15-mark question lab · start here

From daily sunshine minutes to a month-wide report.

Build the solution in a sensible order: decode the six requirements, plan the variables, initialise the totals, test the leap year and update February, read the month, record and validate each day’s sunshine, then convert to hours and minutes and report. Learn the reasoning before revealing the complete program.

1D arraysLeap-year selectionREPEAT … UNTIL validationTotallingDIV/MOD conversion15 marks

Learning targets

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

Lesson overview

01 · Understand

Identify the two given arrays, the question’s own leap-year definition and the six requirements hidden in the bullet list.

02 · Construct

Build the program in order: leap-year selection that updates MonthDays[2], a CASE OF month lookup, and a validated REPEAT … UNTIL loop that stores, totals and tracks each day’s sunshine.

03 · Complete

Convert the total and the highest daily value to hours and minutes with DIV and MOD, and output the year, month, total and highest with suitable messages.

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 leap-year definition, the two given arrays and the “hours and minutes” outputs all affect the structure of your solution.

Question 12 · 15 marks

Cambridge IGCSE Computer Science · 0478/21 · Question 12

A weather monitoring station records the number of minutes of sunshine for each day of a month.

The one-dimensional (1D) array Sunshine[] is used to store the number of minutes of sunshine for each day. The index of the array represents the day.

The one-dimensional (1D) array MonthDays[] contains the number of days in each month. The index of the array represents the month. In this array, 28 is stored in index 2 for the month of February. However, there are 29 days in February in a leap year.

In this program, a leap year is a year that divides by 4 with no remainder. For example, 2026 is not a leap year but 2028 will be a leap year.

Write a program that meets the following requirements:

allows the user to input a year, calculates whether it is a leap year and updates the number of days in February as appropriate
allows a user to input the name of one of the first three months of the year: January, February or March
allows a user to input the number of minutes of sunshine for each day of the month and validates the input
calculates the total amount of sunshine for the month in hours and minutes
finds the highest amount of daily sunshine for the month
outputs the:
- year
- month
- total amount of sunshine for the month in hours and minutes
- the highest amount of daily sunshine for the month.

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

Do not declare any arrays or variables; assume that this has already been done.

All inputs and outputs must contain suitable messages.

Before you move on

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

Mark-scheme map

Input the year, test MOD(Year, 4) = 0 and update MonthDays[2] to 29 or 28 — selection and input.

Input the month name, convert it with LCASE and use CASE OF to set Days from MonthDays[] — selection.

Input each day’s sunshine with REPEAT … UNTIL validation in the range 0 to 1440, store it in Sunshine[] and total the minutes — validation, iteration, totalling.

Track the highest daily sunshine with a strictly-greater maximum test — selection, finding max.

Convert the total and the highest daily to hours and minutes with DIV and MOD, and output the year, month, total and highest — output.

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

Identify the three input phases. The program asks for a year, then a month name, then one sunshine value for every day of that month — three INPUTs that must happen in that order.
Spot the hidden array write. “Updates the number of days in February as appropriate” means writing 29 (or restoring 28) into MonthDays[2] — the only requirement that changes a given array.
Separate per-day work from once-only work. Prompting, validating, storing, totalling and maximum-tracking happen once per day inside a loop; the leap-year test, the month lookup and the three output lines happen once.
Translate the command words. “Validates the input” demands a REPEAT … UNTIL range test; “in hours and minutes” demands DIV and MOD by 60; “highest” demands maximum tracking with a strictly-greater test.
Build the order. initialise → input year, test leap year, update February → input month, set Days → day loop (validate, store, total, max) → convert → output year, month, total and highest.

Inputs: year, month, daily minutesSelection: leap year and maximumIteration: FOR and REPEATOutput: three summary lines

❓ Guiding questions – think before revealing

Why does the first bullet talk about updating an array?
Which parts of the program run once, and which run every day?
What does “in hours and minutes” tell you about the final calculation?

