Explore topic-wise InterviewSolutions in Current Affairs.

a) 1011, 1101 

b) 10101,10111

Answer is

- 1 - √3i

ⁿC_n-7 = ⁿC₆ ⇒ n – 7 = 6 ⇒ n = 13

Two friends have any one of the 366 days on the year 1984 as her birthday. [Year 1984 is a leap year]

Total number of outcomes in which two friends may have their birthday = 366 x 366

Two friends may have same birthday on any one of the 366 Days Of The Year 1984.

Number of favourable outcomes in which Two friends will have same birthday = 366

Probability = Number of favourable outcomes/Total number of outcomes.

Probability that both have same birthday = 366/(366 x 366) = 1/366.

Hence,the  probability  that they both have the same birthday = 1/366.

Correct option is D) in

Relations are also called tables, and sometimes by the older data processing term files. A row is known as a tuple in the relational model, but may also be referred to as a record. Finally, relational model attributes are known as table columns and sometimes as fields.

Equations deal with numerical relationships. A functional dependency deals with the existence of a determinant relationship between attributes, regardless of whether or not there is a numerical relationship between them. Thus, if we know that there is no hot water every Wednesday, No-Hot-Water is functionally dependent on Wednesday. So, if we know it is Wednesday, then we know we will have No-Hot-Water. This is a functional dependency, but not an equation.

A foreign key is used to establish relationships among relations in the relational model. Technically, a foreign key is a column (or columns) appearing in one relation that is (are) the primary key of another table. Although there may be exceptions, the values in the foreign key columns usually must correspond to values existing in the set of primary key values. This correspondence requirement is created in a database using a referential integrity constraint on the foreign key.

A deletion anomaly occurs when, by deleting the facts about one entity, we inadvertently delete facts about another entity; with one deletion, we lose facts about two entities. For example, if we delete the tuple for Student 001289 from a table, we may lose not only the fact that Student 001289 is in Pierce Hall, but also the fact that he has $200 left in his security deposit. An insertion anomaly happens when we encounter the restriction that we cannot insert a fact about one entity until we have an additional fact about another entity. For example, we want to store the fact that the security deposit for Pierce Hall is $300, but we cannot enter this data into the Student relation until a student registers for Pierce Hall.

A relation is in BCNF when every determinant in the relation is a candidate key. This means that any possible primary key can determine all other attributes in the relation. Attributes may not be determined by non-candidate key attributes or part of a composite candidate key. Thus it is said "I swear to construct my tables so that all nonkey columns are dependent on the key, the whole key and nothing but the key, so help me Codd!"

The methods of opening file within C++ program

• Opening a file using constructor.
• Opening a file using member function open() of the class Opening a file using constructor.

The syntax for opening file for output purpose only is ofstream obj(“filename”);
Example:
ofstream out(“results.dat”);

Boyce and Codd form (BCNF):

• When a relation has more than one candidate key, anomalies may result even though the relation is in 3NF.
• 3NFdoes not deal satisfactorily with the case of relation with overlapping candidate keys
• BCNF is based on the concept of a determinant.
• A determinant is any attribute on which some other attribute is fully functionally dependent.

A relation is in BCNF is, and only if, every determinant is a candidate key.

1. A web server is a computer on which a web site is hosted and a program that runs on such a computer. So the term web server refers to both hardware and software.

2. A web site is a collection of web pages stored or “hosted” on a computer that is connected to the internet. Such a computer is known as a Web Server.

3. A web server program is software that runs on the web site hosting Server computer. Its main purpose is serving web pages for requests from web browsers.

4. There are many web server programs available. The most famous and popular of them all is Apache developed by the Apache Foundation. It is free and also available for several operating systems including Windows, Macintosh, and Linux/Unix.

A Functional dependency is denoted by X Y between two sets of attributes X and Y that are subsets of R specifies a constraint on the possible tuple that can form a relation state r of R. The constraint is for any two tuples t1 and t2 in r if t1[X] = t2[X] then they have t1[Y] = t2[Y]. This means the value of X component of a tuple uniquely determines the value of component Y.

It guarantees that the spurious tuple generation does not occur with respect to relation schemas after decomposition.

The domain of attribute must include only atomic (simple, indivisible) values.

1. Partial Key: It is a set of attributes that can uniquely identify weak entities and that are related to same owner entity. It is sometime called as Discriminator. 

2. Alternate Key: All Candidate Keys excluding the Primary Key are known as Alternate Keys. 

3. Artificial Key: If no obvious key, either stand alone or compound is available, then the last resort is to simply create a key, by assigning a unique number to each record or occurrence. Then this is known as developing an artificial key. 

