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(i) - 1 (ii) + 3

(i) Ionization
(ii) Optical
(iii)Coordination isomerism

(i) K₃[Co(C₂O₄)₃]
The central metal ion is Co. Its coordination number is 6. The oxidation state can be given as:
x − 6 = −3 x
= + 3
The d orbital occupation for Co³⁺ is t_2g ⁶eg 0.
(ii) cis-[Cr(en)₂Cl₂]Cl
The central metal ion is Cr. 

The coordination number is 6. The
oxidation state can be given as:
x + 2(0) + 2(−1) = +1
x − 2 = +1 x = +3
The d orbital occupation for Cr3+ is t2g 3.

(iii) (NH₄)₂[CoF_4]
The central metal ion is Co.
The coordination number is 4.
The oxidation state can be given as:
x − 4 = −2 x
= + 2
The d orbital occupation for Co²⁺ is eg ⁴ t_2g 3.

(iv) [Mn(H₂O)₆]SO₄
The central metal ion is Mn.
The coordination number is 6. The
oxidation state can be given as: x +
0 = +2 x = +2
The d orbital occupation for Mn is t_2g ³ eg ^2.

Potassium trioxalatochromate (III)

The IUPAC name for Cr(C_{​​​​​​2}​​​​O₄)₃Cl₃ is TriChloroChromoOxalate

XML stands for eXtensible Markup Language. HTML is used to create web pages / display data whereas XML is used to describe/store/transport data.

George Boole was developed Boolean algebra.

DROP TABLE:-

DROP TABLE statement is used to delete the table and all its data from the database entirely. The syntax for DROP TABLE is DROP TABLE ;

DROP VIEW:-

Removes an existing view from a database. DROP VIEW statement is used to remove a view or an object view from the database. The syntax for DROP VIEW is DROP VIEW ;

The difference between column constraint and table constraint is that column constraint applies only to individual columns, whereas table constraints apply to groups of one or more columns.

Following are the few of database integrity constrains:

• Unique constraint 
• Primary Key constraint
• Default constraint 
• Check constraint

SQL means Structured Query Language. It is the set of commands that is recognized by all RDBMS.

Data Definition Language (DDL)

Data Manipulation Language(DML)

(DML) Data Control Language (DCL)

DDL Commands 

(1) CREATE 

(2) ALTER 

(3) DROP

DML Commands 

(1) INSERT INTO 

(2) DELETE 

(3) UPDATE

DDL is short form of Data Definition Language statements are used to build and modify the structure of database, tables and other objects in the database. When you execute a DDL statement, it takes effect immediately. Some of the commands comprising DDL are CREATE TABLE, DROP TABLE and CREATE INDEX.

DML is abbreviation of Data Manipulation Language. It is used to retrieve, store, modify, delete, insert and update data in database. Examples: SELECT, UPDATE, INSERT statements. 

(b) DROP TABLE CUSTOMER

PEP 8 is a coding convention(a set of recommendations) to write your Python code in order to make it more readable and useful for those after you.

The scope defines the accessibility of the python object. To access the particular variable in the code, the scope must be defined as it cannot be accessed from anywhere in the program. The particular coding region where variables are visible is known as scope. Variables are not visible to the entire code; their visibility can be restricted. Scope verifies which variable can be ‘Seen’. The scope defines the set of rules which tell us how and where a variable can be searched. The variable is searched either to retrieve a value or for assigning value. The namespace is the unique identification of the variable or the method. Namespace tells the python interpreter about the name of the object and the location from where it is trying to access it.

The Namespaces are searched for scope resolution according to the LEGB rule. The LEGB stands for :L: Local, E: Enclosed, G: Global, B: Built-in. The sequence of LEGB is important. The variable is first searched in Local, followed by Enclosed, then global and finally built-in.

