15-mark question lab · start here

A year of rainfall, from 365 daily measurements to a drought verdict.

Build the solution in a sensible order: plan the declarations, zero the array, capture each day’s measurement, total the year and round both statistics, count the dry days and hunt the longest consecutive run, then report every result and decide whether there was a drought. Learn the reasoning before revealing the complete program.

DeclarationsArray initialisationROUND to 2 and 4 dpDry-run streak trackingDrought decision15 marks

Learning targets

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

Lesson overview

01 · Understand

Recognise what the question gives and what it demands: no pre-stored data, a declared Rainfall[] array for 365 days, four statistics at two rounding precisions, and an inclusive drought boundary.

02 · Construct

Build the program pass by pass: declare the array and variables, zero every element, input each day’s measurement, total and round, count the dry days, and scan neighbouring pairs for the longest dry run.

03 · Complete

Output every statistic with clear messages and use selection to deliver the drought verdict when the longest dry run reaches 15 days or more.

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. Nothing is pre-stored in this question — declarations, initialisation and input are all your job — and the two rounding precisions, the consecutive-run scan and the inclusive drought boundary each force a structural decision you should make on paper first.

Question 10 · 15 marks

Cambridge IGCSE Computer Science · 0478/22 · Question 10

A program is required to record and analyse the daily rainfall in millimetres for one year. Assume a year has 365 days.

The one-dimensional (1D) array Rainfall[] is used to store the daily rainfall measurements.

Write a program for the following requirements:

Initialise the array Rainfall[] to zero.
Input the rainfall measurement in millimetres for each day and store it in the array.
Calculate the following statistics after all the inputs have been made for the year:
- the total rainfall for the year in centimetres (1 centimetre = 10 millimetres), rounded to two decimal places
- the mean average daily rainfall in millimetres, rounded to four decimal places
- the total number of days with no rainfall
- the longest number of consecutive days in which there was no rainfall.
Output all the calculated statistics with clear and appropriate messaging.
If the longest number of consecutive days in which there was no rainfall is 15 or more, output a message to indicate there was a drought (a long period when there is little or no rainfall).

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

All arrays, variables or constants used must be declared for this algorithm.

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

Declare the array and variables, initialise Rainfall[] to zero, then input and store the rainfall in millimetres for each of the 365 days — declarations, initialisation, iteration, input.

Input loop stores every daily measurement in Rainfall[] with a suitable prompt — iteration, input, output.

Total the year’s rainfall, convert to centimetres and round to two decimal places; find the mean daily rainfall in millimetres rounded to four decimal places — totalling, rounding.

Count the days with no rainfall and scan for the longest run of consecutive dry days — counting, selection, finding max.

Output every statistic with clear messages and decide, with selection, whether the drought message (15 days or more) is needed — output, selection.

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

Notice what is missing. Unlike papers that pre-store the data, this question supplies only the array name — there is no “already set up” sentence. Your program must declare Rainfall[], zero it and input every day’s measurement itself.
Count the passes. The program runs in clear passes: initialise the array, input 365 values, total the rainfall, count the dry days, scan for the longest dry run, then output the statistics and decide about the drought.
Decode the statistics. “Total … in centimetres … rounded to two decimal places” means divide by 10 then ROUND(…, 2); “mean average … rounded to four decimal places” means divide by 365 then ROUND(…, 4); the other two statistics need counting and a maximum search.
Spot the hidden structure. “Consecutive days” is the hardest requirement: counting the days with no rainfall is a simple count, but the longest run needs a scan that compares each day with the next day.
Plan the running order. declarations → zero the array → input loop → total and convert → count dry days → longest-run scan → outputs → drought decision.

Five passes over the arrayTwo ROUND precisionsPair scan: consecutive daysSelection: drought test

❓ Guiding questions – think before revealing

Which phrases tell you the data is not already stored?
Why are there two different rounding requirements?
What makes the drought requirement a selection problem?

Reveal suggested reasoning

Which phrases tell you the data is not already stored? The question asks you to “initialise the array Rainfall[] to zero” and to “input the rainfall measurement … for each day” — and it insists that “all arrays, variables or constants used must be declared”. Nothing is pre-stored: declaration, initialisation and input are all your job.