4. Compound Key: If no single data element uniquely identifies occurrences within a construct, then combining multiple elements to create a unique identifier for the construct is known as creating a compound key. 

5. Natural Key: When one of the data elements stored within a construct is utilized as the primary key, then it is called the natural key.

It is based on concept of full functional dependency. A functional dependency X Y is full functional dependency if removal of any attribute A from X means that the dependency does not hold any more.

A relation is said to be in DKNF if all constraints and dependencies that should hold on the the constraint can be enforced by simply enforcing the domain constraint and key constraint on the relation.

A Relation schema R is said to be 5NF if for every join dependency {R1, R2, ..., Rn} that holds R, one the following is true 

1. Ri = R for some i. 

2. The join dependency is implied by the set of FD, over R in which the left side is key of R.

A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully functionally dependent on primary key.

A relation schema R is in 3NF if it is in 2NF and for every FD X A either of the following is true 

1. X is a Super-key of R. 

2. A is a prime attribute of R.

In other words, if every non prime attribute is non-transitively dependent on primary key.

A relation schema R is said to be in 4NF if for every Multivalued dependency X Y that holds over R, one of following is true. 

1. X is subset or equal to (or) XY = R. 

2. X is a super key.

Correct option:

(C) (1, –2) 

• Low concentration of sulphur dioxide causes respiratory diseases like asthma, bronchitis etc and irritation to the eyes.
• High concentration of SO₂ leads to stiffness of flower buds. Also it produces acid rain.

Given , 

(Universal Set) U = { 1,2,3,4,5,6,7,8,9 }

A = { 1,2,4,7 }

B = { 1,3,5,7 }

Therefore , A U B = { 1,2,3,4,5,7 }

[ Here U represents Union ( U ) . ]

( Note : Dont Write the repeated numbers multiple times,  only write those numbers 1 time ) 

\(f(x)=\frac{2(x +3)}{3}\)

\(\because\) function f(x) is defined for all real values of x.

\(\therefore\) \(- \infty < x< \infty\)

⇒ \(- \infty < x +3 < \infty\)

⇒ \( - \infty <\frac{2(x +3)}{3} < \infty\)

⇒ \(- \infty < f(x) < \infty\)

Range of function f(x) is \(\mathbb {R}\) or \((-\infty, \infty)\).

If Z1+Z2 = real and Z1.Z2=real

Then Z2 is conjugate of Z1​​​​​

Example 

Let Z1= X+iy  and Z2= A+ ib

If Z1+ Z2 =real them Im(Z1+Z2)=0

Thus y=-b

Similarly for z1.z2= real

A=X

Thus Z1 and Z2 are conjugate

Example 

Z1= 3+ 4i       Z2= 3-4i

Z1+Z2 = 6 real

Z1.Z2= 25  is real  as i²=-1

(-3)^-1 x \((1/3)^{-1}\) = \(\frac1{-3}\times(3^{-1})^{-1}\)

 = \(\frac{-1}3\times3\) = -1

Correct option is (A) All real numbers

\(\frac{2^{x -1} -1}{2^{x + 1} + 1} < 2\)

⇒ \(\frac{2^{x -1} -1}{2^{x + 1} + 1} -2 < 0\)

⇒ \(\frac{2^{x -1} - 1 -2.2^{x + 1} -2}{2^{x +1} + 1}< 0\)

⇒ \(2^{x-1}-2^{x +2}- 3< 0\)   (∵ \(2^{x+1}+1 > 0\) (always))

⇒ \(\frac{2^x}2 - 4.2^x - 3< 0\)

⇒ \(\frac{-7}2.2^x - 3<0\)

⇒ \(\frac 72 . 2^x > -3\)

⇒ \(2^x > \frac{-6}7\)

which is true for all real values of x.

Product of Mean = Product of Extreme

P×21 = 3×77

P×21 = 231

P = 231÷21

P = 11

Solution: we have
Sum of m terms of an A.P. = m/2 [2a + (m -1)d]
Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

m/2 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m² : n²

⇒ [2a + md - d] / [2a + nd - d] = m/n

⇒ 2an + mnd - nd = 2am + mnd - md

⇒ 2an - 2am = nd - md

⇒ 2a (n -m) = d(n - m)

⇒ 2a = d

Ratio of m th term to n th term:

[a + (m - 1)d] / [a + (n - 1)d]

= [a + (m - 1)2a] / [a + (n - 1)2a]

= a [1 + 2m - 2] / a[1 + 2n -2]

= (2m - 1) / (2n -1)

So, the ratio of m th term and the n th term of the arithmetic series is (2m - 1) : (2n -1)

Solution:
Let A be the first term and D be the common difference of A.P.

Given that a,b and c be the sum of the first p,q,r terms respectively.

