NCERT Solutions Class 12 Computer Science Chapter 9 Structured Query Language

Chapter - Structured Query Language

Question 1: Differentiate between delete and drop table command ?
Аnswer: DELETE command is used to remove information from a particular row or rows. If used without any condition, it will delete all row information but not the structure of the table. It is a DML command. DROP table command is used to remove the entire structure of the table and information. It is a DDL command.

Question 2: What is the use of wildcard ?
Аnswer: The wildcard operators are used with the LIKE operator to search a value similar to a specific pattern in a column. There are 2 wildcard operators. % – represents 0,1 or many characters – = represents a single number or character

Question 3: Write SQL query to add a column total price with datatype numeric and size 10, 2 in a table product.
Аnswer: ALTER TABLE product ADD total price number

Question 4: While creating table ‘customer’, Rahul a forgot to add column ‘price’. Which command is used to add new column in the table. Write the command to implement the same.
Аnswer: ALTER TABLE customer ADD price number(10,2)

Question 5: Deepika wants to remove all rows from the table BANK. But he needs to maintain the structure of the table. Which command is used to implement the same ?
Аnswer: DELETE FROM BANK

Question 6: Sonal needs to display name of teachers, who have “0” as the third character in their name.She wrote the following query. Select name from teacher where name = “$$0?”; But the query isn’t producing the result. Identify the problem.
Аnswer: The wildcards are incorrect. The corrected query is SELECT name FROM teacher
WHERE name LIKE’ _ _ 0%’ 

Question 7: Consider the following tables School and Admin and answer this question :
Give the output the following SQL queries :
1. Select Designation Count (*) From Admin Group By Designation Having Count (*) <2;
2. SELECT max (EXPERIENCE) FROM SCHOOL;
3. SELECT TEACHERNAME FROM SCHOOL WHERE EXPERIENCE >12 ORDER BY TEACHER NAME;
4. SELECT COUNT (*), GENDER FROM ADMIN GROUP BY GENDER

Аnswer:
(i) VICE PRINCIPAL      01
(ii) 16
(iii) UMESH
YASH RAJ

Short Answer Type Questions-II 

Question 1: Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables.

- NO is Traveller Number
- KM is Kilometer Travelled
- NOP is number of travellers travelled in vehicle.
- TDATE is Travel Date
1. To display NO, NAME, TDATE from the table TRAVEL in descending order of NO.
2. To display the NAME of all the travelers from the table TRAVEL who are travelling by vehicle with code 101 or 102.
3. To display the NO and NAME of those travelers from the table TRAVEL who travelled between ‘2015-1231’ and ‘2015-04-01’.
4. To display all the details from table TRAVEL for the travelers, who have travelled distance more than 100 KM in ascending order of NOP .
5. SELECT COUNT (*), CODE FROM TRAVEL GROUP BY CODE HAVING COUNT(*)>1;
6. SELECT DISTINCT CODE FROM TRAVEL;
7. SELECT A. CODE,NAME, VTYPE

FROM TRAVEL A,VEHICLE B
WHERE A.CODE=B.CODE AND ‘KM<90;

8.SELECT NAME, KM*PERKM
FROM TRAVEL A,VEHICLE B
WHERE A.CODE=B. CODE AND A.CODE=’105’;
Аnswer:
1. Select NO, Name, TDATE from TRAVEL order by NO desc
2. Select NAME from TRAVEL, where CODE in (101, 102)
3. Select NO, NAME from TRAVEL where TDATE between ’2015-12-31′ and ‘2015-04-01’.
4. Select * from TRAVEL where KM > 100 order by NOP.
5.
COUNT (*)    CODE
2                   101
2                  102

6.
DISTANCE (CODE)
101
103
102
104
105

7. CODE    NAME       VTYPE
    104        Ahmed       khan CAR
    105      Raveena     SUV

8. NAME KM*PERKM
Tarun Ram 14000

Question 2: Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables.
Table :VEHICLE