Reveal suggested reasoning

Why does the first bullet talk about updating an array? February is stored as 28 days in MonthDays[2]. In a leap year the program must change that entry to 29 before the month is chosen, otherwise February would record only 28 days.

Which parts of the program run once, and which run every day? The year input, the leap-year test, the month input and the final outputs run once. The prompt, the validation test, the store, the totalling and the maximum test run once per day inside the FOR loop.

What does “in hours and minutes” tell you about the final calculation? The sunshine is collected in minutes, so it must be split: DIV(Total, 60) gives the whole hours and MOD(Total, 60) the minutes left over. The highest daily value needs the same split.

Reveal Cambridge pseudocode for Stage 1

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Once: read the year, test the leap year, update February
// Once: read the month name and set Days from MonthDays[]

// Every day: prompt, validate, store, total, track the maximum
FOR Count ← 1 TO Days
    // REPEAT ... UNTIL the minutes are in the range 0 to 1440
    // Sunshine[Count] ← Sun, Total ← Total + Sun, test Max
NEXT Count

// Once: convert with DIV and MOD, then output the report

📖 Detailed explanation

“Updates the number of days in February as appropriate” is the requirement that writes to a given array. The leap-year test decides whether MonthDays[2] becomes 29 or is restored to 28.

“Validates the input” demands a loop, not a single test: ask for the day’s minutes, test the range 0 to 1440, and ask again until the value is valid.

“The total amount of sunshine … in hours and minutes” means two numbers in the output, so the minute total must be split with DIV and MOD — and the highest daily value is converted the same way.

“Do not declare any arrays or variables” tells you the answer starts with assignments: Sunshine[] and MonthDays[] already exist and must be used as given.

⚠️ Common mistakes

Starting to code before listing the six bullet requirements in order.
Missing that the leap-year test must write 29 into MonthDays[2] before February’s length is read.
Treating validation as a single IF instead of a REPEAT … UNTIL loop that re-asks.
Forgetting that the outputs must name the year and the month, not just print numbers.

💡 Exam hack / tip

Underline “updates the number of days in February”, “validates the input” and “hours and minutes”. Those three phrases fix the hidden array write, the validation loop and the DIV/MOD conversion — the parts the examiners look for first.

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 array names exactly. Sunshine[] stores each day’s minutes with the day as the index; MonthDays[] stores the length of each month with the month as the index. The mark scheme requires both names as given.
Keep two copies of the month. Month keeps the name exactly as typed for the final output; Month1 holds LCASE(Month) so the CASE OF test matches whatever case the user typed.
Add the loop and input variables. Year holds the year, Days the chosen month’s length, Count the current day and Sun the current day’s minutes.
Add the accumulators. Total collects the month’s minutes and Max remembers the highest daily value seen so far.
Add the conversion variables. Hours and Minutes hold the total split by 60; MaxHours and MaxMinutes hold the same split for the highest day.
Respect the declaration rule. The question says not to declare anything — the plan names what the program uses, and the answer simply assigns them.

Given: Sunshine[] and MonthDays[]Working: Year, Month1, Days, Count, SunAccumulators: Total and MaxResults: Hours and Minutes

❓ Guiding questions – think before revealing

Why do the two arrays use different indexes?
Why keep Month as well as Month1?
Which variables does the day loop update?

Reveal suggested reasoning

Why do the two arrays use different indexes? Sunshine[] is indexed by day, so Sunshine[17] is day 17’s minutes; MonthDays[] is indexed by month, so MonthDays[2] is February’s length. Confusing them stores sunshine in the wrong structure.

Why keep Month as well as Month1? The final output prints the month the user typed. If only the lowercase copy existed the output would read “february”; keeping Month preserves the original for display.

Which variables does the day loop update? Count controls the loop, Sun holds the current input, Sunshine[Count] stores it, and Total and Max accumulate the month’s results across all the days.

