NSWMaths Standard 2Syllabus dot point

How do you evaluate a formula correctly once you replace its pronumerals with numbers?

Substitute numerical values into a formula or algebraic expression and evaluate it, including expressions with powers, fractions and negative numbers

A focused answer to the HSC Maths Standard 2 dot point on substitution. Replacing pronumerals with values, the order of operations, the rule for powers of negatives, brackets and fractions, calculator layout, units and rounding, with worked Australian examples for a tradie's quote, a tank's surface area and stopping distance.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to take a formula or algebraic expression and replace each pronumeral (each letter) with the number given for it. You then evaluate (work out) the result correctly, and state it to the required accuracy with the right units. The arithmetic is mechanical, but almost every lost mark here comes from one of three things: getting the order of operations wrong, mishandling a power of a negative number, or forgetting the units and rounding the question asked for. The real skill is reading a formula as a set of instructions in a fixed order, not just a string of symbols. Read it that way and you can key it into the calculator exactly as written.

The answer

The method, in order

Substitution is four steps, always in this order:

Write the formula exactly as given. Do not start punching the calculator before the formula is on the page; the written line is where the method marks live.
Replace each pronumeral with its value, putting every negative value in brackets. A pronumeral can appear more than once (as $n$ does in the profit formula above) and every copy gets the same value.
Evaluate following the order of operations. This is the step the calculator does, but only if you have entered the brackets and powers faithfully.
State the answer to the accuracy asked for, with the correct units.

The diagram makes step 2 concrete: each given number drops into the slot held by its pronumeral. The fixed call-out fee $90 is the constant (it is there whether or not any hours are worked), the $75 is the coefficient of $h$ (the per-hour amount), and the value $h = 3.5$ is what actually changes from job to job. Substituting $h = 3.5$ into $C = 90 + 75h$ gives $C = 90 + 75 \times 3.5$ , that is $352.50.

Order of operations is the whole game

When you evaluate the substituted line, the operations are not done left to right. They follow the standard priority, often remembered as BIDMAS or BODMAS:

Brackets first,
Indices (powers and roots) next,
Division and Multiplication next, left to right,
Addition and Subtraction last, left to right.

In $C = 90 + 75 \times 3.5$ the multiplication $75 \times 3.5 = 262.5$ happens before the $90$ is added, giving $352.5$ . If you (or a basic calculator) added $90 + 75$ first you would get a meaningless $165 \times 3.5$ . A scientific calculator follows BIDMAS for you. So the danger is not the calculator's arithmetic but what you type into it. Every bracket and power in the formula has to be entered, or the machine works out a different expression from the one written.

Powers of negative numbers: the silent trap

The single most common substitution error in this course is squaring a negative number. The rule is:

$(-2)^2$ means $(-2) \times (-2) = 4$ (a positive answer), but
$-2^2$ means $-(2^2) = -4$ ,

A power binds more tightly than the minus sign in front of it. So when you substitute a negative value into a squared term you must put it in brackets. If a formula contains $c^2$ and you are told $c = -2$ , write $(-2)^2 = 4$ , not $-2^2 = -4$ . The same care applies to a calculator. On many models typing -2 x^2 computes $-(2^2) = -4$ , whereas (-2) x^2 (with the brackets) gives $4$ . Bracket every negative as you substitute and this trap disappears.

Fractions and brackets

A horizontal fraction bar acts as an invisible bracket around its top and around its bottom. Everything on top is grouped, and so is everything on the bottom. When you write a fraction on one line you must put those brackets back. In the trapezium area $A = \dfrac{a + b}{2}\,h$ , the sum $a + b$ is grouped, so on one line it becomes $A = (a + b) \div 2 \times h$ . Drop the bracket and you would wrongly compute $a + \dfrac{b}{2} \times h$ . Likewise in $\text{BMI} = \dfrac{m}{h^2}$ only the $h$ is squared and only the bottom is divided, so square first, then divide.

