15-mark question lab · start here

From a grid of hourly readings to a week of weather reports.

Build the solution in a sensible order: decode the six requirements, plan the counters and sentinels, scan each day’s column for its maximum, minimum and total, output every day’s rounded line, feed the week trackers, then print the week’s three results. Learn the reasoning before revealing the complete program.

Columns are daysNested loopsMax and min sentinelsAverages and roundingCASE OF day names15 marks

Learning targets

By the end, you should be able to explain the algorithm before writing it.

Lesson overview

01 · Understand

Identify the given 2D array Temperatures[] with a column for each day, the six given variables MaxDay, MinDay, AvDay, MaxWeek, MinWeek and AvWeek, and the six requirements hidden in the bullet list.

02 · Construct

Build the program in order: reset the day sentinels inside the day loop, scan each day’s column with the hour loop, total the readings and track the maximum and minimum, then output each day’s name and three rounded values.

03 · Complete

Feed each day’s finished results into MaxWeek, MinWeek and WeekTotal, calculate the rounded weekly average from every reading, and output the week’s three results after the loop.

How Cambridge awards the marks: the mark scheme gives AO2 a maximum of 9 marks and AO3 a maximum of 6 marks. It lists requirements R1–R5, but does not assign a separate fixed mark total to each R requirement. The stage-to-mark links in this guide show alignment, not invented stage marks.

Source question

Read the whole question first

Do not begin coding after the first paragraph. The column-per-day layout, the six given variables and the two levels of maximum-and-minimum tracking all affect the structure of your solution.

Question 10 · 15 marks

Cambridge IGCSE Computer Science · 0478/23 · Question 10

A weather station takes temperature readings once an hour for a week. These temperatures are stored in a two-dimensional (2D) array Temperatures[]

Each column contains 24 readings for a single day. The first temperature is recorded at 00:00 and the final temperature at 23:00. There are seven columns, one for each day of the week, starting with Monday and ending with Sunday.

The variables MaxDay, MinDay and AvDay are used to store the maximum, minimum, and average temperatures for a day. The variables MaxWeek, MinWeek and AvWeek are used to store the maximum, minimum, and average temperatures for the week.

The array has already been set up and the data stored.

Write a program that meets the following requirements:

finds the maximum and minimum temperatures for each day
calculates the average temperature for each day
outputs for each day:
- name of the day, for example Monday
- maximum temperature
- minimum temperature
- average temperature
finds the maximum and minimum temperatures for the week
calculates the average temperature for the week
outputs:
- maximum temperature for the week
- minimum temperature for the week
- average temperature for the week.

All temperatures output must be rounded to two decimal places.

You must use pseudocode or program code and add comments to explain how your code works.

All inputs and outputs must contain suitable messages.

You do not need to declare any arrays or variables; you may assume that this has already been done.

You do not need to initialise the data in the array Temperatures[]

Before you move on

These are the five required techniques identified by the Cambridge mark scheme.

Mark-scheme map

Find the maximum and minimum temperatures for each day — searching each day’s column with strict comparisons.

Calculate the average daily temperature — totalling the day’s readings and dividing by the reading count.

Find the maximum and minimum temperatures for the week — nested searching that compares each day’s finished results with the week trackers.

Calculate the average weekly temperature — totalling every reading of the week and dividing by the number of readings.

Output each day’s name with its rounded maximum, minimum and average, then the week’s rounded maximum, minimum and average — appropriate messages and rounded values throughout.

Use the tabs along the top to work through the seven stages. The complete solution is kept in its own tab, and the interactive visualisation is the final tab.

Decode the requirements

Turn the question wording into a sequence of tasks.

R1 · R2 · R3

🧠 Step-by-step thinking

Read the grid’s orientation first. Temperatures[] stores a column per day: each column holds a single day’s 24 readings, first at 00:00 and last at 23:00, and the seven columns run Monday to Sunday. That is the opposite of the row-per-day grids you may have met elsewhere — and it decides the order of both loop counters.
Decode the six given variables. MaxDay, MinDay and AvDay hold one day’s maximum, minimum and average; MaxWeek, MinWeek and AvWeek hold the week’s. The day variables are re-created for every day; the week variables are set once and survive the whole day loop.
Split the bullet list into per-day and once-only work. Per day: find the maximum and minimum, calculate the average, output the name plus three values. For the week: find the maximum and minimum, calculate the average, output three values — the week’s work rides along inside the day loop and finishes after it.
Notice the rounding demand. “All temperatures output must be rounded to two decimal places” — the averages almost never divide exactly, so ROUND(…, 2) is required, and rounding the maxima and minima too is the literal reading of that line.
Build the order. initialise week sentinels → day loop (reset day sentinels → hour loop (total, max, min) → CASE OF day name → average → four output lines → update week trackers) → weekly average → three output lines.

Column = day, row = hourGiven: six named variablesRound to two decimal placesOnce-only: week outputs

❓ Guiding questions – think before revealing

Why does the column-per-day layout matter so much?
Which variables belong inside the day loop, and which outside?
What does “for example Monday” tell you?

Reveal suggested reasoning

Why does the column-per-day layout matter so much? Because the inner loop must run down a column: Temperatures[HourCounter, DayCounter] with the hour changing fastest visits all 24 readings of one day before the next day begins. Swap the indexes and every comparison mixes seven different days’ readings at the same hour.

