Explore topic-wise InterviewSolutions in Current Affairs.

We have seen that the shaded region and origin are on the opposite side of the line 6x + 2y = 8 

For (0,0) we have 0 + 0 - 8 < 0 . So the shaded region satisfies the inequality 6x + 2y \(\ge\) 8. 

We have seen that the shaded region and origin are on the opposite side of the line x + 5y = 4 

For (0,0) we have 0 + 0 - 4 < 0 . So the shaded region satisfies the inequality x + 5y \(\ge\) 4 . 

We have seen that the shaded region and origin are on the same side of the line x + y = 4 

For (0,0) we have 0 + 0 - 4 < 0 . So the shaded region satisfies the inequality x + y \(\le\) 4 

We have seen that the shaded region and origin are on the same side of the line y = 3 

For (0,0) we have 0 - 3 < 0. So the shaded region satisfies the inequality y \(\le\) 3. 

Thus the linear inequation comprising the given solution set are +2y \(\ge\) 8,x + 5y \(\ge\) 4, x + y \(\le\)4, y \(\le\) 3

Correct option is (A) (3, 2)

\(2^x+3^y=17\)   _________(1)

and \(2^{x+2} - 3^{y+1}=5\)

\(\Rightarrow\) \(2^2.2^x-3.3^y=5\)

\(\Rightarrow\) \(4.2^x-3.3^y=5\)   _________(2)

Let \(2^x=X\;\&\;3^y=Y\)

Then equations (1) & (2) converts to

X + Y = 17    _________(3)

and 4X - 3Y = 5   _________(4)

Put Y = 17 - X from (3) into equation (4), we get

4X - 3 (17 - X) = 5

\(\Rightarrow\) 4X - 51 + 3X = 5

\(\Rightarrow\) 7X = 5+51 = 56

\(\Rightarrow\) X = \(\frac{56}7\) = 8

\(\therefore\) Y = 17 - X

= 17 - 8 = 9    (From (3))

\(\therefore\) X = 8

\(\Rightarrow\) \(2^x=8=2^3\)

\(\Rightarrow\) x = 3

and Y = 9

\(\Rightarrow\) \(3^y=9=3^2\)

\(\Rightarrow\) y = 2

Hence, (x, y) = (3, 2)

Correct option is A) (3, 2)

(D) x + y = 0

Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)]

Hence, the point (a, – a) always lies on the line x + y = 0.

False.

Justification:

We have the equation, x + 2y = 7.

Substituting the values of x = 0 and y = 7 from the point (0, 7) in the equation,

We get,

0 + 2(7) = 14 ≠ RHS

Hence, the graph of the linear equation x + 2y = 7 passes through the point (0, 7).

Correct option is (B) Any real number except 3

For both lines has exactly one solution (unique solution), we have

\(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)

\(\Rightarrow\) \(\frac{2P}2\neq\frac31\)

\(\Rightarrow\) \(P\neq3\)

Hence, for any real value of P except 3, the given pair of linear equations has exactly one solution.

Correct option is B) Any real number except 3

True

If we put x = 0 and y = 3 in LHS of the given equation, we find LHS = 3 x 0 + 4 x 3 = 0 + 12 = 12 = RHS

Hence, (0, 3) lies on the linear equation 3x + 4y = 12.

Correct option is (A) 2

Slope of the line passing through points (2, 3) and (4, 7) is \(m=\cfrac{y_2-y_1}{x_2-x_1}\)

\(=\frac{7-3}{4-2}=\frac{4}{2}=2\)

Correct option is A) 2

Correct option is (D) (2016, 2017)

Point of intersection of the lines x = 2016 and y = 2017 is (2016, 2017).

Correct option is D) (2016, 2017)

Correct option is (A) (- b/a, 0)

Equation of line X-axis is y = 0.

\(\therefore\) For point of intersection of line y = ax+b with X-axis (i.e., y = 0 line), we have

ax + b = 0

\(\Rightarrow\) \(x=\frac{-b}{a}\)

Hence, point of intersection is \(\left(\frac{-b}{a},0\right).\)

Correct option is A) (- b/a,0)

Variations in the off springs is the characteristic of Sexual reproduction.

The Reproduction which involves the formation and fusion of male and female gametes is called sexual reproduction.

There is much wastage of pollen grain in Cross (wind) pollination.

Reproduction is one of the basic characteristics of living organisms by which they continue their progeny.

Special features of wind pollinated flowers are. The flowers are generally small and dull colored. They produce large number of dry and light pollen grain. 

Ex : Sugar cane, Jawara, etc.

Fertilization leads to the formation of Zygote.

