1.1 — Arithmetic

Arithmetic is the foundation of all mathematics in EASA Part 66. This section covers the essential number skills you will use throughout your career in aircraft maintenance — from calculating fuel loads and torque values to converting between measurement systems. Category A requires Level 1 (basic familiarisation); categories B1, B2, B2L, and B3 require Level 2 (general knowledge, able to apply).

Module 1 is examined by multiple-choice questions with three options, of which exactly one is correct, and the pass mark is 75%. That format shapes how you should study arithmetic. The two wrong options are almost never random numbers: they are the answers you get from the specific slips a candidate under time pressure actually makes — dividing by 45 instead of by 0.75, adding numerators and denominators separately, dividing square centimetres by 100 instead of 10,000, or reading a fuel figure in pounds as though it were kilograms. Every section below therefore names the slip as well as the method, because recognising a wrong option for what it is will save you as many marks as knowing the right one.

Four habits do most of the work, and they carry through into the hangar as well as the exam room:

• Read the units before the numbers. Most arithmetic errors in aviation are unit errors wearing an arithmetic disguise. Decide what unit the answer must come out in before you start.
• Estimate first. A rough answer obtained in five seconds tells you the order of magnitude, and the order of magnitude is what a misplaced decimal point destroys.
• Carry full precision, round once at the end. Rounding part-way through and then continuing compounds the error, and on a multi-step conversion it can move the answer far enough to select the wrong option.
• Answer the question that was asked. A calculation can be flawless and still score zero if the question wanted litres and you gave kilograms, or wanted the radius and you gave the diameter.

The arithmetic here is not self-contained knowledge you use once and forget. It is the machinery every later module runs on: mass and balance moments in Module 7, Ohm's law and power in Module 3, stress and strain in Module 2, gear and pulley ratios in that same module, blade angles and areas in Module 17. Getting fluent here means you are never doing two hard things at once later, and it is the reason candidates who are shaky on Module 1 tend to be shaky everywhere.

Arithmetical Terms and Signs

You must know the standard mathematical symbols and the terms associated with each operation.

Symbol	Meaning	Example
\( + \)	Addition (sum)	\( 8 + 5 = 13 \)
\( - \)	Subtraction (difference)	\( 15 - 7 = 8 \)
\( \times \)	Multiplication (product)	\( 6 \times 4 = 24 \)
\( \div \)	Division (quotient)	\( 20 \div 5 = 4 \)
\( = \)	Equals	\( 3 + 4 = 7 \)
\( \neq \)	Not equal to	\( 5 \neq 6 \)
\( > \)	Greater than	\( 10 > 7 \)
\( < \)	Less than	\( 3 < 8 \)
\( \geq \)	Greater than or equal to	\( x \geq 0 \)
\( \leq \)	Less than or equal to	\( T \leq 100 \)
\( \approx \)	Approximately equal to	\( \pi \approx 3.14 \)
\( \pm \)	Plus or minus (tolerance)	\( 50 \pm 2 \) mm

Key terminology: In addition, the numbers being added are addends and the result is the sum. In subtraction, you subtract the subtrahend from the minuend to get the difference. In multiplication, the numbers are factors and the result is the product. In division, the dividend is divided by the divisor to give the quotient and possibly a remainder.

Further Symbols You Will Meet

The symbols above are the arithmetic core. The following appear constantly in maintenance data, formulae and exam stems, and are worth recognising on sight.

Symbol	Meaning	Example
\( \sqrt{a} \)	Square root	\( \sqrt{49} = 7 \)
\( \sqrt[3]{a} \)	Cube root	\( \sqrt[3]{27} = 3 \)
\( a^n \)	Index, exponent or power	\( 4^3 = 64 \)
\( \lvert a \rvert \)	Absolute value (size, sign discarded)	\( \lvert -6 \rvert = 6 \)
\( \% \)	Per cent (parts per hundred)	\( 25\% = 0.25 \)
\( a : b \)	Ratio of \( a \) to \( b \)	\( 20 : 60 = 1 : 3 \)
\( \sum \)	The sum of all the values	\( \sum x = x_1 + x_2 + x_3 \)
\( \propto \)	Is proportional to	\( A \propto r^2 \)
\( \therefore \)	Therefore (introduces a conclusion)	\( \therefore V = 12 \)
\( (\;) \; [\;] \; \{\;\} \)	Brackets: evaluate the contents first	\( 2 \times (3 + 4) = 14 \)

Round, square and curly brackets all mean exactly the same thing. Different shapes are used only so that the eye can pair them up when brackets are nested inside brackets; there is no rule that says one shape is worked before another, and the innermost pair is always evaluated first whatever its shape.

Signed (Directed) Numbers

A number can carry a direction as well as a size. On a number line, positive numbers run to the right of zero and negative numbers to the left, so −20 is smaller than −5 even though 20 is larger than 5. This catches candidates out with temperatures: an OAT of −40 °C is colder than −15 °C, and a limit written as \( T \ge -55\,{}^\circ\text{C} \) permits −50 °C but not −60 °C.

Keep the two uses of the minus sign apart. In \( 7 - 2 \) the sign is an operation: take 2 away. In \( -2 \) it is part of the number: two units to the left of zero. When both appear together the brackets tell you which is which, as in \( 7 + (-2) \).

Situation	Rule	Example
Adding a negative	Becomes a subtraction	\( 7 + (-2) = 5 \)
Subtracting a negative	Becomes an addition	\( (+3) - (-4) = +7 \)
Multiply or divide, signs alike	Result is positive	\( (-4) \times (-3) = +12 \)
Multiply or divide, signs unlike	Result is negative	\( (-4) \times 3 = -12 \)
Negative base, even power	Result is positive	\( (-2)^4 = 16 \)
Negative base, odd power	Result is negative	\( (-2)^3 = -8 \)

The "two like signs give a positive" rule is not an arbitrary convention to be memorised. Read \( (-4) \times (-3) \) as "the opposite of four lots of negative three". Four lots of \( -3 \) is \( -12 \); the opposite of \( -12 \) is \( +12 \). The same reasoning explains subtracting a negative: taking away a debt leaves you better off, so \( (+3) - (-4) \) moves you four units further to the right, to \( +7 \).

The Order of Operations

An expression such as \( 8 + 4 \times 3 \) has only one correct value, because there is a fixed order in which operations are carried out. Work it left to right and you get 36; obey the order and you get 20. The convention, from first to last, is:

• Brackets — innermost pair first, working outwards.
• Orders — powers and roots.
• Division and multiplication — these rank equally, worked left to right.
• Addition and subtraction — these also rank equally, worked left to right.

You will meet this as BODMAS, BIDMAS or PEMDAS. These are three mnemonics for one and the same convention, not three competing ones: they differ only in the word chosen for brackets (Brackets or Parentheses) and for powers (Orders, Indices or Exponents). Any of them, applied properly, gives the same answer.

Three symbols group their contents as though brackets were printed around them, and forgetting this is a reliable source of wrong answers:

• A fraction bar groups the whole numerator and the whole denominator. In \( \frac{6 + 4}{2} \) the addition is completed first, giving \( \frac{10}{2} = 5 \) — not \( 6 + 2 \).
• A radical sign groups everything under it, so \( \sqrt{25 + 144} \) means the root of 169, which is 13. It is not the sum of the two separate roots, a point returned to in the section on roots below.
• A number written directly against a bracket means multiply. \( 2(5 - 3) \) is \( 2 \times 2 = 4 \), and the multiplication cannot happen until the bracket is resolved.

The Laws of Arithmetic

Three laws govern which rearrangements of an expression are legitimate. They matter because rearranging is how you make arithmetic easy enough to do reliably in your head or on paper.

• Commutative law: order does not matter for addition or multiplication. \( a + b = b + a \) and \( a \times b = b \times a \). It does not hold for subtraction or division: \( 7 - 3 = 4 \) but \( 3 - 7 = -4 \), and \( 8 \div 2 = 4 \) but \( 2 \div 8 = 0.25 \).
• Associative law: grouping does not matter for addition or multiplication. \( (a + b) + c = a + (b + c) \). This is why a column of masses can be totalled in any order — useful when you want to pair figures that add to round numbers.
• Distributive law: \( a(b + c) = ab + ac \). Multiplication distributes across a sum, which is the reason long multiplication works at all.