Which variables belong inside the day loop, and which outside? MaxDay, MinDay and the day’s total are reset inside it, once per day. MaxWeek, MinWeek and the week’s total are initialised once, before the loop, and are updated by each day’s finished results — that is what the mark scheme’s “nested searching” means.

What does “for example Monday” tell you? The output must show the day’s name, not its number. The published answer uses CASE OF on the day counter to assign the string “Monday” and outputs it as the first of the day’s four lines.

Reveal Cambridge pseudocode for Stage 1

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Once, before the day loop: initialise the week
// MaxWeek, MinWeek, WeekTotal

FOR DayCounter ← 0 TO Days - 1        // per day (one column):
    // reset MaxDay, MinDay, DayTotal

    FOR HourCounter ← 0 TO Hours - 1  // per reading (down the column):
        // add Temperatures[HourCounter, DayCounter] to DayTotal
        // compare it with MaxDay and MinDay
    NEXT HourCounter

    // CASE OF DayCounter → day name
    // AvDay ← DayTotal / Hours
    // output day name, maximum, minimum, average
    // update MaxWeek, MinWeek, WeekTotal
NEXT DayCounter

// Once, after the loop: AvWeek and the week's three output lines

📖 Detailed explanation

The bullets are two reports in one. Every “for each day” requirement lives inside the day loop; every “for the week” requirement is fed by the day loop and printed after it. One pass over the grid serves both.

The published answer numbers from zero. FOR DayCounter ← 0 TO Days - 1 pairs hour 0 with the 00:00 reading and lets CASE OF map day 0 straight to “Monday” — a natural fit for this question’s layout. Numbering from 1 also works if every index and label moves together.

The rounding line is a requirement, not a footnote. It is the last thing the question says before the standard closing lines, and R3 in the mark scheme names “rounded values” explicitly — plan ROUND(…, 2) into every temperature output from the start.

The closing lines shape the answer. “You do not need to declare any arrays or variables” and “you do not need to initialise the data” mean the algorithm begins straight away with the given names — no input, no setup pass.

⚠️ Common mistakes

Treating each row as a day and scanning across seven different days’ readings at one hour.
Resetting MaxDay and MinDay once, before the day loop, so Tuesday inherits Monday’s extremes.
Printing the week’s three results inside the day loop, so they appear seven times.
Outputting the day’s number instead of its name — the question says “for example Monday”.

💡 Exam hack / tip

Say the layout aloud before writing anything: columns are days, rows are hours. Then underline rounded to two decimal places and the six variable names — those three facts drive every loop, comparison and OUTPUT line in the answer.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · algorithm designR1–R3 · full requirement coverageAO3 · logical orderAO3 · comments explain purpose

This planning stage supports AO2 by identifying the full range of techniques required. It supports AO3 by establishing a logical sequence. These are alignment notes, not separately allocated marks.

Plan the data and identifiers

Choose variables and arrays that store exactly what the algorithm needs.

Required data structures

🧠 Step-by-step thinking

Use the given names exactly. Temperatures[] and the six variables MaxDay, MinDay, AvDay, MaxWeek, MinWeek and AvWeek must appear as given — the mark scheme lists them as the data structures required with names as given in the scenario.
Add the two loop counters. DayCounter walks the seven columns, one day at a time; HourCounter walks the 24 rows inside the current day’s column. The published answer numbers both from 0, so Temperatures[HourCounter, DayCounter] addresses “this hour, this day”.
Add the two totals. DayTotal accumulates one day’s readings so AvDay can divide by the reading count; WeekTotal accumulates every reading of the week for AvWeek. Neither is given by the question — the program declares them itself.
Start the trackers at sentinels. The published answer sets MaxDay and MaxWeek to -1000 and MinDay and MinWeek to 1000 before their scans — values no real temperature can reach, so the first reading always beats them and no special first-reading test is needed.
Add a string for the day name. Day holds the name selected by CASE OF, so the first output line of each day prints “Monday” rather than 0.

Given: Temperatures[] and six variablesCounters: DayCounter, HourCounterTotals: DayTotal, WeekTotalSentinels: -1000 and 1000

❓ Guiding questions – think before revealing

Why HourCounter first in Temperatures[HourCounter, DayCounter] — could it be the other way round?
Why are the sentinels -1000 and 1000 rather than 0?
Do the day results need an array?

Reveal suggested reasoning

Why HourCounter first in Temperatures[HourCounter, DayCounter] — could it be the other way round? The question defines columns as days, so the second index selects the day while the first moves down the hours. Written the other way round the loops would read along a row — seven different days’ readings at one hour — and every maximum and minimum would be wrong.

Why are the sentinels -1000 and 1000 rather than 0? Temperatures can be negative. A MinDay seeded with 0 would never record a freezing reading, and a MaxDay seeded with 0 fails on a day that never rises above zero. Sentinels far outside the data range make the first reading win both comparisons whatever the weather.

Do the day results need an array? No. MaxDay, MinDay and AvDay are printed before the next day begins, and only the week trackers need to survive — three simple variables reused each day are enough, which is exactly how the published answer stores them.

