[Free] 2018(Jan) EnsurePass Pass4sure Oracle 1z0-051 Dumps with VCE and PDF 141-150

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Oracle Database: SQL Fundamentals I

Question No: 141 – (Topic 1)

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?

  1. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS quot;50% Credit Limitquot; FROM customers;

  2. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS quot;50% Credit Limitquot;

    FROM customers;

  3. SELECT DISTINCT cust_income_level #39; #39; cust_credit_limit * 0.50 AS quot;50% Credit Limitquot; FROM customers;

  4. SELECT cust_income_level #39; #39; cust_credit_limit * 0.50 AS quot;50% Credit Limitquot; FROM customers;

Answer: C

Explanation: Duplicate Rows

Unless you indicate otherwise, SQL displays the results of a query without eliminating the duplicate rows.

To eliminate duplicate rows in the result, include the DISTINCT keyword in the SELECT clause immediately after the SELECT keyword.

You can specify multiple columns after the DISTINCT qualifier. The DISTINCT qualifier affects all the selected columns, and the result is every distinct combination of the columns.

Question No: 142 – (Topic 1)

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

  1. A constraint can be disabled even if the constraint column contains data

  2. A constraint is enforced only for the INSERT operation on a table

  3. A foreign key cannot contain NULL values

  4. All constraints can be defined at the column level as well as the table level

  5. A columns with the UNIQUE constraint can contain NULL values

Answer: A,E

Question No: 143 – (Topic 1)

The STUDENT_GRADES table has these columns:

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Which statement finds students who have a grade point average (GPA) greater than 3.0 for the calendar year 2001?

  1. SELECT student_id, gpa FROM student_grades

    WHERE semester_end BETWEEN ’01-JAN-2001′ AND ’31-DEC-2001′

    OR gpa gt; 3.;

  2. SELECT student_id, gpa FROM student_grades

    WHERE semester_end BETWEEN ’01-JAN-2001′ AND ’31-DEC-2001′

    AND gpa gt 3.0;

  3. SELECT student_id, gpa FROM student_grades

    WHERE semester_end BETWEEN ’01-JAN-2001′ AND ’31-DEC-2001′ AND gpa gt; 3.0;

  4. SELECT student_id, gpa FROM student_grades

    WHERE semester_end BETWEEN ’01-JAN-2001′ AND ’31-DEC-2001′ OR gpa gt; 3.0;

  5. SELECT student_id, gpa FROM student_grades

WHERE semester_end gt; ’01-JAN-2001′ OR semester_end lt; ’31-DEC-2001′ AND gpa gt;= 3.0;

Answer: C

Question No: 144 – (Topic 1)

The PART_CODE column in the SPARES table contains the following list of values:

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Which statement is true regarding the outcome of the above query?

  1. It produces an error.

  2. It displays all values.

  3. It displays only the values A%_WQ123 and AB_WQ123 .

  4. It displays only the values A%_WQ123 and A%BWQ123 .

  5. It displays only the values A%BWQ123 and AB_WQ123.

Answer: D Explanation:

Combining Wildcard Characters

The % and _ symbols can be used in any combination with literal characters. The example in the slide displays the names of all employees whose last names have the letter “o” as the second character.

ESCAPE Identifier

When you need to have an exact match for the actual % and _ characters, use the ESCAPE identifier. This option specifies what the escape character is. If you want to

search for strings that contain SA_, you can use the following SQL statement: SELECT employee_id, last_name, job_id

FROM employees WHERE job_id LIKE #39;%SA\_%#39; ESCAPE #39;\#39;;

Question No: 145 – (Topic 1)

View the Exhibit and examine the structure of the PROMOTIONS table. Using the PROMOTIONS table, you need to display the names of all promos done after January 1, 2001, starting with the latest promo. Which query would give the required result? (Choose all that apply.)