Reveal Cambridge pseudocode for Stage 2

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Given by the question — already declared, do not declare again:
//   Sunshine[]   minutes of sunshine, index = day
//   MonthDays[]  days in each month, index = month (28 in index 2)

// Working values the program uses:
Year        // the year entered
Month       // the month name exactly as typed
Month1      // LCASE(Month), used for the CASE OF test
Days        // length of the chosen month, from MonthDays[]
Count       // current day, runs 1 to Days
Sun         // one day's minutes, validated 0 to 1440
Total       // running total of the month's minutes
Max         // highest daily minutes seen so far
Hours, Minutes          // the total as hours and minutes
MaxHours, MaxMinutes    // the highest daily as hours and minutes

📖 Detailed explanation

Meaningful identifiers earn AO3 credit. Total, Max and Days say what they hold, and every name maps back to a phrase in the question.

Two arrays, two jobs. Sunshine[Count] is written once per day; MonthDays[2] is read and written once — the leap-year update is the one write the question demands.

The lowercase copy is a design decision. Converting once with LCASE keeps the CASE OF labels simple and keeps the original month name clean for the output.

No new arrays are needed. The results are consumed as soon as they are calculated, so simple variables are enough — inventing arrays adds risk without credit.

⚠️ Common mistakes

Renaming Sunshine[] or MonthDays[], which the mark scheme treats as not using the supplied data structures.
Printing Month1 in the final output, so “February” appears as “february”.
Inventing arrays for the results when Total, Max and the conversion variables are enough.
Declaring variables when the question explicitly says not to.

💡 Exam hack / tip

Point at each name and ask “which phrase in the question does this store?” Sunshine answers “minutes of sunshine for each day”, MonthDays answers “number of days in each month” and Max answers “highest amount of daily sunshine”.

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 working values

Give every accumulator a known starting value — and declare nothing.

R2 · R3

🧠 Step-by-step thinking

Zero every accumulator. Total ← 0 and Max ← 0 so the first day’s addition and comparison start from a known value.
Zero is a safe seed for Max here. The validation only accepts values from 0 to 1440, so no daily value can be negative and the true maximum is at least 0.
Clear the conversion variables too. The published answer sets Hours, Minutes, MaxHours and MaxMinutes to 0 before calculating them — every value is intentional.
Start Days at 0. Until the month is chosen the program does not know how many days to loop over, so Days ← 0 marks it as not yet set.
Explain why nothing is declared. The question states the arrays and variables already exist, so the answer opens with assignments — a DECLARE would ignore the question’s instruction.

Zeroed: Total and MaxCleared: Hours and MinutesNo declarations

❓ Guiding questions – think before revealing

Why can Max start at 0 for this question?
Why does the published answer not DECLARE anything?
Which values must never be reset inside the day loop?

Reveal suggested reasoning

Why can Max start at 0 for this question? The validation loop only accepts values of at least 0, so the true maximum can never be below 0. A tracker starting at 0 is therefore already no bigger than the real maximum, and the first positive reading takes the lead.

Why does the published answer not DECLARE anything? The question says “Do not declare any arrays or variables; assume that this has already been done.” The mark scheme’s example answer therefore begins with assignments such as Total ← 0.

Which values must never be reset inside the day loop? Total and Max accumulate across the whole month. Resetting either inside the FOR loop would erase the running results and leave only the last day standing.

Reveal Cambridge pseudocode for Stage 3

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Starting values — set once, before any input
Hours ← 0
Minutes ← 0
MaxHours ← 0
MaxMinutes ← 0
Max ← 0
Days ← 0
Total ← 0

// No DECLARE statements — the question says the
// arrays and variables have already been declared

📖 Detailed explanation

“Assume that this has already been done” shapes the whole answer: start with assignments, not declarations. The examiner wants the algorithm, not the declarations.

Zero works for Max only because of the validation. The range test guarantees Sun >= 0, so a seed of 0 can never hide a real maximum. Without that guarantee, seeding the maximum from 0 would be a risk.

Initialisation happens once, before any input. No accumulator is reset mid-month, so every day’s contribution survives to the final report.

⚠️ Common mistakes

Opening the answer with DECLARE statements after the question explicitly says not to.
Resetting Total or Max inside the day loop, so only the final day’s value survives.
Seeding Max with an invented value above 1440, which no valid reading could ever replace.
Forgetting to zero Total and relying on it happening to hold 0.