Why are there two different rounding requirements? The total is reported in centimetres to two decimal places, but the mean daily rainfall stays in millimetres to four decimal places. Each statistic needs its own ROUND — and the mean must be calculated before the total is overwritten with its centimetre value.

What makes the drought requirement a selection problem? The message is conditional: it is needed only when the longest dry run is “15 or more” days. That fixes a boundary test — MaxDryDays >= 15 — and a run of exactly 15 days counts as a drought.

Reveal Cambridge pseudocode for Stage 1

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Declare the array and every variable
// Zero every element of Rainfall[]
// Input 365 daily measurements

// After all the inputs have been made:
// Total the year's rainfall in mm
// Convert: cm total to 2 dp, mean to 4 dp
// Count the days with no rainfall
// Find the longest run of consecutive dry days

// Output the four statistics
// Decide whether there was a drought

📖 Detailed explanation

“All arrays, variables or constants used must be declared” is the clearest signal in the whole question: every identifier you invent needs a DECLARE statement before it is used.

“After all the inputs have been made” tells you the statistics are calculated in later passes — the input loop finishes first, then the analysis loops begin.

“Consecutive days in which there was no rainfall” demands a run, not a count: the scan must compare each day with the next day, because a run is a property of neighbouring elements.

“15 or more” fixes the drought boundary at >= 15 — a longest run of exactly 15 days is a drought.

⚠️ Common mistakes

Skipping the declarations because earlier papers pre-declared the data for you.
Reading “total rainfall … in centimetres” as millimetres and forgetting the ÷ 10 conversion.
Treating “the longest number of consecutive days” as a simple count of dry days.
Missing that the drought test includes 15 itself.

💡 Exam hack / tip

Underline “must be declared”, “two decimal places”, “four decimal places” and “15 or more”. Those four phrases decide the declarations, both ROUND commands and the drought boundary — the highest-value words in the question.

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

Declare the array exactly as demanded. DECLARE Rainfall : ARRAY[1:365] OF REAL — the question names the array and the year length; REAL because millimetre measurements carry decimal places.
This question has no given variables. No “already set up” sentence appears, so every identifier beyond Rainfall[] is yours to declare — the mark scheme’s R1 explicitly credits declaring the variables and the array.
Add the loop counter. DECLARE Index : INTEGER walks the array in every pass — initialisation, input, totalling, counting and the streak scan all reuse it.
Add the statistics variables. TotalRain (REAL) and AverageRain (REAL) hold the rounded values; NoRain (INTEGER) counts the days with no rainfall.
Add the streak pair. DryDays tracks the run of consecutive dry days that is still growing, and MaxDryDays remembers the longest — both INTEGER.

DECLARE Rainfall : ARRAY[1:365] OF REALCounter: IndexStreak pair: DryDays and MaxDryDays

❓ Guiding questions – think before revealing

Why is Rainfall[] declared as REAL rather than INTEGER?
Why does the array run from 1 to 365?
Why does the streak need two variables?

Reveal suggested reasoning

Why is Rainfall[] declared as REAL rather than INTEGER? Rainfall measurements in millimetres can carry decimal places — the statistics are rounded to two and four decimal places, which only makes sense if the stored values are real numbers.

Why does the array run from 1 to 365? The question says to assume a year has 365 days, and Cambridge pseudocode arrays are declared with their bounds — ARRAY[1:365] gives one element per day, indexed from 1 like the loops that follow.

Why does the streak need two variables? DryDays holds the run that is still growing as the scan moves along the year; MaxDryDays keeps the best value seen. Without the pair, the scan would forget a long run the moment it ended.

Reveal Cambridge pseudocode for Stage 2

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

DECLARE Rainfall : ARRAY[1:365] OF REAL
DECLARE Index : INTEGER
DECLARE TotalRain : REAL
DECLARE AverageRain : REAL
DECLARE NoRain : INTEGER
DECLARE DryDays : INTEGER
DECLARE MaxDryDays : INTEGER

📖 Detailed explanation

Declarations are a requirement, not decoration. R1 in the mark scheme credits declaring the array and the variables — an answer that starts straight into loops loses that evidence.

One counter serves every pass. Each pass finishes before the next begins, so Index can be reused from 1 to 365 in all five loops.