In most plants cross pollination is brought about with help of insects. Like butterflies, moths and bees. Such flowers show certain modifications to attract insects. These modifications are “mainly concerned with the color of the petals, some flowers contain glands called nectaries which produce the nectar to attract the insects, that bring about cross pollination, flowers which open during night, usually have a dull coloration, but have an aromatic smell that attracts insects.

The seed is en closed in the fruit. This is a characteristic feature of angiosperms.

1. The following changes take place after the fertilization. 

2. The diploid Zygote develops into an embryo .which later differentiates in to a seed, which later grows in to a new plant. 

3. Surrounding the embryo a nutritive tissue called endosperm is formed. 

4. The entire ovule now becomes the seed. 

5. The coverings to the ovule transform in to the seed coat. 

6. Ovary portion of the carpel gets transformed in to the fruit enclosing the seed. 

7. Petals, sepals and other parts of the flower fall off. 

8. Thus pollination and fertilization result in the formation of seed. Which is enclosed in the fruit.

Correct option is (D) 1

\(\frac{5}{x}+\frac{3}{y} = 6\)

\(\because\) y = 3

\(\therefore\) \(\frac{5}{x}+\frac{3}{3} = 6\)

\(\Rightarrow\) \(\frac{5}{x}=6-1=5\)

\(\Rightarrow\) \(x=\frac55=1\)

Correct option is D) 1

Let r₁ and r₂ be the internal and external base radii of spherical shell.

r₁ = 3 cm, and r₂ = 5 cm

Base radius of solid cylinder, r = 7 cm

Let the height of the cylinder = h

As per given statement:

The hollow spherical shell is melted into a solid cylinder, so

Volume of solid cylinder = Volume of spherical shell

πr²h = 4/3 π(r₂³ – r₁³)

⇨ 49h = 4/3(125 – 27)

or h = 8/3 cm

We know that

Surface area of the sphere = 4 πr²

By substituting the values

4 πr² = 576π

On further calculation

r² = 576/ 4 = 144

By taking square root

r = 12cm

We know that

Volume of the sphere = 4/3 πr³

By substituting the values

Volume of the sphere = 4/3 × π × (12)³

So we get

Volume of the sphere = 2304 π cm³

Therefore, the volume of the sphere is 2304 π cm³.

Let the number of balls formed are n

As per statement,

Volume of metal cone = Total volume of n spherical balls

Volume of cone = n(Volume of any spherical ball)

1/3 π r² h = n (4/3 π r³)

12² x 24 = n x 108

or n = 32

Therefore, 32 spherical balls can be formed.

Correct option is D) cuboid

Radius of the cone = r = 5cm and

Height of the cone = h = 20cm

Let the radius of the sphere = R

As per given statement,

Volume of sphere = Volume of cone

4/3 πR³ = 1/3πr²h

4R³ = 5 × 5 × 20

R = 5 cm

Diameter of the sphere = 2R = 2 x 5 = 10 cm

We know that

Surface area of the sphere = 4 πr²

By substituting the values

4 πr² = 154

On further calculation

4 × (22/7) × r² = 154

So we get

r² = (154 × 7)/ (4 × 22) = 49/4

By taking the square root

r = 7/2 cm

We know that

Volume of the sphere = 4/3 πr³

By substituting the values

Volume of the sphere = 4/3 × (22/7) × (7/2)³

So we get

Volume of the sphere = 179.67 cm³

Therefore, the volume of the sphere is 179.67 cm³

We know that

Volume of the sphere = 4/3 πr³

By substituting the values

606.375 = 4/3 × (22/7) × r³

On further calculation

r³ = (606.375 × 3 × 7)/ 88

So we get

r³ = 144.703125

By taking cube root

r = 5.25m

We know that

Surface area of the sphere = 4 πr²

By substituting the values

Surface area of the sphere = 4 × (22/7) × 5.25²

So we get

Surface area of the sphere = 346.5 m²

Therefore, the surface area of the sphere is 346.5 m².

x + 2y = 5 …………… (1) 

2x-y = 0 ………… (2) 

2x = y ⇒ x = \(\frac{y}{2}\)

This value substitute in equation (1)

\(\frac{y}{2}\) + 2y = 5 ⇒ y + 4y = 10 

5y = 10 ⇒ y = 2 

This value substitute in equation (2) 

2x – 2 = 0 ⇒ 2x = 2 ⇒ x = 1 

∴ x = 1, y = 2.