Units and rounding

A substituted answer is not finished until it carries the right unit and the right accuracy. If the lengths went in as metres, an area comes out in square metres and a volume in cubic metres. If the question says "to the nearest whole metre" or "correct to two decimal places", do the rounding at the very end, never on the values in between. Rounding early and then carrying on lets a small error grow. Keep full precision on the calculator and round only the final line.

How exam questions ask about substitution

The wording varies, but each version maps onto the same four-step method:

"Evaluate / find the value of [expression] when..." This is pure substitution: replace the pronumerals and evaluate. The accuracy is usually stated; if not, give an exact value or a sensible rounding.
"Use the formula to find..." A formula is supplied (often a measurement, finance or NESA-named formula like stopping distance or BAC) and you substitute the given quantities. Write the formula line first.
"Calculate [a quantity] correct to N decimal places / the nearest..." Same substitution, but the marks include the rounding, so keep full precision until the last line and then round.
"...correct to N significant figures" or "...with appropriate units" are explicit reminders that the final presentation is being marked, not just the number.
"What does this value represent / interpret your answer" asks you to read the substituted result in context, including its sign (a negative profit is a loss, a negative change is a decrease).
A value appears more than once in the formula (for example $n$ in a profit model): substitute that same value into every occurrence.

Substituting into three Australian formulae

Three different formulae are evaluated by substitution: a tradie's quote (a multi-term expression with a constant and a coefficient), the surface area of a closed cylinder (a formula with a power and $\pi$ ), and a stopping distance (a formula combining a fraction and a squared term), plus a signed example to drill the negatives rule. Each one follows the same four steps.

Quote a tradie's call-out job

An electrician charges a fixed call-out fee of $90 plus $75 for each hour on the job, so the cost of a job lasting $h$ hours is the formula $C = 90 + 75h$ dollars. Quote a job that takes $h = 3.5$ hours.

Write the formula and substitute. Replace $h$ with $3.5$ , keeping the multiplication explicit:

C = 90 + 75 \times 3.5

Evaluate, multiplication before addition. The order of operations multiplies first:

75 \times 3.5 = 262.5

then adds the fixed fee:

C = 90 + 262.5 = 352.5

State the answer with units. The quote is $C = 352.5$ , that is $352.50. Writing it to two decimal places is the natural form for money even though the raw number is $352.5$ .

Surface area of a closed cylindrical tank

A closed cylinder has surface area $SA = 2\pi r (r + h)$ , where $r$ is the base radius and $h$ is the height. A water tank has radius $r = 0.6$ m and height $h = 1.5$ m. Find its surface area correct to two decimal places.

Write the formula and substitute. Replace $r$ with $0.6$ and $h$ with $1.5$ :

SA = 2 \times \pi \times 0.6 \times (0.6 + 1.5)

Work the bracket first. Brackets have top priority:

0.6 + 1.5 = 2.1

Multiply the rest, left to right. Now the expression is a chain of multiplications:

SA = 2 \times \pi \times 0.6 \times 2.1 \approx 7.9168

Round and add units. $SA \approx 7.92$ m $^2$ . The units are square metres because surface area is an area. A frequent slip is to evaluate $2\pi r \times r + h$ , forgetting the bracket; that gives only $3.76$ , which is far too small for a tank this size, a useful sanity check.

Stopping distance on the freeway

The stopping distance of a car in metres is $d = \dfrac{5Vt}{18} + \dfrac{V^2}{170}$ , where $V$ is the speed in km/h and $t$ is the driver's reaction time in seconds. Find the stopping distance at a freeway speed of $V = 100$ km/h for a tired driver with reaction time $t = 1.5$ s, correct to the nearest metre.

Write the formula and substitute. Replace $V$ with $100$ and $t$ with $1.5$ in both terms:

d = \frac{5 \times 100 \times 1.5}{18} + \frac{100^2}{170}

Evaluate the reaction term. The numerator is $5 \times 100 \times 1.5 = 750$ , then divide by $18$ :

\frac{750}{18} \approx 41.67

Evaluate the braking term. Square first ( $100^2 = 10\,000$ ), then divide by $170$ :

\frac{10\,000}{170} \approx 58.82

Add and round. $d \approx 41.67 + 58.82 = 100.49$ m, which is $100$ m to the nearest metre. At freeway speed a tired driver needs roughly the length of a football field to stop, which is exactly the kind of real-world reading this formula is designed to give.