💡 Exam hack / tip

Say it as a rule: zero what you total, zero what you track, declare nothing. The published answer’s first seven lines do exactly that.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · correct starting stateR2 · R3 · accumulators start at zeroAO3 · follows the question’s instructions

Initialisation supports AO2 because the accumulators can genuinely store the required data, and AO3 because the answer follows the question’s instructions. These are alignment notes, not separately allocated marks.

Read the year and test for a leap year

One input, one MOD test, one write to MonthDays[2].

🧠 Step-by-step thinking

Prompt, then read. The program prints a message asking for the year and then reads Year — the question demands a suitable message on every input.
Apply the question’s own definition. A leap year is “a year that divides by 4 with no remainder”, so the test is exactly MOD(Year, 4) = 0.
Update the given array. The TRUE branch writes MonthDays[2] ← 29; the ELSE branch writes MonthDays[2] ← 28, so the entry is correct either way.
Check the definition against the examples. 2026 gives MOD(2026, 4) = 2 — not a leap year; 2028 gives MOD(2028, 4) = 0 — a leap year. The question’s own examples are your test data.
Do this before reading the month. February’s length must already be 29 when the CASE OF copies MonthDays[2] into Days.

Selection: IF … THEN … ELSETest: MOD(Year, 4) = 0Write: MonthDays[2] ← 29 or 28

❓ Guiding questions – think before revealing

Why MOD and not DIV?
Why does the ELSE branch write 28?
Why must this test run before the month is read?

Reveal suggested reasoning

Why MOD and not DIV? MOD(Year, 4) is the remainder — “no remainder” is exactly MOD(Year, 4) = 0. DIV gives the quotient, which says nothing about divisibility by itself.

Why does the ELSE branch write 28? The array already stores 28 for February. Writing it again is harmless and makes the update explicit for both outcomes — the published answer includes both branches so the selection is complete.

Why must this test run before the month is read? If the user chooses February, MonthDays[2] is copied into Days. The leap-year update has to happen before that copy, or a leap-year February would record only 28 days.

Reveal Cambridge pseudocode for Stage 4

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

OUTPUT "Please enter a year"
INPUT Year

IF MOD(Year, 4) = 0
  THEN
    MonthDays[2] ← 29
  ELSE
    MonthDays[2] ← 28
ENDIF

📖 Detailed explanation

The question defines its own leap year. Only “divides by 4 with no remainder” is wanted here. Adding the real-world century rules answers a different question and risks the logic marks.

This is the one write to a given array. Both branches assign MonthDays[2], so February’s length is correct whether the year is a leap year or not.

The question’s examples are ready-made test data. 2026 must fail the test and 2028 must pass it — check your own condition against both before moving on.

⚠️ Common mistakes

Testing Year / 4 = 0 or DIV(Year, 4) = 0, which checks the wrong thing entirely.
Writing Days ← 29 directly instead of updating MonthDays[2], which breaks the array lookup for February.
Adding century rules such as “divides by 400” that the question’s definition does not ask for.
Running the leap-year test after the month input, so February’s length is read before it is corrected.

💡 Exam hack / tip

Turn the definition into code the moment you read it: divides by 4 with no remainder becomes MOD(Year, 4) = 0. Every 0478 “divides with no remainder” phrase is a MOD test in disguise.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · selection and inputR1 · leap-year test and array updateAO3 · accurate condition

The leap-year stage is the clearest R1 evidence of selection and input for AO2, and the MOD condition is AO3 accuracy evidence. These are alignment notes, not separately allocated marks.

Read the month and set the number of days

Read the name, lowercase it, and map it to Days with CASE OF.

