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Oracle 1Z1-051 Test Passed : Oracle Database: SQL Fundamentals I

1Z1-051 actual test
  • Exam Code: 1Z1-051
  • Exam Name: Oracle Database: SQL Fundamentals I
  • Updated: Sep 19, 2026
  • Q & A: 254 Questions and Answers
  • PDF Demo
  • PC Test Engine
  • Online Test Engine
  • Total Price: $59.99  

About Oracle 1Z1-051 Exam

A Oracle certification puts you in leadership territory; the Oracle Database: SQL Fundamentals I exam is the trailhead. The 254 practice questions at TestPassed are the most direct path up.

Oracle 1Z1-051 Exam Overview:

Certification Vendor:Oracle
Exam Name:Oracle Database: SQL Fundamentals I
Exam Number:1Z0-051
Exam Duration:120 minutes
Certificate Validity Period:N/A (exam retired / replaced)
Exam Format:Multiple choice, Multiple answer (select multiple correct options)
Passing Score:60%
Exam Price:US$95
Available Languages:English
Real Exam Qty:64-66
Related Certifications:Oracle Certified Associate (OCA) Oracle Database 11g Administrator
Oracle Certified Associate (OCA) Oracle PL/SQL Developer
Sample Questions:Free Download real 1Z1-051 test passed rate
Exam Way:Onsite at Pearson VUE or online proctored (historically)
Pre Condition:None formally required; foundational SQL knowledge recommended
Official Syllabus URL:https://education.oracle.com/

Oracle 1Z1-051 Exam Syllabus Topics:

SectionObjectives
Manipulating Data (DML)- INSERT, UPDATE, DELETE rows
Displaying Data from Multiple Tables- Write joins (equijoins, nonequijoins)
- Self-join and Cartesian product
Using DDL Statements- Create and manage tables
- Create other schema objects
Using Single-Row Functions- Character, number, and date functions
Using Set Operators- Combine queries with UNION, INTERSECT, MINUS
Using Subqueries to Solve Problems- Define and write subqueries
Restricting and Sorting Data- Limit rows retrieved by queries
- Sort query results
- Use substitution to restrict/sort output
Conversion Functions and Conditional Expressions- Apply conditional expressions in SELECT
- Use TO_CHAR, TO_NUMBER, TO_DATE
Reporting Aggregated Data Using Group Functions- Include/exclude grouped rows with HAVING
- Group data with GROUP BY
Retrieving Data Using SQL SELECT Statements- List capabilities of SQL SELECT statements
- Execute basic SELECT queries

Oracle 1Z1-051 Exam FAQ: Worry-Free Answers

Oracle Database: SQL Fundamentals I is an official Oracle certification exam, listed under the code 1Z1-051. Passing it earns you the 11g certification, positioned at the Associate level. It also connects to Oracle Certified Associate (OCA) Oracle Database 11g Administrator, Oracle Certified Associate (OCA) Oracle PL/SQL Developer. For anyone aiming at a leadership track in IT, this credential is one of the clearest markers of verified skill.

Passing Oracle Database: SQL Fundamentals I takes 60%, and the official registration fee is US$95. Retakes charge the full US$95 again, which is why accuracy in preparation matters more than volume. Validate your readiness with repeated TestPassed practice scores above the requirement before committing to a date.

The Oracle Database: SQL Fundamentals I exam contains 64-66 questions to answer within 120 minutes. The SOFT engine from TestPassed imitates the real test scene on your computer and uses special methods to help you master questions and answers, so the official time limit becomes a practiced routine rather than a surprise.

None formally required; foundational SQL knowledge recommended

Vendor requirements are revised periodically, so confirm the current conditions before registering via the official exam page.

Three versions share the same verified content: the PDF prints unlimited copies for paper study, the SOFT engine imitates the real test scene on Windows PCs, and the APP version runs on all electronic products, which is why the majority of examinees choose it. A free demo lets you try before deciding, and after purchase updates are free for 365 days, extendable afterward at a 50% discount.

Worry-free shopping means a 100% money-back guarantee with stated conditions. Take the Oracle Database: SQL Fundamentals I exam within 60 days of purchase; if you fail, you may claim a full refund, provided the exam matches your product. Attempts within 3 days of purchase are ineligible, as are downloaded-but-unused products, free materials, and expired orders; the candidate name must match the payer name. Submit a scanned enrollment slip and the official Score Report PDF within 2 days of the exam, and claims are processed within 7 days. Alternatively, exchange for two other exam products of equal value, free, keeping the update service on your original purchase.