The distributive law is the most useful of the three in practice. To find \( 98 \times 7 \), rewrite 98 as \( (100 - 2) \): \( 100 \times 7 = 700 \), \( 2 \times 7 = 14 \), so the answer is \( 700 - 14 = 686 \). To find \( 7 \times 43 \), split the 43: \( 7 \times 40 = 280 \) and \( 7 \times 3 = 21 \), giving \( 301 \). Neither needs paper.

Two special numbers complete the picture. Adding zero or multiplying by one changes nothing, so both can be inserted anywhere they are convenient — a fact used heavily in unit conversion, where you multiply by a fraction whose value is exactly 1. Multiplying by zero always gives zero. Dividing by zero is not defined and has no answer: a calculator will report an error, and in a formula it signals that the quantity in the denominator cannot physically be zero.

Kinds of Number

• Natural numbers: the counting numbers 1, 2, 3, ... — rivets, blades, cylinders.
• Integers: the whole numbers including zero and the negatives: ..., −2, −1, 0, 1, 2, ...
• Rational numbers: anything that can be written as one integer divided by another, such as \( \frac{3}{4} \) or \( \frac{7}{6} \). Every terminating decimal and every recurring decimal is rational.
• Irrational numbers: those that cannot. Their decimal expansion never terminates and never settles into a repeating pattern. \( \pi \) and \( \sqrt{2} \) are the two you will meet most.
• Real numbers: the rationals and irrationals together — every value that can appear on a number line.

This has a practical consequence rather than being a piece of classification for its own sake. Every measurement you can take is rational, because an instrument reads to a finite number of figures. Any answer that involves \( \pi \) or a square root is therefore an approximation the moment you write it down as a decimal, and how far you take it decides how much rounding error you carry forward.

Methods of Multiplication and Division

Long Multiplication

To multiply large numbers, break one factor into its place-value components and add the partial products.

Long Division

Long division breaks a division problem into a sequence of smaller steps: divide, multiply, subtract, bring down.

Estimation and Cross-checking

Always estimate before calculating. Round each number to a convenient nearby value: \( 247 \times 36 \approx 250 \times 40 = 10{,}000 \). Your exact answer of 8,892 is reasonable. If you got 88,920, you would know something was wrong.

Why Long Multiplication Works

Splitting a factor into partial products is not a trick; it is the distributive law applied to place value. The multiplier 36 is worth \( 30 + 6 \), so

Understanding it this way tells you where the zeros come from. The row for the tens digit is shifted one place left because you are multiplying by 30, not by 3; the row for a hundreds digit is shifted two places, because you are multiplying by 300. Candidates who memorise "put a zero on the second line" get the third line wrong the first time they meet a three-digit multiplier.

Multiplying and Dividing by Powers of Ten

Because our number system is base 10, multiplying by \( 10^n \) moves the decimal point \( n \) places to the right and dividing by \( 10^n \) moves it \( n \) places to the left. Nothing else changes: the digits keep their order and only their place value shifts. So \( 4.85 \times 1000 = 4850 \) and \( 4.85 \div 100 = 0.0485 \). Every metric conversion in this module is an application of that one fact, which is why metric arithmetic is so much less error-prone than imperial.

Multiplying Decimals

Ignore the decimal points, multiply the digits as whole numbers, then place the point so that the answer has as many decimal places as the two factors had between them. The rule follows directly from writing the decimals as fractions:

Dividing by a Decimal

Long division is only comfortable when the divisor is a whole number, so make it one. Multiply the divisor and the dividend by the same power of ten — that leaves the quotient unchanged, because you have multiplied top and bottom of the same fraction by the same thing.

Division That Does Not Come Out Exactly

The long-division example above divides exactly. Most real ones do not, and you then have a choice about how to express what is left over — a choice the question usually decides for you.

Short Cuts Worth Knowing

Each of these is the distributive or associative law in disguise, and each turns a written calculation into a mental one.

To do this	Do this instead	Example
\( \times 5 \)	\( \times 10 \), then halve	\( 86 \times 5 = \frac{860}{2} = 430 \)
\( \div 5 \)	Double, then \( \div 10 \)	\( 93 \div 5 = \frac{186}{10} = 18.6 \)
\( \times 25 \)	\( \times 100 \), then \( \div 4 \)	\( 36 \times 25 = \frac{3600}{4} = 900 \)
\( \times 9 \)	\( \times 10 \), then subtract the number	\( 47 \times 9 = 470 - 47 = 423 \)
\( \times 11 \) (two digits)	Split the digits, insert their sum	\( 43 \times 11 = 4\,|\,7\,|\,3 = 473 \)
Awkward pair	Halve one factor, double the other	\( 16 \times 35 = 8 \times 70 = 560 \)

Checking an Answer Three Ways

Estimation, covered above, is the first line of defence. Two further checks cost seconds and between them catch most of what estimation misses.

Bracket the answer. Instead of one estimate, produce a deliberate under-estimate and a deliberate over-estimate, and confirm the exact answer sits between them. For \( 384 \times 27 \), rounding both factors down gives \( 380 \times 20 = 7{,}600 \) and rounding both up gives \( 390 \times 30 = 11{,}700 \). The exact answer, \( 384 \times 27 = 10{,}368 \), lies inside that band, so it is credible. A bracket is stronger than a single estimate because it tells you not only the order of magnitude but whether you are in the right part of it.

Reverse the operation. A product divided by one of its factors must return the other; a quotient multiplied by the divisor must return the dividend. This is the only check that verifies the answer exactly rather than approximately, and it is worth the extra few seconds on any figure you are about to enter on a worksheet.

Cast out nines. Replace each number by its repeated digit sum, do the same arithmetic on those small numbers, and the answers must agree. For \( 247 \times 53 = 13{,}091 \): \( 2+4+7 = 13 \rightarrow 4 \), and \( 5+3 = 8 \); \( 4 \times 8 = 32 \rightarrow 5 \). The product's digits give \( 1+3+0+9+1 = 14 \rightarrow 5 \), which matches. Had you written 13,081 the digit sum would be 4, and the error would be exposed at once.

Fractions and Decimals

Types of Fractions

• Proper fraction: numerator < denominator, e.g. \( \frac{3}{4} \)
• Improper fraction: numerator ≥ denominator, e.g. \( \frac{7}{4} \)
• Mixed number: whole number + fraction, e.g. \( 1\frac{3}{4} \)

Operations with Fractions

Decimals

Decimals are another way of expressing fractions in base 10. The digit positions to the right of the decimal point represent tenths, hundredths, thousandths, etc.

Converting fractions to decimals: Divide the numerator by the denominator. For example, \( \frac{3}{8} = 3 \div 8 = 0.375 \).

Converting decimals to fractions: Write the decimal over the appropriate power of 10 and simplify. For example, \( 0.625 = \frac{625}{1000} = \frac{5}{8} \).

Fraction	Decimal	Fraction	Decimal
\( \frac{1}{2} \)	0.5	\( \frac{1}{8} \)	0.125
\( \frac{1}{3} \)	0.333...	\( \frac{3}{8} \)	0.375
\( \frac{1}{4} \)	0.25	\( \frac{5}{8} \)	0.625
\( \frac{3}{4} \)	0.75	\( \frac{7}{8} \)	0.875
\( \frac{1}{5} \)	0.2	\( \frac{1}{16} \)	0.0625

Decimal Places and Significant Figures

Decimal places (d.p.): the number of digits after the decimal point. For example, 3.142 has 3 d.p.

Significant figures (s.f.): count all digits starting from the first non-zero digit. For example, 0.00456 has 3 s.f.; 3,050 has 3 s.f. (the trailing zero may or may not be significant depending on context).

Equivalent Fractions and Lowest Terms

A fraction is a division waiting to happen, and multiplying or dividing both parts of it by the same number leaves its value untouched — because that is the same as multiplying the whole fraction by 1. So \( \frac{3}{4} = \frac{6}{8} = \frac{9}{12} = \frac{75}{100} \); all four are the same quantity written with different-sized pieces.

Going the other way is cancelling, and dividing top and bottom by their highest common factor puts the fraction in its lowest terms in one move. Take \( \frac{84}{126} \): since \( 84 = 2^2 \times 3 \times 7 \) and \( 126 = 2 \times 3^2 \times 7 \), their highest common factor is \( 2 \times 3 \times 7 = 42 \), and \( \frac{84 \div 42}{126 \div 42} = \frac{2}{3} \). Cancelling in smaller steps reaches the same place, just more slowly. An answer left uncancelled is not wrong, but an exam option list will normally offer only the cancelled form, so you must be able to recognise it.