🧠 Step-by-step thinking

Prompt for one of three names. The question limits the choice to January, February or March, so the program only needs three cases.
Convert before comparing. Month1 ← LCASE(Month) means “February”, “february” and “FEBRUARY” all match the case label “february”.
Map each name to its month length. CASE OF Month1 assigns Days ← MonthDays[1] for January, MonthDays[2] for February and MonthDays[3] for March.
Keep the original for the output. The final line prints Month, not Month1, so the report shows the name as the user typed it.
February reads the updated value. Because Stage 4 has already written 29 — or restored 28 — into MonthDays[2], the copy into Days is automatically leap-year correct.

Conversion: LCASE(Month)Selection: CASE OFLookup: Days ← MonthDays[]

❓ Guiding questions – think before revealing

Why is LCASE needed before CASE OF?
Why CASE OF rather than three IFs?
Where does February’s 29 come from?

Reveal suggested reasoning

Why is LCASE needed before CASE OF? String comparison is case sensitive. If the user types “February” and the case label is “february”, the match fails. Lowercasing the input once makes the match reliable for any capitalisation.

Why CASE OF rather than three IFs? One value tested against several constant alternatives is exactly what CASE OF expresses. Three IF statements also work, but the case structure shows the examiner a cleaner selection and mirrors the published answer.

Where does February’s 29 come from? From MonthDays[2], which the leap-year branch has just updated. This stage never mentions 28 or 29 itself — it only copies from the array.

Reveal Cambridge pseudocode for Stage 5

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

OUTPUT "Please enter a month"
INPUT Month

Month1 ← LCASE(Month)

CASE OF Month1
   "january"  : Days ← MonthDays[1]
   "february" : Days ← MonthDays[2]
   "march"    : Days ← MonthDays[3]
ENDCASE

📖 Detailed explanation

The case labels are lowercase strings. “january”, “february” and “march” match the lowercased input, so the comparison is safe whatever case the user types.

The array does the remembering. February’s 28-or-29 decision was made in Stage 4; this stage simply reads MonthDays[2], whatever it now holds.

Days drives the next loop. The recording loop runs FOR Count ← 1 TO Days, so this lookup fixes how many times the daily input repeats.

⚠️ Common mistakes

Comparing the raw input against lowercase labels, so “February” fails to match.
Hard-coding Days ← 29 for February instead of reading MonthDays[2].
Printing Month1 in the final output, so the month appears lowercased.
Trying MonthDays[Month] — the month is a name, not an index.

💡 Exam hack / tip

The pair LCASE then CASE OF is the standard 0478 recipe for reading a choice by name: lowercase the input, keep the original for display, and let the case labels do the matching.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · selection and inputR1 · month name to DaysAO3 · reliable case matching

The month lookup completes the R1 evidence for AO2, and the LCASE conversion is AO3 reliability evidence. These are alignment notes, not separately allocated marks.

Record, validate and total each day

A FOR loop over the days with a REPEAT … UNTIL inside it.

🧠 Step-by-step thinking

Loop over the days. FOR Count ← 1 TO Days visits every day of the chosen month exactly once.
Validate inside a REPEAT loop. Ask for the day’s minutes, then test Sun >= 0 AND Sun <= 1440. While the test fails, the loop asks again — a day cannot have negative minutes or more than 1440.
Store, then accumulate. Sunshine[Count] ← Sun keeps the day’s value in the given array, then Total ← Total + Sun adds it to the running total.
Track the highest with a strict test. IF Sun > Max updates the champion only when today’s value is strictly greater, so an equal value never displaces an earlier day.
Keep the storage outside the REPEAT. The array write, the totalling and the maximum test happen once per day, after a valid value has been accepted.

Iteration: FOR + REPEATValidation: 0 to 1440 inclusiveTotalling and max tracking

❓ Guiding questions – think before revealing

Why REPEAT … UNTIL rather than WHILE?
Why does the range stop at 1440?
What happens to Total and Max across the days?

Reveal suggested reasoning

Why REPEAT … UNTIL rather than WHILE? The prompt must run at least once — you cannot test a value you have not read. REPEAT … UNTIL reads first and tests afterwards; a WHILE loop would need the input duplicated before the loop.