VCODE	VEHICLETYPE	PERKM
VOl	VOLVO BUS	150
V02 V03	AC DELUXE BUS ORDINARY BUS	125 80
V0’5	SUV	30
V04	CAR	18

Note:
PERKM is Freight Charges per kilometer.
Table : TRAVEL

CNo 101	CNAME K.Niwal	TRAVELDATE 2015-12-13	KM 200	VCODE VOl	NOP 32
103	Fredrick Sym	2016-03-21	120	VO3	45
105 102	Hitesh Jain Ravi Anish	2016-04-23 2016-01-13	450 80	VO2 VO2	42 40
107	John Malina	2015-02-10	65	VO4	2
104	Sahanubhuti	2016-01-28	90	VO5	4
106	Ramesh Jaya	2016-04-06	100	VOl	25

- Km is Kilometers Travelled
- NOP is number of passengers travelled in vehicle.
1.To display CNO, CNAME, TRAVELDATE from the table TRAVEL in descending order of CNO.
2.To display the CNAME of all the customers from the table TRAVEL who are travelling by vehicle with code V01 or V02.
3.To display the CNO and CNAME of those customers from the table TRAVEL who travelled between ‘2015-12-31’ and ‘2015-05-01’.
4.To display all the details from table TRAVEL for the customers, who have travel distance more than 120 KM in ascending order of NOP.
5.SELECT COUNT (*) , VCODE FROM TRAVEL
GROUP BY VCODE HAVING COUNT(*)>1;
6. SELECT DISTINCT VCODE FROM TRAVEL;
7. SELECT A. VCODE, CNAME, VEHICLETYPE
FROM TRAVEL A,VEHICLE B
WHERE A.VCODE=B.VCODE AND KM<90;
8. SELECT CNAME, KM*PERKM FROM TRAVEL A,VEHICLE B
WHERE A.VCODE=B . VCODE AND A.VCODE= ‘V05 ‘ ;
Аnswer:
(i) Select CNO, CNAME, TRAVELDATE from TRAVEL order by CNO desc
(ii) Select CNAME from TRAVEL, where VCODE in (‘VOl’, ‘ V02 ‘)
(iii)Select CNO, CNAME from TRAVEL where TRAVELDATE between ‘2015-12-31’ and ‘2015-05-01 ‘
(iv) Select * from TRAVEL where KM > 120 order by NOP.
(v)

COUNT (*)	VCODE
2 2	V01 V02

DISTANCE (CODE)
V01
V03
V02
V04
V05
(vii)

VCODE	CNAME	VEHICLETYPE
V04	JOHN MALINI	CAR

(viii)
CNAME KM*PERKM
Sahanubhuti 30
Note: PERKM is neither given in query nor in TABLE so no output is also acceptable.

Long Answer Type Questions

Question 1: Consider the following tables FACULTY and COURSES. Write SQL commands for the statements (i) to (v) and give outputs for SQL queries (vi) to (vii)
FACULTY

F_ID     Fname       Lname        Hire_date        Salary
102      Amit           Mishra         12-10-1998     12000
103      Nitin           Vyas            24-12-1994     8000
104      Rakshit      Soni             18-5-2001       14000
105      Rashmi      Malhotra      11-9-2004       11000
106      Sulekha     Srivastava    5-6-2006         10000

COURSES
C_ID   FJD   Cname
C21    102    Grid Computing         40000
C22    106    System Design          16000
C23    104    Computer Security     8000
C24    106    Human Biology          15000
C25    102    Computer Network     20000
C26    105    Visual Basic 6000

(i) To display details of those Faculties whose salary is greater than 12000.
Аnswer:
Select * from faculty
where salry > 12000;

(ii) To display the details of courses whose fees is in th range of 15000 to 50000(both values included).
Аnswer:
Select * from Courses
where fees between 15000 and 50000;

(iii)To increase the fees of all courses by 500 of “System Design” Course.
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Аnswer:
Update courses set fees = fees + 500
where Cname = “System Design”;

(iv)To display details of those courses which are taught by ‘Sulekha’ in descending order of courses.
Аnswer:
Select * from faculty fac, courses cour
where fac.f_id = cour.f_id and fac.fname = ‘Sulekha’ order by cname desc;

(v)Select COUNT (DISTINCT F_ID) from COURSES;
Аnswer: 4

(vi)Select MIN (Salary) from FACULTY, COURSES where COURSES.F_ID = FACULTY.FJD;
Аnswer: 6000

