Explore topic-wise InterviewSolutions in Current Affairs.

Encapsulation and polymorphism are two characteristics of OOP. 

1. Encapsulation: The method of combining data, attributes, and methods in the same entity is called encapsulation.

2. Polymorphism: Polymorphism is a Greek word which means many shapes. In OOP, Polymorphism means the ability to take on many forms. The term is applied both to objects and to operations. 

Characteristics of Stacks: 

•  It is a LIFO data structure 
•  The insertion and deletion happens at one end i.e. from the top of the stack

E learning can be defined as a learning system which makes use of electronic resources mainly computers and the internet. It encompasses everything from traditional classrooms that incorporate basic technology to online schools and universities. 

Merits 

Since in e learning teachers use technology to teach, it is highly engaging. Students can learn anywhere, anytime according to their convenience thus saving a lot of time and money.

E-learning: learning through the courses which are delivered online. 

Benefits : 

1. Self Paced (Learn at your speed). 
2. We can attend classes at any time and anywhere

Hyperlink allows us to link HTML elements (text and/or image) to another document or new section within the current document. 

Name : anchor tag, Syntax<a href ="Link Address> Hyperlink Text</a>

d = √((-3 - 2)² + (7 - 4)² + (2 + 1)²) 

= √(25 + 9 + 9) = √43

Let the present age of the man be x years

And His son's  be y years.  

1st Equation :-

After 5 years man’s age = x + 5  

After 5 years ago son’s age = y + 5  

As per the question  

⇒ x + 5 = 3(y + 5)

⇒ x – 3y = 10 ……………(i)  

2nd Equation

5 years ago man’s age = x – 5  

5 years ago son’s age = y – 5  

As per the question

⇒ x – 5 = 7(y – 5)

⇒ x – 7y = -30 …….(ii)

Subtracting (ii) from (i), we have  

⇒ 4y = 40  

⇒ y = 10  

Putting y value in Eq (i)

⇒ x – 3y = 10

⇒ x – 3 × 10 = 10  

⇒ x = 10 + 30

⇒ x = 40

Hence, man’s present age = 40 years and son’s present age = 10 years.

cot^-1(- 1/√3) = π - cot^-1(- 1/√3)

[∵ cot^-1(-x) = π – cot^-1(x)]

= π - π/3

∴ cot^-1(- 1/√3) = 2π/3

α = - π/3 ∴ G.S. x = 2nπ ± α 

∴ [x = -2nπ ± - π/3]

Given a * b = LCM of a and b ∀ a, b ∈ N 5*7 = 35

R={(x,y): x ∈ A, y ∈ B, 3x - y = 0} 

3(1) - 3 = 0, 3(2) - 6 = 0, 3(3) - 9 = 0, 3(4) - 12 = 0 

∴ R = {(1,3), (2,6),(3,9),(4,12)}

Domain of R = A ⇒ Co-domain of R = A 

Range of R = {3, 6, 9, 12}

Converse: q → p = “Something has low temperature implies it is cold”. 

Contrapositive: = q → -p = something does not have low temperature implies it is not cold.

n(C) = 40, n(T) = ? n( C ∩ T) = 10 

n(C ∪ T) = ? n(T - C) = n(C ∪ T) = n(C) + A(T) - n(C ∩ T) 

40 = 40 + n(T) -10 - 40 - 30 = n(T) 

∴ [N(T) = 10] ⇒ n(T - C) = n(T)n(C ∩ T) = 10 - 10 = 0

Given variation = 60 σ = 21, X̄ = ? 

Take coefficient of variation 

= σ/X x ω₀

60 = 21/X x ω₀ ⇒ X = 21/60 x 100 = 210/6

X̄ = Arithmatic mean = 35.

Given:

Number of followers gained by Harsh in a day = 230

Number of followers gained by Ashish in a day = 240

Standard Deviation for Harsh = 2.3

Standard Deviation for Ashish = 1.2

Number of platforms = 5

Formula used:

Mean = (sum of observations)/(number of observations)

Coefficient of variation = (SD/Mean) × 100 

where, SD = Standard deviation

Calculation:

For Harsh,

Mean = 230/5 = 46

Coefficient of variation = (2.3/46) × 100

⇒ Coefficient of variation for Harsh= 1/20 × 100 = 5

For Ashish,

Mean = 240/5 = 48

Coefficient of variation = (1.2/48) × 100

⇒ Coefficient of variation for Ashish = 1/40 × 100 = 2.5

Difference between their coefficient of variation = 5 – 2.5 = 2.5

∴ Harsh is more consistent and their difference is 2.5 

(1/3 + 4/1 + 4/3) + i(7/3 + 1/3 - 1/1) = ((1 + 12 + 4)/3) + i((7 + 1 - 3)/3) 

= 17/3 + i(4/3) = a + ib

1. (a) x < 0, y < 0 

2. The shaded figure is a rectangle. The side parallel to x axis are y = -1 and y = 1. The side perpendicular to x axis are x = 2 and x = -2. Hence the inequality that represent the shaded region are -2 ≤ x ≤ 2; -1 ≤ y ≤ 1.