Reveal Cambridge pseudocode for Stage 2

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Given by the question — already set up and stored:
//   Temperatures[,]  2D, column = day (Monday to Sunday),
//                    row = hour (00:00 to 23:00)
//   MaxDay, MinDay, AvDay      one day's results
//   MaxWeek, MinWeek, AvWeek   the week's results

// Declared by the program:
DayCounter            // current day, runs 0 to Days - 1
HourCounter           // current hour, runs 0 to Hours - 1
DayTotal              // running total of today's readings
WeekTotal             // running total of the week's readings
Day                   // day name selected by CASE OF

CONSTANT Hours ← 24
CONSTANT Days ← 7

📖 Detailed explanation

The two indexes have different meanings. The first selects the hour (the row), the second selects the day (the column). Reading the cell address aloud as “this hour, this day” catches a swapped index before it costs marks.

The totals exist for the averages. AvDay ← DayTotal / Hours and AvWeek ← WeekTotal / (Days * Hours) — the averages are totals divided by reading counts, so the totals must be accumulated while the scan runs.

Sentinels are a data decision, not an afterthought. Starting values the data can always beat remove any special case for the first reading — AO3 credit comes from the program working for any temperatures, including sub-zero ones.

Meaningful identifiers earn AO3 credit. DayTotal, WeekTotal and Day say what they hold, and the six given names map straight back to the question’s wording.

⚠️ Common mistakes

Renaming Temperatures[] or the six given variables, which the mark scheme treats as not using the supplied data structures.
Swapping the indexes — Temperatures[DayCounter, HourCounter] reads along a row of seven days at one hour.
Seeding the minimum trackers with 0, which no freezing reading can beat.
Storing the day results in an array when three simple variables reset per day are enough.

💡 Exam hack / tip

Point at each name and ask “which phrase in the question does this store?” MaxDay answers “maximum … for a day”, WeekTotal answers “average temperature for the week” once divided, and Day answers “name of the day, for example Monday”.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · appropriate data structuresR1 · R2 · given names reusedAO3 · meaningful identifiers

The data plan supports AO2 because the chosen variables store exactly the required data, and AO3 because every identifier is meaningful. These are alignment notes, not separately allocated marks.

Initialise the sentinels and totals

Two reset rhythms — the week once, each day inside the loop.

R1 · R2

🧠 Step-by-step thinking

Initialise the week before the loop. MaxWeek ← -1000, MinWeek ← 1000 and WeekTotal ← 0 run once, because the week’s trackers must survive all seven days and still hold their values for the final output lines.
Reset the day inside the loop. MaxDay ← -1000, MinDay ← 1000 and DayTotal ← 0 sit at the top of the day loop body — inside it, not before it — so every day starts its scan from the same clean state.
The first reading always wins. Any real temperature is greater than -1000 and less than 1000, so the 00:00 reading sets both MaxDay and MinDay — no IF HourCounter = 0 special case is needed.
Two rhythms, two places. The week’s sentinels are set once for the whole run; the day’s sentinels return to their starting values once per day. Mixing the two rhythms — resetting the week inside the loop, or the day only once — is the classic slip.

Set once: MaxWeek, MinWeek, WeekTotalReset per day: MaxDay, MinDay, DayTotalFirst reading beats both sentinels

❓ Guiding questions – think before revealing

Why must MaxWeek and MinWeek never be reset inside the day loop?
Why are the day resets inside the loop rather than before it?
Could the sentinels be replaced by seeding from the first reading?

Reveal suggested reasoning

Why must MaxWeek and MinWeek never be reset inside the day loop? They carry the week’s running records. Resetting them per day would leave them holding only Sunday’s results, and the final output lines would report the last day’s weather as the week’s.

Why are the day resets inside the loop rather than before it? Each day needs a fresh scan. If MaxDay kept Monday’s 21, Tuesday could never record its own 16 as a maximum, and every day after Monday would report Monday’s extremes.

Could the sentinels be replaced by seeding from the first reading? Yes — an IF HourCounter = 0 test that copies the first reading into both trackers works equally well. The sentinels simply make the first comparison do the seeding for free, which is why the published answer prefers them.

Reveal Cambridge pseudocode for Stage 3

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Once, before the day loop
MaxWeek ← -1000   // initialise max and min temperatures and total for the week
MinWeek ← 1000
WeekTotal ← 0

FOR DayCounter ← 0 TO Days - 1
    MaxDay ← -1000 // initialise max and min temperatures and total for each day
    MinDay ← 1000
    DayTotal ← 0

    // hour loop goes here ...

    // week updates happen here, still inside the day loop
NEXT DayCounter

📖 Detailed explanation

Each reset lives at the level of the loop it serves. The week’s state is set once, outside; the day’s state is reset once per day, inside. That placement is what lets one set of variables serve both reports.

The sentinels make the scan uniform. Every reading — including the first — is compared the same way, so the code has no special cases and works whatever the temperatures turn out to be.

WeekTotal starts at 0 for the same reason. It accumulates a day at a time inside the loop; starting it at anything else would corrupt the weekly average.

⚠️ Common mistakes

Resetting MaxWeek and MinWeek inside the day loop, so the week’s results become Sunday’s.
Resetting MaxDay, MinDay and DayTotal only once, before the day loop, so every day inherits Monday’s extremes and total.
Seeding the minimum sentinels with 0 in a week that could include freezing readings.
Initialising DayTotal but forgetting WeekTotal, leaving the weekly average built on an unset value.