Question 2: Consider the following DEPT and WORKER tables. Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii):
TABLE : DEPT

DCODE	DEPARTMENT	CITY
D01	MEDIA	DELHI
D02	MARKETING	DELHI
D03	INFRASTRUCTURE	MUMBAI
D05	FINANCE	KOLKATA
D04	HUMAN RESOURCE	MUMBAI

TABLE : WORKER

WNO	NAME . Y;	DOJ	DOB	GENDER	DCODE
1001	George K	2013-09-02	1991-09-01	MALE	D01
1002	Ryma Sen	2012-12-11	1990-12-15	FEMALE	D03
1003	Mohitesh	2013-02-03	1987-09-04	MALE	D05
1007	Anil Jha	2014-01-17	1984-10-19	MALE	D04
1004	Manila Sahai	2012-12-09	1986-11-14	FEMALE	D01
1005	RSAHAY	2013-11-18	1987-03-31	MALE	D02
1006	Jaya Priya	2014-06-09	1985-06-23	FEMALE	D05

Note : DOJ refers to date of joining and DOB refers to date of birth of workers.

(i)To display Wno. Name, Gender from the table WORKER in descending order of Wno. Ans.
Аnswer:
SELECT WNO, Name, Gender FROM Worker
ORDER BY Wno DESC;

(ii)To display the Name of all the FEMALE workers from the table WORKER.
Аnswer:
SELECT Name FROM Worker
WHERE gender = ‘FEMALE’;

(iii)To display the Wno and Name of those workers from the table WORKER who are born between
‘1987-01-01’ and ‘1991-12-01’.
Аnswer:
SELECT Wno, Name FROM Worker
WHERE DOB BETWEEN ‘1987-01-01’ AND ‘1991-12-01’;

OR
SELECT Wno, Name FROM worker
WHERE DOB > = 1987-01-01′ AND DOB < = ‘1991-12-01’;
WHERE DOB BETWEEN ‘1987-01-01’ AND ‘1991-12-01’;
OR
WHERE DOB > = ‘1987-01-01’ AND DOB < = ‘1991-12-01’;

(iv)To count and display MALE workers who have joined after ‘1986-01-01’.
Аnswer:
SELECT COUNT (*) FROM Worker
WHERE GENDER = ‘MALE’ AND DOJ > ‘198601-01’;
OR
SELECT * FROM Worker
WHERE GENDER = ‘MALE’ AND DOJ > ‘198601-01’;
(Any valid query for counting and/or displaying for male workers will be awarded 1 mark)

(v) SELECT COUNT (*), DCODE FROM WORKER GROUP BY DCODE HAVING COUNT (*) > 1;
Аnswer:
COUNT (*) DCODE
2 D01
2 D05

(vi)SELECT DISTINCT DEPARTMENT FROM DEPT;
Аnswer:
Department
MEDIA
MARKETING
INFRASTRUCTURE
FINANCE
HUMAN RESOURCE

(viii)SELECT NAME, DEPARTMENT, CITY FROM WORKER W, DEPT D WHERE W DCODE = D.
DCODE AND WNO < 1003;
Аnswer:
NAME                      DEPARTMENT          CITY
George K                 MEDIA                     DELHI
Ryma Sen                infrastructure         MUMBAI

(viii) SELECT MAX (DOJ), MIN (DOB) FROM WORKER;
Аnswer: MAX (DOJ) MIN (DOB)
2014-06-09 1984-10-19
Note : In the output queries, please ignore the order of rows

Question 3: Consider the following DEPT and EMPLOYEE tables. Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii).
TABLE : DEPT

DCODE	DEPARTMENT	LOCATION
D01	INFRASTRUCTURE	DELHI
D02	MARKETING	DELHI
D03	MEDIA	MUMBAI
D05	FINANCE	KOLKATA
D04	HUMAN RESOURCE	MUMBAI

TABLE : EMPLOYE

ENO	NAME	DOJ	DOB	GENDER	DCODE
1001	GEORGE K	2013-09-02	1991-09-01	MALE	D01
1002	Ryma Sen	2012-12-11	1990-12-15	FEMALE	D03
1003	Mohitesh	2013-02-03	1987-09-04	MALE	D05
1007	Anil Jha	2014-01-17	198410-19	MALE	D04
1004	Manila Sahai	2012-12-09	1986-11-14	FEMALE	D01
1005	RSAHAY	2013-11-18	1987-03-31	MALE	D02
1006	JAYA Priya	2014-06-09	1985-06-23	FEMALE	D05

Note :DOJ refers to date of joining and DOB refers to date of Birth of employees.

