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IDS is the first developed Network Data Model.

1. complex to implement 

2. difficult to handle many to many relationships.

The number of entity types involved is known as Degree of relationship.

1. One-Unary, 

2. Two-Binary, 

3. Three -Ternary.

M : N relationship is said to b many to many relationship

(c) Relational Data Base Management System

Commands: 

1. CREATE 

2. DROP

Syntax:

1. CREATE database databasename 

2. DROP database databasename

DML COMMANDS List:

Commands: 

1. INSERT 

2. DELETE

Syntax:

1. Syntax 1: INSERT INTO tablename (column1, column2, column3) VALUES (value 1, value2, value3); 

2. Syntax 2: INSERT INTO tablename VALUES (value1, value2, value3); 

3. DELETE from tablename WHERE columnname= ”value”;

SQL meAnswer Structured Query Language.

SQL: 

1. SQL is a query language

2. SQL is used to query and operate database system.

MySQL: 

1. MySQL is DBMS software. 

2. MySQL allows data handling, storing, modifying, deleting, etc.

Rhombus is the symbol for Relationship in ER Model.

Correct option is B) 16

Area of square = a² = 8² = 64 cm²

\(\therefore\) Area of parallelogram = 64 cm²(Given both areas are equal)

⇒ Base x height = 64

⇒ Base = 64/height = 64/4 = 16 cm

Advantages: 

1. The ability to handle more relationship types,

2. Easy data access, 

3. Data integrity and independence.

Disadvantages: The limitation of network model is difficulty in design and maintenance.

The three database components of Network models are: Network schema, Sub schema and Language. Network schema – schema defines all about the structure of the database.

Sub schema – controls on views of the database for the user. 

Language – basic procedural for accessing the database.

While calculating the area of the parallelogram, we can choose any side as base.

The process of creating, implementing and maintaining the enterprise data in a system is known as Designing of databases.

mysql > use database name test; Database changed mysql>

SQL DCL COMMANDS List Commands: 

1. Grant 

2. Revoke

Description:

1. Used to give permission to specific users on specific database objects like table, view etc 

2. Used to take out permission from specific users on specific database objects like table, view etc.

Functions performed using SQL are listed below:

1. Executes queries against a database. 

2. Retrieves data from database. 

3. Inserts and updates records in a database 

4. Delete records from database. 

5. Creates new databases and new tables in a database.

The three major parts that forms a database are Tables, Queries and Views.

1. Tables – similar to an excel sheet, containing multiple rows and columns. Where each row is a record and each column is an attribute. 

2. Queries – It is a question with multiple conditions posted to the database. The records in the database that satisfies the passed conditions are retrieved. 

3. Views -A set of stored queries.

Database Administrators (DBA’s) takes care of configuration, installation, performance, security and data backup.

DBA’s posses the skills on database design, database queries, RDMS, SQL and networking. The primary task is the creation of new user and providing them with access rights.

The Syntax for inserting record is INSERT INTO table name (Parameterl,Parameter2, Parameter3..) VALUES (Value!, Value2, Value3..).

Data Definition Language (DDL), Data Manipulation Language (DML), Data Query Language (DQL), Action Control Language (TCL), Data Control Language (DCL).

Each instance of the relationship between members of these entity types is called a relationship instance. E.g if ‘B’ is the relationship between the Employee entity and the department entity, then Ram works for Comp. Sc department, Shyam works for Electrical department etc. are relationship instances of the relationship, works for.

A Strong entity is the one which doesn’t depend on any other entity on the schema or database and a strong entity will have a primary key with it (i;e. a unique id which other entities will not have in their attributes). It is represented by one rectangle.

The Query results are listed in Ascending or Descending order using the command ORDER ‘ BY. In some databases the results are sorted by default in Ascending order.

The sub query must follow the below rules: 

1. Subqueries are always written within the parentheses. 

2. Always place the Subquery on the right side of the comparison operator. 

3. ORDER BY clause is not used in sub query, since Subqueries cannot manipulate the results internally.

Deleting Record: The existing record in a table is removed from the table using DELETE , command. Entire record or specified columns in the table can be deleted. If we want to perform delete operation on specific columns, then that condition is given in the WHERE condition. If the condition is not specified, then the entire data will be deleted.

Syntax: DELETE from tablename WHERE columnname= “value”;

Modifying Record: SQL provides us with modifying and updating the existing records in a table using UPDATE command. The age of Krishna in Biodata table is changed using the below Syntax.

