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(C₆H₅)₃P.RhCl

Pentaaquasulphatochromium (III) bromide.

Manufacture of K₂Cr₂O₇ from chromite ore involves the following steps 

Step-1: Conversion of chromite ore into sodium chromate: Chromite ore is heated with Na₂CO₃ in the presence of excess of air to form sodium chromate 

4FeOCr₂O₃ + 8Na₂CO₃ + 7O₂→ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ 

Step-2: Conversion of sodium chromate into sodium dichromate. 

The yellow solution of sodium chromate is filtered and acidified with H₂SO₄ to give a mixture of Na₂Cr₂O₇ and sodium sulphate 

2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O 

The less soluble Na₂SO₄ is removed by filtration. 

Step-3: Conversion of sodium dichromate into potassium dichromate: The solution of sodium dichromate is treated with KCl to give potassium dichromate. 

Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Postulates: 

1. Central metal ion in a complex shows two types of valences primary valence and secondary valence. 

2. The primary alence is ionisable and satisfied by negative ions. 

3. The secondary valence is non ionisable. It is equal to the coordination number of the central metal ion or atom. It is fixed for a metal. Secondary valences are satisfied by negative ions or neural molecules (ligands). 

4. The primary valence is non directional. The secondary valence is directional. Ions or molecules attached to satisfy secondary valences have characteristic spatial arrangements. Secondary valence decides geometry of the complex compound.

Penta ammine bromido cobalt (111) sulphate [CO(NH₃)₅SO₄]Br

(1) 2, 1

4d orbital

Angular nodes = l = 2

Radial nodes = (n - l - 1)

 = 4 - 2 - 1 = 1

(d) ionic solid with [PCl₄]+ tetrahedral and [PCl₆]– octahedra

[Hint : Because down the group E – H bond dissociation enthalpy decreases.]

Let the radius of the base of original cylinder is r. 

And the height of the original cylinder is h. 

∴ The volume of the original cylinder is \(\pi\)r²ℎ. 

Given that radius of base of reduced cylinder is half of radius of original cylinder and height will be same. 

∴ The radius of reduced cylinder is \(\frac{r}2\)

And the height of the reduced cylinder is h. 

∴ The volume of the reduced cylinder is \(\pi\)\((\frac{r}{2})^2\) ℎ. 

Now, the ratio of the volume of reduced cylinder and the volume of original cylinder is

\(\frac{Volume\,of\,reduced\,cylinder}{Volume\,of\,original\,cylinder}\) = \(\frac{\pi(\frac{r}2)^2h}{\pi{r}^2h}\) = \(\frac{\pi{r}^2h}{4\pi{r}^2h}\) = \(\frac{1}4\).

Hence, the required ratio is 1 : 4.

Given:

The ratio of milk and water in the first vessel = 3 : 5

The ratio of milk and water in the second vessel = 3 : 2

The ratio of milk and water in the final mixture = 2 : 3

Concept used:

(New quantity of milk/New quantity of water) = (Quantity of milk taken from first vessel + Quantity of milk taken from the second vessel)/(Quantity of water taken from first vessel + Quantity of water taken from the second vessel)

Calculation:

Let the mixture from the two-vessel be mixed in the ratio x : y.

Quantity of milk taken from the first vessel = 3x/8

Quantity of water taken from the first vessel = 5x/8

Quantity of milk taken from the second vessel = 3y/5

Quantity of water taken from the second vessel = 2y/5

Then, {(3x/8) + (3y/5}/{(5x/8) + (2y/5)} = 2/3

⇒ {(15x + 24y)/40}/{(25x + 16y)/40} = 2/3

⇒ 3(15x + 24y) = 2(25x + 16y)

⇒ 45x + 72y = 50x + 32y

⇒ 50x – 45x = 72y – 32y

⇒ 5x = 40y

⇒ x/y = 8/1

Given:

Total amount which is divided among A, B and C = Rs. 6,300

X : Y = 7 : 5 and Y : Z = 4 : 3

Calculation:

X : Y = 7 : 5     ----(1)

Y : Z = 4 : 3     ----(2)

Multiply by 4 in equation (1) and multiply by 5 in equation (2)

X : Y : Z = 28 : 20 : 15

Ratio of X : Y : Z = 28x : 20x : 15x

According to the question

28x + 20x + 15x = 6,300

⇒ 63x = 6,300

⇒ x = 6,300/63

⇒ x = 100

∴ Share of Y = 20 × 100 = Rs. 2,000

Given

Total sum of money = Rs. 4758

The sum is divided among A, B, C and D in ratio :

\(\frac{1}{2}:1:\frac{1}{4}:\frac{1}{5}\)

Calculation

\(\frac{1}{2}:1:\frac{1}{4}:\frac{1}{5}\)

LCM of 5, 4, and 2 is 20, Multiplying all ratios by 20 we get,

10 : 20 : 5 : 4

So, the equivalent ratios are 10 : 20 : 5 : 4

Total amount = 4758

Share of B = (20/39) × 4758

⇒ Rs. 2440

Given:

A : B : C = 3 : 7 : 6

Calculation:

Sum amount of A, B, C

⇒ 3 + 7 + 6

⇒ 16 units

⇒ 16 units = 768

⇒ 1 unit = 48

Share of B

⇒ 48 × 7

⇒ 336

∴ The share of B is 336.