💡 Exam hack / tip

Say it as a rule: set the week once, reset the day every day, zero every total at the level that accumulates it. Three lines before the loop, three lines inside it — nothing else moves.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · correct starting stateR1 · R2 · sentinels and totalsAO3 · resets at the right loop level

Initialisation supports AO2 because the trackers and totals genuinely start from a known state, and AO3 because the resets sit at the right loop level. These are alignment notes, not separately allocated marks.

Scan each day’s column

An hour loop inside a day loop works down one column at a time.

🧠 Step-by-step thinking

Nest the loops day-outside, hour-inside. FOR DayCounter ← 0 TO Days - 1 wraps FOR HourCounter ← 0 TO Hours - 1, so the program reads Monday’s column from 00:00 to 23:00, then Tuesday’s, and so on — column by column, exactly the order the columns are stored in.
Read each cell once. Temperatures[HourCounter, DayCounter] is this hour’s reading for the current day — one value that then does three jobs: it adds to DayTotal, is compared with MaxDay and is compared with MinDay.
Strict comparisons keep the first occurrence. > updates MaxDay only when a reading is strictly greater; < updates MinDay only when strictly less. A reading that ties a record changes nothing.
The loops run Days × Hours times. 7 × 24 = 168 visits in the real program — every read exactly once, in column-by-column order.
NEXT HourCounter ends the day’s search. By then MaxDay, MinDay and DayTotal hold the day’s maximum, minimum and total, ready for the average, the output lines and the week updates.

Iteration: FOR inside FORAccess: Temperatures[h, d]Searching: strict > and <

❓ Guiding questions – think before revealing

Why is the day loop the outer one?
One reading, three jobs — in which order?
What happens when a reading ties the day’s maximum?

Reveal suggested reasoning

Why is the day loop the outer one? Because a day’s results must be complete — scanned, averaged and printed — before the next day’s resets destroy them. Swapping the loops would compare one hour’s readings across all seven days, which is a different — and wrong — report.

One reading, three jobs — in which order? The published answer adds to DayTotal first, then runs the two comparisons. The order of the three statements does not affect the results; what matters is that all three use the same cell Temperatures[HourCounter, DayCounter].

What happens when a reading ties the day’s maximum? Nothing — = fails both > and <. The earlier reading keeps the record, which is the same rule the week comparison relies on when two days tie.

Reveal Cambridge pseudocode for Stage 4

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

FOR DayCounter ← 0 TO Days - 1
    MaxDay ← -1000
    MinDay ← 1000
    DayTotal ← 0

    FOR HourCounter ← 0 TO Hours - 1
        DayTotal ← DayTotal + Temperatures[HourCounter, DayCounter]
        // update total maximum and minimum
        IF Temperatures[HourCounter, DayCounter] > MaxDay
          THEN
            MaxDay ← Temperatures[HourCounter, DayCounter]
        ENDIF
        IF Temperatures[HourCounter, DayCounter] < MinDay
          THEN
            MinDay ← Temperatures[HourCounter, DayCounter]
        ENDIF
    NEXT HourCounter

    // day name, average, output and week updates follow (Stages 05-06)
NEXT DayCounter

📖 Detailed explanation

Nested iteration and searching are the R1 techniques. The mark scheme names “searching, totalling” — the hour loop walking the column while totalling into DayTotal and tracking both extremes is exactly that.

The scan order matches the storage order. Working down a column means HourCounter changes while DayCounter waits — the same order the highlighted cells fill in the visualisation’s grid.

Both comparisons use the same cell expression. Reading Temperatures[HourCounter, DayCounter] three times looks repetitive, but it shows the examiner one value driving all three updates — and it is how the published answer writes it.

A day of falling temperatures updates MinDay every hour. If each reading is cooler than the last, < is TRUE twenty-four times — that is correct behaviour, not a bug: the minimum keeps moving until the scan ends.

⚠️ Common mistakes

Swapping the indexes — Temperatures[DayCounter, HourCounter] reads along a row of seven days at one hour.
Using >= or <=, so a later tied reading steals the record from the earlier one.
Putting the day resets inside the hour loop, so MaxDay can never exceed the last reading.
Looping HourCounter to Days instead of Hours — the two constants are different numbers.

💡 Exam hack / tip

Read the cell address aloud as “this hour, this day”: Temperatures[HourCounter, DayCounter]. If your sentence sounds like “this day, this hour”, your indexes are the wrong way round.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · searching and totallingR1 · max, min and total per dayAO3 · correct two-index accessAO3 · comments explain purpose

The column scan is the R1 evidence for AO2, and the two-index access with strict comparisons is AO3 accuracy evidence. These are alignment notes, not separately allocated marks.

Output each day’s rounded line

CASE OF selects the name; ROUND delivers two decimal places.

R1 · R3

🧠 Step-by-step thinking

The name comes from CASE OF. The question asks for “name of the day, for example Monday” — CASE OF DayCounter with labels 0 to 6 assigns the string held in Day, which becomes the first of the day’s four output lines.
The average divides the day’s total. AvDay ← DayTotal / Hours — 24 in the real program. The division rarely lands on an exact value, which is why the question insists on rounding.
Four lines, four facts. Day name, maximum, minimum, average — printed inside the day loop, before the next day’s resets overwrite MaxDay, MinDay and DayTotal.
Round every temperature. ROUND(AvDay, 2) is essential; wrapping the maximum and minimum in ROUND(…, 2) as well is the literal reading of “All temperatures output must be rounded to two decimal places”, and it changes nothing because the stored readings are already exact.
Suitable messages. “Maximum temperature 21”, not a bare 21 — the question’s closing line asks for messages on every output.