To decide which of two fractions is larger without converting either to a decimal, cross-multiply: compare \( \frac{5}{8} \) with \( \frac{7}{11} \) by forming \( 5 \times 11 = 55 \) against \( 7 \times 8 = 56 \). The larger product sits above the larger fraction, so \( \frac{7}{11} \) is the bigger of the two. This works because you are effectively putting both over the common denominator \( 8 \times 11 \) without bothering to write it.

Choosing the Denominator When You Add

The addition formula above always works, because \( bd \) is always a common denominator. It is not always the lowest one, so the answer it produces often needs cancelling afterwards. Using the lowest common multiple of the denominators instead keeps the numbers small and usually delivers the answer already in lowest terms.

Mixed Numbers in a Calculation

The safe method with mixed numbers is to convert every term to an improper fraction first, work entirely in improper fractions, and convert back only at the very end. To convert, multiply the whole number by the denominator and add the numerator: \( 4\frac{3}{8} = \frac{(4 \times 8) + 3}{8} = \frac{35}{8} \). To convert back, divide and keep the remainder over the denominator.

Subtraction of mixed numbers is where the converting-first rule pays for itself. \( 5\frac{1}{4} - 2\frac{3}{4} \) cannot be done by taking the fractions apart, because \( \frac{1}{4} \) is smaller than \( \frac{3}{4} \). As improper fractions it is simply \( \frac{21}{4} - \frac{11}{4} = \frac{10}{4} = 2\frac{1}{2} \). If you prefer to keep the whole numbers separate you must borrow one whole from the 5, rewriting it as \( 4\frac{5}{4} \), and then \( 4\frac{5}{4} - 2\frac{3}{4} = 2\frac{2}{4} = 2\frac{1}{2} \). Both reach the same answer; the first route has fewer places to go wrong.

Multiplying and Dividing: Cancel Before You Multiply

Multiplication of fractions needs no common denominator, and cancelling diagonally before multiplying keeps the numbers small. In \( \frac{3}{8} \times \frac{4}{9} \), the 4 and the 8 share a factor of 4, and the 3 and the 9 share a factor of 3, leaving \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \). Multiplying first gives \( \frac{12}{72} \), which cancels to the same \( \frac{1}{6} \) with more effort.

The word "of" means multiply. "Three quarters of 82" is \( \frac{3}{4} \times 82 \), worked in the section above. This is worth fixing in your mind, because most percentage and ratio questions are phrased with "of" rather than with a multiplication sign.

Dividing by a fraction is done by inverting the second fraction and multiplying, and the reason is worth knowing rather than memorising: \( \frac{c}{d} \times \frac{d}{c} = 1 \), so multiplying by \( \frac{d}{c} \) exactly undoes a multiplication by \( \frac{c}{d} \). Read \( \frac{3}{4} \div \frac{2}{5} \) as "how many two-fifths fit into three-quarters"; the answer \( \frac{3}{4} \times \frac{5}{2} = \frac{15}{8} = 1\frac{7}{8} \) says a bit less than two, and the reason is worth being precise about: \( \frac{2}{5} \) is smaller than \( \frac{3}{4} \), so at least one of them fits, but it is also more than half of \( \frac{3}{4} \) — half of three-quarters is \( \frac{3}{8} = 0.375 \), while \( \frac{2}{5} = 0.4 \) — so a second one cannot fit and the answer has to land between one and two.

Terminating and Recurring Decimals

Dividing the numerator by the denominator either stops or repeats — it can never wander on at random, because there are only so many possible remainders. Which of the two happens is completely predictable:

A recurring decimal is marked with a dot over the repeating digit, or over the first and last digits of a repeating group: \( \frac{1}{3} = 0.\dot{3} \), \( \frac{1}{6} = 0.1\dot{6} \), and \( \frac{1}{7} = 0.\dot{1}4285\dot{7} \), whose block of six digits then repeats for ever. A mixed number behaves the same way: \( \frac{7}{6} = 1 + \frac{1}{6} = 1.1\dot{6} \), a whole 1 followed by the recurring sixth. Recurring decimals are still rational numbers — they came from a fraction and can be turned back into one.

Turning a recurring decimal back into a fraction is a short piece of arithmetic rather than a lookup. Call the decimal \( x \), multiply it by whichever power of ten shifts one full repeating block to the left of the point, then subtract the original — the endless tails are identical and cancel exactly. For \( x = 0.\dot{3} \): \( 10x = 3.\dot{3} \), so \( 10x - x = 3 \), giving \( 9x = 3 \) and \( x = \frac{3}{9} = \frac{1}{3} \). For a two-digit block, \( x = 0.\dot{3}\dot{6} \): \( 100x = 36.\dot{3}\dot{6} \), so \( 99x = 36 \) and \( x = \frac{36}{99} = \frac{4}{11} \). Check the second one by dividing 4 by 11, which returns 0.363636... as it should.

Rounding: How and When

Rounding to a stated number of decimal places means looking at the digit one place beyond the last one you are keeping. If it is 5 or more, increase the last kept digit by one; if it is 4 or less, leave it alone. So \( \pi = 3.14159\ldots \) is 3.14 to 2 d.p. and 3.142 to 3 d.p.

Rounding to a stated number of significant figures works identically, except that you start counting from the first non-zero digit rather than from the decimal point. So 0.0045678 is 0.00457 to 3 s.f., and 47,382 is 47,000 to 2 s.f. or 47,400 to 3 s.f. The zeros that replace discarded digits in a whole number are placeholders, not new information.

Simply cutting off the surplus digits is truncation, not rounding, and it gives a different result: 2.678 truncates to 2.67 but rounds to 2.68. When a question asks for an answer "to two decimal places" it means rounding.

Significant Figures as a Statement About Precision

Significant figures are not only a rounding convention; they are how a written number declares how well it is known. The rules for counting them are:

• Leading zeros are never significant. They only position the decimal point, which is why 0.00456 carries 3 s.f. and not 5.
• Zeros between non-zero digits always are. 3.05 has 3 s.f.
• Trailing zeros after a decimal point are significant. 2.50 mm has 3 s.f. and says the measurement was taken to hundredths of a millimetre; 2.5 mm has 2 s.f. and does not.
• Trailing zeros in a whole number are ambiguous unless the context settles it — which is exactly why scientific notation exists.

Writing a number in the form \( a \times 10^n \) removes the ambiguity completely, because only the digits in \( a \) are shown: \( 3.05 \times 10^3 \) is 3 s.f. while \( 3.050 \times 10^3 \) is 4 s.f., and both are the value 3,050. The mechanics of that notation are covered in the final section of this note.

Two conventions govern how precision survives a calculation, and they are not the same rule:

• For multiplication and division, quote the answer to the number of significant figures of the least precise input. \( 12.34 \times 2.1 = 25.914 \), but 2.1 carries only 2 s.f., so the honest answer is 26.
• For addition and subtraction, quote the answer to the number of decimal places of the least precise input. \( 12.34 + 2.1 = 14.44 \), quoted as 14.4.

The practical version of this is that a calculator's ten digits are not ten digits of knowledge. A dimension read from a steel rule and one read from a micrometer are different measurements even when both happen to display as 2.5 mm, and multiplying a rule reading by a micrometer reading cannot produce micrometer-grade accuracy. Quoting more figures than your least precise instrument justifies is a claim you cannot support.

Fractions of an Inch in the Hangar

Imperial drill, reamer and fastener sizes step in halves, quarters, eighths, sixteenths, thirty-seconds and sixty-fourths of an inch, so converting a fraction to a decimal and on to millimetres is routine work rather than an exam exercise. The millimetre column below is exact, because the inch is defined as exactly 25.4 mm — the conversion arithmetic itself is set out in the units section.