(i)To display Eno, Name, Gender from the table EMPLOYEE in ascending order of Eno.
Аnswer: SELECT Eno, Name, Gender FROM Employee ORDER BY Eno;

(ii)To display the Name of all the MALE employees from the table EMPLOYEE.
Аnswer: SELECT Name FROM EMPLOYEE WHERE
Gender = ‘MALE’;

(iii)To display the Eno and Name of those employees from the table EMPLOYEE who are born between ‘1987-01-01’ and ‘1991-12-01’.
Аnswer:

SELECT Eno, Name FROM Employee
WHERE DOB BETWEEN ‘1987-01-01’ AND ‘1991-12-01’;
OR
SELECT Eno, Name FROM Employee
WHERE DOB > = ‘1987-01-01′ AND DOB < =’1991-12-01’;
OR
SELECT Eno, Name FROM Employee WHERE DOB > ‘1987-01-01’ AND DOB < ‘199112-01’;
WHERE DOB BETWEEN ‘1987-01-01’ AND ‘1991-12-01’;
OR
WHERE DOB > = ‘1987-01-01’ AND DOB < = ‘1991-12-01’;
OR
WHERE DOB > ‘1987-01-01’ AND DOB < ‘199112-01’);

(iv)To count and display FEMALE employees who have joined after ‘1986-01-01’;
Аnswer: SELECT count (*) FROM Employee
WHERE GENDER = ‘FEMALE’ AND DOJ > ‘1986-01-01’;
OR
SELECT * FROM Employee
WHERE GENDER = ‘FEMALE’ AND DOJ > ‘1986-01-01’;
(Any valid query for counting and/or displaying for female employees will be awarded 1
mark)
(v)SELECT COUNT (*), DCODE FROM EMPLOYEE
GROUP BY DCODE HAVING COUNT (*) > 1;
Аnswer:
COUNT    DCODE
2            D01
2            D05
(½Mark for correct output)

(vi)SELECT DISTINCT DEPARTMENT FROM DEPT
Аnswer: Department
INFRASTRUCTURE
MARKETING
MEDIA
FINANCE
HUMAN RESOURCE

(vii) SELECT NAME, DEPARTMENT FROM EMPLOYEE E, DEPT D WHERE E. DCODE = D.DCODE
AND ENO <1003;
NAME                  DEPARTMENT
George K             MEDIA
Ryma Sen            infrastructure

(viii) SELECT MAX (DOJ), MIN (DOB) FROM EMPLOYEE;
Аnswer:
MAX (DOJ) MIN (DOB)
2014-06-09 1984-10-19
Note : In the output queries, please ignore the order of rows.

Question 4: Write SQL commands for the queries (i) to (iv) and output for (v) & (viii) based on a table
COMPANY and CUSTOMER

CID    NAME                CITY             PRODUCTNAME
111    SONY                 DELHI           TV
222    NOKIA                MUMBAI        MOBILE
333    ONIDA                DELHI            TV
444    SONY                 MUMBAI        MOBILE
555    BLACKBERRY   MADRAS       MOBILE
666    DELL                  DELHI            LAPTOP

CUSTID    NAME                       PRICE      QTY   CID
101        ROHAN SHARMA        70,000      20      222
102        DEEPAK KUMAR         50,000      10      666
103        MOHAN KUMAR          30,000      5        111
104        SAHIL BANSAL            35,000      3        333
105        NEHA SONI                  25,000      7        444
106        SONAL AGGARWAL    20,000      5        333
107        ARUN SINGH               50,000     15       666

