Expressions and Formulae · Study Notes · Edexcel IGCSE 4MA1 Higher

Substitute carefully, derive formulae from words, evaluate using BIDMAS. Trivial when done right; a constant source of careless errors when done wrong.

At a glance

Substitute — replace variable with value. Use BRACKETS for negatives.
Evaluate — apply BIDMAS strictly.
Derive formulae from word problems: identify structure first.
Show every step — gain method marks even if arithmetic slips.
$(-2)^{2} = 4$ but $-2^{2} = -4$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.2 — Substitute numerical values into expressions and formulae.
2.2 — Evaluate expressions using BIDMAS.
2.2 — Derive simple formulae from word problems.

Substitution — the disciplined approach

▼

Brackets around negatives. BIDMAS strictly. Show working.

Standard procedure:

Write the formula.
Substitute values, using BRACKETS for any negative.
Apply BIDMAS:
- Brackets first.
- Indices (powers).
- Division and Multiplication (left to right).
- Addition and Subtraction.

Example. $V = \pi r^{2} h$ . Find $V$ when $r = 3$ , $h = 7$ . (Use $\pi \approx 3.14$ .)

Substitute: $V = 3.14 \times (3)^{2} \times 7$ .
Compute: $V = 3.14 \times 9 \times 7 = 197.82$ .

Negative values demand brackets.

If $x = -3$ :

$x^{2} = (-3)^{2} = 9$ .
$-x^{2} = -(-3)^{2} = -9$ . (Negative outside the squaring.)
$(-x)^{2} = (3)^{2} = 9$ . (Negate first, then square.)

Brackets decide whether the minus sign is squared — always bracket a substituted negative.

Worked example. Evaluate $3a^{2} - 2ab + b^{2}$ when $a = -2$ and $b = 4$ .

Substitute with brackets: $3(-2)^{2} - 2(-2)(4) + (4)^{2}$

BIDMAS:

Powers: $(-2)^{2} = 4$ . $(4)^{2} = 16$ .
Multiplication: $3 \times 4 = 12$ . $2 \times (-2) \times 4 = -16$ .
Combine: $12 - (-16) + 16 = 12 + 16 + 16 = 44$ .

Edexcel tip. Show every step. Mark schemes give M1 for substitution and A1 for the answer. If your final value is wrong but substitution is right, you still get the M1.

Substitute first.
Brackets for negatives.
BIDMAS strictly.
Show every step.

Deriving formulae from word problems

▼

Identify the structure. Choose letters carefully. Build the equation.

Process:

Read the problem.
Identify the QUANTITIES involved. Define LETTERS for each.
Identify the RELATIONSHIP (sum, product, fraction, percentage).
Write the formula.

Example: Taxi tariff.

Initial fee: £ $2$ .
Per-km rate: £ $1.50$ .
Distance: $d$ km.
Total cost: $C$ .

Structure: fixed + (rate × distance). Formula: $C = 2 + 1.5d$ .

Example: Compound interest.

Principal: $P$ .
Rate: $r\%$ per year.
Years: $n$ .
Final: $A$ .

$A = P \times \left(1 + \dfrac{r}{100}\right)^{n}$ .

Example: Geometric. A rectangle has length $L$ and width $W$ . Find the perimeter.

Perimeter $= L + W + L + W = 2L + 2W = 2(L + W)$ .

Combining formulae.

When two formulae share a variable, you can substitute one into the other.

Example: A cone with radius half the height. Volume?

Cone volume: $V = \dfrac{1}{3}\pi r^{2} h$ .
Constraint: $r = h/2$ , so $h = 2r$ .
Substitute: $V = \dfrac{1}{3}\pi r^{2} \cdot 2r = \dfrac{2}{3}\pi r^{3}$ .

Worked qualitative. A rope is cut into 3 pieces. The first is twice the length of the second; the third is $5$ cm longer than the first. Total length is $35$ cm. Find each piece.

Let second piece = $x$ .
First = $2x$ .
Third = $2x + 5$ .
Sum: $x + 2x + 2x + 5 = 35$ , so $5x = 30$ , $x = 6$ .
Pieces: $6, 12, 17$ cm.