Substitute a set of negative values

Evaluate $T = a - 5b + c^2$ when $a = 3$ , $b = -4$ and $c = -2$ . This one is all about handling the signs.

Write the formula and substitute, bracketing every negative. Put $-4$ and $-2$ in brackets:

T = 3 - 5 \times (-4) + (-2)^2

Apply the power first. Because the negative is inside the bracket, the square is positive:

(-2)^2 = 4

Then the multiplication. A negative times a negative is positive:

-5 \times (-4) = +20

Finally add and subtract, left to right.

T = 3 + 20 + 4 = 27

Had you written $-2^2$ without the bracket you would have got $-4$ instead of $+4$ , and the final answer would have been a wrong $19$ . The bracket is what saves the mark.

Final answers: the tradie's quote is $C = 352.5$ , that is $352.50; the tank's surface area is $SA \approx 7.92$ m $^2$ ; the tired driver's stopping distance at $100$ km/h is about $100$ m; and the signed expression evaluates to $T = 27$ .

Common traps

Adding before multiplying: In $90 + 75 \times 3.5$ the multiplication is done first ( $262.5$ ), then $90$ is added. Working strictly left to right gives the wrong answer.
Squaring a negative without brackets: $(-2)^2 = 4$ , but $-2^2 = -4$ . Always bracket a negative value before applying a power, on paper and on the calculator.
Dropping the fraction's invisible brackets: A fraction bar groups its whole top and whole bottom. Written on one line, $\dfrac{a+b}{2}$ must become $(a + b) \div 2$ , not $a + b \div 2$ .
Forgetting that only part of a term is powered: In $\dfrac{m}{h^2}$ only the $h$ is squared. Square $h$ first, then divide; do not square the whole fraction.
Rounding too early: Round only the final answer. Rounding the intermediate values and then continuing lets the error grow.
Dropping the units or the right power of the unit: An area is in square units, a volume in cubic units. A bare number, or metres where it should be square metres, loses the presentation mark.
Substituting a repeated pronumeral only once: If a letter appears twice in the formula, every occurrence takes the same value.

Exam technique

Always write the formula line before you substitute; that line, plus the substituted line, is where the working marks are awarded even if the final arithmetic slips. Bracket every negative value as you substitute, and bracket the whole top and bottom of any fraction when you flatten it to one line, because that is what your calculator needs to see. Key the expression into a scientific calculator in one go using its bracket and power keys rather than computing pieces and re-entering them, which avoids transcription errors and early rounding. Keep full precision on the display and round only on the last line, to exactly the accuracy the question states (decimal places, significant figures, or "nearest..."), and finish with the correct units. If an answer looks wrong (a tank with $3.76$ m $^2$ of surface, a BMI of $43$ ), check the order of operations and the brackets before anything else.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC-style2 marksThe simple interest earned on an investment is

I = \dfrac{Prn}{100}

, where

P

is the principal in dollars,

r

is the annual rate as a percentage and

n

is the number of years. Find the interest earned when

P = 2500

r = 4.5

and

n = 3

Show worked answer →

Write the formula and substitute. Replace $P$ with $2500$ , $r$ with $4.5$ and $n$ with $3$ :

$I = \dfrac{2500 \times 4.5 \times 3}{100}$

Work the top, then divide. The fraction bar groups the whole numerator, so multiply across the top first: $2500 \times 4.5 \times 3 = 33\,750$ . Then divide by $100$ :

$I = \dfrac{33\,750}{100} = 337.5$

State the answer with units. $I = 337.5$ , that is $337.50 in interest.

Markers reward the substituted line, evaluating the numerator before dividing, and the final value $337.50 (a bare $337.5$ without the dollar sign usually drops the presentation mark).