1. To display those company name which are having prize less than 30000.
2. To display the name of the companies in reverse alphabetical order.
3. To increase the prize by 1000 for those customer whose name starts with S?
4. To add one more column total price with decimal(10,2) to the table customer
5. SELECT COUNTO ,CITY FROM COMPANY GROUP BY CITY;
6. SELECT MIN(PRICE), MAX(PRICE) FROM CUSTOMER WHERE QTY>10;
7. SELECT AVG(QTY) FROM CUSTOMER WHERE NAME LIKE “%r%;
8. SELECT PRODUCTNAME,CITY, PRICE FROM COMPANY, CUSTOMER WHERE
COMPANY.CID=CUSTOMER.CID AND PRODU CTN AME=”MOBILE”;
Аnswer:
1. To display those company name which are having prize less than 30000.
SELECT NAME FROM COMPANY WHERE COMPANY.CID=CUSTOMER. CID AND PRICE < 30000
2.To display the name of the companies in reverse alphabetical order.
SELECT NAME FROM COMPANY
ORDER BY NAME DESC?;
3.To increase the prize by 1000 for those customer whose name starts with “S”
UPDATE CUSTOMER
SET PRICE = PRICE + 1000;
WHERE NAME LIKE ‘S%’;
4.To add one more column total price with decimal(10,2) to the table customer
ALTER TABLE CUSTOMER
ADD TOTALPRICE DECIMAL(10,2);
5.SELECT COUNT(*) ,CITY FROM COMPANY GROUP BY CITY;
3    DELHI
2    MUMBAI
1    MADRAS
6.SELECT MIN(PRICE), MAX(PRICE) FROM
CUSTOMER WHERE QTY> 10;
50000,70000
7.SELECT AVG(QTY) FROM CUSTOMER
WHERE NAME LIKE “%r%;
8.SELECT PRODUCTNAME, CITY, PRICE FROM COMPANY, CUSTOMER WHERE

COMPANY.CID=CUSTOMER.CID AND PRODUCTNAME=”MOBILE”;

MOBILE    MUMBAI   70000
MOBILE    MUMBAI   25000

Question 5: Consider the following tables SCHOOL and ADMIN and answer this question :
Table : SCHOOL

1. To display TEACHERNAME, PERIODS of all teachers whose periods are more than 25.
2. To display all the information from the table SCHOOL in descending order of experience.
3. To display DESIGNATION without duplicate entries from the table ADMIN.
4. To display TEACHERNAME, CODE and corresponding DESIGNATION from tables
SCHOOL and ADMIN of Male teachers.
Аnswer:
1. To display TEACHERNAME, PERIODS of all teachers whose periods are more than 25.
SELECT TEACHERNAME, PERIODS
FROM SCHOOL WHERE PERIODS >25.
2.To display all the information from the table SCHOOL in descending order of experience.
- SELECT * FROM SCHOOL;
3.To display DESIGNATION without duplicate entries from the table ADMIN.
SELECT DISTINCT DESIGNATION FROM ADMIN;
4.To display TEACHERNAME, CODE and corresponding DESIGNATION from tables SCHOOL
and ADMIN of Male teachers.
SELECT TEACHERNAME.CODE
DESIGNATION FROM SCHOOL.CODE = ADMIN.CODE
WHERE GENDER = MALE;

Question 6: Answer the questions (a) and (b) on the basis of the following tables SHOPPE and ACCESSORIES.
Аnswer:
Id        SName                         Area
S001   ABC computronics        CP
S002   All Infotech Media         GKII
S003   Tech Shoppe                CP
S004   Geeks Tecno Soft        Nehru Place
S005   Hitech Tech Store        Nehru Place

No      Name                      Price       Id
A01    Mother Board         12000     SOI
A02    Hard Disk               5000      SOI
A03    Keyboard                500        S02
A04    Mouse                    300         SOI
A05    Mother Board         13000     S02
A06   Keyboard                400         S03
A07   LCD                        6000       S04
T08   LCD                        5500       S05
T09   Mouse                     350         S05
T10   Hard Disk               4500        S03

1. To display Name and Price of all the Accessories in ascending order of their Price.
2. To display Id and SName of all Shoppe located in Nehru Place.
3. To display Minimum and Maximum Price of each Name of Accessories.
4. To display Name, Price of all Accessories and their respective SName where they are available.