Edexcel tip. Always DEFINE variables explicitly. 'Let $x$ be ...' earns the first mark.

Define letters first.
Identify the structure.
Combine via substitution.
Always state 'Let $x$ be ...'.

BIDMAS in formulae

▼

Order of operations. Brackets, Indices, Division/Multiplication, Addition/Subtraction.

The rules:

Brackets — innermost first.
Indices — powers and roots.
Division and Multiplication — left to right (not strict order between these two).
Addition and Subtraction — left to right.

Examples:

Work down the list: Brackets and Indices outrank Multiply/Divide, which outrank Add/Subtract.

$3 + 5 \times 2 = 3 + 10 = 13$ (NOT $8 \times 2 = 16$ ).

$(3 + 5) \times 2 = 8 \times 2 = 16$ .

$2^{3} + 4 = 8 + 4 = 12$ .

$2 \cdot (3 + 4)^{2} = 2 \cdot 49 = 98$ .

Common pitfalls.

$-2^{2} = -4$ but $(-2)^{2} = 4$ .
$1 / 2 \cdot 4 = 2$ (left to right: $0.5 \times 4 = 2$ , NOT $1 / 8$ ).
$2(3) = 2 \cdot 3 = 6$ (juxtaposition is multiplication).

With formulae.

For $V = \dfrac{1}{2}bh$ where $b = 6, h = 8$ :

$V = \dfrac{1}{2} \times 6 \times 8 = 24$ .

For $\dfrac{a + b}{c}$ where $a = 3, b = 5, c = 2$ :

Numerator first: $3 + 5 = 8$ .
Then divide: $8 / 2 = 4$ .

Worked qualitative. Why is $4 + 3 \times 2 \ne 14$ ?

BIDMAS says multiplication BEFORE addition.
$3 \times 2 = 6$ first.
Then $4 + 6 = 10$ .
Calculators handle this automatically — but writing intermediate steps prevents confusion.

Edexcel tip. When a formula has multiple operations, write each step on a new line: substitution, simplification, evaluation. Mark schemes credit each transition.

BIDMAS: Brackets > Indices > D/M > A/S.
Brackets innermost first.
D and M same precedence, left to right.
Indices include powers AND roots.

Quick recap

Substitute first; brackets for negatives.
BIDMAS strictly.
Derive: define letters, identify structure.
Show every step for method marks.
$(-x)^{2} \ne -x^{2}$ in general.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Substitute with brackets for negatives.
BIDMAS order of operations.
Define variables before using.
Show every step — earn method marks.

How it’s examined

Substitution and formulae appear regularly Paper 1H + 2H (3-5 marks). Often combined with geometry, finance, or physical contexts. Examiner reports flag (1) substituting negatives without brackets, (2) BIDMAS errors, (3) skipping the substitution step.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on expressions and formulae, written exactly the way a tutor would explain them at the board.