Why does the range stop at 1440? A day holds 24 × 60 = 1440 minutes, so no daily sunshine total can exceed it. Both boundaries are inclusive: 0 minutes is a valid dull day and 1440 is a completely sunny one.

What happens to Total and Max across the days? Neither is reset inside the loop. Each accepted day adds to Total and gets one attempt at Max, so after NEXT Count they hold the month’s figures.

Reveal Cambridge pseudocode for Stage 6

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

FOR Count ← 1 TO Days
    REPEAT
        OUTPUT "Please enter sunshine for day in minutes ", Count
        INPUT Sun
    UNTIL Sun >= 0 AND Sun <= 1440
    Sunshine[Count] ← Sun
    Total ← Total + Sun
    IF Sun > Max
      THEN
        Max ← Sun
    ENDIF
NEXT Count

📖 Detailed explanation

The two loops do different jobs. The FOR loop moves through the days; the REPEAT loop refuses to release one day until its value is valid.

The prompt sits inside the REPEAT. A rejected value is followed immediately by a fresh ask — exactly what “validates the input” requires, with the day number in the message.

Three updates follow each accepted value. Store in Sunshine[Count], add to Total, test against Max — this one loop body is the iteration, validation, totalling and selection evidence.

The index is the day. Day 1 is stored in Sunshine[1], matching the question’s statement that the index of the array represents the day.

⚠️ Common mistakes

Testing only Sun >= 0 and accepting 1500 minutes — more than a day contains.
Placing Sunshine[Count] ← Sun inside the REPEAT loop, so an invalid attempt is stored before it is tested.
Writing WHILE Sun >= 0 AND Sun <= 1440, which accepts the first invalid value and loops forever on a valid one.
Resetting Total inside the FOR loop, so only the last day is totalled.

💡 Exam hack / tip

Read the boundaries from the question: sunshine cannot be negative and a day has at most 1440 minutes, so the UNTIL condition is Sun >= 0 AND Sun <= 1440. Say it aloud as “ask again until the value fits the day”.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · validation, iteration and totallingR2 · validated inputs, totals and maximumAO3 · correct loop structureAO3 · comments explain purpose

This stage carries the R2 evidence for AO2 — validation, iteration, selection and totalling — and the loop structure is AO3 logic evidence. These are alignment notes, not separately allocated marks.

Convert to hours and minutes and report

DIV and MOD by 60, then the three output lines.

🧠 Step-by-step thinking

Convert the total. Hours ← DIV(Total, 60) gives the whole hours; Minutes ← MOD(Total, 60) gives the minutes left over.
Convert the highest too. The published answer splits Max the same way — MaxHours ← DIV(Max, 60) and MaxMinutes ← MOD(Max, 60) — so both results are reported in hours and minutes.
Output the year and month first. The first output line names the session: the month as originally typed, followed by the year the user entered.
Then the two results. One line for the total, one for the highest daily value, each with a suitable message and both numbers.
Check against the bullet list. Year, month, total in hours and minutes, highest daily — four required pieces of information across three output lines.

Conversion: DIV and MOD by 60Output: three linesCheck: four facts

❓ Guiding questions – think before revealing

Why DIV for the hours and MOD for the minutes?
Why convert the maximum as well?
Which month name goes in the output?

Reveal suggested reasoning

Why DIV for the hours and MOD for the minutes? DIV(Total, 60) is the whole number of hours — the quotient — and MOD(Total, 60) is what remains once those hours are removed. Together they split any minute total exactly.

Why convert the maximum as well? The output asks for the highest daily sunshine alongside the total in hours and minutes; converting both keeps the report consistent, and the mark scheme’s example answer does exactly this.

Which month name goes in the output? Month, the original input. Using Month1 would print the lowercased name the user never typed.