(b)Write the output of the following SQL commands:
1. SELECT DISTINCT NAME FROM ACCESSORIES WHERE PRICE > =5000;
2. SELECT AREA, COUNT(*) FROM SHOPPE GROUP BY AREA;
3. SELECT COUNT (DISTINCT AREA) FROM SHOPPE;
4. SELECT NAME, PRICE*0.05 DISCOUNT FROM ACCESSORIES
Аnswer: (a) (i) SELECT Name, Price FROM ACCESSORIES ORDER BY Prices;
(ii)SELECT ID, SName FROM SHOPPE WHERE Area=”Nehru Place”;
(iii)SELECT Name, max (Price); min(Price) FROM ACCESSORIES, Group By Name;
(iv)SELECT Name, price, Sname FROM ACCESSORIES, SHOPPE WHERE
SHOPPE.ID=ACCESSORIES.ID
(b)(i) Name
Mother Board
Hard Disk
LCD
(ii)
Area                Count
CP                      2
GK II                  1
Nehru Place       2
(iii) count(Distinct Area)
3
(iv) Name
Name             DISCOUNT
600                      600
Hard Disk            250
Key Board           20
Hard Disk            225

Question 7: Answer the questions (a) to (g) on the basics of the following tables APPLICANTS and COURSB.
1. To display name, fee, gender, join year about the applicants, who have joined before 2010.
2. To display names of applicants, who are paying fee more than 30000.
3. To display names of all applicants in ascending order of their join year.
4. To display the year and the total number of applicants joined in each YEAR from the table APPLICANTS.

C_ID   Course
A01     Fashion Design
A02     Networking
A03     Hotel Management
A04     Event Management
A05    Office Management

5.To display the C_ID (i.e., CourselD) and the number of applicants registered in the course from the APPLICANTS and table.
6.To display the applicant’s name with their respective course’s name from the tables APPLICANTS and COURSES.
7.Give the output statements: of following SQL statements :
(i)SELECT Name, Join year FROM APPLICANTS
WHERE GENDER=’F’ and C_ID=’A02′;
(ii) SELECT MIN (Join year) FROM APPLICANTS
(iii)SELECT AVG (Fee) FROM APPLICANTS WHERE C_ID=’A01′ OR C_ID=’A05′;
(iv)SELECT SUM (Fee), C_ID FROM APPLICANTS
GROUP BY C_ID HAVING COUNT(*)=2;
Аnswer: (a) SELECT NAME,FEE,GENDERJOINYEAR FROM APPLICANTS
WHERE JOINYEAR<2000;
(b)SELECT NAME FROM APPLICANTS
WHERE FEE>30000;
(c)SELECT NAME FROM APPLICANTS
ORDERBY JOINYEAR ASC;
(d)SELECT YEAR, COUNTf) FROM APPLICANTS;
(e)SELECT C_ID, COUNT(*) FROM APPLICANTS, COURSES
WHERE APPLICANTS.C_ID=COURSES; C_ID;
(f)SELECT NAME,COURSE FROM
APPLICANTS, COURSES
WHERE APPLICANTS.C_ID=COURSES. C_ID;
(g)(i) Avisha 2009
(ii) 2009
(iii) 67
(iv)55000 A01

Question 8: Write SQL queries for (a) to (g) and write the output for the SQL queries mentioned shown in (hi) to (h4) parts on the basis of table ITEMS and TRADERS :
Table : ITEMS

- To display the details of all the items in ascending order of item names (i.e., INAME).
- To display item name and price of all those items, whose price is in the range of 10000 and 22000 (both values inclusive).
- To display the number of items, which are traded by each trader. The expected output of this query should be:
- To display the price, item name and quantity (i.e., qty) of those items which have quantity more than 150.
- To display the names of those traders, who are either from DELHI or from MUMBAI.
- To display the names of the companies and the names of the items in descending order of company names.
- Obtain the outputs, of the following SQL queries based on the data given in tables