1Substitute values into a formula
Foundation• substitution
▼
Question
Given $A = \dfrac{1}{2}bh$ , find $A$ when $b = 8$ and $h = 5$ .
Step-by-step solution
1. Step 1
  Substitute the values directly:
  $A = \frac{1}{2} \times 8 \times 5$
2. Step 2
  Compute: $A = 20$ .
Answer
$A = 20$
Examiner tip
Always show the substitution step. Mark schemes give M1 for the substitution and A1 for the answer.
2Substitute negative values
Foundation• Adapted from 4MA1/1H May/Jun 2024 Q4• substitution, negatives
▼
Question
Find the value of $3a^{2} - 2ab + b^{2}$ when $a = -2$ and $b = 4$ .
Step-by-step solution
1. Step 1
  Substitute carefully — use BRACKETS for negative values.
  $3(-2)^{2} - 2(-2)(4) + (4)^{2}$
2. Step 2
  Compute each: $3 \times 4 - 2 \times (-8) + 16 = 12 + 16 + 16$ .
3. Step 3
  Sum: $44$ .
Answer
$44$
Examiner tip
ALWAYS use brackets when substituting negative values. $(-2)^{2} = 4$ but $-2^{2} = -4$ . Without brackets, you'd get the wrong sign.
3Derive a formula from a word problem
Higher• formula, deriving
▼
Question
A taxi charges £ $2$ initial fee + £ $1.50$ per km. Write a formula for the total cost £ $C$ for a $d$ -km journey.
Step-by-step solution
1. Step 1
  Identify the structure: fixed fee + (rate × distance).
2. Step 2
  $C = 2 + 1.5d$ .
Answer
$C = 2 + 1.5d$
Examiner tip
Define variables clearly. Use simple letters: $d$ for distance, $C$ for cost. Show the formula derived from the structure.
4Use multiple formulae together
Higher• formulae, multi-step
▼
Question
The volume of a cone is $V = \dfrac{1}{3}\pi r^{2} h$ . The radius is half the height. Find $V$ in terms of $r$ alone.
Step-by-step solution
1. Step 1
  From 'radius half of height': $h = 2r$ .
2. Step 2
  Substitute into $V$ :
  $V = \frac{1}{3}\pi r^{2}(2r) = \frac{2}{3}\pi r^{3}$
Answer
$V = \dfrac{2}{3}\pi r^{3}$
Examiner tip
Multi-step formula problems test substitution AND simplification. Show every step.
5Writing a product in standard notation [1 mark]
Getting started• Paper 1H/2H style — 1 mark• notation, simplifying
▼
Question
Simplify $a \times b \times 3 \times a$ [1]
Step-by-step solution
1. Step 1
  Algebra has conventions: leave out the $\times$ signs, put the number first, write $a \times a$ as $a^{2}$ , and put the letters in alphabetical order.
2. Step 2
  B1 Collect the number, then the letters.
  $3 \times a \times a \times b = 3a^{2}b$
Answer
$3a^{2}b$
Examiner tip
Examiner tip: $a \times a$ is $a^{2}$ , not $2a$ . The index counts how many times the letter is multiplied.

Common mistake: Writing $3 \times 2ab$ or $6ab$ , which treats $a \times a$ as $a + a$ .
6Substituting a negative decimal into a cube [2 marks]
Getting started• Paper 1H/2H style — 2 marks• substitution, negative numbers, powers
▼
Question
$P = 4r^{3} - 5r$

Work out the value of $P$ when $r = -1.5$ [2]
Step-by-step solution
1. Step 1
  M1 Substitute with brackets round the negative decimal.
  $P = 4 \times (-1.5)^{3} - 5 \times (-1.5)$
2. Step 2
  A1 A negative number cubed stays negative, and subtracting a negative is adding.
  $P = 4 \times (-3.375) + 7.5 = -13.5 + 7.5 = -6$
Answer
$P = -6$
Examiner tip
Examiner tip: Odd powers of a negative number are negative, and even powers are positive. $(-1.5)^{3} = -3.375$ .

Common mistake: Writing $-5 \times -1.5$ as $-7.5$ , which gives $P = -21$ . A sign slip like this is the most common way to lose the A mark.
7Formula from a chain of comparisons [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• deriving formulae, writing expressions
▼
Question
Sam has $x$ sweets.

Priya has $7$ fewer sweets than Sam.

Leon has half as many sweets as Priya.

The total number of sweets they have is $N$ .

Write a formula for $N$ in terms of $x$ . Give your answer in its simplest form. [3]
Step-by-step solution
1. Step 1
  M1 Priya's number of sweets.
  $x - 7$
2. Step 2
  M1 Leon has half of Priya's, so halve the whole of $x - 7$ .
  $\frac{x - 7}{2}$
3. Step 3
  A1 Add all three and simplify, writing the formula with $N$ as the subject.
  $N = x + (x - 7) + \frac{x - 7}{2} = \frac{2x + 2x - 14 + x - 7}{2} = \frac{5x - 21}{2}$
Answer
$N = \dfrac{5x - 21}{2}$ (or $N = 2.5x - 10.5$ )
Examiner tip
Examiner tip: Check with a number. If $x = 11$ , Priya has $4$ and Leon has $2$ , a total of $17$ , and $\dfrac{5 \times 11 - 21}{2} = 17$ .