2022 HSC-style3 marksThe volume of a closed cylindrical drum is

V = \pi r^2 h

, where

r

is the base radius and

h

is the height, both in metres. Find the volume of a drum with

r = 0.45

m and

h = 1.2

m, correct to two decimal places.

Show worked answer →

Write the formula and substitute. Replace $r$ with $0.45$ and $h$ with $1.2$ :

$V = \pi \times 0.45^2 \times 1.2$

Apply the power first. Order of operations squares the radius before any multiplication, and only the $r$ is squared:

$0.45^2 = 0.2025$

Multiply the chain, left to right. Now it is a string of multiplications:

$V = \pi \times 0.2025 \times 1.2 \approx 0.7634$

Round and add units. $V \approx 0.76$ m $^3$ . The units are cubic metres because a volume multiplies three lengths.

Markers reward squaring $0.45$ before multiplying (a common slip is $\pi \times 0.45 \times 1.2 = 1.70$ , far too large), the substituted line, and the rounded value $0.76$ m $^3$ with cubic units.

2023 HSC-style4 marksThe kinetic energy of a moving vehicle, in joules, is

E = \dfrac{1}{2}mv^2

, where

m

is the mass in kilograms and

v

is the speed in metres per second. (a) Find the kinetic energy of a

1200

kg car travelling at

v = 15

m/s. (b) The same car speeds up to

v = 20

m/s. Find how much its kinetic energy increases.

Show worked answer →

Part (a) - substitute the first speed. Put $m = 1200$ and $v = 15$ into the formula, keeping the square on the speed only:

$E = \dfrac{1}{2} \times 1200 \times 15^2$

Square before multiplying. Indices come before multiplication, so square first: $15^2 = 225$ . Then multiply the chain:

$E = \dfrac{1}{2} \times 1200 \times 225 = 600 \times 225 = 135\,000$ J

Part (b) - substitute the second speed. Replace $v$ with $20$ , again squaring first ( $20^2 = 400$ ):

$E = \dfrac{1}{2} \times 1200 \times 400 = 600 \times 400 = 240\,000$ J

Find the increase. Subtract the first energy from the second:

$240\,000 - 135\,000 = 105\,000$ J

So the kinetic energy rises by $105\,000$ J. Notice the speed went up by a factor of $\tfrac{20}{15}$ , yet the energy more than doubled, because $v$ is squared in the formula.

Markers reward squaring each speed before multiplying, the two energy values $135\,000$ J and $240\,000$ J with units, and the correct increase of $105\,000$ J.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksThe perimeter of a rectangle is

P = 2L + 2W

. Find

P

when

L = 8.5

cm and

W = 6

cm.

Show worked solution →

Write the formula and substitute. Replace $L$ with $8.5$ and $W$ with $6$ , keeping each multiplication explicit:

P = 2 \times 8.5 + 2 \times 6

Evaluate, multiplication first. Order of operations does the two multiplications before the addition:

P = 17 + 12 = 29

Answer with units. $P = 29$ cm. (Markers want the unit on a perimeter; a bare $29$ usually drops the final mark.)

foundation2 marksEvaluate

E = 5g - 2h

when

g = -3

and

h = 4

Show worked solution →

Substitute, bracketing the negative. Put $-3$ in brackets so the sign is not lost:

E = 5 \times (-3) - 2 \times 4

Multiply each term first. $5 \times (-3) = -15$ and $2 \times 4 = 8$ :

E = -15 - 8

Combine. Subtracting $8$ from $-15$ moves further negative:

E = -23

core3 marksBody mass index is

\text{BMI} = \dfrac{m}{h^2}

, where

m

is mass in kilograms and

h

is height in metres. Find the BMI of a person with

m = 72

kg and

h = 1.68

m, correct to one decimal place.

