Applying Number · Study Notes · Edexcel IGCSE 4MA1 Higher

Real-world number problems: money, time, units, rates and tariffs. Crucial for Paper 2H — and for life beyond exams.

At a glance

Tariffs: fixed charge + (rate × units).
Best buy: lowest unit price wins.
Currency: rate as multiplier; reverse with division.
Time: base 60 (not 100). $0.5$ h $= 30$ min.
Units: convert to consistent before calculating.
Speed-distance-time: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$ .

What you’ll learn

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

1.12 — Solve real-life problems involving money, time, units and rates.
1.12 — Calculate best buys, taxi fares, currency conversions.
1.12 — Use speed-distance-time formula.

Tariffs and rates

▼

Many bills follow: fixed charge + per-unit rate × units used.

Tariff structure.

Total cost = fixed charge + (rate × units used)

Examples:

Taxi: £ $3.50$ initial + £ $1.20$ /km. For $14$ km: $3.50 + 1.20 \times 13 = £19.10$ (note: $13$ additional, not $14$ ).
Electricity: £ $15$ standing charge + $20$ p/kWh. For $300$ kWh: $15 + 0.20 \times 300 = £75$ .
Gym: £ $25$ /month + £ $5$ /visit. For $8$ visits in a month: $25 + 5 \times 8 = £65$ .

Read carefully. Phrases like 'first $X$ included', 'additional $Y$ ' affect counting.

Worked qualitative. A water bill: £ $30$ standing charge + $0.20$ p/litre after the first $1000$ litres. Cost for $1500$ litres?

First $1000$ litres included in standing charge.
Additional $500$ litres at $0.20$ p each: $500 \times 0.20 = £100$ .
Total: $30 + 100 = £130$ .

Edexcel tip. Underline key numbers in the question. Identify: fixed cost, rate, units used. Compute step by step.

Total = fixed + rate × units.
Read 'first X' and 'additional Y' carefully.
Show working step by step.

Currency and units

▼

Multiply for one direction, divide for the other. Always identify the direction.

Currency conversion.

If £1.25$:

£ → 1.25$.
1.25$.

Worked example. Convert £ $240$ to USD, then 400$ back to GBP.

£300$.
400 / 1.25 = £320$.

Sense check. £ $1$ buys 1.25 $(more than$ 1 $). So £ →$ $$ should give a BIGGER number. Multiply confirms.

Unit conversions.

Memorise:

$1$ km = $1000$ m. $1$ m = $100$ cm. $1$ cm = $10$ mm.
$1$ kg = $1000$ g. $1$ g = $1000$ mg.
$1$ L = $1000$ ml = $1000$ cm³.
$1$ tonne = $1000$ kg.
$1$ hour = $60$ min = $3600$ s.
$1$ day = $24$ h.

Time arithmetic.

Time is BASE 60, NOT base 10.

$0.5$ hours = $30$ minutes (NOT $50$ ).
$1$ h $30$ min + $2$ h $45$ min = $3$ h $75$ min = $4$ h $15$ min.

Worked qualitative. A train leaves at $10{:}45$ and arrives at $14{:}20$ . Journey length?

Method 1 — split:

$10{:}45$ → $11{:}00$ = $15$ min.
$11{:}00$ → $14{:}00$ = $3$ h.
$14{:}00$ → $14{:}20$ = $20$ min.
Total: $3$ h $35$ min.

Method 2 — convert to minutes from start of day:

$14{:}20 = 14 \times 60 + 20 = 860$ min.
$10{:}45 = 10 \times 60 + 45 = 645$ min.
Difference: $215$ min $= 3$ h $35$ min.

Edexcel tip. Show the conversion explicitly. 'Converting both to minutes: ...' earns the M1 mark.

Currency: multiply or divide by rate.
Sense check the direction.
Memorise unit equivalents.
Time = base 60.

Best buys and unit prices

▼

Find unit price for each option. Lowest wins.

Step 1. Pick a common unit (g, ml, item).

Step 2. Compute total cost ÷ quantity for each option.

Step 3. Compare. Lowest unit price = best buy.

Worked example. Brand A: $500$ g for £ $3.00$ . Brand B: $750$ g for £ $4.20$ .

A: $300$ p / $500$ g = $0.60$ p/g.
B: $420$ p / $750$ g = $0.56$ p/g.

B is cheaper.

Reduce each option to one common unit; the lowest unit price wins.

Bonus offers.

Sometimes packages have 'extra free'. Calculate the EFFECTIVE quantity.