Common mistake: Writing Leon's number as $\dfrac{x}{2} - 7$ , which halves only the $x$ . It needs a bracket or a full fraction line.
8Substituting into a formula with a cube root [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• substitution, roots, accuracy
▼
Question
The radius, $r$ cm, of a sphere with volume $V$ cm³ is given by

$r = \sqrt[3]{\frac{3V}{4\pi}}$

Work out the value of $r$ when $V = 500$ . Give your answer correct to 3 significant figures. [3]
Step-by-step solution
1. Step 1
  M1 Substitute the value.
  $r = \sqrt[3]{\frac{3 \times 500}{4\pi}} = \sqrt[3]{\frac{1500}{12.566\ldots}}$
2. Step 2
  M1 Evaluate the fraction under the root.
  $\frac{1500}{12.566\ldots} = 119.366\ldots$
3. Step 3
  A1 Cube root, then round at the very end.
  $r = \sqrt[3]{119.366\ldots} = 4.9237\ldots = 4.92 \text{ (3 s.f.)}$
Answer
$r = 4.92$ (3 s.f.)
Examiner tip
Examiner tip: On the calculator, put the whole denominator $4\pi$ in brackets, or work out the fraction first and write it down. Check: $\frac{4}{3}\pi \times 4.9237^{3} \approx 500$ .

Common mistake: Typing $1500 \div 4 \times \pi$ , which multiplies by $\pi$ instead of dividing, and gives $r = 10.6$ . Taking a square root instead of a cube root gives $10.9$ . Both are far too big for a sphere of volume $500$ cm³.
9Two-part hire charge, then solving [4 marks]
Stretch• Paper 1H/2H style — 4 marks• deriving formulae, linear equations
▼
Question
A company hires out vans. It charges £ $55$ per day, plus $30$ p for every kilometre driven over the first $100$ km.

(a) Write a formula for the total charge, £ $C$ , for hiring a van for $d$ days and driving $m$ kilometres, where $m > 100$ . [2]

(b) Jack hires a van for $3$ days. The total charge is £ $231.60$ . Work out how many kilometres Jack drove. [2]
Step-by-step solution
1. Step 1
  M1 (a) Only the kilometres over $100$ are charged, which is $m - 100$ km. $30$ p is £ $0.30$ .
  $0.3(m - 100)$
2. Step 2
  A1 Add the daily charge.
  $C = 55d + 0.3(m - 100)$
3. Step 3
  M1 (b) Substitute $d = 3$ and $C = 231.6$ , then solve.
  $231.6 = 165 + 0.3(m - 100) \;\Rightarrow\; 0.3(m - 100) = 66.6 \;\Rightarrow\; m - 100 = 222$
4. Step 4
  A1 Add back the first $100$ km.
  $m = 322$
Answer
(a) $C = 55d + 0.3(m - 100)$ oe $\quad$ (b) $322$ km
Examiner tip
Examiner tip: Keep everything in pounds, because $C$ is in pounds. Check (b): $165 + 0.3 \times 222 = 231.6$ .

Common mistake: Using $0.3m$ instead of $0.3(m - 100)$ , which charges for the free kilometres as well. It gives $m = 222$ in part (b), which is the number of charged kilometres, not the total.
10Triangle and rectangle with equal perimeters [5 marks]
Stretch• Paper 1H/2H style — 5 marks• deriving formulae, perimeter, linear equations
▼
Question
An isosceles triangle has sides of length $(3x - 1)$ cm, $(3x - 1)$ cm and $(2x + 4)$ cm.

A rectangle has length $(2x + 5)$ cm and width $x$ cm.

(a) Find an expression, in terms of $x$ , for the perimeter of the triangle. Give your answer in its simplest form. [1]

(b) The triangle and the rectangle have the same perimeter. Work out the area of the rectangle. [4]
Step-by-step solution
1. Step 1
  B1 (a) Add the three sides.
  $2(3x - 1) + (2x + 4) = 8x + 2$
2. Step 2
  M1 (b) Perimeter of the rectangle.
  $2(2x + 5) + 2x = 6x + 10$
3. Step 3
  M1 Set the perimeters equal and solve.
  