Meaningful identifiers earn AO3 credit. NoRain, DryDays and MaxDryDays say exactly what they store — an examiner can map every name back to a phrase in the question.

⚠️ Common mistakes

Starting to code without any DECLARE statements, which the question explicitly forbids.
Declaring Rainfall as ARRAY[0:364], which fights the 1-based indexing used throughout Cambridge pseudocode.
Using one variable for the current run and the longest run, so the maximum is lost when a run ends.
Declaring TotalRain as INTEGER when the total must be rounded to two decimal places.

💡 Exam hack / tip

Borrow the mark scheme’s own names: TotalRain, AverageRain, NoRain, DryDays, MaxDryDays. Each name compresses a phrase from the question, and the declarations take ten seconds to write once the list exists.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · appropriate data structuresR1 · array and variables declaredAO3 · meaningful identifiers

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

Zero the array before anything else

The question lists initialisation as a requirement — and safe accumulation needs it too.

🧠 Step-by-step thinking

Zero every element with a loop. FOR Index ← 1 TO 365 … Rainfall[Index] ← 0 … NEXT Index — there is no “clear whole array” command in Cambridge pseudocode.
Understand why the question demands it. Bullet 1 explicitly requires the initialisation: it is a listed requirement, part of R1, not an optional safety step.
Initialise each accumulator before its own pass. TotalRain ← 0 before the totalling loop and NoRain ← 0 before the counting loop — each accumulator starts from a known value.
Leave the streak pair for its own stage. DryDays and MaxDryDays are set inside the NoRain > 0 guard, because their starting values depend on whether any dry day exists at all.

Loop: 1 to 365Every element ← 0Accumulators start at 0

❓ Guiding questions – think before revealing

Why does zeroing the array need its own loop?
Why does the input loop not make the zeroing redundant?
Where do TotalRain and NoRain get their starting values?

Reveal suggested reasoning

Why does zeroing the array need its own loop? The array has 365 separate elements and a single assignment clears only one. A FOR loop from 1 to 365 sets every Rainfall[Index] ← 0 — and because bullet 1 asks for exactly this, the loop is marked work, not housekeeping.

Why does the input loop not make the zeroing redundant? Every element will indeed be overwritten by the input loop. The question still lists initialisation as a requirement — and the zeroed array guarantees that no leftover value can ever reach the statistics.

Where do TotalRain and NoRain get their starting values? Immediately before the loops that use them: TotalRain ← 0 before the totalling pass and NoRain ← 0 before the counting pass. Each accumulator must begin from a known value before anything is added to it.

Reveal Cambridge pseudocode for Stage 3

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Requirement 1 — initialise the array
FOR Index ← 1 TO 365
    Rainfall[Index] ← 0
NEXT Index

// Accumulators — zeroed before their own passes
TotalRain ← 0
...
NoRain ← 0
...

📖 Detailed explanation

“Initialise the array Rainfall[] to zero” is requirement number one in the question and appears word-for-word in R1 — the zeroing loop is marked work.

Initialisation is element by element. Cambridge pseudocode has no whole-array assignment, so the FOR loop is the expected idiom.

Known starts make accumulators trustworthy. TotalRain ← 0 and NoRain ← 0 guarantee the statistics begin from zero rather than from whatever the memory happened to hold.

⚠️ Common mistakes

Omitting the zeroing loop because the input loop fills every element anyway.
Writing Rainfall[] ← 0, which is not valid pseudocode — one assignment clears one element.
Zeroing TotalRain inside the totalling loop, so the total never grows past a single day.
Initialising DryDays outside the NoRain > 0 guard, where a year with no dry days would leave the streak values meaningless.

💡 Exam hack / tip

Say it as a rule: an array is initialised element by element, an accumulator is initialised immediately before its loop. The zeroing loop takes three lines and satisfies a bullet on its own.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · correct starting stateR1 · Rainfall[] zeroedAO3 · reliable logic

Initialisation supports AO2 because the array and accumulators can genuinely store the required data, and AO3 because the logic starts from a reliable state. These are alignment notes, not separately allocated marks.

Capture the year of measurements

One loop prompts for and stores each day’s rainfall in millimetres.