Brand C: $750$ g + $20\%$ extra free for £ $4.20$ .

Effective quantity: $750 \times 1.2 = 900$ g.
Unit price: $420 / 900 = 0.467$ p/g.

Worked qualitative. Why are unit prices the universal yardstick?

They normalise for package size.
Easy to compare different brands.
Many countries require unit prices on shop labels by law.

Edexcel tip. State which is the best buy CLEARLY. 'Brand B at $0.56$ p/g, lower than Brand A at $0.60$ p/g.' Mark schemes credit the comparison reasoning.

Unit price = cost ÷ quantity.
Lowest wins.
Check for 'extra free' adjustments.
State the comparison clearly.

Speed, distance, time

▼

$d = vt$ . Watch units and time conversions.

The triangle:

       D
      ---
      v|t

So:

$D = v \times t$ .
$v = D / t$ .
$t = D / v$ .

Cover the quantity you need: distance on top, speed and time multiply below.

Worked example. Car travels $135$ km in $2$ hours $15$ min. Find the average speed in km/h.

Convert time to hours: $2$ h $15$ min = $2.25$ h.
$v = 135 / 2.25 = 60$ km/h.

Worked example. A train travels at $90$ km/h for $40$ minutes. Distance?

Convert time to hours: $40/60 = 2/3$ h.
$D = 90 \times 2/3 = 60$ km.

Compound trips.

Often two parts at different speeds.

Example: $30$ km at $60$ km/h, then $20$ km at $40$ km/h. Average speed for the WHOLE trip?

Time for first leg: $30/60 = 0.5$ h.
Time for second leg: $20/40 = 0.5$ h.
Total distance: $50$ km. Total time: $1$ h.
Average speed: $50$ km/h. (NOT $50$ km/h average of speeds — that would be wrong!)

Worked qualitative. Why isn't average speed the same as average of speeds?

Speeds are weighted by TIME, not by speeds.
$60$ km/h for $0.5$ h covers more ground than $40$ km/h for $0.5$ h.
The 'simple average' of $60$ and $40$ is $50$ , but only correct if equal TIMES are spent at each speed.
For equal DISTANCES, use harmonic mean: $\dfrac{2}{1/60 + 1/40} = \dfrac{2}{0.0167 + 0.025} = 48$ .

Edexcel tip. Always convert time to consistent units (hours OR minutes — not mixed). Speed has units (km/h, m/s) — get them right.

$D = vt$ .
Convert time consistently.
Average speed = total distance / total time.
NOT simply the average of two speeds.

Quick recap

Tariffs: fixed + rate × units.
Currency: multiply or divide by rate.
Time: base 60.
Best buy: lowest unit price.
$D = vt$ for speed problems.
Always check + state units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Tariff: total = fixed + rate × units.
Unit price = cost / quantity.
Currency: multiply (or divide) by exchange rate.
Time: $1$ h $= 60$ min, $1$ min $= 60$ s.
Speed: $D = vt$ .

How it’s examined

Applying number appears Paper 2H (4-6 marks): tariffs, best buys, currency, speed problems. Examiner reports flag (1) wrong direction in currency conversion, (2) treating time as base 10, (3) double-charging in two-part tariffs.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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

