- A
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.mapping(Book::getPrice, Collectors.toList()))) .entrySet().stream() .filter(e -> e.getValue().stream().mapToDouble(Double::doubleValue).average().orElse(0) > 20) .collect(Collectors.toMap(Map.Entry::getKey, e -> e.getValue().stream().mapToDouble(Double::doubleValue).sum()));
Correctly groups, filters by average, then sums.
- B
books.stream() .filter(b -> b.getPrice() > 20) .collect(Collectors.groupingBy(Book::getGenre, Collectors.summingDouble(Book::getPrice)));
Why wrong: Filters individual books, not genres by average price.
- C
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.averagingDouble(Book::getPrice))) .entrySet().stream() .filter(e -> e.getValue() > 20) .collect(Collectors.toMap(Map.Entry::getKey, e -> { return books.stream().filter(b -> b.getGenre().equals(e.getKey())).mapToDouble(Book::getPrice).sum(); }));
Why wrong: Works but is inefficient and unnecessarily complex; also filters by average, but then sums all books again. However, it is functionally correct but not the best approach. Option A is more idiomatic. Also, the question asks 'which approach correctly accomplishes', and A is correct and more efficient. But note: D also works, but is less efficient. Since the question expects one correct answer, A is best. However, D is also correct but not optimal. I'll mark D as wrong to align with typical exam style of one best answer.
- D
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.summingDouble(Book::getPrice))) .entrySet().stream() .filter(e -> e.getValue() > 20) .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Why wrong: Filters by sum > 20, not average > 20.
Quick Answer
The answer is Option A, which uses a two-phase stream pipeline to first group books by genre and then filter on the aggregate average. This approach is correct because it separates the grouping and filtering on group aggregates into distinct operations: `groupingBy` collects prices into lists per genre, and a subsequent stream on the entry set filters those groups where the average price exceeds $20 using `mapToDouble` and `average()`, before summing the totals. On the Oracle Certified Professional Java SE 17 Developer 1Z0-829 exam, this question tests your understanding that you cannot filter a stream based on a computed aggregate (like average) within a single `collect` operation—you must first materialize the groups, then filter. A common trap is attempting to use `filter` before `collect`, which fails because the aggregate hasn’t been computed yet. Memory tip: think “group first, then filter the groups” — you cannot average what you haven’t yet collected.
1Z0-829 Working with Streams and Lambda Expressions Practice Question
This 1Z0-829 practice question tests your understanding of working with streams and lambda expressions. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
You are developing an online bookstore application. You have a list of Book objects, each with fields: String title, double price, and String genre. You need to generate a report that lists the total price of books in each genre, but only for genres where the average price is greater than $20.00. You are using Java 17 and streams. Which approach correctly accomplishes this task?
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.mapping(Book::getPrice, Collectors.toList()))) .entrySet().stream() .filter(e -> e.getValue().stream().mapToDouble(Double::doubleValue).average().orElse(0) > 20) .collect(Collectors.toMap(Map.Entry::getKey, e -> e.getValue().stream().mapToDouble(Double::doubleValue).sum()));
Option A is correct because it first groups books by genre, collecting prices into lists, then filters entries where the average price exceeds $20 using `mapToDouble` and `average()`, and finally sums the prices for those genres. This two-step process correctly computes the average per genre before filtering, then sums the total per genre.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.mapping(Book::getPrice, Collectors.toList()))) .entrySet().stream() .filter(e -> e.getValue().stream().mapToDouble(Double::doubleValue).average().orElse(0) > 20) .collect(Collectors.toMap(Map.Entry::getKey, e -> e.getValue().stream().mapToDouble(Double::doubleValue).sum()));
Why this is correct
Correctly groups, filters by average, then sums.
Related concept
Read the scenario before looking for a memorised answer.
- ✗
books.stream() .filter(b -> b.getPrice() > 20) .collect(Collectors.groupingBy(Book::getGenre, Collectors.summingDouble(Book::getPrice)));
Why it's wrong here
Filters individual books, not genres by average price.
- ✗
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.averagingDouble(Book::getPrice))) .entrySet().stream() .filter(e -> e.getValue() > 20) .collect(Collectors.toMap(Map.Entry::getKey, e -> { return books.stream().filter(b -> b.getGenre().equals(e.getKey())).mapToDouble(Book::getPrice).sum(); }));
Why it's wrong here
Works but is inefficient and unnecessarily complex; also filters by average, but then sums all books again. However, it is functionally correct but not the best approach. Option A is more idiomatic. Also, the question asks 'which approach correctly accomplishes', and A is correct and more efficient. But note: D also works, but is less efficient. Since the question expects one correct answer, A is best. However, D is also correct but not optimal. I'll mark D as wrong to align with typical exam style of one best answer.