Internal commands: The internal commands are those commands that are automatically loaded in the memory. Some commonly used DOS internal commands are: -

1. Cls

Syntax :- C :/> Cls

2. Dir 

Syntax:- C : / > dir.

a. C : / > dir/p – It displays the list of directories or files page wise

     b. C: / > dir/w- It displays the list of directories or files width wise

     c. C : / > dir/d: –It display list of directories or files in drive D

     d. C : / > dir filename . extension – It displays the information of specified file.

     e. C : / > dir file name with wild cards.

3. Date 

Syntax: - C : /> date

Current date is: sat 3-09-2021

Enter of new date (mm-dd-yy):-

4. Time

Syntax : - C : / > time

Current time is 12 : 39 - 48 : 36 p

Enter new time : -

5. Ver

Syntax : - C : / > Ver

MS – Dos version 6 : 20

Copy card .

2. External commands: -  This commands are not permanent part of the memory. To execute or run this commands an external file is required.

Example : [.] Dot  exe, bat.

Some commonly used DOS external commands are .

1. CHKDSK

Syntax : - C : / > CHKDSK drive name

Eg:- C : / > CHKDSK   e :

If drive name is not mentioned by default current drive is considered.

2. Diskcopy

Syntax : - C : / > disk copy < source path > < destination path >

Ex: - c : / > diskcopy A : B :

NOTE: - This command is used after diskcopy command to ensure that disk is copied successfully.

3. Format

Syntax : - C : / > format drive name

Ex : C : / > format  A:

4. Label 

Syntax : C : / > label drive name

Ex : C : / > label  A:

5.Scandisk

Syntax : - C : / > scandisk < drive names >

 C : / > Scandisk  A :

Correct answer is

(d) EJKNFTGP

(c) Brass

Brass is least like the other words in the group.

Correct answer is

(d) South-West

Considering the given series

51, 52, 56, 83, 99, 224, 259, 603

The logic of the given series can be explained as

51 + 1³ = 52

52 + 2² = 56

56 + 3³ = 83

83 + 4² = 99

99 + 5³ = 224

224 + 6² = 260

260 + 7³ = 603

∴ Wrong term in given number series is 259

Given:

The number is 25⁹

Concept used:

Factors which are perfect cube have their power as multiple of 3.

Calculation:

To simplify 25⁹ 

⇒ 25⁹ = (5²)⁹ 

⇒ 25⁹ = 5¹⁸

⇒ 259 = (5³)⁶ 

According to question we have to find the number of factors which are perfect cube

⇒ Factors of 259 which are perfect cube are 5³, 5⁶, 5⁹, 5¹², 5¹⁵, 5¹⁸ and 1

⇒ Factors of 259 which are perfect cube are 53, 253, 1253, 6253, 31253, 156253, and 1

∴ Number of factors of 25⁹ which are perfect cube are 7.

First of all subtract 382 by 4, 509 by 5 and 636 by 6.

382-4 = 378

509-5 = 504

636-6 = 630

Then, take out the HCF of 378, 504 and 630

378 and 504

504 = 378 x 1 + 126

378 = 126 x 3 + 0

HCF (378 and 504) = 126

Now, HCF (126 and 630) = 126

The greatest number that divided these number is 126.

First of all subtract 382 by 4, 509 by 5 and 636 by 6.

382 - 4 = 378

509 - 5 = 504

636 - 6 = 630

Factor of 378, 504 and 630 is

378 = 2 x 3 x 3 x 3 x 7

504 = 2 x 2 x 2 x 3 x 3 x 7

630 = 2 x 3 x 3 x 5 x 7

Common factor = 2 x 3 x 3 x 7 = 126

H.C.F. = 126

126 is the greatest number that will divide 382, 509 and 636 leaving remainders 4, 5 and 6 respectively.

The smallest number which is divided by 35, 56 and 91 is LCM (35, 56, 91). 

Write 35, 56 and 91 in the prime factorization. 

35 = 5 × 7 

56 = 2 × 2 × 2 × 7 = 2³ × 7 

91 = 7 × 13. 

∴ LCM (35, 56, 91) = 5× 7 × 2³ × 13 = 35 × 8 × 13 = 3640. 

Hence, the smallest number which is divided by 35, 56 and 91 is 3640. 