Let a – d, a, a + d are the roots of the given equation.

Now, sum of the roots a – d + a + a + d = 24/4

3a = 6 a = 2

Product of the roots (a – d)a(a + d)= −18/4

a(a² – d²) = – 9/2

= 2(4 – d²) = – 9/2

4(4 – d²) = – 9

16 – 4d² = – 9

4d² = 25

d = ±5/2

∴ roots are  -1/2, 2  and 9/2

Answer: twice

TSA - CSA

= (2πrh + 2πr²) - (2πrh)

= 2πr²

Radius of cylinder = 2*Height

Find - Relation between C.S.A. and base area.

Solution - C.S.A. refers to Curved Surface Area. The formula to represent Curved Surface Area is - 2πrh.

In the equation, r represents radius and h represents height.

Base area of cylinder is given by - πr².

Keeping the values in equation-

As, R = 2h

Curved Surface Area is - 2π(2h)h

Curved Surface Area is - 4πh²

New base area is - π(2h)²

New base area is - 4πh².

Hence, Curved Surface Area and base area is - 4πh². Therefore, Curved Surface Area will be equal to base area.

Correct answer is c. B2:F15 

Answer: perpendicular

(i) It can be accomplished with function overloading. 

void repeat(int n, char c);

void repeat();

void repeat(char c);

void repeat(int n);

(ii) It can be accomplished with function overloading.

int average(int a,int b);

double average(double a,double b);

(iii) It can be accomplished with default argument. 

double mass (density d, volume v);

double mass(density d, volume v=1.0);

(iv) It can be accomplished with function overloading. 

int average(int a,int b);

double average(int c,int d);

(v) It can be accomplished with function overloading. 

char handle(char a);

void handle(char a);

The program contains two functions namely exchange and swap which call the exchange function. In main function swap function is called, which actually call the exchange function. In call of exchange function first argument is a address of variable and second argument is value of the variable. In call of the swap function both argument contains the address of the variable. After the execution of program the value of ‘a’ is 15 and value of ‘b’ is 5. 

int themax(int a)

{

return a;

}

int themax(int a,int b)

{

if(a>b)

return a;

else return b;

}

int themax(int a[ ])

Answer: Sharpen

The easy and best technique for sharpening uses Photoshop to create a high pass layer and then layer masks to control where and how much sharpening is applied. 

A high pass layer works in the same way an un sharp mask works on film: the high pass layer creates slightly blurry halos around edges in the original image, thus increasing the contrast at edges. The difference between high pass sharpening and most other methods of sharpening is that high pass sharpening does not actually adjust or change any pixels in your original image. Also, because high pass exists as a separate layer, you can adjust the layer’s Opacity and Blending Modes to control the strength of sharpening over the entire image. You can then use masks to control where sharpening is applied within the image and to make localized changes in the strength of sharpening.

Another advantage to sharpening with a high pass layer is you can save the layer with the Photoshop file and go back later and change the settings and areas where sharpening has been applied. This is a huge advantage when working with an image that you might finish for web viewing and printing, or printing on different papers. Process:

1. Evaluate the Image: First we need to evaluate the image we are going to sharpen like., where the sharpening is needed and the amount of sharpening is needed. 

2. Stamp New Layer or Duplicate Your Image Layer: Sharpening is usually the last or almost the last thing you do when preparing an image, and you want to be sure that you are applying sharpening to a finished, complete image. Therefore, before making your sharpening layer, ensure you have a duplicate layer of your completed image. To Stamp New Layer, check that all layers you want included in your final image are visible (turned on). Click on (select) your topmost layer and then stretch your fingers to use Command-Alt Shift-E/Control-Alt-Shift-E to activate the Stamp Visible command and make that stamp a new layer. 

3. Desaturate the High Pass Layer: A disadvantage to sharpening with a high pass layer is the potential for increasing or adding noise to a photo. With other sharpening tools such as Un sharp Mask, you can control noise problems with adjustments to the different values set in the tool. With the high pass option, you control noise by doing three things: Ensure your image capture is as clean as possible. This means using a tripod if not shooting at a high shutter speed, choosing the lowest ISO possible, and using the correct exposure. 

• Deal with any noise while processing your image. 