Selection: CASE OF day namesCalculation: DayTotal / HoursOutput: four lines per dayRounding: ROUND(…, 2)

❓ Guiding questions – think before revealing

Why CASE OF rather than seven IF statements?
When does rounding actually change a value?
Why must these lines print inside the day loop?

Reveal suggested reasoning

Why CASE OF rather than seven IF statements? One counter is tested against seven constant alternatives — exactly what CASE OF expresses. Seven IFs also work, but the case structure shows the examiner a clean selection and mirrors the published answer.

When does rounding actually change a value? Whenever the division does not terminate at two decimal places — 76 ÷ 6 = 12.666…, so ROUND(AvDay, 2) outputs 12.67. When it divides exactly, such as 84 ÷ 6 = 14, the rounded output is simply 14.00.

Why must these lines print inside the day loop? MaxDay, MinDay and AvDay are overwritten by the next day’s scan. Printing immediately — while the values are still this day’s — is what makes reusing the same three variables safe.

Reveal Cambridge pseudocode for Stage 5

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// the hour loop has just finished for this day
CASE OF DayCounter    // select message for day
  0 : Day ← "Monday"
  1 : Day ← "Tuesday"
  2 : Day ← "Wednesday"
  3 : Day ← "Thursday"
  4 : Day ← "Friday"
  5 : Day ← "Saturday"
  6 : Day ← "Sunday"
ENDCASE

AvDay ← DayTotal / Hours    // output results for day
OUTPUT Day
OUTPUT "Maximum temperature ", ROUND(MaxDay, 2)
OUTPUT "Minimum temperature ", ROUND(MinDay, 2)
OUTPUT "Average temperature ", ROUND(AvDay, 2)

📖 Detailed explanation

Output with rounded values is half of R3. The mark scheme’s R3 names “output with appropriate messages and rounded values” — the CASE OF name, the four messages and the ROUND wrappers are that evidence, for the day half of the question.

The CASE OF runs after the scan, before the output. The day’s results are final at NEXT HourCounter; selecting the name there means Day is ready for the first OUTPUT line.

The labels match the loop’s numbering. With DayCounter running 0 to 6, label 0 is Monday. If you number the days from 1 instead, the labels must shift with them — a mismatch prints the wrong name for every day.

Rounding never changes the maxima and minima here. They are stored readings, already exact — but the ROUND wrapper satisfies the question’s closing line at no cost.

⚠️ Common mistakes

Printing the day’s results after the day loop, so only Sunday’s survive.
Forgetting the day name, or printing DayCounter instead of Day.
Outputting AvDay unrounded — 12.666666… is not two decimal places.
CASE OF labels that disagree with the loop’s numbering, printing “Tuesday” for Monday’s column.

💡 Exam hack / tip

Match your four OUTPUT lines to the question’s sub-bullets — name, maximum, minimum, average — and wrap every temperature in ROUND(…, 2). Tick the four sub-bullets and the day half of R3 is safe.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · output with suitable messagesR1 · R3 · name and rounded values per dayAO3 · correct CASE OF and ROUND

The day outputs complete the R1 evidence and deliver the day half of R3 for AO2, and the CASE OF labels with ROUND wrappers are AO3 accuracy evidence. These are alignment notes, not separately allocated marks.

Feed the week trackers

Each finished day updates MaxWeek, MinWeek and WeekTotal.

🧠 Step-by-step thinking

Update the week inside the day loop. The published answer places the three updates straight after the day’s output lines, still inside FOR DayCounter — each day’s finished MaxDay, MinDay and DayTotal feed the week’s records.
MaxDay > MaxWeek. Strictly greater — a day that merely ties the week’s best changes nothing, and the earlier day keeps the record.
MinDay < MinWeek. Strictly less. (The printed mark scheme reads “IF MinDay > MinWeek” — an obvious typesetting slip: with >, MinWeek could never fall below its 1000 seed and the week’s minimum would output as 1000.)
WeekTotal gathers every reading. WeekTotal ← WeekTotal + DayTotal accumulates the grand total of all 168 readings — the sum the weekly average divides.
These three never reset. Initialised once before the loop, updated seven times inside it, and still holding the week’s results when the loop ends.

Selection: > and < testsUpdate: MaxWeek, MinWeek, WeekTotalOnce per day, inside the loop

❓ Guiding questions – think before revealing

Why compare after the day’s outputs rather than inside the hour loop?
What happens when two days tie for the week’s maximum?
Why does WeekTotal add DayTotal and not AvDay?

Reveal suggested reasoning

Why compare after the day’s outputs rather than inside the hour loop? A day’s maximum is only final once its whole column has been scanned. Comparing hourly partial results would test half-finished days against the week’s records; comparing the finished MaxDay and MinDay tests exactly seven complete days.

What happens when two days tie for the week’s maximum? The test is strictly >, so the later day changes nothing — the first day to reach the value keeps the record. The tie is simply a FALSE comparison, which is why the visualisation includes one deliberately.

Why does WeekTotal add DayTotal and not AvDay? Because WeekTotal totals readings. Adding the averages would also give the right weekly mean here — every day has the same number of readings — but the grand total is the direct partner of the Days * Hours divisor, and it is what the published answer accumulates.