  A1 $x = 4$
  $8x + 2 = 6x + 10 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4$
4. Step 4
  A1 The rectangle is $13$ cm by $4$ cm.
  $(2 \times 4 + 5) \times 4 = 13 \times 4 = 52$
Answer
(a) $8x + 2$ $\quad$ (b) $52$ cm²
Examiner tip
Examiner tip: Check the shapes are real. With $x = 4$ the triangle is $11$ , $11$ , $12$ and both perimeters are $34$ cm.

Common mistake: Stopping at $x = 4$ , or giving the perimeter $34$ . The question asks for the area, so the last mark needs $52$ cm².

Question 1
Paper 1H/2H style — 1 mark1 mark
Here are two statements.

A $\quad 5n - 3 = 17$

B $\quad P = 5n - 3$

In one statement $n$ is an unknown. In the other $n$ is a variable.

Write down the letter of the statement in which $n$ is a variable. [1]
Model answer
In A, $n$ has one fixed value ( $n = 4$ ), so it is an unknown.

In B, $n$ can take any value, and $P$ changes with it, so $n$ is a variable.

Answer: B
Why this scores
Mark scheme:
- B1 for B
Why this scores full marks: An equation (A) is solved to find its unknown. A formula (B) describes how one quantity varies as another changes, so its letters are variables.
Question 2
Paper 1H/2H style — 2 marks2 marks
$T = \dfrac{3}{4}m^{2} - \sqrt{n}$

Work out the value of $T$ when $m = -\dfrac{2}{3}$ and $n = 0.36$

Give your answer as a fraction in its simplest form. [2]
Model answer
$T = \frac{3}{4} \times \left(-\frac{2}{3}\right)^{2} - \sqrt{0.36}$

$= \frac{3}{4} \times \frac{4}{9} - 0.6 = \frac{1}{3} - \frac{3}{5}$

$= \frac{5}{15} - \frac{9}{15}$

Answer: $-\dfrac{4}{15}$
Why this scores
Mark scheme:
- M1 for $\dfrac{3}{4} \times \left(-\dfrac{2}{3}\right)^{2} - \sqrt{0.36}$ oe, or $\dfrac{1}{3}$ and $0.6$ oe both seen
- A1 for $-\dfrac{4}{15}$ cao
Why this scores full marks: The negative fraction goes into a bracket before it is squared, so $\left(-\dfrac{2}{3}\right)^{2} = +\dfrac{4}{9}$ . Without the bracket a calculator gives $-\dfrac{4}{9}$ , and the answer becomes $-\dfrac{1}{3} - \dfrac{3}{5} = -\dfrac{14}{15}$ . The answer is left as a fraction, as the question asks; $-0.2\dot{6}$ does not score the A1.
Question 3
Paper 1H/2H style — 2 marks2 marks
A cuboid measures $x$ cm by $2x$ cm by $5$ cm.

Write a formula for the volume, $V$ cm³, of the cuboid in terms of $x$ . Give your answer in its simplest form. [2]
Model answer
$V = x \times 2x \times 5$

Answer: $V = 10x^{2}$
Why this scores
Mark scheme:
- M1 for $x \times 2x \times 5$ oe
- A1 for $V = 10x^{2}$
Why this scores full marks: It multiplies the numbers ( $2 \times 5 = 10$ ) and the letters ( $x \times x = x^{2}$ ) separately, and writes a formula with $V =$ , not just the expression $10x^{2}$ . Adding the edges to get $3x + 5$ is the usual wrong method.
Question 4
Paper 1H/2H style — 2 marks2 marks
Show that $4(3x - 2) - 3(2x - 5) \equiv 6x + 7$ [2]
Model answer
$4(3x - 2) - 3(2x - 5) = 12x - 8 - 6x + 15$

$= 6x + 7$

The left-hand side simplifies to the right-hand side for all $x$ , so $4(3x - 2) - 3(2x - 5) \equiv 6x + 7$ .
Why this scores
Mark scheme:
- M1 for expanding both brackets with at least three of the four terms correct
- A1 for $12x - 8 - 6x + 15$ simplified to $6x + 7$ , with no errors
Why this scores full marks: The $-3$ multiplies both terms in the second bracket, so $-3 \times -5 = +15$ . Writing $-15$ is a sign error examiners regularly see when a negative multiplies a bracket, and it leads to $6x - 23$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
The area, $A$ , of a triangle with sides $a$ , $b$ and $c$ is given by

$A = \sqrt{s(s - a)(s - b)(s - c)} \quad \text{where } s = \frac{a + b + c}{2}$

A triangle has sides $7$ cm, $8$ cm and $9$ cm.