Given:

\(\frac{8x + 3y}{4x + 5y} = \frac{25}{23}\)

Calculation:

\(\frac{8x + 3y}{4x + 5y} = \frac{25}{23}\)

⇒ 23 (8x + 3y) = 25 (4x + 5y)

⇒ 184x + 69y = 100x + 125y

⇒ 184x – 100x = 125y – 69y

⇒ 84x = 56y

⇒ 3x = 2y

\( ⇒ \frac{x}{y} = \;\frac{2}{3}\)

⇒ x : y = 2 : 3

∴ x : y is 2 : 3

Given:

Vessel A ratio of Milk and water is 3 : 4

Vessel B ratio of Milk and water is 2 : 3

vessel A and B are mixed in the ratio of 1 : 2 respectively.

Calculation:

Let the ratio of milk and water in vessel A be 3x : 4x

So, the total quantity of vessel A is 7x

And the ratio of milk and water in vessel B be 2x : 3x

So, the total quantity of vessel B is 5x

To Mix into the ratio first we need to make quantity equal

So, multiply vessel A by 5 and multiply vessel B by 7

⇒ Vessel A = 15x : 20x

⇒ Vessel B = 14x : 21x

Now mix them in the ratio of 1 and 2

The final ratio of milk and water will be

⇒ milk : water = 15x + 28x : 20x + 42x

⇒ milk : water = 43 : 62

∴ The ratio of milk and water in the final mixture is 43 : 62.

Given: 

\(\frac{a}{b} = \frac{7}{9}, \frac{b}{c} = \frac{3}{5}\) 

Calculation:

a/b = 7/9

Let the value of a be 7x and b be 9x.

⇒ b/c = 3/5

Let the value of b be 3y and c be 5y.

We know that value of b will be same.

⇒ 9x = 3y

⇒ x/y = 1/3

Put x = 1 and y = 3

⇒ a = 7x =7, b = 9x = 9, c = 5y = 15

∴ The value of a : b : c = 7 : 9 : 15

Given:

B's share = (2/3) of C's share

C's share = (9/13) of D's share

Ratio of share of A to B = 8 : 9

Concept used:

A's share + B's share + C's share + D's share = 4200

Calculation:

Let the share of D is x. 

Then, C's share = 9x/13

B's share = 18x/39

A's share = {(18x/39)/9} × 8

⇒ 16x/39

According to question, (16x/39) + (18x/39) + (9x/13) + x = 4200

⇒ (16x + 18x + 27x + 39x)/39 = 4200

⇒ 100x/39 = 4200

⇒ 100x = 163800

⇒ x = 1638

A's share = (16 × 1638)/39

⇒ 26208/39

⇒ 672

B's share = (18 × 1638)/39

⇒ 29484/39

⇒ 756

C's share = (9 × 1638)/39

⇒ 14742/39

⇒ 378

D's share = 1638

Difference between B's share  and D's share = 1638 - 756

⇒ 882

∴ The difference between D's share and B's share is Rs. 882.

Given:

2/9 × A = 4/9 × B = 6/11 × C

Calculation:

 A = 2 × B, C = 22/27 × B

A : B : C = 2B : B : 22/27B

A : B : C = 54 :  27 : 22

Now,

A : B : C = 54 :  27 : 22

⇒ A = 54p, B = 27p, C = 22p

Difference between  A's share and C's share is = (8755/103p) × (54p - 22p)

⇒ 2720

∴ The required result will be 2720.

Given:

A and B work together can do a piece of work in 12 days

B alone can finish it in 30 days

Calculation:

Work done by (A + B)'s in 1 day = 1/12 days

Work done by B's in 1 day = 1/30 days

Work done by A's in 1 day = (1/12 – 1/30)

⇒ (5 – 2)/60 = 1/20

∴ A alone can finish the work in 20 days.

In your day-to-day life, you must have seen several kinds of tables consisting of numbers, figures, names, etc. These tables provide ‘Data’. Data is a collection of numbers gathered to give some information. They are collected from a variety of sources.

Data representation and Data interpretation: Data Interpretation should be done honestly and integrally without any biased manipulation of data.  This means that you need to report your research honestly and that this applies to your methods, your data, your results, and whether you have previously published any of it. One should not make up any data, including extrapolating unreasonably from some of your results or do anything which could be construed as trying to mislead anyone. It is better to undersell than exaggerate your findings. You should aim to avoid bias in any aspect of your research, including design, data analysis, interpretation, and peer review. 

