exam_draft_module2.md

Mock Exam: Module 2 (Iteration, Lists, Strings)

~15 minutes, 4 questions. Intended for ELMS online quiz.

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Question 1: Code Tracing

You're writing a script to clean URLs for privacy. What is the output of the following code?

urls = [
    "https://shop.com/item?utm_source=twitter",
    "http://news.org/article?ref=homepage",
    "https://blog.io/post?utm_campaign=sale",
    "https://docs.dev/page"
]
cleaned = []
tracker_count = 0

for url in urls:
    url = url.replace("http://", "https://")
    if "?" in url:
        base, param = url.split("?")
        if param.startswith("utm"):
            cleaned.append(base)
            tracker_count += 1
        else:
            cleaned.append(url)
    else:
        cleaned.append(url)

print(len(cleaned))
print(tracker_count)
print(cleaned[1])

• A) 4 then 2 then https://news.org/article?ref=homepage
  • Note: Correct! Put in rationale from below.
• B) 2 then 2 then https://blog.io/post
  • Note: we're appending all urls to cleaned (this is a transform pattern), so len(cleaned) is 4 (same as input). blog.io/post is the 3rd item in cleaned (cleaned[2])
• C) 4 then 2 then http://news.org/article?ref=homepage
  • Note: Correct number of cleaned and tracked, and the correct url, but it should be https instead of http since we're doing url.replace("http://", "https://")
• D) 4 then 3 then https://news.org/article
  • Note: Correct number of cleaned, but not tracked (only two utm_ params); so news.org url (correct position, cleaned[1]) should have the ?ref=homepage param since it's not a tracker param
• E) 3 then 2 then https://news.org/article?ref=homepage
  • Note: Correct number of tracked, and correct url, but not len(cleaned)

Correct answer: A

Rationale: Walk through each URL:

1. "https://shop.com/item?utm_source=twitter" — has ?, param starts with utm → append base "https://shop.com/item", tracker_count = 1
2. "http://news.org/article?ref=homepage" — first replaced to https://..., has ?, param is ref=homepage which does NOT start with utm → append the full url "https://news.org/article?ref=homepage"
3. "https://blog.io/post?utm_campaign=sale" — has ?, param starts with utm → append base "https://blog.io/post", tracker_count = 2
4. "https://docs.dev/page" — no ? → append as-is

All 4 URLs end up in cleaned (len = 4). Two had trackers (tracker_count = 2). cleaned[1] is the news URL with https:// and ?ref=homepage preserved.

Key distractors:

• B tests whether students think non-tracker URLs are excluded
• C tests whether students miss the http → https replacement
• D tests whether students think ref=homepage starts with utm
• E tests whether students think URLs without ? are excluded

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Question 2: Choosing the Right Loop

For each of the following tasks, select whether a for loop or a while loop is the best (most suitable, idiomatic) choice.

(2a) Process every item in a list of 100 student grades and count how many are above 90.

• A) for loop
• B) while loop
  • Note: We know the exact collection to iterate through (the list of grades) and we need to visit every item. This is the textbook use case for a for loop, now a while: you could solve with while, but it's unnecessarily clunkier.

Correct answer: A

Rationale: We know the exact collection to iterate through (the list of grades) and we need to visit every item. This is the textbook use case for a for loop.

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(2b) Keep asking the user for a password until they enter the correct one.

• A) for loop
  • Note: We don't know in advance how many attempts it will take. We need to keep going until a condition is met (correct password entered). This requires a while loop.
• B) while loop

Correct answer: B

Rationale: We don't know in advance how many attempts it will take. We need to keep going until a condition is met (correct password entered). This requires a while loop.

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(2c) Search through a list of names and stop as soon as you find "Joel".

• A) for loop (with break)
  • Note: You could use a for loop with break to stop early, OR a while loop with an index that advances until the name is found or you reach the end. Both are reasonable approaches. The for + break approach is slightly simpler since you don't need to manage the index yourself.
• B) while loop
  • Note: You could use a for loop with break to stop early, OR a while loop with an index that advances until the name is found or you reach the end. Both are reasonable approaches. The for + break approach is slightly simpler since you don't need to manage the index yourself.
• C) Either would work well

Correct answer: C

Rationale: You could use a for loop with break to stop early, or a while loop with an index that advances until the name is found or you reach the end. Both are reasonable approaches. The for + break approach is slightly simpler since you don't need to manage the index yourself.

