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Given details are,

Dimensions of log of wood = 3m × 75cm × 50cm

Side of cubical block = 25cm

We know that,

Number of cubical block that can be made from wooden log =

Volume of wooden block / volume of cubical block

= (300 × 75 × 50) / (25 × 25 × 25)

= 72 blocks

We know that, 

Volume of cuboid = Length × Breadth × Height 

Therefore, 

Total volume of the block = (500 × 70 × 32) 

= 1120000 cm³ 

Total volume of each plank = 200 × 25 × 8 

= 40000 cm³ 

Hence,

Total number of planks that can be made = \(\frac{Total\,volume\,of\,the\,block}{Volume\,of\,each\,plank}\)

= \(\frac{1120000}{40000}\)

= 28 planks

Given details are,

Length of water tank = 3m

Breadth of water tank = 2m

Height of water tank = 1m

So, Dimensions of water tank is = 3m × 2m × 1m

Volume the water tank can hold = l × b × h

= 3 × 2 × 1

= 6m³

= 6000 litres

∴ The water tank can hold 6000 litres of water.

The two types of overloading is operator overloading and function overloading.

Inheritance is the process of creating new classes, called derived class, from existing or base classes. The derived class inherits all the capabilities of the base class. Using Inheritance some qualities of the base classes are added to the newly derived class, apart from its own features. Inheritance permits code reusability.

Correct option is C. Rs. 160

Given,

Length of wall = 8m

Breadth of wall = 4 m

Thickness of wall = 20 cm = \(\cfrac{20}{100}m\)

Volume of wall = \(l\times b\times h\) = 8 \(\times4\times\cfrac{20}{100}\) = \(\cfrac{32}5m^3\)

∵ cost of construction of 1m³ wall = Rs. 25

∴ cost of construction of \(\cfrac{32}5\)m³ wall = \(\cfrac{32}5\times25\) = Rs. 160

Given,

Weight of 1m³ marble = 0.25 kg

Weight of marble 28 cm wide and 5 cm thick = 112 kg

Let length of block = \(l\) cm

Then volume of block = \(l\) × 28 × 5 cm³ = 140 \(l\) cm³

Weight of block = 140 \(l\) × 0.25

= 112 = 140 \(l\) × 0.25

\(=​​​​l=\cfrac{112\times100}{140\times0.25}\) = 3.2 cm

Given details are,

Length of beam = 5 m

Width of beam = 40 cm = 0.4 m

Volume of wood in beam = 0.6 m³

Let thickness of beam be ‘h’ m

We know that,

Volume = l × b × h

h = volume/(l × b)

= 0.6/(5×0.4)

= 0.3m

∴ Thickness of the beam is 0.3m

Given details are,

Dimensions of one plank = 3m × 15cm × 5cm = 300cm × 15cm × 5cm

Dimensions of wooden block = 6m × 75cm × 45cm = 600cm × 75cm × 45cm

We know that,

Number of planks that can be prepared = volume of wooden block / volume of one plank

= (600 × 75 × 45) / (300 × 15 × 5)

= 90 planks

∴ 90 planks are required to prepare the block.

1. OOP provides a dear modular structure for programs. Large problems can be reduced to smaller and more manageable problems.
2. In OOP, data can be made private to a class such that only member functions of the class can access the data. This principle of data hiding helps the programmer to build a secure program.
3. Implementation details are hidden from other modules and other modules has a clearly defined interface.
4. It is easy to maintain and modify existing code as new objects can be created with small differences to existing ones.
5. With the help of polymorphism, the same function or same operator can be used for different purposes. This helps to manage software complexity easily.
6. In OOP, programmer not only defines data types but also deals with operations applied for data structures.
7. It is easy to model a real system as real objects are represented by programming objects in OOP. With the help of inheritance, we can reuse the existing class to derive a new class such that the repetition of code is eliminated and the use of existing class is extended. This saves time and cost of program.

The different types of inheritance are single inheritance, multilevel inheritance, multiple inheritances, hierarchical inheritance and hybrid inheritance.

The object is a collection of data members and member functions. Objects occupy space in memory and have address associated with them. Objects interact by sending messages to one another.

Output stream is a sequence of characters from the program to an output device like monitor, printer.

The few disadvantages of OOP are:

1. Size:

Object-Oriented programs are much larger than other programs. In the early days of computing, space on hard drives, floppy drives and in memory was at a premium. Today we do not have these restrictions.