सही विकल्प है (D) 06

सही विकल्प है (C) हेनरी फेयोल

Given l ∥m and AC is transversal. 

z = 40° (Alternate interior angles) 

l ∥m and AB is transversal and x + (y + z) = 180° (co-interior angles are supplementary) 

x + y + z = 180° 

(x + y) + 40° = 180° (co-interior angles are supplementary) (∵ z = 40°) 

x + y + 40° = 180° 

x + y + 40°-40° = 180° – 40° 

∴ x + y = 140° 

But x = y (given)

2x = 140°

∴ x = 140°/2 = 70° = y 

So, x = 70°, y = 70° and z = 40°

Glycolysis is strongly regulated by the complex interplay between ATP consumption, NADH₂ regeneration and regulation of various glycolytic enzymes like hexokinase, PFK-1, pyruvate kinase, etc. Besides, it is also controlled by hormones like glucagon, epinephrine and insulin.

Glycolysis takes place in the cytoplasm of a cell.

1. SELECT NO, NAME, TDATE FROM TRIP ORDER BY NO;

2. SELECT NAME FROM TRIP WHERE TCODE = 101 OR TCODE = 103;

3. SELECT NO AND NAME FROM TRIP WHERE ‘2015-02-10’ < TDATE < ‘2015-04-01’;

4. SELECT NO, NAME, TDATE, KM, TCODE FROM TRIP WHERE KM >100 ORDER BY NOP;

5. TO DISPLAY THE MORE THAN ONE COUNT OF TCODE FROM THE TABLE TRIP

6. TO DISPALY SEPERATE TCODE OF TABLE TRIP

7. TO DISPAY THE NAME AND CODE OF THOSE TRANS PORTERS, WHO HAVE TRAVELLED MORE THAN 90 KMS.

8. TO DISPLAY THE NAME AND EXPENDITARE OF A TRANSPORTER WHO HAVE TCODE AS 105.

1. SELECT INAME FROM ITEMS ORDER BY INAME ASC;

2. SELECT INAME, PRICE FROM ITEMS WHERE PRICE => 10000 AND PRICE =< 22000; (c) SELECT TCODE, COUNT (CODE) FROM ITEMS GROUP BY TCODE;

3. SELECT PRICE, INAME, QTY FROM ITEMS WHERE (QTY> 150);

4. SELECT TNAME FROM TRADERS WHERE (CITY = “DELHI”) OR (CITY = “MUMBAI”)

5. SELECT COMPANY, INAME FROM ITEMS ORDER BY COMPANY DESC;

6. (hi) 38000 

1200 

(h2)1075000 

(h3)T01 

T02 

TO3 

(h4) LED SCREEN 40 DISP HOUSE INC CAR GPS SYSTEM ELECTRONICS SALES

1. Primary Key of CARDEN = C code CARDEN 

Alternate Key = CarName: 

Primary key of Customer = Code 

Alternate Key of Customer = Cname 2

2. SELECT CarName From CARDEN

(i)  WHERE Color = “SILVER”;

(ii) SELECT CarName, Make, Capacity From 

 CARDEN ORDER BY Capacity DESC;

(iii) SELECT MAX(Charges) Frm CARDEN;

(iv) ELECT Cname, CarName From CARDEN, CUSTOMER WHERE CARDEN. Ccode = CUSTOMER. Ccode;

3. (i) 4 

  (ii) MAX(Charges) MIN (Charges) 35 112 

  (iii) 5 

  (iv) SX4 C Class

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

1. SELECT * FROM PRODUCTS ORDER BY PNAME ASC;

2. SELECT PNAME, PRICE FROM PRODUCTS WHERE ((PRICE => 10000) AND (PRICE = < 15000));

3. SELECT SUPCODE, COUNT (PID) FROM PRODUCTS GROUP BY SUPCODE;

4. SELECT PRICE, PNAME, QTY FROM PRODUCTS WHERE (QTY > 100);

5. SELECT SNAME FROM SUPPLIERS WHERE ((CITY = “DELHI”) OR (CITY = “CHENNAI”));

6. SELECT COMPANY, PNAME FROM PRO-DUCTS ORDER BY COMPANY DESC; 4

7. SOI1 

(gl)s02 

s03 

(g2) 28000 

1100 

(g3) 550000 

(g4) PNAME SNAME Vi 

DIGITAL CAMERA 14 X GETALL INC 

PENDRIVE16 GB GETALL INC

ALTER TABLE CUSTOMER ADD PRICE NUMBER (10, 2).