ITEMS and TRADERS above.
(h1)SELECT MAX (PRICE), MIN (PRICE) FROM ITEMS;
(h2) SELECT PRICE*QTY
FROM ITEMS WHERE CODE-1004;
(h3) SELECT DISTINCT TCODE FROM ITEMS;
(h4) SELECT INAME, TNAME
FROM ITEMS I, TRADERS T WHERE I.TCODE=T.TCODE AND
QTY<100;
Аnswer:
(a) SELECT INAME FROM ITEMS ORDER BY INAME ASC;
(b)SELECT INAME, PRICE FROM ITEMS WHERE PRICE => 10000 AND PRICE =< 22000;
(c)SELECT TCODE, COUNT (CODE) FROM
ITEMS GROUP BY TCODE;
(d)SELECT PRICE, INAME, QTY FROM ITEMS
WHERE (QTY> 150);
(e)SELECT TNAME FROM TRADERS
WHERE (CITY = “DELHI”) OR (CITY = “MUMBAI”)
ORDER BY COMPANY DESC;
(g) (h1) 38000
1200
(h2) 1075000
(h3) T01
T02
T03
(h4) LED SCREEN 40 DISPHOUSE INC
CAR GPS SYSTEM ELECTRONICS
SALES

Question 9: Write SQL queries for (a) to (f) and write the outputs for the SQL queries mentioned shown in (hi) to (h4) parts on the basis of tables PRODUCTS and SUPPLIERS
Table : PRODUCTS

(a)To display the details of all the products in ascending order of product names (e., PNAME).
(b)To display product name and price of all those products, whose price is in the range of 10000 and 15000 (both values inclusive).
(c)To display the number of products, which are supplied by each supplier, i.e., the expected output should be;
2
2
1

(d)To display the price, product name and quantity (i.e., qty) of those products which have quantity more than 100.
(e)To display the names of those suppliers, who are either from DELHI or from CHENNAI.
(f)To display the name of the companies and the name of the products in descending order of company names.
(g)Obtain the outputs of the following SQL queries based on the data given in tables
PRODUCTS and SUPPLIERS above.
(gl) SELECT DISTINCT SUPCODE FROM PRODUCTS;
(g2) SELEC MAX (PRICE), MIN (PRICE) FROM PRODUCTS
(g3) SELECT PRICE*QTY
FROM PRODUCTS WHERE PID = 104;
(g4) SELECT PNAME, SNAME
FROM PRODUCTS P, SUPPLIERS S WHERE P SUPCODE = S. SUPCODE AND QTY>100;
Аnswer:
(a) SELECT * FROM PRODUCTS ORDER. BY PNAME ASC;
(b)SELECT PNAME, PRICE FROM PRODUCTS WHERE ((PRICE => 10000) AND (PRICE = < 15000));
(c)SELECT SUPCODE, COUNT (PID) [Yz] FROM PRODUCTS GROUP BY SUPCODE;
(d)SELECT PRICE, PNAME, QTY FROM PRODUCTS WHERE (QTY > 100);
(e)SELECT SNAME FROM SUPPLIERS WHERE ((CITY = “DELHI”) OR (CITY = “CHENNAI”));
(f)SELECT COMPANY, PNAME FROM PRODUCTS ORDER BY COMPANY DESC;
(g) SOI
(g1) S02
S03
(g2) 28000
1100
(g3) 550000
(g4) PNAME SNAME
DIGITAL CAMERA 14 X GET ALL INC
PENDRIVE 16 GB GET ALL INC

Question 10: Give a suitable example of a table with sample data and illustrate Primary and Alternate Keys in it. Consider the following tables CARDEN and CUSTOMER and answer (b) and (c) parts of this question :
Table : CARDEN

Ceode    CarName    Make        Colour    Capacity   Charges
501       A-Star         Suzuki      RED          3Q          14
503       Indigo        Tata           SILVER     3            12
502       Innova       Toyota       WHITE       7           15
509      SX4            Suzuki       SILVER      4           14
510      C Class       Mercedes   RED           4           35

Table : CUSTOMER
CCode        Cname             Ceode
1001          Hemant Sahu    501
1002          Raj Lai              509
1003          Feroza Shah      503
1004          Ketan Dhal        502