🧠 Step-by-step thinking

Prompt, then input, then store. The question insists inputs contain suitable messages: OUTPUT the day number, then INPUT Rainfall[Index] — the measurement lands directly in the array.
The loop runs from day 1 to day 365. One pass per day, reusing Index; by NEXT Index the whole year is stored.
No validation is demanded. The question asks only for input and storage — adding range checks is not required and costs time you should spend on the statistics.
Note the order of the passes. Input finishes before any statistic is calculated — the question says the statistics come “after all the inputs have been made for the year”.

OUTPUT then INPUTStore directly in the array365 passes

❓ Guiding questions – think before revealing

Why does the prompt include the day number?
Why store straight into Rainfall[Index]?
Could the statistics be calculated inside this loop?

Reveal suggested reasoning

Why does the prompt include the day number? “All inputs and outputs must contain suitable messages” is a closing requirement of the question. Outputting Index inside the prompt tells the user exactly which day’s measurement is expected.

Why store straight into Rainfall[Index]? The array is the question’s named storage for the daily measurements. Reading into a temporary variable and copying adds a line without adding any credit.

Could the statistics be calculated inside this loop? The total and the dry-day count could accumulate here, but the question explicitly asks for the statistics “after all the inputs have been made” — separate passes are the expected shape and keep each requirement visible.

Reveal Cambridge pseudocode for Stage 4

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Requirement 2 — input each day's measurement
FOR Index ← 1 TO 365
    OUTPUT "Input the rainfall measurement for day ", Index
    INPUT Rainfall[Index]
NEXT Index

📖 Detailed explanation

The prompt is part of the requirement. R1 credits iteration, input and output together — a bare INPUT with no message loses the output evidence.

One loop, one job. Because initialisation already happened, this pass only gathers data; every analysis pass comes later and reads the finished array.

Index doubles as the prompt’s day number. The same counter that selects the element is printed in the message — no second variable needed.

⚠️ Common mistakes

INPUT without a prompting OUTPUT, breaking the “suitable messages” requirement.
Hard-coding 365 separate INPUT statements instead of using a loop.
Starting the statistics before this loop finishes, against the question’s “after all the inputs” wording.
Prompting with a fixed message that never names the day.

💡 Exam hack / tip

The pattern is a drumbeat: prompt, read, store — 365 times. If your loop body says which day it wants and where the value lands, R1’s input evidence is complete.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · iteration and inputR1 · 365 measurements storedAO3 · clear messages

The capture loop is AO2 evidence for iteration, input and output, and the day-numbered prompts are AO3 communication evidence. These are alignment notes, not separately allocated marks.

Total the year and round both statistics

Millimetres to centimetres at 2 dp; the mean at 4 dp.

🧠 Step-by-step thinking

Total the raw millimetres first. TotalRain ← TotalRain + Rainfall[Index] over all 365 days gives the year’s total in millimetres.
Compute the mean before overwriting the total. AverageRain ← ROUND(TotalRain / 365, 4) — the mean is in millimetres, so it must divide the millimetre total while it still exists.
Then convert the total to centimetres. TotalRain ← ROUND(TotalRain / 10, 2) — one centimetre is ten millimetres, and the question wants the total reported in centimetres to two decimal places.
Let ROUND do the rounding. The question says “rounded to two decimal places” and “rounded to four decimal places” — the pseudocode ROUND(value, places) command expresses both directly.

Totalling loopROUND(x, 4) then ROUND(x / 10, 2)Order matters: mean first

❓ Guiding questions – think before revealing

Why must the mean be calculated before the total is converted to centimetres?
Why divide by 365 and not 364?
How does the ROUND command know how many places?

Reveal suggested reasoning

Why must the mean be calculated before the total is converted to centimetres? The mean is required in millimetres. Once TotalRain ← ROUND(TotalRain / 10, 2) has run, the millimetre total is gone — dividing the centimetre value by 365 would produce a mean a tenth of the correct size.

Why divide by 365 and not 364? The mean averages the daily rainfall across the whole assumed year — every one of the 365 stored measurements contributes, so the divisor is 365.

How does the ROUND command know how many places? Its second argument gives the precision directly: ROUND(TotalRain / 10, 2) rounds to two decimal places and ROUND(TotalRain / 365, 4) to four — no manual multiplying and adding of 0.5 is needed.