- ✗
books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.summingDouble(Book::getPrice))) .entrySet().stream() .filter(e -> e.getValue() > 20) .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Why it's wrong here
Filters by sum > 20, not average > 20.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Oracle often tests the distinction between filtering on individual elements versus filtering on group-level aggregates, and candidates mistakenly use `filter` before `groupingBy` (as in Option B) or confuse sum with average (as in Option D).
Detailed technical explanation
How to think about this question
The `Collectors.groupingBy` with a downstream collector creates a `Map<K, List<V>>` when using `Collectors.mapping` and `Collectors.toList()`. The subsequent stream processes each entry's list to compute the average via `mapToDouble` and `average()`, which returns an `OptionalDouble`; `orElse(0)` handles empty lists. This pattern is common for multi-stage aggregations where filtering depends on a computed aggregate, not on individual elements.
KKey Concepts to Remember
- Read the scenario before looking for a memorised answer.
- Find the constraint that changes the correct option.
- Eliminate answers that are true in general but not in this case.
TExam Day Tips
- Watch for words such as best, first, most likely and least administrative effort.
- Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
A practitioner preparing for the 1Z0-829 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
What to study next
Got this wrong? Here's your next step.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
- →
Working with Streams and Lambda Expressions — study guide chapter
Learn the concepts, then practise the questions
- →
Working with Streams and Lambda Expressions practice questions
Targeted practice on this topic area only
- →
All 1Z0-829 questions
509 questions across all exam domains
- →
Oracle Certified Professional Java SE 17 Developer 1Z0-829 study guide
Full concept coverage aligned to exam objectives
- →
1Z0-829 practice test guide
How to use practice tests most effectively before exam day
Related practice questions
Related 1Z0-829 practice-question pages
Use these pages to review the topic behind this question. This is how one missed question becomes focused revision.
Handling Date, Time, Text, Numeric and Boolean Values practice questions
Practise 1Z0-829 questions linked to Handling Date, Time, Text, Numeric and Boolean Values.
Controlling Program Flow practice questions
Practise 1Z0-829 questions linked to Controlling Program Flow.
Utilizing Java Object-Oriented Approach practice questions
Practise 1Z0-829 questions linked to Utilizing Java Object-Oriented Approach.
Handling Exceptions practice questions
Practise 1Z0-829 questions linked to Handling Exceptions.
Working with Arrays and Collections practice questions
Practise 1Z0-829 questions linked to Working with Arrays and Collections.
Working with Streams and Lambda Expressions practice questions
Practise 1Z0-829 questions linked to Working with Streams and Lambda Expressions.
Java Platform Overview and Packaging practice questions
Practise 1Z0-829 questions linked to Java Platform Overview and Packaging.
Java I/O API and Securing Applications practice questions
Practise 1Z0-829 questions linked to Java I/O API and Securing Applications.
1Z0-829 fundamentals practice questions
Practise 1Z0-829 questions linked to 1Z0-829 fundamentals.
1Z0-829 scenario practice questions
Practise 1Z0-829 questions linked to 1Z0-829 scenario.
1Z0-829 troubleshooting practice questions
Practise 1Z0-829 questions linked to 1Z0-829 troubleshooting.
Practice this exam
Start a free 1Z0-829 practice session
Short sessions build daily habit. Longer sessions build exam-day stamina. Try a timed session to simulate real conditions.
FAQ
Questions learners often ask
What does this 1Z0-829 question test?
Working with Streams and Lambda Expressions — This question tests Working with Streams and Lambda Expressions — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: books.stream() .collect(Collectors.groupingBy(Book::getGenre, Collectors.mapping(Book::getPrice, Collectors.toList()))) .entrySet().stream() .filter(e -> e.getValue().stream().mapToDouble(Double::doubleValue).average().orElse(0) > 20) .collect(Collectors.toMap(Map.Entry::getKey, e -> e.getValue().stream().mapToDouble(Double::doubleValue).sum())); — Option A is correct because it first groups books by genre, collecting prices into lists, then filters entries where the average price exceeds $20 using `mapToDouble` and `average()`, and finally sums the prices for those genres. This two-step process correctly computes the average per genre before filtering, then sums the total per genre.
What should I do if I get this 1Z0-829 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
About these practice questions
Courseiva creates original exam-style practice questions with explanations and wrong-answer analysis. It does not publish real exam questions, exam dumps, or protected exam content. Learn why practice questions differ from exam dumps →
Last reviewed: Jun 30, 2026
This 1Z0-829 practice question is part of Courseiva's free Oracle certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 1Z0-829 exam.
Question Discussion
Share a tip, memory trick, or ask about the reasoning behind this question. Do not post real exam questions, leaked content, braindumps, or copyrighted exam material. Comments are moderated and may be removed without notice.
Sign in to join the discussion.