Reveal Cambridge pseudocode for Stage 6

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// still inside FOR DayCounter, after the day's output lines

IF MaxDay > MaxWeek    // update total maximum and minimum
  THEN
    MaxWeek ← MaxDay
ENDIF

IF MinDay < MinWeek
  THEN
    MinWeek ← MinDay
ENDIF

WeekTotal ← WeekTotal + DayTotal   // update total for week

NEXT DayCounter

📖 Detailed explanation

Nested searching is the R2 technique. The mark scheme names “nested searching, totalling” for the week — the week’s maximum and minimum are found by comparing each day’s finished results, a search nested inside the day loop, and WeekTotal is the totalling.

Strict tests keep the first occurrence. A day that equals the record does not take it — the same rule the hour scan uses, applied one level up.

The updates are three statements, not a second grid scan. The week’s results are built from the day’s results while they are still available — no second pass over the 168 readings is needed.

The minimum comparison must point the right way. For a minimum, the update fires when the day’s value is smaller: MinDay < MinWeek. Writing > leaves the 1000 seed untouched and outputs an absurd weekly minimum.

⚠️ Common mistakes

Writing IF MinDay > MinWeek — as printed in the mark scheme — so MinWeek stays at 1000 and the week’s minimum is wrong.
Updating the week trackers inside the hour loop, testing half-finished days against the week’s records.
Using >= or <=, so a later tied day steals the week’s record.
Resetting WeekTotal inside the day loop, so the weekly average is built from Sunday alone.

💡 Exam hack / tip

Write the two week tests as a pair and read them aloud: greater than the week’s best, less than the week’s worst. If either comparison sounds backwards, the record it guards can never change hands.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · nested searching and totallingR2 · week records and running totalAO3 · strict tests inside the day loop

The week updates are the R2 evidence for AO2, and their strict tests at the right loop level are AO3 logic evidence. These are alignment notes, not separately allocated marks.

Finish with the week’s results

One average and three output lines, printed once after the loop.

R2 · R3

🧠 Step-by-step thinking

The weekly average divides by the reading count. AvWeek ← WeekTotal / (Days * Hours) — 7 × 24 = 168 readings. (The printed mark scheme divides by Days alone, which averages the daily totals; the requirement is the average temperature for the week, so the divisor is the number of readings.)
Three lines, once each. “Maximum temperature for week”, “Minimum temperature for week” and “Average temperature for week” — printed after NEXT DayCounter, when all seven days have fed the trackers.
ROUND closes the report. ROUND(AvWeek, 2) — 516 ÷ 42 = 12.285… becomes 12.29 in the visualisation’s sample week; the maximum and minimum are wrapped too, honouring the question’s rounding line.
Every requirement now has its output. Seven day blocks of four lines plus three week lines: the six bullets are complete, and the program ends.

Calculation: WeekTotal / (Days * Hours)Output: three final linesOnce-only, after NEXT DayCounter

❓ Guiding questions – think before revealing

Why divide by Days * Hours and not by Days?
Where do the week’s output lines belong?
Which values do the final lines print?

Reveal suggested reasoning

Why divide by Days * Hours and not by Days? Because WeekTotal is the sum of every reading. Dividing by 7 gives the average daily total; dividing by 168 gives the average temperature — which is what the question asks for.

Where do the week’s output lines belong? After the day loop, executed once. Inside the loop they would print seven times, once per day, with the records still mid-update.

Which values do the final lines print? MaxWeek and MinWeek — the records that survived all seven comparisons — and AvWeek, computed from WeekTotal only after the loop ends. None of the day variables appears in them.

Reveal Cambridge pseudocode for Stage 7

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

NEXT DayCounter

// once, after the day loop
AvWeek ← WeekTotal / (Days * Hours)   // mean of every reading

OUTPUT "Maximum temperature for week ", ROUND(MaxWeek, 2)
OUTPUT "Minimum temperature for week ", ROUND(MinWeek, 2)
OUTPUT "Average temperature for week ", ROUND(AvWeek, 2)

📖 Detailed explanation

Calculation and output together complete R2 and R3. The weekly average is the last calculation (R2) and the three final lines are the week half of R3’s “output with appropriate messages and rounded values”.

The divisor must match the total. Totals of readings divide by counts of readings. Keeping the pair consistent — DayTotal / Hours, WeekTotal / (Days * Hours) — is the AO3 accuracy check for both averages.

The closing lines mirror the question’s final bullet. Maximum, minimum and average for the week, in the order the bullet lists them, each with its message and its rounding.

⚠️ Common mistakes

Dividing WeekTotal by Days alone, reporting the average daily total as a temperature.
Printing the week’s lines inside the day loop, seven times over.
Forgetting ROUND(AvWeek, 2) on the final average.
Outputting MaxDay instead of MaxWeek, reporting Sunday’s weather as the week’s.

💡 Exam hack / tip

Before writing the last three lines, say what each prints: the records that survived seven days, and the total divided by every reading. If your sentence mentions a day variable, the line is in the wrong place.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · average and outputR2 · R3 · week results output onceAO3 · correct divisor and clean closure

The weekly average and final outputs complete the R2 and R3 evidence for AO2, and the matching divisor is AO3 accuracy evidence. These are alignment notes, not separately allocated marks.