∴  a = Sp

a = ( p/2)[ 2A + ( p -1)D ]

a / p = (1/2)[ 2A + ( p -1)D ] --------------(1)

∴  b = Sq

b = ( q/2)[ 2A + ( q -1)D ]

b / q = (1/2)[ 2A + ( q -1)D ] --------------(2)

∴  c = Sr

c = ( r/2)[ 2A + ( r -1)D ]

c / r = (1/2)[ 2A + ( r -1)D ] --------------(3)

Now a/p(q - r) + b/q (r- p ) + c/r (p - q)

⇒ (1/2)[ 2A + ( p -1)D ](q - r) + (1/2)[ 2A + ( q -1)D ](r- p ) + (1/2)[ 2A + ( r -1)D ](p - q)

⇒ (1/2)( 2A)[ q - r + r- p + p - q)  + (1/2)D [ ( p -1)(q - r) + ( q -1)(r- p ) + ( r -1)(p - q) ]

⇒ (1/2)( 2A) (0) + (1/2)D [ pq - pr - q + r + qr - pq - r + p + rp - rq - p + q]

⇒ (0) + (1/2)D (0)

= 0

Formula Used:

nPr =n!/(n-r)!

nCr = n!/[(n-r)! r! ]

ⁿP_r/ⁿC_r = r!

Calculation:

Given:

nPr = 720

nCr = 6

nPr/nCr = r!

r! = 720/6

r! = 120

r! = 5 × 4 × 3 × 2 × 1

r = 5 

∴ The value of r! = 5! 

Let required distance = x kms 

At 4 kmph, time taken = X/4

At 5 kmph, time taken = X/5 

Given X/4 = 5/60 = X/5 + 25/60s

Getting x = 2.5 km

Required arrangement = ⁷P₄ = 840

Duplicate ratio of 5 : 3 = 52 : 32 = 25 : 9

The numbers are in the ratio 3 : 5 

Let the common ratio be x. 

Therefore the numbers are 3x and 5x. 

If 5 is added to both the numbers the ratio become 2:3. 

∴ (3x + 5):(5x + 5)= 2 : 3 

∴ 6x + 15 = 10x + 5 

∴ x = 1 

Thus, the numbers are 3x = 3 × 1 = 3 and 5x = 5 × 1 = 50.

1^st tap can fill 1/12 th tank in 1 min 

2^nd tap can fill 1/15th tank in 1 min Drain pipe drain out 1/20th tank in 1 min 

In 1 min = (1/12 + 1/15 - 1/20)th = (1/10)th of tank will get filled

∴ the tank will get filled in 10 minutes.

2 men and 4 women can do a work in 33 days.

66 men and 132 women can do a work in 1 day

Similarly 3 men and 5 women can do a work in 24 days

72 men and 120 women can do a work in 1 day

∴ 66 men + 132 women = 72 men + 120 women 

∴ 1 man = 2 women 5 men + 2 women = 10 women + 2 women =12 women

Let the required number of days be x

2 men + 4 women = 4 women + 4 women = 8 women

8W : 12W = x : 33

12x = 33 × 8

∴ x = 22 days

5x = 60 ⇒ x = 12

ⁿP₄ = n(n – 1)(n – 2)(n – 3) = 6.5.4.3 = 360 

∴ n = 6

Answer is 2,4,6,8,10,

Answer is 1,3,5,7,

The difference between the 15^th term and the 5^th term = 40

i.e., ten times an of a common difference = 40

Common difference = 40/10 = 4

When we add 8 x common difference we will get 20^th term.

The difference between the 12^th term and the 20^th term = (20 – 12) × common difference

= 8 x 4 = 32

The difference between the first term and the 21^th term = (21 – 1) × Common difference

= 20 × = 80

a. 23rd term = 32

35th term = 104

x₂₃ + 12d = x₃₅

12d = 104 - 32 = 72

d = 6

b. Middle term offirst 35 terms = \(\sqrt{4}\)

c. 18 th term = X₁₉ = X₂₃ – 5d = 32 -5 × 6 = 32 – 30 = 2

Sum of 35 terms = 18^th term × 35 = 70

i. Sum of first n natural numbers = n(n + 1)/2

∴ Sum of first 20 natural numbers

= 20 x (20 + 1)/2 = (20 x 21)/2

= 420 / 2 = 210

ii. The algebric form = 6n + 5

Common difference 'd' = coefficient of n = 6

First term 'x₁' = (coefficient of n) + (constant term) = 6 + 5 = 11

Sum of first 20 terms

= \(\frac{1}{2}\) an(n + 1) + nb

= \(\frac{1}{2}\) x 6 x 20 x 21 + 20 x 5

= 3 x 20 x 21 + 100

= 1260 + 100 = 1360