∴ the smallest number that when divided by 35, 56 and 91 leaves the same remainder 7 

in each case = 3640 + 7 = 3647.

Given:

Numbers have exactly 3 factors

Concept used:

If the number can be expressed as xa.yb.zc, where x, y, and z are prime numbers and a, b, and c are whole numbers,

Then total number of factors of the number = (a + 1) × (b + 1) × (c + 1)

Numbers having only 3 factors can be expressed as x², here x is a prime number.

Calculation:

As the numbers have only three factors 

⇒ Numbers are of the form x², here x is a prime number

⇒ Total number of factors of the number = 2 + 1 = 3, i.e. x, x², and 1

⇒ 2² = 4, 32 = 9, 52 = 25, 72 = 49, 112 = 121, 132 = 169

⇒ Natural number which are less than 200 having 3 factors are 4, 9, 25, 49, 121, 169

∴ There are 6 natural numbers which are lass than 200 and have exactly 3 factors.

GIVEN:

The number is 24, 36, and 40

CONCEPT USED:

Prime factorization of Number is the multiplication of the Least prime number

CALCULATION:

The number is 24, 36, and 40

⇒ 24 = 2 × 2 × 2 × 3

⇒ 36 = 2 × 2 × 3 × 3

⇒ 40 = 2 × 2 × 2 × 5

⇒ Least Common Multiple = 2 × 2 × 2 × 3 × 3 × 5 = 360

∴ The L.C.M of 24, 36, and 40 is 360

Given: 

The number is 165, 246, 300, and 408

The remainder is 3 in each case

Concept: 

First, subtract 3 from each number then find the HCF of the obtained number after subtracting.

Calculation: 

165 - 3 = 162, 246 - 3 = 243, 300 - 3 = 297, 408 - 3 = 405

Now, find the HCF of 162, 243, 297, and 405

162 = 27 × 6

243 = 27 × 9

297 = 27 × 11

405 = 27 × 15

Now, we see that nothing is common among (6, 9, 11, 15) so highest common factor is 27.

∴ The required greatest number is 27.

Concept Used: 

Greatest divisor = HCF[(a – b), (b – c), (a – c)]; where a > b > c

Calculation:

⇒ greatest number = HCF [(99 – 53), (99 – 30), (53 – 30)]

⇒ greatest number = HCF [46, 69, 23] = 23

∴ Required greatest number that will divide 99, 53 and 30 so as to leave the same remainder in each case is 23

Given:

After subtraction of a certain number from the number 6846, the result will be completely divisible by 6, 8, and 12. We have to find a certain number.

Concept Used:

Concept of LCM

Calculation:

6 = 2 × 3

8 = 2 × 2 × 2

12 = 2 × 2 × 3

LCM of 6, 8 and 12 is 2 × 2 × 3 × 2 = 24

Now, dividing 6846 by 24 get the quotient = 285 and the remainder = 6

Thus, it is clear that if we subtract 6 from the number 6846 the result will be completely divisible by 24

⇒ (6846 - 6) = 6840 will be completely divisible by 24

⇒ 6840 will be completely divisible by 6, 8 and 12

∴ The required least number is 6.

GIVEN:

N = 125⁴ × 8³⁴ × 4³ × 25⁴

CONCEPT:

The number of zeroes is the number of power of 10.

CALCULATION:

N = 125⁴ × 8³⁴ × 4³ × 25⁴

N = 5¹² × 2¹⁰² × 2⁶ × 5⁸

N = 5²⁰ × 2¹⁰⁸

N = 10²⁰ × 2⁸⁸

∴ The number of zeroes of the expression 125⁴ × 8³⁴ × 4³ × 25⁴ is 20.

GIVEN:

N = 274¹²³ + 972⁷¹⁶ + 123⁴⁵⁶

CALCULATION:

N = 274¹²³ + 972⁷¹⁶ + 123⁴⁵⁶

To find the unit digit –

⇒ 274123 + 972716 + 123456

⇒ (4⁴)³⁰ ×  4³ + (2⁴)¹⁷⁹ + (3⁴)¹¹⁴

⇒ 4 + 6 + 1

⇒ 11

⇒ Unit digit, 1

∴ The unit digit in 274¹²³ + 972⁷¹⁶ + 123⁴⁵⁶ is 1.