Delivery is immediate: files unlock for download at payment and are emailed to you within one minute. If nothing arrives within 2 hours, check spam and contact our 7/24 customer attendants, who reply within two hours. Installation is unlimited, and credit card payment keeps your money safe.

Oracle Database: SQL Fundamentals I is structured into 10 official domains. The leading ones are Using Subqueries to Solve Problems, Manipulating Data (DML), and Reporting Aggregated Data Using Group Functions. The full topic breakdown appears above; accurate preparation starts with an accurate map.

Oracle Database: SQL Fundamentals I Sample Questions:

Question #1

The EMPLOYEES table contains these columns:
EMPLOYEE_ID NUMBER(4)
ENAME VARCHAR2 (25)
JOB_ID VARCHAR2(10)
Which SQL statement will return the ENAME, length of the ENAME, and the numeric position of the letter "a" in the ENAME column, for those employees whose ENAME ends with a the letter "n"?

  • A. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, ,-1,1) FROM EMPLOYEES WHERE SUBSTR(ENAME, -1, 1) = 'n';
  • B. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, 'a') FROM EMPLOYEES WHERE SUBSTR(ENAME, -1, 1) = 'n';
  • C. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR(ENAME, -1, 1) = 'n';
  • D. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR(ENAME, 1, 1) = 'n';
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

Explanation: Only visible for TestPassed members. You can sign-up / login (it's free).

Question #2

What are two reasons to create synonyms? (Choose two.)

  • A. Your tables have difficult names.
  • B. You want to work on your own tables.
  • C. You want to use another schema's tables.
  • D. You have too many columns in your tables.
  • E. You have too many tables.
  • F. Your tables names are too long.
Reveal Solution  Discussion  0

Correct Answer: A,F  🗳️

Explanation: Only visible for TestPassed members. You can sign-up / login (it's free).

Question #3

Which CREATE TABLE statement is valid?

  • A. CREATE TABLE ord_details
    (ord_no NUMBER(2) UNIQUE, NOT NULL,
    item_no NUMBER(3),
    ord_date DATE DEFAULT SYSDATE NOT NULL);
  • B. CREATE TABLE ord_details (ord_no NUMBER(2) PRIMARY KEY, item_no NUMBER(3) PRIMARY KEY, ord_date DATE NOT NULL);
  • C. CREATE TABLE ord_details
    (ord_no NUMBER(2) ,
    item_no NUMBER(3),
    ord_date DATE DEFAULT NOT NULL,
    CONSTRAINT ord_uq UNIQUE (ord_no),
    CONSTRAINT ord_pk PRIMARY KEY (ord_no));
  • D. CREATE TABLE ord_details
    (ord_no NUMBER(2),
    item_no NUMBER(3),
    ord_date DATE DEFAULT SYSDATE NOT NULL,
    CONSTRAINT ord_pk PRIMARY KEY (ord_no, item_no));
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

Explanation: Only visible for TestPassed members. You can sign-up / login (it's free).

Question #4

Which two statements are true regarding constraints? (Choose two.)

  • A. A constraint can be disabled even if the constraint column contains data.
  • B. A foreign key cannot contain NULL values.
  • C. A column with the UNIQUE constraint can contain NULL values.
  • D. A constraint is enforced only for the INSERT operation on a table.
  • E. All constraints can be defined at the column level as well as the table level.
Reveal Solution  Discussion  0

Correct Answer: A,C  🗳️

Explanation: Only visible for TestPassed members. You can sign-up / login (it's free).

Question #5

Examine the data in the PROMO_BEGIN_DATE column of the PROMOTIONS table:
PROMO_BEGIN _DATE 04-jan-00
10-jan-00
15-dec-99
18-oct-98
22-aug-99
You want to display the number of promotions started in 1999 and 2000.
Which query gives the correct output?

  • A. SELECT COUNT(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '99' THEN 1 ELSE 0 END) "1999", COUNT(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '00' THEN 1 ELSE 0 END) "2000" FROM promotions;
  • B. SELECT COUNT(DECODE(SUBSTR(TO_CHAR(promo_begin_date,'yyyy'), 8), '1999', 1, 0))
    "1999", COUNT(DECODE(SUBSTR(TO_CHAR(promo_begin_date,'yyyy'), 8),'2000', 1,
    0)) "2000"
    FROM promotions;
  • C. SELECT SUM(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '99' THEN 1 ELSE 0 END)
    "1999",SUM(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '00' THEN 1 ELSE 0 END)
    "2000"
    FROM promotions;
  • D. SELECT SUM(DECODE(SUBSTR(promo_begin_date,8),'00',1,0)) "2000",
    SUM(DECODE(SUBSTR
    (promo_begin_date,8),'99',1,0)) "1999"
    FROM promotions;
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

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