Fraction of an inch	Decimal inches	Millimetres
\( \frac{1}{16} \)	0.0625	1.5875
\( \frac{3}{16} \)	0.1875	4.7625
\( \frac{5}{16} \)	0.3125	7.9375
\( \frac{7}{16} \)	0.4375	11.1125
\( \frac{9}{16} \)	0.5625	14.2875
\( \frac{11}{16} \)	0.6875	17.4625
\( \frac{13}{16} \)	0.8125	20.6375
\( \frac{15}{16} \)	0.9375	23.8125

The odd sixteenths are listed because the even ones reduce to eighths, quarters and halves that you already have. Notice how the pattern is generated: one sixteenth is 0.0625 in, so the whole scale can be rebuilt by repeated addition if you forget it. The same arithmetic run backwards identifies the nearest imperial size to a metric one. A 6 mm dimension is \( 6 \div 25.4 = 0.2362 \) in, which falls between \( \frac{7}{32} \) (0.21875 in) and \( \frac{1}{4} \) (0.25 in) and is closest to \( \frac{15}{64} \) (0.234375 in) — a difference of under 0.05 mm. That arithmetic tells you which imperial size is nearest; whether a near-size may actually be substituted for the specified one is a maintenance-data question, and the answer is found in the manual, not in the sum.

Factors and Multiples

A factor of a number divides into it exactly with no remainder. A multiple is the product of that number and any positive integer.

Prime Numbers and Prime Factorisation

A prime number has exactly two factors: 1 and itself. The first primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... Note that 1 is not prime, and 2 is the only even prime.

Any integer can be written as a product of prime factors. For example: \( 60 = 2^2 \times 3 \times 5 \).

Highest Common Factor (HCF)

The HCF is the largest number that divides exactly into two or more numbers. To find it, identify the prime factors common to both numbers and multiply them together.

Lowest Common Multiple (LCM)

The LCM is the smallest number that is a multiple of two or more numbers. Take the highest power of each prime factor present in either number.

Divisibility Tests

Before you can factorise anything you have to spot what divides into it, and these tests do that by inspection rather than by trial division.

Divisor	Test	Example
2	Last digit is even	3,174
3	Digit sum divides by 3	4,713 (digits sum to 15)
4	Last two digits divide by 4	7,316 (16 divides by 4)
5	Last digit is 0 or 5	2,485
6	Passes the tests for both 2 and 3	5,214 (even; digits sum to 12)
8	Last three digits divide by 8	12,536 (536 divides by 8)
9	Digit sum divides by 9	6,633 (digits sum to 18)
10	Last digit is 0	4,730
11	Alternately subtract and add the digits	8,151 (\( 1 - 5 + 1 - 8 = -11 \))

The tests for 4 and 8 work because 100 and 1,000 are themselves divisible by 4 and 8, so everything above the last two or three digits can be ignored. The tests for 3 and 9 are the same idea used in casting out nines, which is why that check exists at all.

Finding Every Factor Without Missing One

Factors come in pairs: if 3 divides into 84 then so does \( 84 \div 3 = 28 \), and the two are found together. Work upwards from 1, writing down both members of each pair, and stop as soon as the two members meet — that is, once you pass \( \sqrt{84} \approx 9.2 \). Anything beyond that point has already been written down as the partner of a smaller factor.

Prime Factorisation in Practice

The reliable method is repeated division by the smallest prime that will go, carrying on until you reach 1. For 360: \( 360 \div 2 = 180 \), \( \div 2 = 90 \), \( \div 2 = 45 \); 2 no longer divides, so move to 3: \( 45 \div 3 = 15 \), \( \div 3 = 5 \); then \( 5 \div 5 = 1 \). Collecting the divisors gives \( 360 = 2^3 \times 3^2 \times 5 \). Every whole number greater than 1 has exactly one such factorisation, which is the reason the whole method is trustworthy — and the reason 1 is excluded from the primes, since allowing it would let you write any number in endlessly many ways.

Once you have the prime factorisation you can count the factors without listing them: add one to each index and multiply. For \( 84 = 2^2 \times 3 \times 7 \) that gives \( 3 \times 2 \times 2 = 12 \), matching the twelve found above. For 24 it gives \( 4 \times 2 = 8 \), matching the eight listed in the example earlier in this section. For \( 360 = 2^3 \times 3^2 \times 5 \) it gives \( 4 \times 3 \times 2 = 24 \) factors, which is exactly why a circle of 360° can be divided evenly so many ways — into 2, 3, 4, 5, 6, 8, 9, 10, 12 and more equal parts, all landing on whole degrees. Seven is not among the factors of 360, and that is why seven equally spaced fasteners around a circle come out at an awkward 51.43° apart rather than a whole number.

The same square-root limit tests whether a number is prime at all. To check 197, you need only try the primes up to \( \sqrt{197} \approx 14 \): it is odd, its digits sum to 17 so 3 does not divide it, it does not end in 0 or 5, and 7, 11 and 13 all leave remainders. There is nothing left to test, so 197 is prime.

Euclid's Method for the HCF

Prime factorisation is the clearest route to the HCF, but it becomes slow with large or awkward numbers. Euclid's method avoids factorising anything: divide the larger number by the smaller, then replace the pair with the divisor and the remainder, and repeat until the remainder is zero. The last non-zero remainder is the HCF.

The Link Between HCF and LCM

Test it on the pair worked above: \( 12 \times 144 = 1728 \), and \( 36 \times 48 = 1728 \). The identity gives you a free cross-check on both answers at once, and it also gives you the second quantity as soon as you have the first — having found the HCF by Euclid, the LCM is \( \frac{a \times b}{\text{HCF}} \).

Where Factors and Multiples Are Actually Used

These are not classroom exercises. Each one solves a recurring practical problem.

• The LCM tells you when repeating events coincide. If one task falls due every 100 flight hours and another every 150, the two fall together every \( \text{LCM}(100,150) = 300 \) hours — useful for grouping work into a single visit rather than two.
• The HCF tells you the largest equal share. Making up identical kits from 84 rivets and 126 washers, the HCF of 42 says you can make 42 kits, each holding 2 rivets and 3 washers, with nothing left over. That is the same 42 that reduced \( \frac{84}{126} \) to \( \frac{2}{3} \) in the fractions section — cancelling a fraction and dividing stock into equal shares are the same arithmetic.
• The HCF also predicts wear patterns in a gear train. If a 20-tooth pinion drives a 60-tooth gear, the HCF is 20, so every tooth on the pinion meets the same three gear teeth for the life of the assembly and any local wear or damage is repeatedly hammered on the same partners. Choose tooth counts with no common factor — a 23-tooth pinion against a 60-tooth gear, HCF 1 — and the pattern does not repeat until the pinion has turned \( \text{LCM}(23,60) \div 23 = 60 \) times, by which point every tooth has met every other and wear is spread across the whole set. This is the reasoning behind a hunting-tooth ratio.

Weights, Measures and Conversion Factors

Aviation uses a mixture of SI (metric) and imperial/US customary units. You must be able to convert between them confidently.

Length

Conversion	Factor
1 inch (in)	25.4 mm (exact)
1 foot (ft) = 12 in	0.3048 m
1 yard (yd) = 3 ft	0.9144 m
1 mile (statute)	1.609 km
1 nautical mile (NM)	1.852 km (exact)

Mass

Conversion	Factor
1 pound (lb)	0.4536 kg
1 kilogram (kg)	2.205 lb
1 US ton (short ton)	907.2 kg
1 Imperial ton (long ton)	1,016 kg
1 metric tonne	1,000 kg

Volume

Conversion	Factor
1 US gallon	3.785 litres
1 Imperial gallon	4.546 litres
1 litre	1,000 cm³ = 0.001 m³

Pressure

Conversion	Factor
1 atmosphere (atm)	101.325 kPa = 14.696 psi
1 bar	100 kPa = 14.504 psi
1 psi	6.895 kPa
1 atm	29.92 inHg = 760 mmHg

Temperature

SI Base Units and Prefixes

The SI system is built on seven base units — the metre (length), kilogram (mass), second (time), ampere (electric current), kelvin (thermodynamic temperature), mole (amount of substance) and candela (luminous intensity). Everything else is derived from them: the newton is \( \text{kg}\cdot\text{m}/\text{s}^2 \), the pascal is \( \text{N}/\text{m}^2 \), the volt is \( \text{W}/\text{A} \). Prefixes then scale any unit by a power of ten, so no new conversion factors are needed.

