
4. Drop everything after the line.

$4.6382$ to 2 d.p.: keep $4.63|82$. The next digit is 8, so round up to $4.64$.

Why look at only one digit? You are deciding whether the number is nearer to 4.63 or to 4.64. The halfway point is 4.635. The digit straight after the line tells you which side of halfway you are on; the digits after it cannot change that.

Zeros you must keep. $1.9963$ to 2 d.p. is $2.00$, not 2. Rounding up the 9 makes it 10, which carries into the next place: $1.99\to 2.00$. The two zeros show the accuracy asked for, and leaving them off loses the mark.

Rounding to significant figures

Significant figures (s.f.) count from the first non-zero digit, wherever it is. In $0.007251$ the first significant figure is 7; the zeros in front only show the place value.

The method is the same as for decimal places once you have found where to start counting.

• $0.007251$ to 2 s.f.: count two digits from the 7, $0.0072|51$. The next digit is 5, so round up to $0.0073$.
• $27486$ to 3 s.f.: $274|86$. The next digit is 8, so the 4 becomes a 5. But the number must stay the same size, so the dropped digits become zeros: $27500$, not 275.

This is the most important point about significant figures: rounding never changes the size of a number. $27486$ is about twenty-seven thousand, and so is $27500$. The zeros at the end of $27500$ hold the place value and are not significant; zeros in the middle, as in $3.05$, always are.

$3.0496$ to 3 s.f. is $3.05$. The zero after the point is significant because it comes after the first non-zero digit.

Significant figures or decimal places? Decimal places count from the decimal point, so they suit money or measurements in fixed units. Significant figures count from the first real digit, so they work for any size of number, from $0.0004$ to $4000000$. That is why the exam's default accuracy for a calculator answer is 3 significant figures.

Rounding to the nearest 10, 100 or 1000

"Correct to the nearest thousand" means find the multiple of 1000 that the number is closest to. Look at the hundreds digit: 5 or more rounds up.

The syllabus example: 5764 correct to the nearest thousand. The hundreds digit is 7, so round up: 6000. (To the nearest hundred it would be 5800, and to the nearest ten 5760.)

Notice that 5764 to the nearest thousand and 5764 to 1 significant figure are both 6000. That is not a coincidence: for a four-digit number, the first significant figure is the thousands digit.

Estimating a calculation

An estimate is a quick approximate answer. You make one by rounding every number to 1 significant figure, then doing the simpler calculation in your head or on paper.

Estimates matter for two reasons. On the non-calculator papers you may be asked for one directly. On the calculator papers an estimate is your check: if your calculator says 4.8 and your estimate is 480, a bracket or a decimal point has gone wrong.

Show the rounded numbers. In an estimate question the method mark is for writing each number correct to 1 significant figure. An answer with no working, even a good one, may score only part of the marks.

The answer to an estimate is the value of the rounded calculation, exactly. Do not work out the real answer and round it: that is not what the question asks, and it will not earn the marks.

Dividing by a small number. Rounding often leaves you dividing by something like 0.5 or 0.2. Dividing by 0.5 is the same as multiplying by 2, because there are two halves in every whole. Dividing by 0.2 is multiplying by 5. Use the method from the four operations lesson: $\frac{30}{0.2}=\frac{300}{2}=150$.

Estimates with quantities and measurements

The same idea works for real quantities: lengths, areas, costs, journeys. Round each measurement to 1 significant figure and work with the rounded values, keeping the units.

It is often worth asking whether the estimate is too big or too small. If every number in a multiplication has been rounded up, the estimate is bigger than the true value (an overestimate). If they have all been rounded down it is an underestimate. In a division the bottom number works the other way: rounding the divisor down makes the answer bigger.

When some numbers go up and some go down you cannot tell without working out the real value, and a question will not ask you to.

Rounding answers to a reasonable degree of accuracy

When a question gives a context and says "give your answer to a sensible (or reasonable) degree of accuracy", the context decides.

• Money goes to the nearest cent: 2 decimal places of a dollar. A calculator showing 34.9044 means 34.90 dollars.
• Things that must be whole cannot be fractions, and the question decides whether to round up or down.
• "How many coaches are needed for 213 people, 45 per coach?" $213÷45=4.73\dots$, and 4 coaches leave people behind, so round up to 5.
• "How many 3.85-dollar bags can you buy with 50 dollars?" $50÷3.85=12.98\dots$, and you cannot afford a 13th bag, so round down to 12.
• Measurements should not be given more accurately than the data. If lengths were measured to the nearest centimetre, an area such as $15.8194{\text{m}}^2$ is better given as $15.8{\text{m}}^2$ (3 s.f.).

Without any context, the exam rule is: give a non-exact answer correct to 3 significant figures, and angles to 1 decimal place.

Round only at the end. Keep every digit in your calculator until the final answer. Rounding halfway through, and then rounding again, can change the last digit. $54.549\dots$ to 1 d.p. is $54.5$; but if you first round to $54.55$ and then round that to 1 d.p., you get $54.6$, which is wrong.