Unit digit of a number is digits in the one's place of the number.

It is rightmost of the number. like 523417, 7 is the unit digit.

Given:

When 29, 43 and 71 divided by the number, it leaves the remainder 5, 7 and 11 respectively.

Concept Used:

Concept of HCF

Calculation:

29 - 5 = 24

43 - 7 = 36

71 - 11 = 60

Now,

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

60 = 2 × 2 × 3 × 5

HCF of 24, 36 and 60 is 2 × 2 × 3 = 12

∴ The required greatest number is 12.

GIVEN:

HCF of two numbers = 8

Their product = 5632

CALCULATION:

Since HCF of two numbers is 8.

Let numbers are ‘8a’ and ‘8b’ respectively. (‘a’ and ‘b’ are co-prime.)

Product of number = 8a × 8b = 5632

ab = 88

Possible values of ‘a’ and ‘b’ is 11 and 8 in any order.

Value of A = 8a = 88

Value of B = 8b = 64

Required sum = 88 + 64

= 152

Calculation:

Difference between first number and second number= 91 - 43 = 48

Difference between second number and third number = 183 - 91 = 92

Difference between third number and first number = 183 - 43 = 140

HCF of 48, 92, 140 = 4

∴ The greatest number that will divided 43, 91 and 183 so as to leave the same remainder in each case is 4.

GIVEN:

N = (652)¹⁹² + (183)²⁰¹ × (174)⁸²⁴

CALCULAION:

N = (652)¹⁹² + (183)²⁰¹ × (174)⁸²⁴

We have to find the unit digit –

N = (652)⁴ + (183)¹ × (174)⁴

Unit digit of N = (2)⁴ + (3)¹ × (4)⁴

Unit digit of N = 6 + 3 × 6

Unit digit of N = 6 + 18

Unit digit of N = 24

Unit digit of N = 4

Given:

A number is divided by 126 leaves remainder 43

Concept used:

Number = Divisor × Dividend + Remainder

Calculation:

Let the number be 126x + 43

126x + 43 when divided by 21

⇒ (126x + 43)/21

⇒ 126x/21 + 43/21

⇒ 6x + (42 + 1)/21

⇒ 6x + 42/21 + 1/21

⇒ 6x + 2 + 1/21

Remainder = 1

∴ If we divide 126x + 43 by 21, we get 1 as a remainder.

Given:

Mitesh’s profit = 20%

Ritesh’s profit = 10%

Ritesh’s C.P = Rs. 3960

Calculations:

Let Mitesh’s C.P be x.

∴ According to the question,

⇒ x × 120/100 × 110/100 = 3960

⇒ x = (3960 × 10 × 10)/(12 × 11)

⇒ x = 3000

Given:

Percent profit on 20% of commodity = 25%

Percent profit on 10% of commodity = 40%

Percent profit on 70% of commodity = 30%

Total profit = 180

Concept used:

Percent profit / loss = (profit / loss) / C.P × 100

Calculation:

Profit and loss is always calculated on C.P

Let value of the commodity = Rs. X

C.P of 20% of it = 20% of X = X / 5

C.P of 10% of it = 10% of X = X / 10

C.P of 70% of it = 70% of X = 7X / 10

Profit on selling 20% of it = 25% of X / 5 = X / 20,

Similarly, profit on 10% = 40% of X / 10 = X / 25

And profit on 70% = 30% of 7X / 10 = 21X / 100

Total profit = X / 20 + X / 25 + 21X / 100 = 30X / 100

∵ 30X / 100 = 180

⇒ X = 180 × 100 / 30 = 600

GIVEN:

Two – third of a number P is equal to the four – seventh of another number Q.

R is nine – fourteenth of Q.

Difference of P and R is 12.