1Convert units in a multi-step problem
Foundation• units, conversion
▼
Question
A car uses $6.5$ litres of fuel per $100$ km. The fuel costs £ $1.40$ per litre. How much will it cost in fuel for a $250$ km journey?
Step-by-step solution
1. Step 1
  Fuel for $250$ km: $\dfrac{250}{100} \times 6.5 = 16.25$ litres.
2. Step 2
  Cost: $16.25 \times 1.40 = £22.75$ .
Answer
£ $22.75$
Examiner tip
Multi-step problems get partial credit if intermediate steps are clear. Always show working — even if final answer is wrong, method marks add up.
2Convert time and combine
Foundation• time
▼
Question
A train leaves at $10{:}45$ and arrives at $14{:}20$ . How long is the journey?
Step-by-step solution
1. Step 1
  From $10{:}45$ to $14{:}45$ : $4$ hours.
2. Step 2
  But arrival is $14{:}20$ , which is $25$ minutes EARLIER than $14{:}45$ .
3. Step 3
  $4$ hours $- 25$ minutes $= 3$ hours $35$ minutes.
Answer
$3$ hours $35$ minutes
Examiner tip
Time arithmetic can be tricky. Convert to total minutes if confused: $14{:}20 = 14 \times 60 + 20 = 860$ min from start of day; $10{:}45 = 645$ min. Difference: $215$ min $= 3$ h $35$ min.
3Money calculation with rate
Foundation• Adapted from 4MA1/2H May/Jun 2024 Q3• money, rate
▼
Question
A taxi charges £ $3.50$ for the first km and £ $1.20$ for each additional km. Calculate the cost of a $14$ km journey.
Step-by-step solution
1. Step 1
  First km: £ $3.50$ .
2. Step 2
  Additional km: $14 - 1 = 13$ km at £ $1.20$ each $= 13 \times 1.20 = £15.60$ .
3. Step 3
  Total: $3.50 + 15.60 = £19.10$ .
Answer
£ $19.10$
Examiner tip
Tariff problems use a fixed charge plus a per-unit rate. Always identify the structure first: 'fixed + (rate × units)'.
4Compare best buys with bonus offers
Higher• best buy
▼
Question
Brand A: $500$ g at £ $3.00$ . Brand B: $750$ g at £ $4.20$ , but with $20\%$ extra free. Which is cheaper per gram?
Step-by-step solution
1. Step 1
  Brand A: cost per gram $= \dfrac{300\,p}{500\,\text{g}} = 0.6$ p per g.
2. Step 2
  Brand B: $20\%$ extra → $750 \times 1.2 = 900$ g. Cost per g $= \dfrac{420\,p}{900\,\text{g}} = 0.467$ p per g.
3. Step 3
  Brand B is cheaper.
Answer
Brand B (about $0.47$ p/g vs $0.60$ p/g for A).
Examiner tip
Always reduce to a unit price (per gram, per litre, per item). When 'extra free' is offered, calculate the EFFECTIVE quantity first.
5Currency conversion
Foundation• currency
▼
Question
1.25 $. Convert (a) £$ 320 $to USD, (b)$ $500$ to GBP.
Step-by-step solution
1. Step 1
  (a) £→1.25$.
  $320 \times 1.25 = \$400$
2. Step 2
  (b) →£ $: divide by$ 1.25$.
  $\frac{500}{1.25} = £400$
Answer
(a) 400 $(b) £$ 400$
Examiner tip
Determine the rate's direction. £1.25 $means: £→$ $ $multiply by$ 1.25 $.$ $→£ $divide by$ 1.25$.
6Decimal hours to hours and minutes [1 mark]
Getting started• Paper 1H/2H style — 1 mark• time, conversion
▼
Question
Write $2.4$ hours in hours and minutes. [1]
Step-by-step solution
1. Step 1
  B1 Keep the $2$ whole hours and change the decimal part into minutes by multiplying by $60$ .
  $0.4 \times 60 = 24 \text{ minutes}$
Answer
$2$ hours $24$ minutes
Examiner tip
Examiner tip: An hour has $60$ minutes, not $100$ , so the digits after the decimal point are not the minutes.

Common mistake: Writing $2$ hours $40$ minutes.
7How many pieces from a sheet — area units [2 marks]
Getting started• Paper 1H/2H style — 2 marks• area, units
▼
Question
A sheet of card has an area of $0.84\ \text{m}^{2}$ . Labels are cut from the sheet. Each label has an area of $120\ \text{cm}^{2}$ . There is no waste.

Work out the number of labels that can be cut from the sheet. [2]
Step-by-step solution
1. Step 1
  M1 Put both areas in the same unit: $1\ \text{m}^{2} = 10\,000\ \text{cm}^{2}$ .
  $0.84 \times 10\,000 = 8400\ \text{cm}^{2}$
2. Step 2
  A1 Divide.
  $8400 \div 120 = 70$
Answer
$70$ labels
Examiner tip
Examiner tip: An answer smaller than $1$ label (from $0.84 \div 120$ ) is a clear sign the units were not matched.

Common mistake: Using $100$ as the area factor, giving $84 \div 120 = 0.7$ .
8Distance from a speed in m/s [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• speed, compound measures
▼
Question
A car travels at a constant speed of $25$ m/s for $18$ minutes.

Work out the distance the car travels. Give your answer in kilometres. [3]
Step-by-step solution
1. Step 1
  M1 Change the time into seconds to match the speed.
  $18 \times 60 = 1080 \text{ s}$
2. Step 2
  M1 Distance $=$ speed $\times$ time.
  $25 \times 1080 = 27\,000 \text{ m}$
3. Step 3
  A1 Change metres to kilometres.
  $27\,000 \div 1000 = 27$
Answer
$27$ km
Examiner tip
Examiner tip: Match the time unit to the speed unit before multiplying: metres per second needs seconds.