Cambridge assessment criteria

What does a high-level response look like?

The mark scheme assesses the answer across AO2 and AO3. Use these descriptors to check the whole program, not just one isolated code section.

9 AO2 + 6 AO3 = 15

AO2 · Apply knowledge and understandingMaximum 9 marks · programming techniques and data structures

Band	What the descriptor means
0	No creditable response.
1–3	At least one programming technique is used. Some data is stored, but not appropriately. Any use of iteration, selection, counting, totalling, input or output may be credited.
4–6	Some programming techniques are appropriate to the problem, with more than one technique applied. Some data structures are appropriate and store some of the required data.
7–9	The full range of techniques needed by the scenario is used appropriately. The chosen data structures are appropriate and store all the required data.

AO3 · Provide solutions to problemsMaximum 6 marks · comments, identifiers, logic and accuracy

Band	What the descriptor means
0	No creditable response.
1–2	Program seen without relevant comments. Some identifier names and some data-structure names are appropriate. The solution may be illogical or inaccurate in many places, but attempts at least one requirement.
3–4	Program seen with some relevant comments. The majority of identifiers and most data-structure names are meaningful. Parts may be illogical or inaccurate, but most requirements are met.
5–6	The program is fully commented. Suitable identifiers are meaningful throughout, and all data structures have meaningful names. The program is logical and accurate and meets all requirements; minor syntax errors are ignored.

How the seven stages contribute

These are links to the assessment criteria, not separate mark allocations.

R1–R5

Stage	Mark-scheme link	AO2 evidence	AO3 evidence
1 · Decode	R1–R3	Identifies the full set of required techniques, including the column-per-day layout and the rounding demand.	Establishes logical order and comments the plan.
2 · Data	Required data structures	Uses Temperatures[] and the six given variables as given, with two counters, two totals and sentinel starts.	Meaningful identifiers throughout.
3 · Initialise	R1, R2	Week sentinels and WeekTotal set once; day sentinels and DayTotal reset per day.	Resets sit at the right loop level.
4 · Day Stats	R1	Nested iteration, searching and totalling down each day’s column.	Accurate two-index access and strict comparisons.
5 · Day Output	R1, R3	CASE OF day names, average from the day total, four rounded output lines per day.	Correct labels, messages and ROUND(…, 2).
6 · Week Stats	R2	Each day’s finished results update MaxWeek, MinWeek and WeekTotal.	Strict tests at the right point inside the day loop.
7 · Summary	R2, R3	Weekly average from every reading; the week’s three results output once.	Divisor matches the total and the report closes cleanly.

Consolidated Cambridge pseudocode

Complete solution

The full answer is presented in one place after the seven stages. Try to construct it yourself before opening this tab during revision.

15-mark exemplar style

Read this carefully: this mirrors the official published mark scheme answer for 0478/23 October/November 2023 Question 10. The array Temperatures[] and the variables MaxDay, MinDay, AvDay, MaxWeek, MinWeek and AvWeek are supplied by the question and are used exactly as given. Three slips in the printed answer are normalised here: the weekly minimum comparison is corrected to IF MinDay < MinWeek (the printed answer reads IF MinDay > MinWeek, which would leave MinWeek stuck at its 1000 seed); the averages are stored in the declared variables AvDay and AvWeek rather than the undeclared DayAverage and WeekAverage the print uses; and the weekly average divides WeekTotal by Days * Hours so it is the true mean of all 168 readings (the printed answer divides by Days alone, which averages the daily totals). The maximum and minimum lines are also wrapped in ROUND(…, 2) to honour the question’s closing rounding line. As published, the loop counters number from 0 — hour 0 is the 00:00 reading and day 0 is Monday — and the sentinels -1000 and 1000 let the first reading set both records.

QUESTION 10 · PUBLISHED MARK SCHEME ANSWERCambridge 0478 style · zero-based counters, as published

// meaningful identifier names and appropriate data structures to store the data required
DECLARE DayCounter, HourCounter : INTEGER
DECLARE AvDay, AvWeek, MaxDay, MinDay, MaxWeek, MinWeek : REAL
DECLARE DayTotal, WeekTotal : REAL
DECLARE Day : STRING

CONSTANT Hours ← 24
CONSTANT Days ← 7

MaxWeek ← -1000    // initialise max and min temperatures and total for the week
MinWeek ← 1000
WeekTotal ← 0

FOR DayCounter ← 0 TO Days - 1
    MaxDay ← -1000    // initialise max and min temperatures and total for each day
    MinDay ← 1000
    DayTotal ← 0
    FOR HourCounter ← 0 TO Hours - 1
        DayTotal ← DayTotal + Temperatures[HourCounter, DayCounter]
        // update total maximum and minimum
        IF Temperatures[HourCounter, DayCounter] > MaxDay
          THEN
            MaxDay ← Temperatures[HourCounter, DayCounter]
        ENDIF
        IF Temperatures[HourCounter, DayCounter] < MinDay
          THEN
            MinDay ← Temperatures[HourCounter, DayCounter]
        ENDIF
    NEXT HourCounter
    CASE OF DayCounter    // select message for day
      0 : Day ← "Monday"
      1 : Day ← "Tuesday"
      2 : Day ← "Wednesday"
      3 : Day ← "Thursday"
      4 : Day ← "Friday"
      5 : Day ← "Saturday"
      6 : Day ← "Sunday"
    ENDCASE
    AvDay ← DayTotal / Hours    // output results for day
    OUTPUT Day    // Results from a day
    OUTPUT "Maximum temperature ", ROUND(MaxDay, 2)
    OUTPUT "Minimum temperature ", ROUND(MinDay, 2)
    OUTPUT "Average temperature ", ROUND(AvDay, 2)
    IF MaxDay > MaxWeek    // update total maximum and minimum
      THEN
        MaxWeek ← MaxDay
    ENDIF
    IF MinDay < MinWeek
      THEN
        MinWeek ← MinDay
    ENDIF
    WeekTotal ← WeekTotal + DayTotal    // update total for week