Reveal Cambridge pseudocode for Stage 5

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Total the year's rainfall in millimetres
TotalRain ← 0
FOR Index ← 1 TO 365
    TotalRain ← TotalRain + Rainfall[Index]
NEXT Index

// Mean daily rainfall in mm, to 4 dp
AverageRain ← ROUND(TotalRain / 365, 4)
// Total rainfall in cm, to 2 dp
TotalRain ← ROUND(TotalRain / 10, 2)

📖 Detailed explanation

Two statistics, one total. Both rounded values come from the single millimetre total — the order of the two assignments is what keeps them both correct.

The conversion is a division by 10. The question defines 1 centimetre = 10 millimetres, so the centimetre total is exactly the millimetre total divided by ten, then rounded.

ROUND’s second argument is the precision. Rounding to two and to four decimal places are two different requirements; writing the places in the command shows the examiner both were read.

⚠️ Common mistakes

Converting TotalRain to centimetres first, then dividing by 365 for a millimetre mean.
Dividing by 10 twice, or multiplying instead of dividing, when converting to centimetres.
Rounding the total to two decimal places while it is still in millimetres.
Dividing by 364, or by the loop counter, instead of the fixed 365.

💡 Exam hack / tip

Write the two ROUND lines together and read them aloud: mean in millimetres by 365 first, total in centimetres by 10 second. That order is the whole trick of this stage.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · totalling and roundingR2 · cm at 2 dp and mean at 4 dpAO3 · correct calculation order

The totalling pass is AO2 evidence for totalling and rounding, and calculating the mean before the conversion is AO3 accuracy evidence. These are alignment notes, not separately allocated marks.

Count dry days and find the longest run

A simple count, then a pair-by-pair scan of neighbouring days.

🧠 Step-by-step thinking

Count the dry days one by one. IF Rainfall[Index] = 0 THEN NoRain ← NoRain + 1 inside its own 1-to-365 loop — the total number of days with no rainfall.
Guard the streak scan with NoRain > 0. If not a single day was dry, there is no run to find: the guard’s ELSE sets MaxDryDays ← 0 and DryDays ← 0, so the scan only runs when at least one dry day exists.
Compare each day with the next. The scan runs FOR Index ← 1 TO 364 and tests Rainfall[Index] = 0 AND Rainfall[Index + 1] = 0 — a run grows only when two neighbouring days are both dry, which is why the loop stops one day early: day 365 has no day 366 to pair with.
Grow the run in the THEN branch. DryDays ← DryDays + 1, and if the growing run beats the best so far, MaxDryDays ← DryDays.
Reset to 1, not 0, in the ELSE. When the pair is not both dry, the old run is over — but the next dry day already begins a brand-new run of length 1, so DryDays ← 1 keeps the count correct.

Count: Rainfall[Index] = 0Pairs: Index and Index + 1ELSE resets DryDays to 1

❓ Guiding questions – think before revealing

Why does the scan loop run to 364 rather than 365?
Why does the ELSE branch reset DryDays to 1 rather than 0?
Why is the NoRain > 0 guard needed?

Reveal suggested reasoning

Why does the scan loop run to 364 rather than 365? Each pass tests a pair of neighbouring days: Rainfall[Index] and Rainfall[Index + 1]. When Index reaches 364 the pair is days 364 and 365 — the last two days of the year. Running to 365 would ask for Rainfall[366], which does not exist.

Why does the ELSE branch reset DryDays to 1 rather than 0? A failed pair means the old run is over — but a dry Index + 1 is already the first day of a new run of length 1. Resetting to 0 would undercount every run by one: the next TRUE pair would give a two-day run a length of 1.

Why is the NoRain > 0 guard needed? If no day was dry, DryDays and MaxDryDays starting at 1 would claim a one-day dry run that never happened. The guard’s ELSE sets both to 0, so a completely wet year correctly reports a longest run of zero days.