Common mistake: Working out $25 \times 18 = 450$ and giving $0.45$ km.
9Comparing two phone tariffs [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• tariffs, money
▼
Question
Two phone plans:

Plan X: £ $12$ per month, plus $5$ p for every minute over $300$ minutes in that month

Plan Y: £ $21$ per month, with unlimited minutes

Ahmed uses $460$ minutes every month.

Work out which plan is cheaper for Ahmed, and how much he saves in one year by using it. [4]
Step-by-step solution
1. Step 1
  M1 Minutes charged on Plan X.
  $460 - 300 = 160$
2. Step 2
  M1 Monthly cost of Plan X.
  $12 + 160 \times 0.05 = 12 + 8 = 20$
3. Step 3
  M1 Yearly saving: difference per month times $12$ .
  $(21 - 20) \times 12 = 12$
4. Step 4
  A1 Plan X, saving £ $12$ .
Answer
Plan X; he saves £ $12$ a year
Examiner tip
Examiner tip: Only the minutes over $300$ are charged on Plan X. Change $5$ p to £ $0.05$ before adding to the pounds.

Common mistake: Charging all $460$ minutes, $12 + 460 \times 0.05 = 35$ , and so choosing Plan Y.
10Catching the right bus from a timetable [5 marks]
Stretch• Paper 1H/2H style — 5 marks• time, timetables, speed
▼
Question
Buses leave Northgate every $12$ minutes from 06:50. Each bus takes $38$ minutes to reach the college stop.

Sam walks $9$ minutes from home to Northgate. From the college stop he walks $6$ minutes to his class, which starts at 08:15.

(a) Work out the latest time Sam can leave home and still arrive by 08:15. [3]

The bus route is $23.5$ km long.

(b) Work out the average speed of the bus in km/h. Give your answer correct to $3$ significant figures. [2]
Step-by-step solution
1. Step 1
  M1 (a) Sam must reach the college stop by 08:15 $-$ 6 min $=$ 08:09, so the bus must leave Northgate by 08:09 $-$ 38 min $=$ 07:31.
2. Step 2
  M1 Bus times: 06:50, 07:02, 07:14, 07:26, 07:38. The latest one at or before 07:31 is 07:26 (arriving 08:04).
3. Step 3
  A1 Leave home $9$ minutes earlier.
  $07{:}26 - 9 \text{ min} = 07{:}17$
4. Step 4
  M1 (b) $38$ minutes $= \frac{38}{60}$ hour.
  $23.5 \div \frac{38}{60}$
5. Step 5
  A1 Round at the end.
  $37.105\ldots \approx 37.1$
Answer
(a) 07:17 (b) $37.1$ km/h
Examiner tip
Examiner tip: Work backwards from the deadline, then pick a bus that actually runs — 07:31 is not a bus time.

Common mistake: Dividing by $0.38$ hours in (b), giving $61.8$ km/h.

Question 1
Paper 1H/2H style — 1 mark1 mark
Change $7.2$ km into centimetres. [1]
Model answer
$7.2 \times 1000 \times 100 = 720\,000$

$720\,000$ cm
Why this scores
Mark scheme:
- B1 for $720\,000$ cao
Why this scores full marks: The candidate goes km to m ( $\times 1000$ ) then m to cm ( $\times 100$ ) in one line, rather than guessing a single factor.
Question 2
Paper 1H/2H style — 2 marks2 marks
A bag contains $2.5$ kg of rice. One portion uses $65$ g of rice.

Work out the greatest number of full portions that can be made from the bag. [2]
Model answer
$2.5 \text{ kg} = 2500 \text{ g}$

$2500 \div 65 = 38.46\ldots$

$38$ full portions
Why this scores
Mark scheme:
- M1 for $2500 \div 65$ oe
- A1 for $38$ cao
Why this scores full marks: The answer is a count of full portions, so $38.46\ldots$ is rounded down, not to the nearest whole number.
Question 3
Paper 1H/2H style — 2 marks2 marks
A school day starts at 08:40. There are $6$ lessons, each lasting $50$ minutes, a break of $15$ minutes and a lunch break of $45$ minutes. There are no other gaps.

