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Exam Code: 1Z1-051

Exam Name: Oracle (Oracle Database: SQL Fundamentals I)

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NO.1 Which three statements are true regarding the data types in Oracle Database 10g/11g? (Choose
three.)
A. Only one LONG column can be used per table.
B. A TIMESTAMP data type column stores only time values with fractional seconds.
C. The BLOB data type column is used to store binary data in an operating system file.
D. The minimum column width that can be specified for a VARCHAR2 data type column is one.
E. The value for a CHAR data type column is blank-padded to the maximum defined column width.
Answer: ADE

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NO.2 You need to produce a report where each customer's credit limit has been incremented by $1000. In
the output, t he customer's last name should have the heading Name and the incremented credit limit
should be labeled New Credit Limit. The column headings should have only the first letter of each word in
uppercase .
Which statement would accomplish this requirement?
A. SELECT cust_last_name Name, cust_credit_limit + 1000
"New Credit Limit"
FROM customers;
B. SELECT cust_last_name AS Name, cust_credit_limit + 1000
AS New Credit Limit
FROM customers;
C. SELECT cust_last_name AS "Name", cust_credit_limit + 1000
AS "New Credit Limit"
FROM customers;
D. SELECT INITCAP(cust_last_name) "Name", cust_credit_limit + 1000
INITCAP("NEW CREDIT LIMIT")
FROM customers;
Answer: C

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NO.3 Which two statements are true regarding single row functions? (Choose two.)
A. They a ccept only a single argument.
B. They c an be nested only to two levels.
C. Arguments can only be column values or constants.
D. They a lways return a single result row for every row of a queried table.
E. They c an return a data type value different from the one that is referenced.
Answer: DE

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NO.4 Using the CUSTOMERS table, you need to generate a report that shows 50% of each credit
amount in each income level. The report should NOT show any repeated credit amounts in each income
level.
Which query would give the required result?
A. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50
AS "50% Credit Limit"
FROM customers;
B. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50
AS "50% Credit Limit"
FROM customers;
C. SELECT DISTINCT cust_income_level ' ' cust_credit_limit * 0.50
AS "50% Credit Limit"
FROM customers;
D. SELECT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit"
FROM customers;
Answer: C

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NO.5 View the Exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
A. listing of customers who do not have a credit limit and were born before 1980
B. finding the number of customers, in each city, whose marital status is 'married'
C. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D. listing of those customers whose credit limit is the same as the credit limit of customers residing in the
city 'Tokyo'
E. finding the number of customers, in each city, whose credit limit is more than the average credit limit of
all the customers
Answer: DE

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NO.6 Evaluate the following query:
SQL> SELECT promo_name q'{'s start date was }' promo_begin_date
AS "Promotion Launches"
FROM promotions;
What would be the outcome of the above query?
A. It produces an error because flower braces have been used.
B. It produces an error because the data types are not matching.
C. It executes successfully and introduces an 's at the end of each promo_name in the output.
D. It executes successfully and displays the literal " {'s start date was } " for each row in the output.
Answer: C

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NO.7 Which statement is true regarding the INTERSECT operator?
A. It ignores NULL values.
B. Reversing the order of the intersected tables alters the result.
C. The names of columns in all SELECT statements must be identical.
D. The number of columns and data types must be identical for all SELECT statements in the query.
Answer: D

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NO.8 View the E xhibit and examine the data in the EMPLOYEES table.
You want to generate a report showing the total compensation paid to each employee to date.
You issue the following query:
SQL>SELECT ename ' joined on ' hiredate
', the total compensation paid is '
TO_CHAR(ROUND(ROUND(SYSDATE-hiredate)/365) * sal + comm)
"COMPENSATION UNTIL DATE"
FROM employees;
What is the outcome?
A. It generates an error because the alias is not valid.
B. It executes successfully and gives the correct output.
C. It executes successfully but does not give the correct output.
D. It generates an error because the usage of the ROUND function in the expression is not valid.
E. It generates an error because the concatenation operator can be used to combine only two items.
Answer: C