Reveal Cambridge pseudocode for Stage 6

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Count the days with no rainfall
NoRain ← 0
FOR Index ← 1 TO 365
    IF Rainfall[Index] = 0
      THEN
        NoRain ← NoRain + 1
    ENDIF
NEXT Index

// Longest run of consecutive dry days
IF NoRain > 0
  THEN
    MaxDryDays ← 1
    DryDays ← 1
    FOR Index ← 1 TO 365 - 1
        IF Rainfall[Index] = 0 AND Rainfall[Index + 1] = 0
          THEN
            DryDays ← DryDays + 1
            IF DryDays > MaxDryDays
              THEN
                MaxDryDays ← DryDays
            ENDIF
          ELSE
            DryDays ← 1
        ENDIF
    NEXT Index
  ELSE
    MaxDryDays ← 0
    DryDays ← 0
ENDIF

📖 Detailed explanation

Counting and run-finding are different problems. NoRain totalises the dry days wherever they fall; MaxDryDays needs the days to be neighbours, which is why only the scan compares Index with Index + 1.

The AND is what enforces “consecutive”. A pair of dry neighbours extends the run; anything else ends it. One IF with one AND carries the entire meaning of the requirement.

The inner IF updates the champion. Every time the run grows, DryDays > MaxDryDays is retested, so the maximum is never missed even inside a single long run.

The guard protects the edge case. A year with zero dry days skips the scan entirely and reports a longest run of 0 — the ELSE of the guard keeps the output honest.

⚠️ Common mistakes

Scanning to 365 and reading Rainfall[366], which is outside the array.
Resetting DryDays to 0 in the ELSE, undercounting every run by one.
Testing only Rainfall[Index + 1] = 0, which counts every dry day after the first as a new run.
Running the streak scan without the NoRain > 0 guard, so a fully wet year reports a run of 1.

💡 Exam hack / tip

Trace a tiny example by hand: days [5, 0, 3, 0, 0]. The pairs fail, fail, fail, then succeed once — DryDays goes 1, 1, 1, 1, 2 and MaxDryDays ends at 2. If your ELSE resets to 0, the same data gives 1 — and you know instantly which version is right.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · counting, selection and finding maxR2 · dry-day count and longest runAO3 · correct pair logic and reset

The streak scan is the strongest AO2 evidence in the program — counting, selection, iteration and maximum tracking in one structure — and the pair logic is AO3 accuracy evidence. These are alignment notes, not separately allocated marks.

Report the statistics and the drought verdict

Four clear messages, then one boundary selection.

🧠 Step-by-step thinking

Output all four statistics with messages. The centimetre total, the millimetre mean, the dry-day count and the longest run — each in its own OUTPUT with wording that says what the number means.
Test the drought boundary with >=. “15 or more” includes 15 itself: IF MaxDryDays >= 15 decides whether the drought message is needed.
Say something in both branches. The published answer outputs a drought message in the THEN branch and a no-drought message in the ELSE — the user learns the verdict either way.
Use the units in the messages. The total is in centimetres and the mean in millimetres — printing “cm” and “mm” inside the output strings is the clearest messaging of all.

Four OUTPUT linesBoundary: MaxDryDays >= 15Both branches speak

❓ Guiding questions – think before revealing

Why >= 15 and not > 15?
What makes an output message “clear and appropriate”?
Where does the drought decision sit in the program?

Reveal suggested reasoning

Why >= 15 and not > 15? The question says “15 or more”. A longest run of exactly 15 days is a drought, so the boundary value must pass the test — MaxDryDays >= 15.

What makes an output message “clear and appropriate”? It names the statistic and its unit: “The total rainfall for the year was … cm” tells the user what the number is, while a bare number on screen tells them nothing.

Where does the drought decision sit in the program? After the four statistics and after MaxDryDays is final — the selection is the program’s last act, using the finished value from the streak scan.

Reveal Cambridge pseudocode for Stage 7

CODE PLAN · STRUCTURE ONLYBuild this before filling in the details

// Requirement 4 — output the statistics with messages
OUTPUT "The total rainfall for the year was ", TotalRain, " cm"
OUTPUT "The average daily rainfall was ", AverageRain, " mm"
OUTPUT "The total number of days with no rainfall was ", NoRain, " day(s)"
OUTPUT "The longest period with no rainfall was ", MaxDryDays, " day(s)"

// Requirement 5 — the drought decision (15 or more)
IF MaxDryDays >= 15
  THEN
    OUTPUT ""
    OUTPUT "There was a drought this year because the longest ",
        "period with no rainfall was 15 days or longer"
  ELSE
    OUTPUT ""
    OUTPUT "There was no drought this year, because the longest ",
        "period with no rainfall was shorter than 15 days"
ENDIF

📖 Detailed explanation

Four statistics, four messages. Each OUTPUT pairs a label with its value — exactly the “clear and appropriate messaging” the bullet demands.