NEXT DayCounter

AvWeek ← WeekTotal / (Days * Hours)    // mean of every reading of the week
OUTPUT "Maximum temperature for week ", ROUND(MaxWeek, 2)    // output results for week
OUTPUT "Minimum temperature for week ", ROUND(MinWeek, 2)
OUTPUT "Average temperature for week ", ROUND(AvWeek, 2)

Why this is a complete response

It covers the three mark-scheme requirements. R1: the nested loops scan every day’s column, total the readings into DayTotal and track MaxDay and MinDay with strict comparisons, and AvDay divides each day’s total by its reading count. R2: each day’s finished results update MaxWeek, MinWeek and WeekTotal inside the day loop, and AvWeek divides the grand total by the number of readings. R3: CASE OF supplies each day’s name, and every temperature — each day’s maximum, minimum and average, then the week’s maximum, minimum and average — is output with a suitable message and rounded to two decimal places.

What to check before submitting

Check the two-index accesses Temperatures[HourCounter, DayCounter], the sentinel resets at the right loop level, the CASE OF labels 0 to 6, the ROUND(…, 2) wrappers on every temperature output, the strict > and < tests in both the hour scan and the week updates, the Days * Hours divisor, and the comments explaining the program.

Remember: the mark scheme awards AO2 a maximum of 9 marks and AO3 a maximum of 6 marks, giving 15 in total, judged by best fit across the whole response. R1, R2 and R3 are the technique requirements the examiners look for, not a fixed per-item mark split.

Interactive execution trace

Visualisation: watch the program work.

Step through the sample week. Watch the sentinels reset for each day, the hour scan work down one day’s column at a time, the maxima and minima change hands, every day’s four rounded lines appear, and the week trackers take the records.

Last tab · Trace player

Sample input → expected results

The real question stores 24 readings per day (00:00 to 23:00) — 168 cells, far too many to trace one by one. This demonstration keeps the week’s seven day columns but shrinks each day to six readings, the 06:00–11:00 window, so Hours = 6 in the trace and 24 in the real program: every daily average divides by 6 here and by 24 in the examination answer, and the weekly average divides by 7 × 6 = 42 here and 7 × 24 = 168 there. The week deliberately tests the boundary cases: the maximum lands on the very first reading (Monday 06:00) and is tied — never beaten — by Wednesday, testing the strictly-greater rule twice; Friday peaks one degree short; the minimum lands late, on Saturday’s final reading; and Monday’s average divides exactly (84 ÷ 6 = 14.00) while Tuesday’s rounds up (76 ÷ 6 = 12.67). This is the authentic 0478/23 October/November 2023 Question 10. These are illustrative values, not data supplied by the examination question.

Input array Temperatures[] — one column per day

Time	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday	Sunday

Each column is one day, exactly as the question defines; HourCounter 0–5 addresses this demo’s 06:00–11:00 window — the real program’s hours run 00:00–23:00 with 24 readings per day.

Reveal expected records and the week summary

Expected readings and records After all calculations

Time	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday	Sunday	Week role

Execution trace

Step 1 of 1

Initialisation

The program begins.

Program output so far

(No output yet)

Live variable values

Day and record counters

Ready

Tip: use Next step slowly and watch HourCounter and DayCounter. The hour counter climbs 0 to 5 down each day’s column while the day counter waits; MaxWeek and MinWeek never reset — Monday breaks both records at once and Saturday breaks the minimum.

Trace colours: blue row = current time slot; amber cell = current reading; green cells = readings already processed — they fill column by column, because the hour loop works down each day’s column before the day counter moves on.

Rounding example: Tuesday’s readings total 76, so AvDay ← 76 / 6 gives 12.666… and ROUND(AvDay, 2) outputs 12.67; Monday’s 84 divides exactly to 14.00. The tie: Wednesday’s 07:00 reading of 21 equals the week’s maximum, but 21 > 21 is FALSE, so Monday — which reached 21 first — keeps the record.

From a grid of hourly readings to a week of weather reports.

Learning targets

01 · Understand

02 · Construct

03 · Complete

Read the whole question first

Before you move on

Decode the requirements

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Plan the data and identifiers

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Initialise the sentinels and totals

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Scan each day’s column

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Output each day’s rounded line

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Feed the week trackers

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Finish with the week’s results

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

What does a high-level response look like?

How the seven stages contribute

Final self-check

Complete solution

Why this is a complete response

What to check before submitting

Visualisation: watch the program work.

Sample input → expected results

Input array Temperatures[] — one column per day

Expected readings and records After all calculations

Execution trace

Initialisation

Live variable values

Day and record counters