Prefix	Symbol	Multiplier	Typically seen in
giga	G	\( 10^{9} \)	GHz (radio frequency)
mega	M	\( 10^{6} \)	MΩ (insulation resistance), MPa
kilo	k	\( 10^{3} \)	kg, kN, kPa
—	—	\( 10^{0} = 1 \)	m, s, A, N
centi	c	\( 10^{-2} \)	cm
milli	m	\( 10^{-3} \)	mm, mA, mV
micro	µ	\( 10^{-6} \)	µm (surface finish), µF
nano	n	\( 10^{-9} \)	nF, ns
pico	p	\( 10^{-12} \)	pF

The Unit-Factor Method

Guessing whether to multiply or divide is the single largest source of conversion errors, and there is a method that removes the guess entirely. Every conversion factor can be written as a fraction whose value is exactly 1. Since 1 inch is 25.4 mm, both of these are equal to one:

This also answers the question the exam actually asks, which is usually about the direction rather than the arithmetic: to convert inches to millimetres you multiply by 25.4, and to convert inches to centimetres you multiply by 2.54. Pounds to kilograms is a multiplication by 0.4536, and kilograms to pounds a multiplication by 2.205 — each pair being the two ways up of the same fact.

Exact Definitions Behind the Rounded Factors

Several of the factors in the tables above are not measurements at all but definitions, agreed internationally and exact by construction. Knowing which is which tells you how far you may safely round.

Quantity	Exact definition	Common working value
Inch	1 in = 25.4 mm exactly	25.4 mm
Pound (avoirdupois)	1 lb = 0.45359237 kg exactly	0.4536 kg
Kilogram to pounds	Reciprocal of the above	2.20462 lb
US gallon	1 US gal = 3.785411784 L exactly	3.785 L
Imperial gallon	1 imp gal = 4.54609 L exactly	4.546 L
Nautical mile	1 NM = 1852 m exactly	1.852 km
Statute mile	1 mile = 1609.344 m exactly	1.609 km

Two Gallons, and Why It Matters

The US gallon and the imperial gallon are genuinely different volumes, not two names for one thing. At 4.54609 litres the imperial gallon is about 20% larger than the US gallon's 3.785411784 litres. Ordering 500 gallons of fuel therefore means 1,893 litres if the figure is American and 2,273 litres if it is British — a difference of 380 litres, which is not a rounding matter. Whenever a volume in gallons appears in maintenance or fuelling data, establish which gallon it is before converting anything.

Fuel: Volume Becomes Mass Only Through Density

Fuel is uplifted by volume, because that is what a bowser meter measures, but it is planned and flown by mass, because that is what affects the aircraft's performance and centre of gravity. Density is the bridge between the two, and it is not a constant to be memorised: it depends on the fuel grade and changes with temperature, and for aviation turbine fuel it typically lies somewhere in the region of 0.78 to 0.84 kilograms per litre at 15 °C. The figure you use must come from the fuelling documentation or from the density actually measured at uplift.

Areas and Volumes: the Factor Gets Squared or Cubed

This is the most reliably examined trap in the whole of unit conversion. A linear conversion factor of \( k \) becomes \( k^2 \) for areas and \( k^3 \) for volumes, because an area is a length multiplied by a length and a volume is a length multiplied by a length multiplied by a length. Since 1 m = 100 cm:

• \( 1\ \text{m}^2 = 100^2 = 10{,}000\ \text{cm}^2 \)
• \( 1\ \text{m}^3 = 100^3 = 1{,}000{,}000\ \text{cm}^3 \)
• \( 1\ \text{in}^2 = 25.4^2 = 645.16\ \text{mm}^2 \)
• \( 1\ \text{in}^3 = 25.4^3 = 16{,}387.064\ \text{mm}^3 \)

One consequence of the cubed factor is worth stating on its own, because it is used constantly and rarely explained: a litre is defined as a cubic decimetre, and since a decimetre is 10 cm, that is \( 10^3 = 1000 \) cubic centimetres. The abbreviation "cc" that appears on engine displacement figures is simply cm³, so an 11,946 cc engine and an 11.9 litre engine are the same engine. A cubic metre is 1,000 litres by the same reasoning.

Speed and Rate of Climb

A knot is one nautical mile per hour, and because the nautical mile (1.852 km) is longer than the statute mile (1.609344 km), a speed in knots is always a smaller number than the same speed in miles per hour. Dividing one definition by the other gives the factor: \( 1.852 \div 1.609344 = 1.15078 \), so 1 knot = 1.15078 mph. Converting to kilometres per hour is easier still, because 1 knot is 1.852 km/h by definition.

Pressure: Gauge Against Absolute

A pressure figure is incomplete until you know what it is measured against. Most workshop instruments — tyre gauges, hydraulic gauges, oxygen trolley gauges — read gauge pressure, which takes the surrounding atmosphere as zero. Absolute pressure takes a vacuum as zero, so at any positive gauge reading the absolute figure is the larger of the two:

A tyre gauge reading 200 psi therefore indicates 200 psig, which is about 214.7 psia at sea level. The distinction is not academic: gas-law calculations and anything involving compression ratios must use absolute pressures, because a gas at "zero" gauge pressure is not a gas that has ceased to exist. Note also that the hectopascal and the millibar are the same size, so 1013.25 hPa and 1013.25 mbar are two names for one pressure.

Temperature: Scales, Intervals and Absolute Zero

Celsius and Fahrenheit differ in two ways at once — the size of the degree and the position of zero — which is why their conversion needs both a factor and an offset. Kelvin shares the Celsius degree size but places zero at absolute zero, so only an offset is needed. Kelvin is written without a degree symbol and never takes a negative value.

Setting the two Celsius-to-Fahrenheit expressions equal shows there is exactly one temperature where the scales read the same number: it is −40, so −40 °C is −40 °F. That is a genuinely useful memory anchor, and it is also a fast check on any conversion — below −40 the Fahrenheit figure is the more negative of the two, above it the Fahrenheit figure is the larger.

Force, Mass and Torque

Mass and weight are different quantities and, in SI, take different units: mass is in kilograms, and weight is the force gravity exerts on that mass, in newtons. With \( g \approx 9.81\ \text{m/s}^2 \), a 100 kg component has a weight of \( 100 \times 9.81 = 981 \) N. The imperial system uses the word "pound" for both ideas, which is why you will see lb for mass and lbf where a force is specifically meant.

Torque is a force multiplied by the perpendicular distance from the axis, so its units are a force unit times a length unit — N·m, lb·ft or lb·in. Because a foot is twelve inches, 1 lb·ft is 12 lb·in, and 1 lb·ft is 1.356 N·m.

Ratio and Proportion

A ratio compares two quantities of the same kind. It is written as \( a : b \) and can also be expressed as the fraction \( \frac{a}{b} \).

Direct Proportion

Two quantities are in direct proportion if when one increases, the other increases by the same factor. If \( y \) is directly proportional to \( x \), then \( y = kx \) for some constant \( k \).

Inverse Proportion

Two quantities are in inverse proportion if when one increases, the other decreases by the same factor. If \( y \) is inversely proportional to \( x \), then \( y = \frac{k}{x} \).

Reading a Ratio Correctly

Two rules govern every ratio question, and almost every wrong answer comes from breaking one of them.

First, both quantities must be in the same unit before the ratio is formed. A 50 cm link and a 2 m rod are not in the ratio 50 : 2. Converting the rod to 200 cm gives 50 : 200, which simplifies to 1 : 4. A ratio is a pure number with no units of its own, and it only becomes one once the units have been made to match.

Second, decide whether the ratio compares a part with another part, or a part with the whole. The ratio 1 : 4 describes a mixture in which one component is a quarter of the size of the other. That same mixture has five parts in total, so the first component is one fifth — 20% — of the whole mixture, while being 25% of the second component. Both statements are true and they are different numbers; which one a question wants depends entirely on whether it gives you a component quantity or a total quantity.

Simplifying and Scaling a Ratio

Multiplying or dividing every term of a ratio by the same number leaves it unchanged — the same rule that governs equivalent fractions. So \( 6 : 5 \) is equally \( 12 : 10 \), \( 18 : 15 \) or \( 24 : 20 \). To test whether a given ratio is equivalent to another, reduce both to their lowest terms and compare: \( 10 : 16 \) reduces to \( 5 : 8 \) and \( 20 : 25 \) reduces to \( 4 : 5 \), so neither is a form of \( 6 : 5 \), while \( 24 : 20 \) divides through by 4 to give exactly \( 6 : 5 \).

The Two Question Shapes

Ratio questions arrive in exactly two forms, and reading which one you have been given is most of the work.

• You are told one component. Scale directly from that component's own share of the ratio. If \( A : B = a : b \) and you know \( A \), then \( B = A \times \frac{b}{a} \). The total plays no part.
• You are told the total. Add the terms to get the number of parts, divide the total by that number to find one part, then scale each component — the method used in the example above.