Work out the time the school day ends. [2]
Model answer
$6 \times 50 + 15 + 45 = 360 \text{ minutes} = 6 \text{ hours}$

08:40 + 6 hours = 14:40
Why this scores
Mark scheme:
- M1 for $6 \times 50 + 15 + 45$ ( $= 360$ ) oe
- A1 for 14:40 or 2:40 pm
Why this scores full marks: Totalling everything in minutes first, then changing to hours, avoids adding clock times one step at a time.
Question 4
Paper 1H/2H style — 3 marks3 marks
A lorry travels at a constant speed of $81$ km/h.

Work out the time, in seconds, the lorry takes to travel $450$ metres. [3]
Model answer
$81 \text{ km/h} = \frac{81\,000}{3600} = 22.5 \text{ m/s}$

$450 \div 22.5 = 20$

$20$ seconds
Why this scores
Mark scheme:
- M1 for $81 \times 1000 \div 3600$ oe ( $= 22.5$ )
- M1 for $450 \div 22.5$ oe, or $0.45 \div 81 \times 3600$
- A1 for $20$ cao
Why this scores full marks: The speed is changed to m/s so it matches the distance in metres, and the answer is in seconds as asked.
Question 5
Paper 1H/2H style — 3 marks3 marks
A stone has a mass of $1.35$ kg. The density of the stone is $2.7\ \text{g/cm}^{3}$ .

$\text{density} = \frac{\text{mass}}{\text{volume}}$

(a) Work out the volume of the stone. [2]

The stone is put into a measuring jug containing $800$ ml of water. The stone is completely under the water.

(b) Write down the new reading on the jug, in litres. [1]
Model answer
(a) $1.35 \text{ kg} = 1350 \text{ g}$

$1350 \div 2.7 = 500$

$500\ \text{cm}^{3}$

(b) $800 + 500 = 1300$ ml $= 1.3$ litres
Why this scores
Mark scheme:
- M1 for $1350 \div 2.7$
- A1 for $500\ \text{cm}^{3}$
- B1 for $1.3$ litres ft ( $800 +$ their volume) $\div 1000$
Why this scores full marks: Mass is in grams before dividing by a density in $\text{g/cm}^{3}$ , and $1\ \text{cm}^{3} = 1$ ml links the two parts.
Question 6
Paper 1H/2H style — 4 marks4 marks
A car park charges £ $1.80$ for the first hour, then £ $0.90$ for each extra $30$ minutes or part of $30$ minutes.

Mai parks from 10:47 until 14:05.

Work out how much Mai pays. [4]
Model answer
10:47 to 14:05 is $3$ hours $18$ minutes.

After the first hour: $2$ hours $18$ minutes $= 138$ minutes.

$138 \div 30 = 4.6 \text{ so } 5 \text{ extra periods}$

$1.80 + 5 \times 0.90 = 6.30$

£ $6.30$
Why this scores
Mark scheme:
- M1 for $3$ h $18$ min or $198$ minutes
- M1 for $5$ extra periods (from $138 \div 30$ rounded up)
- M1 for $1.80 + n \times 0.90$ where $n$ is their number of extra periods
- A1 for £ $6.30$
Why this scores full marks: "Part of $30$ minutes" means rounding up — the candidate charges $5$ periods, not $4.6$ or $4$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
Leila has a weight of $600$ N.

$\text{pressure} = \frac{\text{force}}{\text{area}}$

When she wears snowshoes, each snowshoe has an area of $0.075\ \text{m}^{2}$ in contact with the snow. When she wears boots, each boot has an area of $150\ \text{cm}^{2}$ in contact with the snow. She stands on both feet.

How many times greater is the pressure she exerts in boots than in snowshoes? [4]
Model answer
Snowshoes: $600 \div (2 \times 0.075) = 600 \div 0.15 = 4000 \text{ N/m}^{2}$

Boots: $150\ \text{cm}^{2} = 0.015\ \text{m}^{2}$

$600 \div (2 \times 0.015) = 600 \div 0.03 = 20\,000 \text{ N/m}^{2}$

$20\,000 \div 4000 = 5$

$5$ times greater
Why this scores
Mark scheme:
- M1 for $150\ \text{cm}^{2} = 0.015\ \text{m}^{2}$ , or $0.075\ \text{m}^{2} = 750\ \text{cm}^{2}$
- M1 for a pressure with two feet, e.g. $600 \div 0.15$
- M1 for both pressures in the same units and a division
- A1 for $5$ cao
Why this scores full marks: Both areas are put in the same unit before comparing — a comparison of $0.075$ with $150$ is meaningless.
Question 8
Paper 1H/2H style — 4 marks4 marks
A flat roof is a rectangle $12$ m by $8.5$ m. During a storm $18$ mm of rain falls on the roof. All of the rain drains into a tank.