Reveal Cambridge pseudocode for Stage 7

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

Hours ← DIV(Total, 60)
Minutes ← MOD(Total, 60)
MaxHours ← DIV(Max, 60)
MaxMinutes ← MOD(Max, 60)

OUTPUT "The sunshine for ", Month, " in the year ", Year
OUTPUT "Total sunshine was ", Hours, " hours and ", Minutes, " minutes"
OUTPUT "The highest daily sunshine was ", MaxHours, " hours and ", MaxMinutes, " minutes"

📖 Detailed explanation

DIV and MOD work as a pair. Every minute is either inside a whole hour (DIV) or left over (MOD). Division alone loses the remainder; MOD alone loses the hours.

The conversion happens after the loop. Total and Max are final only once every day has been recorded, so the four conversion lines sit after NEXT Count.

Suitable messages are part of the marks. The question’s closing line — “All inputs and outputs must contain suitable messages” — is checked here: each output line says what it is printing.

⚠️ Common mistakes

Printing the total in minutes only, ignoring “in hours and minutes”.
Swapping DIV and MOD, so the remainder is reported as hours.
Leaving the highest daily value in minutes while the total is converted.
Printing Month1 instead of Month, or forgetting the year entirely.

💡 Exam hack / tip

Memorise the split as DIV = the whole part, MOD = the leftover. Every “X hours and Y minutes” output in 0478 is the pair DIV(…, 60) and MOD(…, 60).

Reveal the Cambridge mark-scheme and rubric connection

AO2 · arithmetic and outputR3 · conversion and complete outputsAO3 · accurate expressions and messages

The conversion and the three outputs are the R3 evidence for AO2, and the correct DIV and MOD expressions support AO3 accuracy. 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 hidden array update.	Establishes logical order and comments the plan.
2 · Data	Required data structures	Uses Sunshine[] and MonthDays[] as given, with appropriate working variables.	Meaningful identifiers throughout.
3 · Initialise	R2, R3	Accumulators and conversion variables start at zero; nothing is declared.	Follows the question’s instructions with a clean start.
4 · Leap	R1	Inputs the year, tests MOD(Year, 4) = 0 and updates MonthDays[2].	Accurate selection with both branches written.
5 · Month	R1	Inputs the month, converts with LCASE and sets Days through CASE OF.	Reliable case matching; original month kept for output.
6 · Record	R2	Validated daily inputs with REPEAT … UNTIL, storing, totalling and maximum tracking.	Correct loop nesting and strictly-greater maximum test.
7 · Report	R3	DIV/MOD conversion of the total and the maximum, and the three output lines.	Accurate expressions and suitable messages.

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 0478/21 May/June 2026 Question 12. The arrays Sunshine[] and MonthDays[] are supplied by the question and are used exactly as given, and nothing is declared because the question says this has already been done. The answer keeps the user’s original Month for the final output and works from a lowercase copy Month1, and it converts both the total and the highest daily value to hours and minutes before printing the three output lines.

QUESTION 12 · PUBLISHED MARK SCHEME ANSWERCambridge 0478 style · 1-based indexing

// programming techniques of iteration, selection, totalling and output are used
Hours ← 0
Minutes ← 0
MaxHours ← 0
MaxMinutes ← 0
Max ← 0
Days ← 0
Total ← 0

// R1 — input the year and update February for leap years
OUTPUT "Please enter a year"
INPUT Year

IF MOD(Year, 4) = 0    // updating array if leap year
  THEN
    MonthDays[2] ← 29
  ELSE
    MonthDays[2] ← 28
ENDIF

// entering a month and assigning the number of days to a variable
OUTPUT "Please enter a month"
INPUT Month

Month1 ← LCASE(Month)    // converting to lowercase for the case statement — keeping the original for output

// R2 — input each day's sunshine, validated, and total it
CASE OF Month1
   "january"  : Days ← MonthDays[1]
   "february" : Days ← MonthDays[2]
   "march"    : Days ← MonthDays[3]
ENDCASE

FOR Count ← 1 TO Days    // inputting sunshine for each day and validating in a valid range
    REPEAT
        OUTPUT "Please enter sunshine for day in minutes ", Count
        INPUT Sun
    UNTIL Sun >= 0 AND Sun <= 1440
    Sunshine[Count] ← Sun    // storing in array
    Total ← Total + Sun      // totalling sunshine
    IF Sun > Max
      THEN
        Max ← Sun    // updating highest daily
    ENDIF
NEXT Count