CALCULATION:

Two – third of a number P is equal to the two – seventh of another number Q. R is three – fourth of P

2P/3 = 4Q/7

7P = 6Q

P : Q = 6 : 7

R = 9Q/14

Q : R = 14 : 9

P : Q : R = (6 × 14) : (7 × 14) : (7 : 9) = 12 : 14 : 9

Let P = 12k, Q = 14k and R = 9k

12k – 9k = 12

k = 4

Now,

= (2/7) × (12k + 14k + 9k)

= (2/7) × 140

= 40

Given:

S.P = Rs. 6800

Loss% = 75%

Formula used:

C.P = S.P × [100/(100 – Loss%)]

Calculation:

C.P = Rs. 6800 × (100/25)

⇒ Rs. 27,200

∴ The cost price of article is Rs. 27,200

GIVEN:

Three statements.

CALCULATION:

A:

One root = (5 + 2i)

We know that complex root always exists in pair.

Another root = (5 – 2i)

Product of roots = (5 + 2i) (5 – 2i) = (5)² – (2i)² = 25 – 4i² = 25 + 4 = 29

B:

523 = 30 × 17 + 13

Remainder when 523 is divided by 17 = 13

C:

Sum of fraction (11/7) and its reciprocal = (11/7) + (7/11)

= (121 + 49)/77

= 170/49

Hence, none are TRUE. 

Given :

A sold a watch to B at a profit of 20%.

B sold it to C at 30% profit

C sold it to D at 10% loss

B's profit is Rs. 80 more than that of A

To find :

At what price D bought.

Formula used :

Selling price = cost price + profit

Selling price = cost price - loss

Solution :

Let A purchased at Rs. 100

⇒ A sold to B at a 20% profit = 100 × (120/100)

⇒ A sold to B at a 20% profit = 120

⇒ A's profit = 120 - 100

⇒ A's profit = 20     ----(1)

⇒ B sold to C at a 30% profit = 120 × (130/100)

⇒ B sold to C at 30% profit = 156

⇒ B's profit = 156 - 120

⇒ B's profit = 36     ----(2)

⇒ C sold to D at a 10% loss = 156 × (90/100)

⇒ C sold to D at a 10% loss = 140.4

From (1) and (2)

The difference in profit of B and A

⇒ B's profit - A's profit = 36 - 20

⇒ B's profit - A's profit = 16

So. 16x = 80

⇒ x = 5

⇒ D bought it for 140.4 × 5

∴ D bought it for Rs. 702.

Given:

Number of articles brought = 6

The cost price of 6 articles = Rs. 8

Number of articles sold = 10

The selling price of 10 articles = Rs. 12

Concept:

First, equal the number of articles sold and brought so that the cost price of the number of articles sold can be determined. Then calculate the profit or loss percentage.

Formula used:

Loss % = [(CP - SP)/CP] × 100

Calculation:

Taking LCM of the number of articles sold and brought;

We get 30.

CP of 30 articles = 8 × 5 = Rs 40

SP of 30 articles = 12 × 3 = Rs. 36

Loss % = [(40 - 36)/40] × 100

⇒ (4/40) × 100

⇒ 100/10

⇒ 10% 

∴ His percentage of loss is 10%

Given:

The cost price of an article = Rs. 400

Loss% of one article is 10%

Gain % in whole transaction = 20% 

Concept Used:

Selling price = Cost price + Profit

Selling price = Cost price - loss

Calculations:

Selling price of an article with 10% loss = 400 - 400 × 10%

⇒ 400(1 -10%) = 400(1 - 10/100)

⇒ 400(100 - 10)/100 = 4 × 90 

⇒ 360

The total cost price of 3 articles = 3 × 400

⇒ 1200

Overall transaction gain = 20%

⇒ Overall transaction selling price should be (120/100) × 1200 = 1440

Selling price of other two articles should be (1440 - 360)/2 = 540

Gain from other two article individually is 540 - 400 = 140

Gain percentage = (140/400) × 100 = 35%

∴ The required gain percentage on two article is 35%

(d) Discount = 6272 – 5958.4 = 313.6