Work out its area. Give your answer correct to 3 significant figures. [3]
Model answer
$s = \frac{7 + 8 + 9}{2} = 12$

$A = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720}$

$= 26.832\ldots$

Answer: $26.8$ cm²
Why this scores
Mark scheme:
- M1 for $s = 12$
- M1 for $\sqrt{12 \times 5 \times 4 \times 3}$ or $\sqrt{720}$
- A1 for awrt $26.8$
Why this scores full marks: The two-stage formula is used in order: $s$ first, then each bracket. Writing $\sqrt{720}$ before evaluating secures the method marks even if the decimal were mistyped.
Question 6
Paper 1H/2H style — 3 marks3 marks
The total surface area, $A$ cm², of a cylinder of radius $r$ cm and height $h$ cm is given by

$A = 2\pi r^{2} + 2\pi rh$

A cylinder has radius $4$ cm and total surface area $400$ cm².

Work out the height of the cylinder. Give your answer correct to 3 significant figures. [3]
Model answer
$400 = 2\pi \times 4^{2} + 2\pi \times 4 \times h$

$400 = 32\pi + 8\pi h$

$h = \frac{400 - 32\pi}{8\pi} = 11.915\ldots$

Answer: $11.9$ cm
Why this scores
Mark scheme:
- M1 for $400 = 2\pi \times 4^{2} + 2\pi \times 4 \times h$
- M1 for $h = \dfrac{400 - 32\pi}{8\pi}$ oe
- A1 for awrt $11.9$
Why this scores full marks: Keeping $32\pi$ and $8\pi$ exact until the last line avoids premature rounding. Only the $r$ is squared: $2\pi \times 4^{2}$ , not $(2\pi \times 4)^{2}$ .
Question 7
Paper 1H/2H style — 3 marks3 marks
$\dfrac{ax - 12}{4} + \dfrac{x}{2} \equiv 3x + b$

Find the value of $a$ and the value of $b$ . [3]
Model answer
Multiply every term by $4$ :

$ax - 12 + 2x \equiv 12x + 4b$

$(a + 2)x - 12 \equiv 12x + 4b$

$x$ terms: $a + 2 = 12$ , so $a = 10$

Constant terms: $4b = -12$ , so $b = -3$

Answer: $a = 10$ , $b = -3$
Why this scores
Mark scheme:
- M1 for $ax - 12 + 2x = 12x + 4b$ oe, or $\dfrac{a}{4}x - 3 + \dfrac{1}{2}x$ seen
- A1 for $a = 10$
- A1 for $b = -3$
Why this scores full marks: Because $\equiv$ holds for every $x$ , the script clears the fractions and then matches the $x$ terms and the constant terms separately. Every term is multiplied by $4$ , including $b$ on the right. Leaving it as $b$ gives $b = -12$ , the most common slip. Check: $\dfrac{10x - 12}{4} + \dfrac{x}{2} = 2.5x - 3 + 0.5x = 3x - 3$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
Aisha drives for $t$ hours at an average speed of $64$ km/h.

She then walks for $20$ minutes at an average speed of $w$ km/h.

The total distance she travels is $D$ km.

(a) Write a formula for $D$ in terms of $t$ and $w$ . [2]

(b) Aisha drives for $1.5$ hours and travels $97.5$ km in total. Work out her walking speed. [2]
Model answer
(a) Distance $=$ speed $\times$ time. $20$ minutes $= \dfrac{1}{3}$ hour.