Telling Direct from Inverse

There is a simple test that distinguishes the two kinds of proportion from a table of values, and it is worth applying rather than judging by feel.

Relationship	What stays constant	Form	Effect of doubling \( x \)
Direct proportion	The quotient \( y \div x \)	\( y = kx \)	\( y \) doubles
Inverse proportion	The product \( x \times y \)	\( y = \frac{k}{x} \)	\( y \) halves

In direct proportion the graph is a straight line through the origin, and the practical route to an answer is the unitary method: find the value for one, then multiply. If seven identical brackets have a total mass of 2.8 kg, one bracket is \( 2.8 \div 7 = 0.4 \) kg, so twelve of them are \( 12 \times 0.4 = 4.8 \) kg. Note that direct proportion requires the line to pass through the origin: doubling the number of brackets doubles the mass because zero brackets have zero mass. Quantities that merely increase together are not necessarily in proportion.

Gear and Pulley Ratios

Meshing gears turn at speeds inversely proportional to their tooth counts, because every tooth that leaves one gear must have entered the other:

Take the 20-tooth driver and 60-tooth driven gear described above. The driven gear turns at \( \frac{20}{60} = \frac{1}{3} \) of the driver's speed — a reduction drive. Because power is torque multiplied by rotational speed, and an ideal gear train neither creates nor destroys power, the torque available at the slower shaft is correspondingly higher: three times the input torque, less whatever friction takes. Speed down, torque up. A drive arranged the other way round, with a large gear driving a small one, does the reverse: the output shaft turns faster and its torque is lower.

Proportion at the Torque Wrench

Adding an adaptor that lengthens a torque wrench along its own axis changes the lever arm at the fastener but not the arm the wrench itself is measuring. The torque delivered to the fastener is therefore greater than the wrench indicates, in direct proportion to the lengths involved, so the wrench must be set to a lower figure than the specification if the fastener is to receive the specified torque.

Scale Drawings

A drawing scale is a ratio of drawing size to real size, so 1 : 50 means every millimetre on the paper stands for 50 mm on the aircraft. A line measuring 24 mm on such a drawing represents \( 24 \times 50 = 1200 \) mm, or 1.2 m. A detail drawn at 2 : 1 is the other way round — drawn twice life size, so a 24 mm feature on paper is a 12 mm feature in reality.

Areas on a scale drawing do not scale by the same factor. They scale by the square of the linear ratio, for exactly the reason set out for area conversion factors in the units section above: at 1 : 50, one square centimetre of drawing represents \( 50^2 = 2500 \) square centimetres, which is 0.25 m².

Averages and Percentages

Averages

• Mean (arithmetic average): sum of all values divided by the number of values. $$ \text{Mean} = \frac{\sum x}{n} $$
• Median: the middle value when data is arranged in order. For an even count of values, take the mean of the two middle values.
• Mode: the most frequently occurring value.

Percentages

A percentage is a fraction expressed out of 100. The symbol is %.

Choosing Between the Three Averages

The three averages are not interchangeable, and each answers a different question about the same data. The mean uses every value and is the one wanted whenever a total matters — total fuel, total mass, total time. The median reports the middle of the distribution and ignores how extreme the extremes are. The mode reports what happens most often, and is the only one of the three that works on data which are not numbers at all, such as the most frequently recurring defect on a fleet.

The practical difference shows up as soon as one reading is wrong. Take the five temperature readings above and suppose a faulty probe had added a sixth reading of 90 °C. The mean would leap from 21 °C to \( \frac{105 + 90}{6} = 32.5 \) °C — higher than every genuine reading in the set. The median would be recomputed as the mean of the two middle values of 18, 19, 22, 22, 24, 90 — still 22 °C. One rogue value drags the mean a long way and leaves the median almost untouched, which is exactly why a median is preferred when a data set may contain a spurious reading.

The Weighted Mean

Averaging a set of averages by adding them up and dividing is wrong unless each of them was based on the same amount of data. The correct method multiplies each value by its weight — the number of items, hours or units it represents — before dividing by the total weight.

Alongside the average it is worth quoting the range, the difference between the largest and smallest values, because two sets can share a mean and be nothing alike. The five temperature readings run from 18 °C to 24 °C, a range of 6 °C; a set of readings all at 21 °C would have the same mean and a range of zero. When you are looking for a drifting sensor or an unstable system, the spread is often the more informative number.

Average Speed and Average Rate

An average rate is always the total quantity divided by the total of whatever it is per — never the average of the individual rates. Speed is the case that catches people out most reliably.

Moving Between Fractions, Decimals and Percentages

A percentage is just a fraction with 100 fixed as the denominator, so all three notations describe the same quantity and converting between them needs no new arithmetic:

• Fraction to percentage: multiply by 100. \( \frac{9}{20} \times 100 = \frac{900}{20} = 45\% \), and \( \frac{5}{8} \times 100 = \frac{500}{8} = 62.5\% \).
• Decimal to percentage: multiply by 100, which moves the point two places right. \( 0.45 \rightarrow 45\% \).
• Percentage to decimal: divide by 100, moving the point two places left. \( 7.5\% \rightarrow 0.075 \).
• Percentage to fraction: write it over 100 and cancel. \( 45\% = \frac{45}{100} = \frac{9}{20} \).

Converting a percentage to a decimal before calculating with it is almost always the quickest route, because "12% of" then becomes a single multiplication by 0.12 rather than a multiplication and a division.

Increase, Decrease, and Getting Back to the Original

An increase of \( p\% \) multiplies a quantity by \( \left(1 + \frac{p}{100}\right) \) and a decrease multiplies it by \( \left(1 - \frac{p}{100}\right) \). Handling it as one multiplication rather than as "work out the change, then add it on" removes a step and the error that goes with it.

A related distinction is worth getting right in words as well as in figures. If a pass rate rises from 60% to 75%, it has risen by 15 percentage points, but the relative increase is \( \frac{15}{60} \times 100\% = 25\% \). Both descriptions are correct and they are different numbers; saying "a 15% rise" when you mean 15 percentage points misstates the change by a wide margin.

Percentages as Tolerances and Wear Limits

A great deal of maintenance data expresses acceptability as a percentage rather than as an absolute figure, because the same rule then applies across a range of sizes. Turning that percentage into the two numbers you can actually measure against is straightforward, but it must be done from the correct base.

Areas and Volumes

Area Formulae (2D Shapes)

Volume Formulae (3D Shapes)

Perimeter and Circumference

Area answers "how much surface", perimeter answers "how far round". They are different quantities with different units — square millimetres against millimetres — and confusing them is the most heavily exploited error in this part of the syllabus.

The circle formulae are where the marks are lost. \( 2\pi r \) is the distance round the rim; \( \pi r^2 \) is the surface enclosed. And because the diameter is twice the radius, \( d^2 = (2r)^2 = 4r^2 \) — so putting the diameter into the area formula in place of the radius produces an answer exactly four times too large. If a question gives you a diameter, halve it before it goes anywhere near \( \pi r^2 \).

Working Backwards from the Circumference

A circle is fully described by any one of its radius, diameter, circumference or area, so any of them can be recovered from any other. Given the circumference, rearrange \( C = 2\pi r \) to \( r = \frac{C}{2\pi} \), and substituting that into \( A = \pi r^2 \) gives a direct route:

Rings, Sectors and Shapes with Holes in Them

Real components are rarely one clean shape. Almost all of them can be handled by adding and subtracting simple ones.

An annulus — the cross-section of a tube, a washer, a bearing race — is the difference of two circles, \( A = \pi(R^2 - r^2) \). For an outer radius of 30 mm and an inner radius of 22 mm, that is \( \pi(900 - 484) = \pi \times 416 = 1307 \) mm². The difference of two squares gives an equivalent form, \( \pi(R - r)(R + r) = \pi \times 8 \times 52 \), which reaches the same 416π and is easier to do in your head.

A sector is the fraction of a circle cut off by an angle, so both its area and its arc length are simply that fraction of the whole:

A composite shape is handled by subtraction. A rectangular doubler 200 mm by 150 mm with a 60 mm diameter lightening hole has an area of \( (200 \times 150) - (\pi \times 30^2) = 30{,}000 - 2827 = 27{,}173 \) mm². Note the halving of the 60 mm diameter to a 30 mm radius before squaring. Using 60 as the radius would subtract \( \pi \times 60^2 = 11{,}310 \) mm² instead of 2,827 mm², leaving a doubler of 18,690 mm² — an answer over 30% adrift that looks entirely plausible on the page, which is what makes this particular slip worth guarding against.