Calculate the number of litres of water that drain into the tank. [4]
Model answer
$18$ mm $= 0.018$ m

$12 \times 8.5 \times 0.018 = 1.836 \text{ m}^{3}$

$1.836 \times 1000 = 1836$

$1836$ litres
Why this scores
Mark scheme:
- M1 for $18$ mm $= 0.018$ m or $1.8$ cm
- M1 for $12 \times 8.5 \times$ depth in consistent units
- M1 for converting to litres, $\times 1000$ from $\text{m}^{3}$ or $\div 1000$ from $\text{cm}^{3}$
- A1 for $1836$ cao
Why this scores full marks: The rain is treated as a thin cuboid, all three lengths are in metres, and $1\ \text{m}^{3} = 1000$ litres is used once.
Question 9
Paper 1H/2H style — 5 marks5 marks
The exchange rates are £ $1 = 1.17$ euros and £ $1 = 1.27$ US dollars.

(a) Work out how many US dollars $1$ euro is worth. Give your answer correct to $3$ decimal places. [2]

A camera costs $540$ euros in Germany. The same camera costs $585$ US dollars in the USA.

(b) Work out in which country the camera is cheaper, and by how much in pounds. [3]
Model answer
(a) $1.17 \text{ euros} = 1.27 \text{ US dollars}$

$1.27 \div 1.17 = 1.0854\ldots$

$1.085$ US dollars

(b) Germany: $540 \div 1.17 = 461.538\ldots$ £ $461.54$

USA: $585 \div 1.27 = 460.629\ldots$ £ $460.63$

$461.54 - 460.63 = 0.91$

The camera is cheaper in the USA, by £ $0.91$ .
Why this scores
Mark scheme:
- M1 for $1.27 \div 1.17$
- A1 for $1.085$ cao
- M1 for $540 \div 1.17$ or $585 \div 1.27$
- M1 for both prices in the same currency
- A1 for USA and £ $0.90$ to £ $0.91$
Why this scores full marks: Both prices are changed to pounds, the currency the question asks for, and kept to two decimal places. The difference is small, so the candidate does not use the rounded rate from (a).
Question 10
Paper 1H/2H style — 5 marks5 marks
A room has two walls measuring $4.2$ m by $2.6$ m and two walls measuring $3.6$ m by $2.6$ m. The door has an area of $1.9\ \text{m}^{2}$ and the window has an area of $2.4\ \text{m}^{2}$ . These are not painted.

The walls need two coats of paint. One litre of paint covers $12\ \text{m}^{2}$ . Paint is sold in $2.5$ litre tins costing £ $18.40$ each.

Work out the cost of the paint needed. [5]
Model answer
$2 \times 4.2 \times 2.6 + 2 \times 3.6 \times 2.6 = 21.84 + 18.72 = 40.56 \text{ m}^{2}$

$40.56 - 1.9 - 2.4 = 36.26 \text{ m}^{2}$

Two coats: $36.26 \times 2 = 72.52 \text{ m}^{2}$

$72.52 \div 12 = 6.043\ldots \text{ litres}$

$6.043\ldots \div 2.5 = 2.41\ldots \text{ so } 3 \text{ tins}$

$3 \times 18.40 = 55.20$

£ $55.20$
Why this scores
Mark scheme:
- M1 for the total wall area, $40.56$
- M1 for subtracting $4.3$ and doubling ( $= 72.52$ )
- M1 for dividing by $12$ ( $= 6.04\ldots$ )
- M1 for dividing by $2.5$ and rounding up to a whole number of tins
- A1 for £ $55.20$ cao
Why this scores full marks: Each step is labelled, and the candidate rounds up to $3$ tins — you cannot buy $2.4$ tins, and $2$ tins would not be enough.

4MA1 Applying Number

At a glance

What you’ll learn

Tariffs and rates

Currency and units

Best buys and unit prices

Speed, distance, time

Quick recap

Memorise this

How it’s examined

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

1Convert units in a multi-step problem

2Convert time and combine

3Money calculation with rate

4Compare best buys with bonus offers

5Currency conversion

6Decimal hours to hours and minutes [1 mark]

7How many pieces from a sheet — area units [2 marks]

8Distance from a speed in m/s [3 marks]

9Comparing two phone tariffs [4 marks]

10Catching the right bus from a timetable [5 marks]

Model AnswersFull mark-scheme-style answers

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Question 10