1. SELECT NAME,FEE,GENDER,JOINYEAR 

FROM APPLICANTS 

WHERE J OINYE AR <2010

2. SELECT NAME FROM APPLICANTS WHERE FEE >30000

3. SELECT NAME FROM APPLICANTS ORDERBY JOINYEAR ASC

4. SELECT YEAR, COUNT]*) FROM 

APPLICANTS GROUP BY YEAR;

5. SELECT C_ID, COUNT]*) FROM 

APPLICANTS, COURSES GROUP BY ID 

WHERE APPLICANTS.C_ID=COURSES. C_ID

6. SELECT NAME,COURSE FROM APPLICANTS, COURSES 

WHERE APPLICANTS. C_ID=COURSES. C_ID

• Avisha 2009
• 2009
• 67
• 55000 A01

A static method has to be defined outside a class. It can be called without an object. An instance method is defined within a class and has to be invoked on an object.

Data hiding can be defined as the mechanism of hiding the data of a class from the outside world or to be precise, from other classes. Data hiding is achieved by making the members of the class private. Access to private members is restricted and is only available to the member functions of the same class. However, the public part of the object is accessible outside the class.

1. Data Abstraction

2. Encapsulation

3. Object

• Capability to express closeness with the real- world models.
• Reusability-allows addition of new features to an existing one.
• Transitivity-changes in one class get automatically reflected across.

A given name can only be associated with one function at a time, so cannot overload a function with multiple definitions. If you define two or more functions with the same name, the last one defined is used. 

However, it is possible to overload a function, or otherwise genericized it. You simply need to create a dispatcher function that then dispatches to your set of corresponding functions. Another way to genericized a function is to make use of the simple generic module which lets you define simple single-dispatch generic

functions.

def test(): #function 1

print “hello”

def test(a, b): #function 2

return a+b

def test(a, b, c): #function 3

return a+b+c

If you run the code of three test functions, the second test() definition will overwrite the first one. Subsequently, the third test() definition will overwrite the second one. That means if you give the function call test (20,20), it will flash an error stating, “Type Error: add() takes exactly 3 arguments (2 givens)”. This is because Python understands the latest definition of the function test() which takes three arguments.

• Classes tend to be overly generalized.
• Relationship among classes might become artificial.
• Program design is tricky and complicated.
• More skills and thinking in terms of objects is required.

Abstraction and Encapsulation are complementary concepts. Through encapsulation only we are able to enclose the components of the object into a single unit and separate the private and public members. It is through abstraction that only the essential behaviors of the objects are made visible to the outside world. 

So, we can say that encapsulation is the way to implement data abstraction. For example in class Student, only the essential information like roll no, name, date_of_birth, course, etc. of the student will be visible. The secret information like calculation of grades, allotment of examiners etc. will be hidden.

When several function declarations are specified for a single function name in the same scope, the function is said to be overloaded. In other languages, the same function name can be used to define multiple functions with different number and type of arguments, def test(): 

#function 1 print “hello”

def test(a, b): #function 2

return a+b

def test(a, b, c): #function 3

return a+b+c 2

Overriding Methods: It is a method to access, the parameterized constructor with the same name but having different parameters. For example,

Class emp:

def_init_(self, n):

self, a = n def_init_(self):

self, a = 100

Type Error: sum() takes exactly 3 arguments (2 given)

Concept: Polymorphism [Function Overloading]

from functools import wraps

import math

def overloaded(func):

@wraps(func)

def overloaded_func(*args, **kwargs):

for f in overloaded_func.overloads:

try:

return f(*args, **kwargs)

except TÿpeError:

pass

else:

# it will be nice if the error message prints a list of

# possible signatures here

raise TÿpeError(“No compatible signatures”)

def overload_with(func):

overloaded_func.overloads.append(func)

return overloaded_func

overloadedjunc.overloads = [func]

overloaded_func.overload_with = overload_with

return overloaded_func

#############

©overloaded

def area():

print ‘Area’

pass

@area.overload_with

def _(a):

# note that, like property(), the function’s name in

# the “def _(n):” line can be arbitrary, the important

# name is in the “@overloads(a)” line

print ‘Area of square=’,a*a

pass

@area.overload_with

def _(a,b):

# note that, like property(), the function’s name in

# the “def _(nl,n2):” line can be arbitrary, the important t

# name is in the “@overloads(a)” line

print Area of rectangle=’,a*b

pass

@area.overload_with

def _(a.b,c):

s= (a+b+c)/2

print s

print Area of triangle=’, math.sqrt(s*(s-a)* (s-b)*(s-c))

pass

choice=input(“Enter 1-square 2-rectangle 3- tr-iangle”)

if choice==1:

side = input(“Enter side”) area(side)

elif choice ==2:

length = input(“Enter length”)

breadth = input(“Enter breadth”)

area(length,breadth) elif choice==3:

a = inputfEnter sidel”)

b = inputfEnter side2”)

c = inputfEnter side3”) area(a,b,c)

else:

print “Invalid choice”