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Question 3: Pattern Recognition

Consider this code that filters a list based on a condition:

prices = [12.50, 3.99, 25.00, 8.75, 45.00, 2.50]
expensive = []
for price in prices:
    if price > 10:
        expensive.append(price)

Which of the following programs uses the SAME computational pattern?

• A)

words = ["hello", "hi", "hey", "howdy"]
total = 0
for word in words:
    total += len(word)

Note: The original code uses the filter pattern: iterate through a list, check a condition, and collect only the items that pass into a new list. A is an accumulator pattern (summing values, no conditional collection), not a filter.

• B)

temps = [72, 85, 68, 91, 77, 95, 60]
hot_days = []
for temp in temps:
    if temp > 90:
        hot_days.append(temp)

Note: Correct! B is a filter (collect temps > 90) — same pattern, different data and condition.

• C)

names = ["Joel", "Sarah", "John"]
upper_names = []
for name in names:
    upper_names.append(name.upper())

Note: The original code uses the filter pattern: iterate through a list, check a condition, and collect only the items that pass into a new list. D is a transform/map pattern (doubling every item with no filtering), not a filter.

• D)

nums = [1, 2, 3, 4, 5]
doubled = []
for num in nums:
    doubled.append(num * 2)

Note: The original code uses the filter pattern: iterate through a list, check a condition, and collect only the items that pass into a new list. C is a transform/map pattern (applying a transformation to every item with no filtering), not a filter.

Correct answer: B

Rationale: The original code uses the filter pattern: iterate through a list, check a condition, and collect only the items that pass into a new list.

• B is a filter (collect temps > 90) — same pattern, different data and condition.
• A is an accumulator pattern (summing values, no conditional collection).
• C is a transform/map pattern (applying a transformation to every item with no filtering).
• D is also a transform/map pattern (doubling every item with no filtering).

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Question 4: Find the Bug

A student wrote the following program to clean up a list of course codes by converting them to uppercase. But it doesn't work — the output is still the original list with mixed casing. Which line has the bug, and why?

codes = ["inst126", "CMSC132", "Inst201", "math140"]
for i in range(len(codes)):
    codes[i].upper()          
print(codes)

• A) The range(len(codes)) should just be range(codes) — you can't use len() with range().
  • Note: range(len(codes)) is perfectly valid syntax! The issue is that strings are immutable — .upper() returns a new string but does not modify the original. We need to reassign: codes[i] = codes[i].upper().
• B) Line A calls .upper() but doesn't save the result. It should be codes[i] = codes[i].upper() because strings are immutable.
• C) The for loop should use for code in codes: instead of index-based iteration — that's why the .upper() doesn't work.
  • Note: index-based iteration is fine! The issue is that strings are immutable — .upper() returns a new string but does not modify the original. We need to reassign: codes[i] = codes[i].upper(). Plus, a for code in codes loop would have the same problem since code = code.upper() wouldn't modify the list either!
• D) .upper() doesn't work on strings that are already partially uppercase like "CMSC132".
  • Note: .upper() works on any string regardless of existing casing

Correct answer: B

Rationale: Strings are immutable — .upper() returns a new string but does not modify the original. The student needs to reassign: codes[i] = codes[i].upper().

Key distractors:

• A is wrong because range(len(codes)) is perfectly valid syntax
• C is wrong because index-based iteration is fine; the issue is immutability, not the loop style (and actually, a for code in codes loop would have the same problem since code = code.upper() wouldn't modify the list either!)
• D is wrong because .upper() works on any string regardless of existing casing

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Notes for instructor

Concepts tested across questions:

Concept	Q1	Q2	Q3	Q4
for loops	x	x	x	x
while loops		x
String methods (.replace, .startswith, .split)	x			x
in operator (substring check)	x
List building (.append)	x		x
Conditionals inside loops	x		x
String immutability				x
Filter pattern			x
Transform pattern			x
Accumulator pattern	x		x
for vs while reasoning		x

Bloom's taxonomy levels:

• Q1: Apply (trace execution)
• Q2: Analyze (choose appropriate construct for a scenario)
• Q3: Analyze (recognize structural similarity across different problems)
• Q4: Evaluate (diagnose a bug and identify the root cause)

ELMS implementation notes:

• Q1: Single-answer multiple choice
• Q2: Three separate sub-questions, each single-answer MC (could be grouped as one ELMS question with parts, or three quick standalone questions)
• Q3: Single-answer multiple choice
• Q4: Single-answer multiple choice