2. Effort:
Object-Oriented programs require a lot of work to create. Specifically, a great deal of planning goes into an object-oriented program well before a single piece of code is ever written. Initially, this early effort was felt by many to be a waste of time. In addition, because the programs were larger (see above) coders spent more time actually writing the program.

3. Speed:
Object-Oriented programs are slower than other programs, partially because of their size. Other aspects of Object-Oriented Programs also demand more system resources, thus slowing the program down.

4. Not suitable for all types of problems:
There are problems that lend themselves well to functional-programming style, logic -programming style, or procedure-based programming style, and applying object-oriented programming in those situations will not result in efficient programs.

5. Not all programs can be modeled accurately by the objects model. If you just want to read in some data, do something simple to.it and write it back out, you have no need to define classes and objects. However, in some OOP languages, you may have to perform this extra step.

6. The objects often require extensive documentation.

Difference between Procedure Oriented Programming (POP) & Object Oriented Programming (OOP)

The four high level languages that follow object oriented programming approach are

• C++
• C#
• Java
• Python
• Visual Basic

1. OOP provides useful features like Inheritance, polymorphism, and encapsulation, not available in earlier programming methods.

2. Once an object is created, knowledge of its implementation is not necessary for its use. In older programs, coders needed understand the details of a piece of code before using it in this or another program.

The godown has its length (l₁) as 40 m, breadth (b₁) as 25 m, height (h₁) as 10 m, while the wooden crate has its length (l₂) as 1.5 m, breadth (b₂) as 1.25 m, and height (h₂) as 0.5 m.

Therefore, volume of godown 

= l₁ × b₁ × h₁ = (40 × 25 × 10) m³

= 10000 m³ = (1.5 × 1.25 × 0.5) m³

= 0.9375 m³

Let n wooden crates can be stored in the godown.

Therefore, volume of n wooden crates = Volume of godown 0.9375 × n = 10000

n = 10000/0.9375 = 10666.66

Therefore, 10666 wooden crates can be stored in the godown.

The function write() is used for string output.

The declaration of string is
char arrayname[ size];

Some of the real-life applications of object-oriented programming are given below;

• Computer graphic applications
• CAD/CAM software
• Object-oriented database
• User interface design such as windows
• Real time systems
• Simulation and modeling
• Artificial intelligence and expert systems.

1. I/O Functions.
2. Mathematical Functions – math.h

Predefined functions are functions that are built into C++ Language to perform some standard operations that are stand-alone and used for general purposes.

Object-oriented programming (OOP) is a programming language model organized around “objects” rather than “actions” and data rather than logic of the program.

Syntax: cout << expression or manipulator

Syntax:
<data-type> <array-name> [<size1>] [<size2>];

The member of structure is accessed using dot (.) operator.

The structure definition acts as a blueprint for the creation of variables.

The list of all structure members is called a template.

First Series:

Coefficient of variation: 30%

s = 15

x̄ = ?

Coefficient of variation = \(\frac{s}{\bar x}\) × 100

∴ 30 = \(\frac{15}{\bar x}\) × 100

∴ 30x̄ = 1500

∴ x̄ = \(\frac{1500}{30}\)

∴x̄ = 50

Second series:

Coefficient of variation: 25%

s = 9

x̄ = ?

Coefficient of variation = \(\frac{s}{\bar x}\) × 100

∴ 25 = \(\frac{9}{\bar x}\) × 100

∴ 25x̄ = 900

∴ x̄ = \(\frac{900}{25}\)

∴x̄ = 36

The general syntax for declaration of structure variable is Structurename variable;

The iomanip.h header file contains various function and macros for i/o manipulations for creating parameterized manipulations.

We have, 

r₁ = r₂ 

let , 

l₁ =4x, 

l₂ = 3x

\(\frac{S_1}{S_2}=\frac{\pi r_1l_1}{\pi r_2l_2}\) \(=\frac{l_1}{l_2}=\frac{4}{3}\)

Answer is (C) Rubber

Mixed cropping is the growing of two or more crops simultaneously intermingled with each other without a definite row pattern with the object of minimising risk of crop failure. 

Mixed farming is a method by which the farmer uses some part of his land for additional activities like cattle rearing, poultry or fishing in order to earn additional income and minimise risks associated with agriculture. 

Mixed cropping is the growing of two or more crops simultaneously intermingled with each other without a definite row pattern with the object of minimising risk of crop failure. 