The blank OUTPUT separates the verdict. An empty OUTPUT "" line leaves a gap before the drought message, a small touch the published answer uses to make the verdict stand out.

The decision uses the finished MaxDryDays. By this point the streak scan is complete, so the selection reads a final value — no ordering risk remains.

⚠️ Common mistakes

Using > 15, which wrongly reports “no drought” for a run of exactly 15 days.
Printing bare numbers with no labels, against the “suitable messages” requirement.
Forgetting the units — the total is in centimetres, the mean in millimetres.
Placing the drought selection inside the streak scan, where MaxDryDays is still changing.

💡 Exam hack / tip

End-of-question checklist: four labelled statistics, units on show, >= 15 boundary, both branches speaking. Four ticks in the margin secure R3 completely.

Reveal the Cambridge mark-scheme and rubric connection

AO2 · output and selectionR3 · statistics and drought verdictAO3 · clear messages and >= boundary

The report stage is the R3 evidence — every statistic output once and the drought decided by selection — and the message quality is AO3 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 passes, both ROUND precisions and the drought boundary.	Sets a logical order and a commented plan.
2 · Data	Required data structures	DECLARE statements for the array and all six variables, as the question demands.	Meaningful identifiers throughout.
3 · Initialise	R1	Every element of Rainfall[] zeroed by a loop; accumulators set before their passes.	Clear comments and a reliable starting state.
4 · Capture	R1	Iteration prompts for and stores all 365 daily measurements.	Day-numbered prompts with suitable messages.
5 · Totals	R2	Millimetre total, centimetre conversion at 2 dp, mean at 4 dp.	Mean calculated before the total is overwritten.
6 · Streaks	R2	Dry-day counting plus the pair scan for the longest consecutive run.	Correct AND test, 364 bound and ELSE reset to 1.
7 · Report	R3	Four statistics output with clear messages.	Drought decision with >= 15 and both branches speaking.

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

QUESTION 10 · FULL PSEUDOCODEMirrors the published 0478/22 May/June 2026 mark scheme answer

// array and variable declaration
DECLARE Rainfall : ARRAY[1:365] OF REAL
DECLARE Index : INTEGER
DECLARE TotalRain : REAL
DECLARE AverageRain : REAL
DECLARE NoRain : INTEGER
DECLARE DryDays : INTEGER
DECLARE MaxDryDays : INTEGER

// initialising array
FOR Index ← 1 TO 365
    Rainfall[Index] ← 0
NEXT Index

// daily input of rainfall
FOR Index ← 1 TO 365
    OUTPUT "Input the rainfall measurement for day ", Index
    INPUT Rainfall[Index]
NEXT Index

// finding the total and average rainfall
TotalRain ← 0
FOR Index ← 1 TO 365
    TotalRain ← TotalRain + Rainfall[Index]
NEXT Index

// average daily rainfall to 4dp
AverageRain ← ROUND(TotalRain / 365, 4)
// total rainfall in cm to 2dp
TotalRain ← ROUND(TotalRain / 10, 2)

// counting the number of days with no rain
NoRain ← 0
FOR Index ← 1 TO 365
    IF Rainfall[Index] = 0
      THEN
        NoRain ← NoRain + 1
    ENDIF
NEXT Index

// finding the longest number of consecutive rain free days
IF NoRain > 0
  THEN
    MaxDryDays ← 1
    DryDays ← 1
    FOR Index ← 1 TO 365 - 1
        IF Rainfall[Index] = 0 AND Rainfall[Index + 1] = 0
          THEN
            DryDays ← DryDays + 1
            IF DryDays > MaxDryDays
              THEN
                MaxDryDays ← DryDays
            ENDIF
          ELSE
            DryDays ← 1
        ENDIF
    NEXT Index
  ELSE
    MaxDryDays ← 0
    DryDays ← 0
ENDIF