• Desaturate the high pass layer. Even when working on a colour image, the colour information in a high pass layer is irrelevant, so we’re going to remove that information right from the top to ensure that extra information doesn’t add noise. 

4. Apply the High Pass Filter: To apply the high pass filter to your sharpening layer, go to Filter > Other > High Pass. This will bring up a dialogue box with a Radius slider. You want to increase the Radius slider (increase the number of pixels affected) until the details in the image just begin to pop. You’ll find you need more Radius when you’re working with high resolution images. Don’t be surprised if you need around 10 to 20 pixels or more. Select OK 

5. Set the Layer’s Blending Mode: Change the layer’s blending mode in the blending options drop down menu to Soft Light. As you become more familiar with using high pass sharpening, experiment with Hard Light and Overlay blending modes as well 

6. Adjust the Layer Opacity: You will likely find that you now have a bit more sharpening than you need. That’s okay; starting with 100%, adjust the Layer Opacity down to get the amount of sharpening you need. Aim for an opacity that gives you the right amount of sharpening in the area of your image that needs the most sharpening. Add a layer mask by clicking on the Layer Mask icon at the bottom of your Layers window, or by going to Layer > Layer Mask.

Tuning is adjustment of control parameters to the optimum values for the desired control response. Stability is a basic requirement. However, different systems have different behavior, different applications have different requirements, and requirements may conflict with one another.

1. a large beer cask. 2. a measure of capacity, usually equal to 252 wine gallons.

This method was introduced by John G. Ziegler and Nathaniel B. Nichols in the 1940s. The Ziegler‐ Nichols’ closed loop method is based on experiments executed on an established control loop (a real system or a simulated system).  

double power(double n,int p=2)

{

double res=pow(n,p);

return res;

}

int power(int n,int p=2)

{

int res=pow(n,p);

return res;

}

In order to find the best possible match, the compiler follows the following steps: 1. Search for an exact match is performed. If an exact match is found, the function is invoked. For example, 2. If an exact match is not found, a match trough promotion is searched for. Promotion means conversion of integer types char, short, enumeration and int into int or unsigned int and conversion of float into double. 3. If first two steps fail then a match through application of C++ standard conversion rules is searched for. 4. If all the above mentioned steps fail, a match through application of user-defined conversions and built-in conversion is searched for.

For example,

void afunc(int);

void afunc(char);

void afunc(double);

afunc(471); //match through standard conversion. Matches afunc(int)

The compiler will follow the following steps to interpret more than one definitions having same name:

(i) if the signatures of subsequent functions match the previous function’s, then the second is treated as a redeclaration of the first.

(ii) if the signature of the two functions match exactly but the return type differ, the second declaration is treated as an erroneous re-declaration of the first and is flagged at compile time as an error.

(iii) if the signature of the two functions differ in either the number or type of their arguments, the two functions are considered to be overloaded.

Correct option: 2. Cobalt Blue

Correct option: 4. All of this

Correct option: 1. Face Powder Brush

When Romeo first sees Juliet at the Capulet ball, he is immediately enraptured by her beauty. He immediately asks the servingman at the party who she is. He says that Juliet stands out against the darkness like a jeweled earring in an Ethiopian. He says her "beauty too rich for use, for earth to dear." Compared to the other women at the party, Juliet is like a white dove among crows. He famously says he "ne'er saw true beauty till this night." 

In Act II, during the famous balcony scene, Romeo glorifies Juliet's beauty by saying Juliet is the sun. He says even the moon is jealous of Juliet's beauty. He compares her eyes to the two brightest stars who had to go away but "do entreat in her eyes." He says her eyes are so bright and if they were stars that "that birds would sing and think it were not night." He refers to Juliet as "bright angel," "a winged messenger of heaven." 

Romeo glorifies Juliet’s beauty saying that her beauty was so brilliant that it could inspire a torch to burn more brightly meaning that even torchlight seems pale compared to Juliet’s beauty. He tells us that her beauty was like twilight, soft and radiant, and shines just like a glittering jewel worn by a dark-skinned woman, an Ethiopian woman. Her beauty according to Romeo was priceless and too rich for use and very expensive for humans, meaning that everybody cannot afford to have a beauty like Juliet. A white pigeon in the midst of black crows can be admired easily, similarly, Juliet was more beautiful than all the other beautiful girls.

Mary Kom won a medal in each of the six World Boxing Championships she attended. There were a number of other international level Boxing Championships in Taiwan, Vietnam Denmark and so on. But it was retaining her world title in 2006 by defeating Steluta Duta of Romania 22-7 at the fourth World Championship in New Delhi that she considered her greatest achievement in life because she was able to win at home.

Dona Laura claims to have a neighbour’s right to criticize Don Gonzalo.