Inter cropping is growing of two or more crops simultaneously in the same field in a definite row pattern with the object of increasing the productivity per unit area.

India produces about of 1/6 the world’s vegetables.

Nitrogen is an essential element for plant growth. Leguminous plants are instrumental in restoring the nitrogen supply of the soil. Therefore leguminous plants are alternated with other crops, reducing fertiliser needs. 

Pulses is a leguminous crop

Given details are,

Length of a cuboid = 2m

Breadth of a cuboid = 4m

Height of a cuboid = 5m

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm²

= 2 (2×4 + 4×5 + 5×2)

= 2 (8 + 20 + 10)

= 2 (38)

= 76 m²

Given, 

Area of 3 adjacent faces of cuboid = x= 8 cm² , y = 18 cm² , z = 25 cm²

From previous question we get,

= v² =xyz

= v²= 8 × 18 × 25 = 3600

= v = \(\sqrt{3600}\) = 60 cm³

Given,

Area of 3 adjacent faces of a cuboid = x, y, z

V = volume of cuboid

Let  a, b, c are respectively length, breadth, height of each faces of cuboid

So, x = ab

= y = bc

= z = ca

V = abc

Hence , xyz = ab × bc × ca = (abc)² = v² (v=abc)

= v² = xyz Proved.

Let us consider,

Areas of three faces of cuboid as x,y,z

So, Let length of cuboid be = l

Breadth of cuboid be = b

Height of cuboid be = h

Let, x = l×b

y = b×h

z = h×l

Else we can write as

xyz = l² b² h²….. (i)

If ‘V’ is volume of cuboid = V = lbh

V² = l² b² h² = xyz …… from (i)

∴ V² = xyz

Hence proved.

Let a, b and c be the length, breadth, and height of the cuboid. 

Then, x = ab, y = bc and z = ca [Since areas of three adjacent faces of a cuboid are x, y and z (Given)] 

And xyz = ab x bc x ca = (abc)² ……(1) 

We know, Volume of a cuboid ( V ) = abc …..(2) 

From equation (1) and (2), we have 

V² = xyz 

Hence proved.

Surface area of a hemisphere = 2πr² 

Given, 

diameter of the hemisphere bowl is 10.5 cm 

Surface area of the bowl = 2 × (22/7) × (10.5/2)2 = 173.25 cm² 

Given, 

cost of the plating it on the inside at the rate of Rs. 4 per 100 cm²

Cost of plating the hemisphere bowl = \(\frac{4}{100}\times\) 173.25 

= Rs. 6.93

125x³ – 27y³ – 225x²y + 135xy² 

Above expression can be written as (5x)³−(3y) ³−3(5x)²(3y) + 3(5x)(3y)² 

Using: a³ − b³ − 3a²b + 3ab² = (a−b)³ 

= (5x − 3y)³

p³ + 27 

= p³ + 3³   [a³ + b³ = (a + b)(a² –ab + b²)] 

= (p + 3)(p² – 3p – 9) 

Therefore, p³ + 27 = (p + 3)(p² – 3p – 9).

8x³ + 27y³ + 36x²y + 54xy² 

Above expression can be written as (2x)³ + (3y)³ + 3 x (2x)² x 3y + 3 x (2x)(3y)² 

Which is similar to [a³ + b³ + 3a²b + 3ab² = (a + b)³ ] 

Here a = 2x and b = 3y 

= (2x+3y)³ 

Therefore, 8x³ + 27y³ + 36x²y + 54xy² = (2x+3y)³

Inner radius of the hemisphere ‘r’ = 5 cm 

Outer radius of the hemisphere ‘R’ 

= inner radius + thickness 

= (5 + 0.25) cm = 5.25 cm 

Ratio of areas = 3πR² : 3πr² 

= R² : r² 

= (5.25)² : 5² 

= 27.5625 : 25 

= 1.1025:1 

= 11025 : 10000 

= 441 : 400 

Volume of a cuboid of measures 12 cm × 9 cm × 6 cm

V₁ = l₁× b₁ × h₁ = 12 × 9 × 6

Volume of the smaller cuboid of measures 4 cm × 3 cm × 2 cm

V₂ = l₂b₂h₂ = 4 × 3 × 2

∴ No. of cuboids are made

= V1/V2 = (12 × 9 × 6)/(4 × 3 × 2) = 27