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NO.9 View the Exhibit to examine the description for the SALES table.
Which views can have all DML operations performed on it? (Choose all that apply.)
A. CREATE VIEW v3
AS SELECT * FROM SALES
WHERE cust_id = 2034
WITH CHECK OPTION;
B. CREATE VIEW v1
AS SELECT * FROM SALES
WHERE time_id <= SYSDATE - 2*365
WITH CHECK OPTION;
C. CREATE VIEW v2
AS SELECT prod_id, cust_id, time_id FROM SALES
WHERE time_id <= SYSDATE - 2*365
WITH CHECK OPTION;
D. CREATE VIEW v4
AS SELECT prod_id, cust_id, SUM(quantity_sold) FROM SALES
WHERE time_id <= SYSDATE - 2*365
GROUP BY prod_id, cust_id
WITH CHECK OPTION;
Answer: AB

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NO.10 View the Exhibit and examine the data in the CUSTOMERS table.
Evaluate the following query:
SQL> SELECT cust_name AS "NAME", cust_credit_limit/2 AS MIDPOINT,MIDPOINT+100 AS "MAX
LOWER LIMIT"
FROM customers;
The above query produces an error on execution.
What is the reason for the error?
A. An alias cannot be used in an expression.
B. The a lias NAME should not be enclosed with in double quotation marks .
C. The MIDPOINT+100 expression gives an error because CUST_CREDIT_LIMIT contains NULL
values.
D. The a lias MIDPOINT should be enclosed with in double quotation marks for the
CUST_CREDIT_LIMIT/2 expression .
Answer: A

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NO.11 Which SQL statements would display the value 1890.55 as $1,890.55? (Choose three .)
A. SELECT TO_CHAR(1890.55,'$0G000D00')
FROM DUAL;
B. SELECT TO_CHAR(1890.55,'$9,999V99')
FROM DUAL;
C. SELECT TO_CHAR(1890.55,'$99,999D99')
FROM DUAL;
D. SELECT TO_CHAR(1890.55,'$99G999D00')
FROM DUAL;
E. SELECT TO_CHAR(1890.55,'$99G999D99')
FROM DUAL;
Answer: ADE

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NO.12 You need to extract details of those products in the SALES table where the PROD_ID column
contains the string '_D123'.
Which WHERE clause could be used in the SELECT statement to get the required output?
A. WHERE prod_id LIKE '%_D123%' ESCAPE '_'
B. WHERE prod_id LIKE '%\_D123%' ESCAPE '\'
C. WHERE prod_id LIKE '%_D123%' ESCAPE '%_'
D. WHERE prod_id LIKE '%\_D123%' ESCAPE '\_'
Answer: B

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NO.13 Examine the structure of the PROMOTIONS table:
name Null Type
PROMO_ID NOT NULL NUMBER(6)
PROMO_NAME NOT NULL VARCHAR2(30)
PROMO_CATEGORY NOT NULL VARCHAR2(30)
PROMO_COST NOT NULL NUMBER(10,2)
The management wants to see a report of unique promotion costs in each promotion category.
Which query would achieve the required result?
A. SELECT DISTINCT promo_cost, promo_category FROM promotions;
B. SELECT promo_category, DISTINCT promo_cost FROM promotions;
C. SELECT DISTINCT promo_cost, DISTINCT promo_category FROM promotions;
D. SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1;
Answer: D

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NO.14 Which two statements are true regarding the USING and ON clauses in table joins? (Choose two.)
A. Both USING and ON clauses can be used for equijoins and nonequijoins.
B. A maximum of one pair of columns can be joined between two tables using the ON clause.
C. The ON clause can be used to join tables on columns that have different names but compatible data
types.
D. The WHERE clause can be used to apply additional conditions in SELECT statements containing the
ON or the USING clause.
Answer: CD

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NO.15 View the Exhibit and examine the structure of the PRODUCTS table.
All products have a list price.
You issue the following command to display the total price of each product after a discount of 25% and a
tax of 15% are applied on it. Freight charges of $100 have to be applied to all the products.
SQL>SELECT prod_name, prod_list_price -(prod_list_price*(25/100))
+(prod_list_price -(prod_list_price*(25/100))*(15/100))+100
AS "TOTAL PRICE"
FROM products;
What would be the outcome if all the parenthese s are removed from the above statement?
A. It produces a syntax error.
B. The result remains unchanged.
C. The total price value would be lower than the correct value.
D. The total price value would be higher than the correct value.
Answer: B