Show worked solution →

Substitute into the fraction. Replace $m$ with $72$ and $h$ with $1.68$ :

\text{BMI} = \frac{72}{1.68^2}

Apply the power before dividing. The square is on the denominator only, so square $1.68$ first:

1.68^2 = 2.8224

Divide. Now divide the mass by that result:

\text{BMI} = \frac{72}{2.8224} \approx 25.5102

Round to one decimal place. $\text{BMI} \approx 25.5$ . (A common slip is dividing $72$ by $1.68$ without squaring, which gives $42.86$ and is plainly too large for a BMI.)

core3 marksThe area of a trapezium is

A = \dfrac{a + b}{2} \, h

. Find

A

when

a = 6.4

b = 9.2

m and

h = 5

m, correct to two decimal places.

Show worked solution →

Substitute, keeping the bracket. The whole sum $a + b$ is over $2$ , so write it in a bracket:

A = \frac{(6.4 + 9.2)}{2} \times 5

Work inside the bracket first. Brackets come before the division:

6.4 + 9.2 = 15.6

Divide, then multiply. Halve the bracket, then multiply by the height:

\frac{15.6}{2} = 7.8, \qquad 7.8 \times 5 = 39

Answer with units. $A = 39.00$ m $^2$ . The units are square metres because an area multiplies two lengths.

exam4 marksThe stopping distance of a car (in metres) is modelled by

d = \dfrac{5Vt}{18} + \dfrac{V^2}{170}

, where

V

is the speed in km/h and

t

is the reaction time in seconds. (a) Find the stopping distance for an alert driver travelling at

V = 60

km/h with

t = 0.75

s, correct to the nearest metre. (b) A distracted driver at the same speed has

t = 1.5

s. Find the extra stopping distance, to the nearest metre.

Show worked solution →

Part (a) - substitute the alert driver's values. Put $V = 60$ and $t = 0.75$ into each term:

d = \frac{5 \times 60 \times 0.75}{18} + \frac{60^2}{170}

Evaluate the reaction term. Numerator $5 \times 60 \times 0.75 = 225$ , then divide by $18$ :

\frac{225}{18} = 12.5

Evaluate the braking term. Square first: $60^2 = 3600$ , then divide by $170$ :

\frac{3600}{170} \approx 21.18

Add and round. $d \approx 12.5 + 21.18 = 33.68$ m, which is $34$ m to the nearest metre.

Part (b) - only the reaction time changes. The braking term stays $\frac{3600}{170} \approx 21.18$ . Recompute the reaction term with $t = 1.5$ :

\frac{5 \times 60 \times 1.5}{18} = \frac{450}{18} = 25

so $d \approx 25 + 21.18 = 46.18$ m, which is $46$ m to the nearest metre.

Extra distance. $46.18 - 33.68 = 12.5$ m further, i.e. about $12$ to $13$ m of extra stopping distance from the slower reaction alone. (Doubling the reaction time doubled only the reaction term, not the whole answer - a good check that you treated the two terms separately.)

exam4 marksA market stall's daily profit is

P = 14n - 250 - 0.5n

dollars, where

n

is the number of items sold. (a) Find the profit when

n = 18

. (b) Find the profit when

n = 40

. (c) Interpret the sign of each answer.

Show worked solution →

Part (a) - substitute $n = 18$ . Replace every $n$ with $18$ :

P = 14 \times 18 - 250 - 0.5 \times 18

Evaluate the products first. $14 \times 18 = 252$ and $0.5 \times 18 = 9$ :

P = 252 - 250 - 9 = -7

so $P = -7$ , a profit of negative $7.

Part (b) - substitute $n = 40$ . Replace every $n$ with $40$ :

P = 14 \times 40 - 250 - 0.5 \times 40 = 560 - 250 - 20 = 290

so $P = 290$ , that is $290.

Part (c) - interpret the signs. At $n = 18$ the profit is negative (a value of $-7$ , i.e. negative $7), so selling only $18$ items makes a $7 loss that day. At $n = 40$ the profit is positive ($290), a genuine gain. The sign of a substituted answer carries meaning: a negative result here is a loss, not an error to be ignored.

What this dot point is asking

The answer

The method, in order

Order of operations is the whole game

Powers of negative numbers: the silent trap

Fractions and brackets

Units and rounding

How exam questions ask about substitution

Exam-style practice questions

Practice questions

Related dot points