Triangles: Which Height?

In \( A = \frac{1}{2} \times b \times h \), the height must be measured perpendicular to the base you have chosen, not along a sloping side. In a right-angled triangle the two legs are already perpendicular to each other, so either can serve as the base and the other is then the height — which is why the area of a right-angled triangle is simply half the product of its two shorter sides.

When all three sides are known but no perpendicular height is, Heron's formula gives the area without needing one. With \( s \) as half the perimeter:

Curved Surface Against Total Surface

A surface-area question is only answerable once you know which surfaces are wanted, and the cone is where this bites. Its curved (lateral) surface is \( \pi r l \), where \( l \) is the slant height measured up the sloping side — not the vertical height. Adding the circular base, \( \pi r^2 \), gives the total surface \( \pi r(r + l) \). If a question asks only for the curved area, including the base makes the answer too large; if it asks for the total and you omit the base, too small.

The slant height is rarely given directly, but it follows from the vertical height and the base radius, since they form a right-angled triangle: \( l^2 = r^2 + h^2 \). A cone of base radius 30 mm and vertical height 40 mm therefore has \( l = \sqrt{900 + 1600} = \sqrt{2500} = 50 \) mm, so its curved surface is \( \pi \times 30 \times 50 = 4712 \) mm², its base is \( \pi \times 30^2 = 2827 \) mm², and its total surface is 7,539 mm² — which \( \pi r(r+l) = \pi \times 30 \times 80 \) confirms directly.

The cylinder decomposes the same way, and its formula stops being something to memorise once you see it. Unroll the curved surface and it is a rectangle whose width is the circumference and whose height is the cylinder's height, giving \( 2\pi r h \); add the two circular ends, \( 2 \times \pi r^2 \), and you have \( 2\pi r(r + h) \).

What Happens When You Scale a Shape

Double every linear dimension of a cube of side 1: the surface area goes from 6 to 24 and the volume from 1 to 8. That is the same \( k^2 \) and \( k^3 \) rule set out for conversion factors in the units section, now applied to a shape rather than to a unit. Its physical consequence is the part worth carrying: volume grows faster than surface area, so a large tank holds far more per unit of skin than a small one, which is why large volumes lose heat comparatively slowly and why doubling a duct's dimensions has a much bigger effect on the air it carries than on the material it is made from.

Capacity, Displacement and Compression

Volume calculations become capacity calculations as soon as the units change: 1 litre is 1,000 cm³, and 1 m³ is 1,000 litres.

Prisms, Tapers and Awkward Tank Shapes

Real tanks and ducts are rarely boxes or perfect cylinders, but most of them are prisms — solids of constant cross-section — and every prism obeys one rule regardless of how complicated that cross-section is:

The cuboid and the cylinder are just the two commonest cases of it: \( l \times w \times h \) is a rectangular cross-section times a length, and \( \pi r^2 h \) is a circular one. So a tank whose section is a trapezium, a composite shape, or a rectangle with a rounded corner is handled by finding that one area and multiplying — there is no separate formula to learn.

Solids that taper to a point rather than keeping their section are the other family, and they follow a single rule too: a pyramid or a cone occupies exactly one third of the prism or cylinder that shares its base and height. That is why the cone formula in the table above is \( \frac{1}{3}\pi r^2 h \) — the cylinder of the same base radius and height holds three times as much. A cone of base radius 30 mm and vertical height 40 mm therefore holds \( \frac{1}{3} \times \pi \times 900 \times 40 = 37{,}699\ \text{mm}^3 \), against the 113,097 mm³ of the cylinder that encloses it.

Areas in Sheet-Metal and Pipework

Two everyday uses of area close this section, and both turn a geometric answer into something you can act on.

Mass of a sheet. Area multiplied by thickness gives volume, and volume multiplied by density gives mass. A piece of 2 mm aluminium-alloy sheet 1.5 m by 1.0 m has a volume of \( 1.5 \times 1.0 \times 0.002 = 0.003\ \text{m}^3 \), which is 3,000 cm³. Airframe aluminium alloys have a density of roughly 2.7 grams per cubic centimetre, so the sheet weighs about \( 3000 \times 2.7 = 8100 \) g, or 8.1 kg. The same three-step chain — area, then volume, then mass — estimates the weight of any repair patch or doubler before it is cut.

Bore and flow velocity. The cross-sectional area of a pipe is \( A = \frac{\pi d^2}{4} \), which is just \( \pi r^2 \) written in terms of the bore. Because the diameter is squared, doubling the bore quadruples the area: a 10 mm bore gives \( \pi \times 5^2 = 78.5\ \text{mm}^2 \), while a 20 mm bore gives \( \pi \times 10^2 = 314\ \text{mm}^2 \). For a given volumetric flow, velocity is flow divided by area, so that same doubling of the bore reduces the flow velocity to a quarter of what it was. A small change in a pipe or restrictor size has a much larger effect than its dimensions suggest.

Squares, Cubes, Square and Cube Roots

Squaring a number means multiplying it by itself: \( n^2 = n \times n \). The square root reverses this: \( \sqrt{n^2} = n \).

Cubing means raising to the power of 3: \( n^3 = n \times n \times n \). The cube root reverses it: \( \sqrt[3]{n^3} = n \).

\( n \)	\( n^2 \)	\( n^3 \)	\( \sqrt{n} \)	\( \sqrt[3]{n} \)
1	1	1	1.000	1.000
2	4	8	1.414	1.260
3	9	27	1.732	1.442
4	16	64	2.000	1.587
5	25	125	2.236	1.710
10	100	1,000	3.162	2.154

Perfect Squares and Perfect Cubes Worth Recognising

Recognising a perfect square or cube on sight turns a root question into a recall question. The table above covers the smallest cases; these are the others you will meet most often.

\( n \)	6	7	8	9	11	12	15	20
\( n^2 \)	36	49	64	81	121	144	225	400
\( n^3 \)	216	343	512	729	1,331	1,728	3,375	8,000

Note that 64 appears as both a square and a cube, being \( 8^2 \) and \( 4^3 \), and that 729 is \( 9^3 \) as well as \( 27^2 \). A number can be a perfect square and a perfect cube at the same time, which is worth knowing before you assume a question has only one sensible reading.

Two Square Roots, but One Radical Sign

Every positive number has two square roots, one positive and one negative, because both \( 5 \times 5 \) and \( (-5) \times (-5) \) give 25. The radical sign is defined to mean the principal root — the non-negative one — so \( \sqrt{25} = 5 \) and not \( \pm 5 \). But solving the equation \( x^2 = 25 \) is a different question, and its answer is \( x = \pm 5 \), because both values satisfy it.

The distinction matters as soon as you take a square root while rearranging a formula. Recovering a radius from an area, or a velocity from a kinetic energy, produces two mathematical answers of which only the positive one is physically meaningful — there is no such thing as a radius of −30 mm. Discarding the negative root is a decision you make about the physics, not something the arithmetic does for you.

Roots of Negative Numbers

Squaring a negative gives a positive, so no real number squared can produce a negative — hence there is no real square root of a negative number. Cube roots behave differently, and this is not an exception to the rule but a consequence of it. Cubing a negative keeps it negative, since \( (-2) \times (-2) \times (-2) = -8 \), so every negative number has exactly one real cube root and it is negative: \( \sqrt[3]{-8} = -2 \) and \( \sqrt[3]{-125} = -5 \).

The general pattern is that an even root of a negative number has no real value, while an odd root always does. If you are ever unsure, cube the candidate and see whether it reproduces the original.

Estimating a Root Without a Calculator

Bracket first. To find \( \sqrt{50} \), note that \( 7^2 = 49 \) and \( 8^2 = 64 \), so the answer lies between 7 and 8 and must be very close to 7, since 50 is barely above 49. That alone is enough to reject a wrong option in a multiple-choice question.

To sharpen it, use the divide-and-average method: divide the number by your current guess, then average the guess with the result. Each pass roughly doubles the number of correct digits.