// R3 — converting to hours and minutes for the total and the highest daily
Hours ← DIV(Total, 60)
Minutes ← MOD(Total, 60)
MaxHours ← DIV(Max, 60)     // converting the highest daily too
MaxMinutes ← MOD(Max, 60)

// outputting results
OUTPUT "The sunshine for ", Month, " in the year ", Year
OUTPUT "Total sunshine was ", Hours, " hours and ", Minutes, " minutes"
OUTPUT "The highest daily sunshine was ", MaxHours, " hours and ", MaxMinutes, " minutes"

Why this is a complete response

It covers the three mark-scheme requirements. R1: the year is input, the leap-year test MOD(Year, 4) = 0 updates MonthDays[2], and the month name is input and converted with LCASE for the CASE OF lookup. R2: each day’s minutes are validated with REPEAT … UNTIL in the range 0 to 1440, stored in Sunshine[], totalled, and the highest is tracked with a strictly-greater test. R3: the total and the highest daily value are converted to hours and minutes with DIV and MOD, and the year, month, total and highest are output with suitable messages.

What to check before submitting

Check the MOD(Year, 4) = 0 condition and both MonthDays[2] branches, the LCASE conversion before CASE OF, the UNTIL condition Sun >= 0 AND Sun <= 1440, the strictly-greater maximum test, the four DIV and MOD conversions, 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 one complete sample session for February 2028. Watch the leap-year test update February to 29 days, the REPEAT … UNTIL loop reject −10, 1500 and 1441 and re-ask, the running total climb day by day, and the DIV and MOD splits prepare the three output lines.

Last tab · Trace player

Sample input → expected results

The sample session enters the year 2028 — MOD(2028, 4) = 0, so February is updated to 29 days — and the month February, then supplies 29 daily sunshine values. The data deliberately tests both boundaries of the valid range: 0 minutes on day 1 and exactly 1440 minutes on day 17 are accepted, while −10 and 1500 on day 4 and 1441 on day 17 are rejected and re-asked. The leading ordinary reading, 726 minutes, lands early on day 3, so day 22’s close 719 — one minute short — never takes the lead; only the upper-boundary 1440 itself, on day 17, overtakes it and stands as the month’s highest. The month’s total does not divide by 60 exactly, so the minutes remainder is non-zero. The walkthrough shows one complete run of the authentic question, from the year input to the three output lines. These are illustrative values, not data supplied by the examination question.

Sample session inputs February 2028 · leap year

Day	Sunshine (min)

The day number is the index used by Sunshine[] — day 1 is stored in Sunshine[1], day 29 in Sunshine[29].

Reveal expected session results

Expected session results After all calculations

Day	Validation	Stored (min)	Highest so far

Execution trace

Step 1 of 1

Initialisation

The program begins.

Program output so far

(No output yet)

Live variable values

Session counters

Ready

Tip: use Next step slowly and watch Count, Total and Max. Count advances one day at a time to 29, Total grows by each accepted value, and Max changes only on a strictly greater day — watch it jump to 726 on day 3 and to 1440 on day 17.

Trace colours: blue row = current day; amber cell = current input; orange cells = days already processed.

Boundary example: day 17’s first attempt, 1441, is rejected because the valid range is 0 to 1440 inclusive — 1441 <= 1440 is FALSE — and the re-ask accepts exactly 1440, a fully sunny day. That value splits exactly: DIV(1440, 60) = 24 hours and MOD(1440, 60) = 0 minutes. The month’s total does not split exactly: DIV(12706, 60) = 211 hours and MOD(12706, 60) = 46 minutes.

From daily sunshine minutes to a month-wide report.

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 working values

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Read the year and test for a leap year

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Read the month and set the number of days

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Record, validate and total each day

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Convert to hours and minutes and report

🧠 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

Sample session inputs February 2028 · leap year

Expected session results After all calculations

Execution trace

Initialisation

Live variable values

Session counters