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  1. SELECT promo_name, promo_begin_date FROM promotions WHERE promo_begiii_date gt; #39;01-JAN-01#39; ORDER BY 2 DESC;

  2. SELECT promo_name. promo_begiii_date FROM promotions

    WHERE promo_begin_date gt; #39;01-JAN-01#39; ORDER BY promo_name DESC:

  3. SELECT promo_name. promo_begin_date FROM promotions WHERE promo_begin_date gt; #39;01-JAN-01#39; ORDER BY 1DESC:

  4. SELECT promo_name, promo_begin_date quot;START DATEquot; FROM promotions WHERE promo_begin_date gt; #39;01-JAN-01#39; ORDER BY quot;START DATEquot; DESC;

Answer: A,D

Question No: 146 – (Topic 1)

See the exhibit and examine the structure of the CUSTOMERS and GRADES tables:

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You need to display names and grades of customers who have the highest credit limit. Which two SQL statements would accomplish the task? (Choose two.)

A.

SELECT custname, grade FROM customers, grades

WHERE (SELECT MAX(cust_credit_limit)

FROM customers) BETWEEN startval and endval;

B.

SELECT custname, grade FROM customers, grades

WHERE (SELECT MAX(cust_credit_limit)

FROM customers) BETWEEN startval and endval AND cust_credit_limit BETWEEN startval AND endval; C.

SELECT custname, grade FROM customers, grades

WHERE cust_credit_limit = (SELECT MAX(cust_credit_limit) FROM customers)

AND cust_credit_limit BETWEEN startval AND endval;

D.

SELECT custname, grade FROM customers , grades

WHERE cust_credit_limit IN (SELECT MAX(cust_credit_limit) FROM customers)

AND MAX(cust_credit_limit) BETWEEN startval AND endval;

Answer: B,C

Question No: 147 – (Topic 1)

Evaluate the following SQL statement:

SQLgt; SELECT cust_id, cust_last_name quot;Last Namequot; FROM customers

WHERE country_id = 10 UNION

SELECT cust_id CUST_NO, cust_last_name FROM customers

WHERE country_id = 30;

Which ORDER BY clause are valid for the above query? (Choose all that apply.)

  1. ORDER BY 2,1

  2. ORDER BY CUST_NO

  3. ORDER BY 2,cust_id

  4. ORDER BY quot;CUST_NOquot;

  5. ORDER BY quot;Last Namequot;

Answer: A,C,E Explanation:

Using the ORDER BY Clause in Set Operations

  • The ORDER BY clause can appear only once at the end of the compound query.

  • Component queries cannot have individual ORDER BY clauses.

  • The ORDER BY clause recognizes only the columns of the first SELECT query.

  • By default, the first column of the first SELECT query is used to sort the output in an ascending order.

Question No: 148 – (Topic 1)

Examine the statement:

Create synonym emp for hr.employees;

What happens when you issue the statement?

  1. An error is generated.

  2. You will have two identical tables in the HR schema with different names.

  3. You create a table called employees in the HR schema based on you EMP table.

  4. You create an alternative name for the employees table in the HR schema in your own schema.

Answer: D

Question No: 149 – (Topic 1)

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You created an ORDERS table with the following description: Exhibit:

You inserted some rows in the table. After some time, you want to alter the table by creating the PRIMARY KEY constraint on the ORD_ID column.

Which statement is true in this scenario?

  1. You cannot add a primary key constraint if data exists in the column

  2. You can add the primary key constraint even if data exists, provided that there are no duplicate values

  3. The primary key constraint can be created only a the time of table creation

  4. You cannot have two constraints on one column

Answer: B

Question No: 150 – (Topic 1)

Which is an iSQL*Plus command?

  1. INSERT

  2. UPDATE

  3. SELECT

  4. DESCRIBE

  5. DELETE

  6. RENAME

Answer: D

Explanation: Explanation:

The only SQL*Plus command in this list: DESCRIBE. It cannot be used as SQL command. This command returns a description of tablename, including all columns in that table, the datatype for each column and an indication of whether the column permits storage of NULL values.

Incorrect answer:

A INSERT is not a SQL*PLUS command B UPDATE is not a SQL*PLUS command C SELECT is not a SQL*PLUS command E DELETE is not a SQL*PLUS command F RENAME is not a SQL*PLUS command

Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 7

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