(b)Write SQL commands for the following statements:
1. To display the names of all the silver coloured cars.
2. Tp display names of car, make and capacity of cars in descending order of their sitting
capacity.
3. To display the highest charges at which a vehicle can be hired from CARDEN.
4. To display the customer name and the corresponding name of the cars hired by them.
(c)Give the output of the following SQL queries:
(i)SELECT COUNT(DlST!NCT Make) FROM CARDEN;
(ii)SELECT MAX(Charges), MIN (Charges) FROM CARDEN;
SELECT COUNTO, Make FROM CARDEN;
(iv) SELECT CarName FROM CARDEN WHERE Capacity=4;
Аnswer:
(a) Primary Key of CARDEN = Ceode of CARDEN
Alternate Key = CarName:
Primary key of Customer = Ceode
Alternate Key of Customer = Cname
(b) (i) SELECT CarName From CARDEN
WHERE Color = “SILVER”;
(ii)SELECT CarName, Make, Capacity
From CARDEN ORDER BY Capacity DESC;
(iii)SELECT MAX(Charges) From CARDEN;
(iv)SELECT Cname, CarName From
CUSTOMER, CARDEN WHERE CARDEN. Ccode = CUSTOMER. Ccode;
(c) (i) 4
(ii) MAX(Charges)    MIN(Charges)
35                              12
(iii)5
(iv)SX4
C Class

Question 11: (a) Give a suitable example of a table with sample data and illustrate Primary agd Candidate Keys in it. Consider the following tables CABHUB and CUSTOMER and answer (b) and (c) parts of this question :
Table : CABHUB

(b) Write SQL commands for the following statements:
1. To display the names of all the white coloured vehicles.
2. To display name of vehicle name and capacity of vehicles in ascending order of their sitting capacity.
3. To display the highest charges at which a vehicle can be hired from CABHUB.
4. To display the customer name and the corresponding name of the vehicle hired by them.
(c)Give the output of the following SQL queries :
1. SELECT COUNT (DISTINCT Make) FROMCABHUB;
2. SELECT MAX(Charges), MIN(Charges) FROM CABHUB;
3. SELECT COUNT (*) Make FROM CABHUB;
4. SELECT Vehicle FROM CABHUB WHERE Capacity=4;
Аnswer:
(a) Primary Key of CABHUB = Vcode
Alternate key of CABHUB = Vehicle Name.

Primary Key of Customer = Ccode
Alternate Key of CUSTOMER = Cname.
(b) (i) SELECT VehicleName FROM CABHUB WHERE Colour =”WHITE”;
(ii)SELECT VehicleName, Capacity From CABHUB ORDER BY Capacity ASC;
(iii)SELECT MAX(Charges) FROM CABHUB;
(iv)SELECT Cname,VehicleName FROM CABHUB, CUSTOMER WHERE CUSTOMER.
Vcode=CABHUB. Vcode;
(c)4
(ii)Max(Charges)   Min(Charges)
35                           12
(iii)5
(iv)SX4
C Class

Long Answer Type Question – II

Question 1:

1. To display all the details of those watches whose name ends with ‘Time’
2. To display watch’s name and price of those watches which have price range in between 5000-15000.
3. To display total quantity in store of Unisex type watches.
4. To display watch name and their quantity sold in first quarter.
5. select max(price), min(qty_store) from watches;
6. select quarter, sum(qty_sold) from sale group by quarter;
7. select watch_name,price, type from watches w, sale s where w.watchid!=s.watchid;
8. select watch_name, qty_store, sum(qty_sold), qty_store-sum(qty_sold) “Stock” from watches w, sales where w. watchid=s. watchid group by s.watchid;
Аnswer:
(i) Select*from watches where watch_name like’Time’
(ii)select watchjname, price from watches where price between 5000 and 15000;
(iii)select sum(qty_store) from watches where type like ‘Unisex’;
(iv)select watch name, qty_sold from watches w,sale s where w.watchid=s.watchid and quarter=l;
(v)

max(price)    min(qty_store)
25000           100
(vi)
quarter        suxn(qty_sold)
1                 15
2                 30
3                 45
4                 15
(vii)
Watch_name     price     type
HighFashion      7000     Unisex
(viii)
Watch_name    qty_store    qty_sold   Stock
HighTime          100             25              75
LifTime             150             40             110
Wave                200             30            170
GoldenTime     100             10              90