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DA0-002 Data Acquisition and Preparation Practice Question

Exhibit

SELECT customer_id, COUNT(*) as order_count FROM orders GROUP BY customer_id HAVING COUNT(*) > 5;

Refer to the exhibit. What does the query return?

⚠ Common exam trap

CompTIA often tests the distinction between 'at least' (>=) and 'more than' (>) in HAVING clauses, and candidates may misread the condition as including exactly 5 orders.

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

✓

Customers who have placed more than 5 orders.

The query uses a HAVING clause with COUNT(*) > 5, which filters groups (by customer ID) to only those with more than 5 orders. The GROUP BY customer ID ensures the count is per customer, so the result is customers who have placed more than 5 orders. Option C is correct because the condition is strictly greater than 5, not at least 5.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

✗
All orders grouped by customer ID.
Why it's wrong here
The query returns aggregated counts, not individual orders.
✗
Customers who have placed at least 5 orders.
Why it's wrong here
HAVING >5 excludes those with exactly 5 orders.
✓
Customers who have placed more than 5 orders.
Why this is correct
HAVING COUNT(*) > 5 ensures only customers with more than 5 orders are included.
✗
All customers who have placed orders.
Why it's wrong here
The query filters on count > 5, not all customers.

Go deeper

DA0-002 Data Acquisition and Preparation Practice Question

Exhibit

Why each option matters

Option-by-option breakdown

Related to this question