Syntax: UPDATE tablename SET column1= “new value” Where column2= “value2”; Example: mysql>UPDATE Biodata SET age=13 WHERE firstname= “Krishna”;

(b) 72 cm²

Because,

From the question is given that,

Base of the parallelogram = 16 cm

Height of the parallelogram = 4.5 cm

∴area of the parallelogram = base × height

= 16 × 4.5

= 72 cm²

Area of parallelogram ABCD = AB × AE = 8 × 4 cm² = 32 cm² 

Let altitude corresponding to AD be h. Then, 

h × AD = 32 

or h × 6 = 32 

or h = 32/6 = 16/3

Thus, altitude corresponding to AD is 16/3 cm.

Area of school playground is 160 m × 80 m = 12800 m² 

Area kept for swings = 80 m × 80 m = 6400 m²

Area of path parallel to the width of playground = 80 m × 1.5 m = 120 m² 

Area of path parallel to the remaining length of playground = 80 m × 1.5 m = 120 m². 

Area common to both paths = 1.5 m × 1.5 m = 2.25 m². [since it is taken twice for measuerment it is to be subtracted from the area of paths] 

Total area covered by both the paths 

= (120 + 120 – 2.25) m² 

= 237.75 m². 

Area covered by grass = Area of school playground – (Area kept for swings + Area covered by paths) 

= 12800 m² – [ 6400 + 237.75] m² 

= (12800 – 6637.75) m² 

= 6162.25 m².

Given length of the floor of the room = 9.68 m

Breadth of the floor of the room = 6.2 m

We know that area of rectangle = length x breadth

Area of the floor = 9.68 m x 6.2 m

= 60.016 m²

Given that length of the tile = 22 cm

Breadth of the tile = 10 cm

Area of one tile = 22 cm x 10 cm

= 220 cm² 

= 0.022 m² [Since 1 m² = 10000 c m²]

Thus, Number of tiles = 60.016 m²/0.022 m²

= 2728

We have cost of one tile = Rs. 2.50

Total cost = Number of tiles x Cost of one tile

= (2728 x 2.50)

= Rs. 6820

Correct option is  B) 19 cm

Given length of the corridor = 8 m

Also breadth of the corridor = 6 m

Area of the corridor of a school = Length x Breadth

= (8 m x 6 m)

= 48 m²

We have length of the canvas sheet = 2 m

Breadth of the canvas sheet = 1 m

Area of one canvas sheet = Length x Breadth

= (2 m x 1 m)

= 2 m²

Thus, Number of canvas sheets required to cover the corridor = 48/2 = 24

Given Cost of one canvas sheet = Rs. 8

Total cost of the canvas sheets = number of canvas sheet x cost of one canvas sheet

= (24 x 8) = Rs. 192

Given that length of lane = 180 m

Breadth of the lane = 5 m

Area of a lane = Length x Breadth

= 180 m x 5 m

= 900 m²

Length of the brick = 20 cm

Breadth of the brick = 15 cm

Area of a brick = Length x Breadth

= 20 cm x 15 cm

= 300 cm² 

= 0.03 m² [Since 1 m² = 10000 cm²]

Required number of bricks = 900/0.03 = 30000

Cost of 1000 bricks = Rs. 750

Total cost of 30,000 bricks = 750×30,000/1000

= Rs. 22,500

(a) 43.75 cm²

Area of the shaded part = Area of rectangle CDEF + Area of trapezium ABCD 

= 2.5 cm x 5 cm + \(\frac12\) x (2.5 cm + 10 cm) x 5 cm 

= 12.5 cm² + 31.25 cm² = 43.75 cm².

(a) Area of the parallelogram = Base × height

Base = 20 cm

Area = 246 sq cm

Height = ?

246 = 20 × h

∴ Height  = \(\frac{246}{20} = 12.3cms\)

Height = 12.3 cms

(b) Height = 15 cm

Area = 154.5 cm²

Base = ?

Area of the parallelogram = Base × height

154.5 = B × 15

∴ B = \(\frac{154.5}{15} = 10.3cms\)

∴ Base = 10.3 cms

(c) Height = 8.4 cm

Area = 48.72 cm²

Base = ?