$D = 64t + \frac{w}{3}$

(b)

$97.5 = 64 \times 1.5 + \frac{w}{3} = 96 + \frac{w}{3}$

$\frac{w}{3} = 1.5$

Answer: $w = 4.5$ km/h
Why this scores
Mark scheme:
- M1 (a) for $64t$ or $\dfrac{w}{3}$ oe seen
- A1 for $D = 64t + \dfrac{w}{3}$ oe
- M1 (b) for $97.5 = 96 + \dfrac{w}{3}$ oe (ft their formula)
- A1 for $4.5$ cao
Why this scores full marks: The time is changed to hours before it is used, because the speed is in km/h. Writing $20w$ is the usual error and gives an impossible walking speed of $0.075$ km/h.
Question 9
Paper 1H/2H style — 5 marks5 marks
The stopping distance, $d$ metres, of a car travelling at $v$ km/h is given by

$d = 0.2v + 0.006v^{2}$

(a) Work out the stopping distance at $50$ km/h. [2]

(b) Show that the stopping distance at $100$ km/h is more than twice the stopping distance at $50$ km/h. [3]
Model answer
(a)

$d = 0.2 \times 50 + 0.006 \times 50^{2} = 10 + 15$

$d = 25 \text{ m}$

(b)

$d = 0.2 \times 100 + 0.006 \times 100^{2} = 20 + 60 = 80 \text{ m}$

Twice the stopping distance at $50$ km/h is $2 \times 25 = 50$ m.

$80 > 50$ , so the stopping distance at $100$ km/h is more than twice the stopping distance at $50$ km/h.
Why this scores
Mark scheme:
- M1 (a) for $0.2 \times 50 + 0.006 \times 50^{2}$
- A1 for $25$
- M1 (b) for $0.2 \times 100 + 0.006 \times 100^{2}$
- A1 for $80$
- A1 for comparing $80$ with $2 \times 25 = 50$ and a correct conclusion
Why this scores full marks: Only $v$ is squared, so $0.006 \times 100^{2} = 60$ , not $(0.006 \times 100)^{2} = 0.36$ . The comparison is written out with both numbers and a conclusion. In a "Show that", the answer $80$ alone does not finish the argument.
Question 10
Paper 1H/2H style — 5 marks5 marks
A coat has a normal price of £ $p$ .

In a sale, all prices are reduced by $20\%$ . Kim also uses a voucher for £ $5$ off the sale price.

Kim pays £ $K$ for the coat.

(a) Show that $K = 0.8p - 5$ [2]

(b) Kim pays £ $43$ for the coat.

Ben buys the same coat when there is no sale. He uses a voucher for $15\%$ off the normal price.

Work out how much more Ben pays than Kim. [3]
Model answer
(a) A reduction of $20\%$ leaves $80\%$ of the price, so the sale price is

$p - 0.2p = 0.8p$

The £ $5$ voucher is then taken off the sale price:

$K = 0.8p - 5$

(b)

$0.8p - 5 = 43 \;\Rightarrow\; 0.8p = 48 \;\Rightarrow\; p = 60$

Ben pays $0.85 \times 60 = 51$

$51 - 43 = 8$

Answer: £ $8$
Why this scores
Mark scheme:
- M1 (a) for $0.8p$ or $p - 0.2p$ oe
- A1 for $K = 0.8p - 5$ , with the $20\%$ reduction applied before the £ $5$ is subtracted
- M1 (b) for $0.8p - 5 = 43$ oe
- M1 for $0.85 \times 60$ oe (ft their $p$ )
- A1 for $8$ cao
Why this scores full marks: Part (a) follows the order in the words: the $20\%$ comes off first, then the voucher. Taking the £ $5$ off first gives $0.8(p - 5) = 0.8p - 4$ , which is not the printed formula. Part (b) uses the formula to find the normal price, then answers the question actually asked, the difference in what they pay. Stopping at $p = 60$ or at Ben's price of £ $51$ loses the final mark, and $0.15 \times 60 = 9$ is the discount, not Ben's price.

4MA1 Expressions and Formulae

At a glance

What you’ll learn

Substitution — the disciplined approach

Deriving formulae from word problems

BIDMAS in formulae

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Substitute values into a formula

2Substitute negative values

3Derive a formula from a word problem

4Use multiple formulae together

5Writing a product in standard notation [1 mark]

6Substituting a negative decimal into a cube [2 marks]

7Formula from a chain of comparisons [3 marks]

8Substituting into a formula with a cube root [3 marks]

9Two-part hire charge, then solving [4 marks]

10Triangle and rectangle with equal perimeters [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10