Splitting a Root into Factors

For non-negative numbers, a root distributes across a product and a quotient:

Pulling out a perfect-square factor simplifies awkward roots. Since \( 50 = 25 \times 2 \), \( \sqrt{50} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \); using \( \sqrt{2} = 1.414 \) from the table above, that gives \( 5 \times 1.414 = 7.07 \), agreeing with the hand calculation. The same idea makes \( \sqrt{0.04} \) easy: it is \( \sqrt{\frac{4}{100}} = \frac{2}{10} = 0.2 \).

Squaring Numbers That End in 5

Multiply the leading digits by the next whole number up, then write 25 on the end. So \( 35^2 \): \( 3 \times 4 = 12 \), giving 1225. \( 65^2 \): \( 6 \times 7 = 42 \), giving 4225. \( 85^2 \): \( 8 \times 9 = 72 \), giving 7225. It follows from the distributive law rather than being a coincidence, and it is quick enough to be worth using as a cross-check on a calculator entry.

Units Are Squared and Cubed Along with the Numbers

Squaring a quantity squares its unit as well as its number, which is why an area comes out in mm² and a volume in mm³. That single fact is the origin of the area and volume conversion factors listed in the units section. Taking a root goes the other way and halves or thirds the unit: the square root of an area in mm² is a length in mm.

This gives you a free check on any rearranged formula. If you have taken the square root of something and the units that come out are not a length, the rearrangement is wrong.

Why Squares and Cubes Dominate Engineering Numbers

Squares and cubes are not decoration in engineering formulae; they describe how quantities really behave, and each of the following is a direction you should be able to state without hesitating.

• Area against diameter. Because \( A = \frac{\pi d^2}{4} \), halving a bolt's diameter leaves a quarter of the cross-section. For the same applied load the stress, \( \frac{\text{force}}{\text{area}} \), is therefore four times higher — which is why an undersized fastener is a far more serious substitution than its dimensions suggest.
• Kinetic energy against speed. Since \( E_k = \frac{1}{2}mv^2 \), doubling the speed does not double the energy that has to be absorbed, it quadruples it. A touchdown at 1.4 times the intended speed arrives with roughly twice the kinetic energy to be dissipated.
• Volume against bore. The same squared diameter met under bore and flow velocity above also sits inside swept volume, \( \pi r^2 h \): for a given stroke, doubling the bore quadruples a cylinder's displacement.
• Intensity against distance. Energy radiating outwards spreads over a spherical surface of area \( 4\pi r^2 \), so at twice the distance the same energy covers four times the area and the intensity falls to a quarter. This inverse-square behaviour applies to radiated signal strength and to radiation dose alike.

Squares and cubes are the special cases \( n = 2 \) and \( n = 3 \) of a general system of indices, and roots are the same system with fractional indices — \( \sqrt{a} \) is \( a^{1/2} \) and \( \sqrt[3]{a} \) is \( a^{1/3} \). The rules that govern all of them together are the subject of the final section.

Powers and Roots (Laws of Indices)

An index (plural: indices), also called an exponent or power, indicates how many times a base is multiplied by itself: \( a^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ times}} \)

Where the Index Laws Come From

The three laws in the table are not conventions to be taken on trust. Writing the powers out shows why they must hold: \( a^3 \times a^4 \) is three \( a \)s multiplied by four \( a \)s, which is seven \( a \)s, so the indices add. Division cancels \( a \)s from top and bottom, so the indices subtract. A power of a power repeats the multiplication, so the indices multiply.

The remaining four entries look arbitrary until you notice that each is forced by the first three — they are the only definitions that keep the laws working.

• Zero index. \( \frac{a^m}{a^m} \) is obviously 1. The division law says it is \( a^{m-m} = a^0 \). Both must be true, so \( a^0 = 1 \) for every \( a \) except zero.
• Negative index. \( \frac{a^0}{a^n} \) is \( \frac{1}{a^n} \). The division law says it is \( a^{0-n} = a^{-n} \). Hence \( a^{-n} = \frac{1}{a^n} \) — a negative index means a reciprocal, never a negative number.
• Fractional index. By the power-of-a-power law, \( \left(a^{1/n}\right)^n = a^{n \times \frac{1}{n}} = a \). Something which gives \( a \) when raised to the power \( n \) is by definition the \( n \)th root, so \( a^{1/n} = \sqrt[n]{a} \).
• General fractional index. Combining the last two, \( a^{m/n} \) is the \( n \)th root raised to the power \( m \), in either order.

What the Laws Do Not Say

Every index law above concerns a product or quotient of powers of the same base. Reading them more widely than that is the standard way to lose a mark here.

• Different bases do not combine by adding indices. \( 2^3 \times 5^2 \) cannot be simplified to any single power; it is just \( 8 \times 25 = 200 \).
• But equal indices do combine across different bases. \( a^m \times b^m = (ab)^m \), which is why \( 2^3 \times 5^3 = 10^3 = 1000 \). The matching pair is \( \frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m \), so \( \frac{6^2}{3^2} = 2^2 = 4 \).
• Sums of powers do not combine at all. \( 2^3 + 2^4 = 8 + 16 = 24 \), which is not \( 2^7 \). The addition law is for multiplication, not addition.
• A power does not distribute across a bracketed sum. \( (3 + 4)^2 = 49 \), not \( 3^2 + 4^2 = 25 \).

Standard Form

A number is in standard form (also called scientific notation) when it is written as \( a \times 10^n \) with exactly one non-zero digit before the decimal point — that is, with \( 1 \le a \lt 10 \) — and \( n \) a whole number. To convert, count how many places the point has to move to leave one digit in front of it: moving it left gives a positive exponent, moving it right gives a negative one.

• \( 4{,}730{,}000 = 4.73 \times 10^{6} \) — the point moved six places left.
• \( 0.0000413 = 4.13 \times 10^{-5} \) — the point moved five places right.

The sign of the exponent is therefore a reliable size check on its own: a positive exponent means a number of 10 or more, a negative exponent means a number below 1, and \( 10^0 \) covers everything between. If you have written a very small number with a positive exponent, the point went the wrong way.

Calculating in Standard Form

Standard form turns awkward arithmetic into two easy pieces: handle the leading numbers, and handle the powers of ten with the index laws.

Operation	Method	Example
Multiply	Multiply the leading numbers, add the exponents	\( (3 \times 10^4)(2 \times 10^{-7}) = 6 \times 10^{-3} \)
Divide	Divide the leading numbers, subtract the exponents	\( \frac{4.8 \times 10^{6}}{1.6 \times 10^{2}} = 3 \times 10^{4} \)
Raise to a power	Raise the leading number, multiply the exponent	\( (2 \times 10^{3})^{4} = 16 \times 10^{12} = 1.6 \times 10^{13} \)
Add or subtract	Make the exponents equal first	\( 3.2 \times 10^{3} + 0.45 \times 10^{3} = 3.65 \times 10^{3} \)

After multiplying or dividing, the leading number may fall outside the 1-to-10 range and the answer then has to be renormalised. \( (6 \times 10^4)(5 \times 10^3) = 30 \times 10^7 \), and since \( 30 = 3 \times 10^1 \), that becomes \( 3 \times 10^8 \). The addition row is the one candidates skip: \( 3.2 \times 10^3 + 4.5 \times 10^2 \) cannot be added as it stands, because the two terms are not counting the same thing. Rewriting \( 4.5 \times 10^2 \) as \( 0.45 \times 10^3 \) makes them comparable, and only then do the leading numbers add.

Engineering Notation and the SI Prefixes

Engineering notation is standard form with one extra restriction: the exponent must be a multiple of three, so the leading number sits anywhere from 1 up to 1,000. The point of the restriction is that every allowed exponent corresponds to an SI prefix, so a value written this way can be read straight off in the units an engineer actually uses.

• \( 4.7 \times 10^{-5} \) farads is \( 47 \times 10^{-6} \) F, read as 47 µF.
• \( 2.2 \times 10^{6} \) ohms is already at a multiple of three, and reads as 2.2 MΩ.
• \( 1.5 \times 10^{9} \) hertz reads as 1.5 GHz.
• \( 8.2 \times 10^{-11} \) farads becomes \( 82 \times 10^{-12} \) F, or 82 pF.

This is why component markings, test-set ranges and maintenance data step in thousands rather than in arbitrary powers, and it is worth being fluent in both directions — a resistance quoted as 2,200,000 Ω and one quoted as 2.2 MΩ are the same reading, and an insulation-test limit is far easier to check against a meter when both are expressed the same way.