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NO.16 View the Exhibit; e xamine the structure of the PROMOTIONS table.
Each promotion has a duration of at least seven days .
Your manager has asked you to generate a report, which provides the weekly cost for each promotion
done to l date.
Which query would achieve the required result?
A. SELECT promo_name, promo_cost/promo_end_date-promo_begin_date/7
FROM promotions;
B. SELECT promo_name,(promo_cost/promo_end_date-promo_begin_date)/7
FROM promotions;
C. SELECT promo_name, promo_cost/(promo_end_date-promo_begin_date/7)
FROM promotions;
D. SELECT promo_name, promo_cost/((promo_end_date-promo_begin_date)/7)
FROM promotions;
Answer: D

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NO.17 View the Exhibit and examine the structure of the PRODUCTS table.
You need to generate a report in the following format:
CATEGORIES
5MP Digital Photo Camera's category is Photo
Y Box's category is Electronics
Envoy Ambassador's category is Hardware
Which two queries would give the required output? (Choose two.)
A. SELECT prod_name q'''s category is ' prod_category CATEGORIES
FROM products;
B. SELECT prod_name q'['s ]'category is ' prod_category CATEGORIES
FROM products;
C. SELECT prod_name q'\'s\' ' category is ' prod_category CATEGORIES
FROM products;
D. SELECT prod_name q'<'s >' 'category is ' prod_category CATEGORIES
FROM products;
Answer: CD

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NO.18 Evaluate the following query:
SELECT INTERVAL '300' MONTH,
INTERVAL '54-2' YEAR TO MONTH,
INTERVAL '11:12:10.1234567' HOUR TO SECOND
FROM dual;
What is the correct output of the above query?
A. +25-00 , +54-02, +00 11:12:10.123457
B. +00-300, +54-02, +00 11:12:10.123457
C. +25-00 , +00-650, +00 11:12:10.123457
D. +00-300 , +00-650, +00 11:12:10.123457
Answer: A

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NO.19 Examine the structure of the SHIPMENTS table:
name Null Type
PO_ID NOT NULL NUMBER(3)
PO_DATE NOT NULL DATE
SHIPMENT_DATE NOT NULL DATE
SHIPMENT_MODE VARCHAR2(30)
SHIPMENT_COST NUMBER(8,2)
You want to generate a report that displays the PO_ID and the penalty amount to be paid if the
SHIPMENT_DATE is later than one month from the PO_DATE. The penalty is $20 per day.
Evaluate the following two queries:
SQL> SELECT po_id, CASE
WHEN MONTHS_BETWEEN (shipment_date,po_date)>1 THEN
TO_CHAR((shipment_date - po_date) * 20) ELSE 'No Penalty' END PENALTY
FROM shipments;
SQL>SELECT po_id, DECODE
(MONTHS_BETWEEN (po_date,shipment_date)>1,
TO_CHAR((shipment_date - po_date) * 20), 'No Penalty') PENALTY
FROM shipments;
Which statement is true regarding the above commands?
A. Both execute successfully and give correct results.
B. Only the first query executes successfully but gives a wrong result.
C. Only the first query executes successfully and gives the correct result.
D. Only the second query executes successfully but gives a wrong result.
E. Only the second query executes successfully and gives the correct result.
Answer: C

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NO.20 View the Exhibit and examine the structure of the SALES, CUSTOMERS, PRODUCTS, and TIMES
tables.
The PROD_ID column is the foreign key in the SALES table, which references the PRODUCTS table.
Similarly, the CUST_ID and TIME_ID columns are also foreign keys in the SALES table referencing the
CUSTOMERS and TIMES tables, respectively.
Evaluate the following CREATE TABLE command:
CREATE TABLE new_sales(prod_id, cust_id, order_date DEFAULT SYSDATE)
AS
SELECT prod_id, cust_id, time_id
FROM sales;
Which statement is true regarding the above command?
A. The NEW_SALES table would not get created because the DEFAULT value cannot be specified in the
column definition.
B. The NEW_SALES table would get created and all the NOT NULL constraints defined on the specified
columns would be passed to the new table.
C. The NEW_SALES table would not get created because the column names in the CREATE TABLE
command and the SELECT clause do not match.
D. The NEW_SALES table would get created and all the FOREIGN KEY constraints defined on the
specified columns would be passed to the new table.
Answer: B

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