// outputting the statistics with messages
OUTPUT "The total rainfall for the year was ", TotalRain, " cm"
OUTPUT "The average daily rainfall was ", AverageRain, " mm"
OUTPUT "The total number of days with no rainfall was ", NoRain, " day(s)"
OUTPUT "The longest period with no rainfall was ", MaxDryDays, " day(s)"

// checking if there was a drought
IF MaxDryDays >= 15
  THEN
    OUTPUT ""
    OUTPUT "There was a drought this year because the longest period with no rainfall was 15 days or longer"
  ELSE
    OUTPUT ""
    OUTPUT "There was no drought this year, because the longest period with no rainfall was shorter than 15 days"
ENDIF

Why this is a complete response

It meets the three mark-scheme requirements: R1 — the array and variables are declared, Rainfall[] is initialised to zero and every daily rainfall in millimetres is input with a prompt and stored; R2 — the total is converted to centimetres and rounded to two decimal places, the mean daily rainfall is rounded to four decimal places, the days with no rainfall are counted and the longest consecutive dry run is found; R3 — every statistic is output with a clear message and selection decides and reports whether there was a drought.

What to check before submitting

Check that every identifier is declared, the zeroing loop runs before the input loop, the mean is calculated before TotalRain is overwritten with the centimetre value, the streak scan stops at day 364 and resets DryDays to 1 (not 0) in the ELSE branch, and the drought test uses >= 15.

Remember: the marks for this question are awarded as a maximum of 9 marks for AO2 and a maximum of 6 marks for AO3 — 15 in total — each judged by best fit across the whole answer. The R1–R3 labels list the technique requirements the examiners look for; they are not a fixed per-item split of the marks.

Interactive execution trace

Visualisation: watch the program work.

Step through the first 20 days of a sample year. Watch the array zeroed and filled, the millimetre total rounded to centimetres and a four-decimal mean, the dry-day counter climb, and the pair-by-pair scan find the dry run that decides the drought.

Last tab · Trace player

Sample input → expected results

This question is the authentic Cambridge IGCSE 0478/22 May/June 2026 Paper 22 Question 10. A real trace of 365 days would be unwatchable, so the walkthrough shrinks the year to a 20-day window — the real program loops to day 365 and divides the mean by 365. The window deliberately tests four cases: an isolated dry day 2 that checks the streak reset, a 16-day dry spell (days 5 to 20) that triggers the drought message, a centimetre total of 24.163 ÷ 10 = 2.4163 that rounds up to 2.42 at two decimal places, and a mean of 24.163 ÷ 20 = 1.20815 whose fifth decimal rounds up to 1.2082 at four. These are illustrative values, not data supplied by the examination question.

Sample window inputs First 20 days of the year

Day	Rainfall (mm)

The row number is the day index used by Rainfall[]. The full program continues this input loop to day 365 and divides the mean by 365.

Reveal expected statistics and drought verdict

Expected statistics For the 20-day window

Day	Rainfall (mm)	Classification	Dry run so far

Execution trace

Step 1 of 1

Initialisation

The program begins.

Program output so far

(No output yet)

Live variable values

Drought counters

Ready

Tip: use Next step slowly and watch Index, DryDays and MaxDryDays. The same Index counter serves every pass, the streak grows only on a dry pair, and the drought test runs once, after the scan has finished.

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

Boundary examples: the window’s total is 24.163 mm, so 24.163 ÷ 10 = 2.4163 and ROUND(2.4163, 2) produces 2.42 — the third decimal rounds the second up. The mean, 24.163 ÷ 20 = 1.20815, rounds up to 1.2082 at four decimal places. The drought boundary is inclusive: a longest run of exactly 15 days still passes MaxDryDays >= 15.

A year of rainfall, from 365 daily measurements to a drought verdict.

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

Zero the array before anything else

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Capture the year of measurements

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Total the year and round both statistics

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Count dry days and find the longest run

🧠 Step-by-step thinking

❓ Guiding questions – think before revealing

📖 Detailed explanation

⚠️ Common mistakes

💡 Exam hack / tip

Report the statistics and the drought verdict

🧠 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 window inputs First 20 days of the year

Expected statistics For the 20-day window

Execution trace

Initialisation

Live variable values

Drought counters