Area of the parallelogram = Base × height

48.72 = B × 8.4

∴ B = \(\frac{48.72}{8.4} = 5.8cms\)

∴ Base = 5.8 cms

d) Base = 15.6 cm

Area = 16.38 sq cm

Height = ?

Area of the parallelogram = Base × height

16.38 = 15.6 × h

∴ height = \(\frac{16.38}{15.6} = 1.05cms\)

∴ Height = 1.05 cms

i) Perimeter = 4a = 4(15) = 60cm 

ii) Perimeter = 4a = 88

a = 88/4 = 22

Side = 22cm

Area = a² = (22)² = 484 cm²

Given that length of playground = 62 m 60 cm

= 62.6 m [Since 10 cm = 0.1 m]

Breadth of a playground = 25 m 40 cm

= 25.4 m

Area of a playground = Length x Breadth

= 62.6 m x 25.4 m

= 1590.04 m²

Given rate of turfing = Rs. 2.50/ m² 

Total cost of turfing = (1590.04 x 2.50)

= Rs. 3975.10

Perimeter of a rectangular field = 2(Length + Breadth)

= 2(62.6 + 25.4) = 176 m

Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field

= 3 x 176 m = 528 m

The man walks at the rate of 2 m/sec.

Speed of man walks = 2 x 60 m/min = 120 m/min

Thus, required time to cover a distance of 528 m = 528/120

= 4.4 min

= 4 minutes 24 seconds [Since 0.1 minutes = 6 seconds]

1. a × a (or) a²

2. 48 sq. cm

3. 11 m

Correct option is B) 180

We have l = 15 cm and b = 12 cm

\(\therefore\) Area of rectangle is A = lb = 15 x 12 = 180cm²

1.36 cm

2.22 cm

3. 30 sq.cm

Yes. 

The area of the pathway can be found by dividing it into rectangles and adding the areas of these rectangles.

Length of rectangle 1 = 30 + 1.5 + 1.5 = 33 m

Breadth of rectangle 1 = 1.5 m 

∴ Area of rectangle 1 = 33 x 1.5 

= 49.5 sq. m 

Area of rectangle 4 = Area of rectangle 1 

= 49.5 sq. m. 

Length of rectangle 2 = 65 m 

breadth of rectangle 2 = 1.5 m 

∴ Area of rectangle 2 = 65 x 1.5 

= 97.5 sq. m. 

Area of rectangle 3 = area of rectangle 2 

= 97.5 sq. m. 

∴ Area of pathway = Sum of area of the 4 rectangles 

= 49.5 + 49.5 + 97.5 + 97.5 

= 294 sq. m.

Equal are the opposite sides in a rectangle.

Side of the square piece of cloth = 1 m 

∴ Side of each napkin = 0.5 m 

Length of lace that will be required for 1 napkin = perimeter of the napkin 

= 4 x side = 4 x 0.5 = 2 m 

∴ Perimeter of 4 napkins = 4 x 2 = 8 m 

∴ 8 metre long lace will be required to trim all four napkins.

Area of a rectangle = length × breadth 

= 15 × 5

= 75 sq. cm. 

∴ The area of the rectangle is 75 sq. cm.

Let us assume the base of the parallelogram be x cm.

and height of the parallelogram will be (1/3)x cm.

From the question it is given that the area of the parallelogram is 108 cm².

∴Area of the parallelogram = base × height

= 108 = (x) × ((1/3)x)

= 108 = (1/3)x²

= x² = (108 × 3)

= x² = 324

= x = √324

= x = 18

Then,

The base of the parallelogram be x is 18 cm

Height of the parallelogram will be (1/3)x = (1/3) × 18 = 6 cm

1. 9 cm

2. πd

3. 120

Area = 1/2 d₁d₂

d₁ = 5 + 5 = 10cm

d₂ = 2 + 2 = 4cm

A = 1/2 × 10 × 4 = 20 cm²

d₁ = 3 + 3 = 6 cm

d₂ = 4 + 4 = 8cm

Area = 1/2 d₁d₂

= 1/2 × 6 × 8

= 24 cm²

ABCD is the rectangle formed by the edges of the mobile. PQRS is the rectangle formed by leaving a 1.5 cm wide edge alongside AB, BC, and DC, and a 2 cm edge alongside DA. 

Length of rectangle PQRS = 9.5 cm 

Breadth of rectangle PQRS = 4 cm

Area of screen = Area of rectangle PQRS = 9.5 x 4 = 38 sq .cm