The following techniques can be beneficial for data representation and data interpretation-

• Direct Personal Investigation-
  • know the local conditions, customs, and tradition
• Indirect Oral Investigation-
  • information is collected hm a witness or from a third party who are directly or indirectly related to the problem and possess sufficient knowledge.
• Use of Local Reports-
  • use of local newspaper, magazines, and journals by the investigators
• Questionnaire method- 
  • It involves the preparation of a list of questions relevant to the inquiry and presenting them in the form of a booklet, often called a questionnaire

Thus from the above-mentioned points, it is clear that Debate is not an appropriate activity for 'data representation and data interpretation'.

Given:

X/Y = 12/9

Calculations:

Let the value of X be 12k and the value of Y be 9k, where k is constant.

(X² + Y²)/(X² – Y²)

⇒ k²(12² + 9²)/k²(12² – 9²)

⇒ (144 + 81)/(144 - 81)

⇒ 225/63

∴ (X² + Y²)/(X² – Y²) is 225/63 

Many rules taught in mathematics classrooms “expire” when students develop knowledge that is more sophisticated, such as using new number systems. For example, in elementary grades, students are sometimes taught that “addition makes bigger” or “subtraction makes smaller” when learning to compute with whole numbers, only to find that these rules expire when they begin computing with integers.

More recently, a fifth rule was proposed by Simundza in a laboratory course for precalculus. This is the fifth rule, Experiential is in the words of Simudza, a "direct sensory experience of qualitative phenomena". 

Basic Arithmetic Operations: The four basic arithmetic operations in Maths are Addition, Subtraction, Multiplication, Division. The fifth rule of mathematics is the rule for the application of general knowledge.

• Addition: The addition is a mathematical process of adding things together. The addition process is denoted by the ‘+’ sign. It involves combining two or more numbers into a single term
• Addition Rules
  • The addition of two positive integers is a positive integer
  • The addition of two negative integers is a negative integer
  • While adding positive and negative integer, subtract the integers and use the sign of the largest integer number
• Subtraction: It gives the difference between two numbers. Subtraction is denoted by the ‘-‘ sign. It is almost similar to addition but is the conjugate of the second term.
• Subtraction Rules
  • If both the signs of the integers are positive, the answer will be the positive integer
  • If both the signs of the integers are negative, the answer will be the negative integer
  • If the signs of the integers are different, subtract the values, and take the sign from the largest integer value.
• Multiplication: It is known as repeated addition. It is denoted by ‘×’ or ‘*’. It also combines with the two or more values to result in one a single value
• Multiplication Rules
  • The product of two positive integers is a positive integer.
  • The product of two negative integers is a positive integer.
  • The product of positive and negative integers is a negative integer.
• Division: The division is usually denoted by ‘÷‘ and is the inverse of multiplication. It constitutes two terms dividend and divisor, where the dividend is divided by the divisor to give a single term value.
• Division Rules
  • The division of two positive integers is a positive integer
  • The division of two negative number is a positive integer
  • The division of integers with different signs results in the negative integer

Thus from the above-mentioned points, it is clear that the fifth rule of mathematics is the rule for the application of general knowledge.

Given:

A’s time = 2 × time taken by B = 3 × time taken by C

And together they complete work in 1 day

Concept used:

Days is inversely proportional to efficiency

If A does a work in x days and B does a work in y days, the ratio of days = x : y and

The ratio of efficiency = y : x

If a person does a work in ‘n’ days then one day work by a person will be 1/n part of total work

Calculation:

A’s time = 2 × taken by B

⇒ A’s time/B’s time = 2/1      ----(i)

A’s time = 3 × time taken by C

⇒ A’s time/ time taken C = 3/1     ----(ii)

After multiplying equation(i) by 3 and equation (ii) by 2 we get, Ratio of time is

⇒ A : B : C = 6 : 2 : 3

And Efficiency ratio = (1/6) : (1/2) : (1/3)

⇒ Efficiency ratio = 1 : 3 : 2

Total work = sum of efficiency × days

⇒ (1 + 3 + 2) × 1

⇒ 6 × 1

⇒ 6

A alone can complete the work in= total work/ efficiency of A

⇒ 6/1

⇒ 6 days.

∴ A can alone complete the work in 6 days.

Given:

(a + b - c) : (b + c - a) : (a + c - b) = 6 : 5 : 7

Concept used:

Ratio and proportion

Calculation:

Let,

a + b - c = 6k

b + c - a = 5k

a + c - b = 7k

Adding all the above three equations we get,

⇒ a + b + c = 18k

Adding (a + b + c) + (a + b - c) = 18k + 6k = 24k

⇒ a + b = \(\frac{{24k}}{2}\) = 12k

Adding (a + b + c) + (b + c - a) = 18k + 5k = 23k

⇒ b + c = \(\frac{{23k}}{2}\)

We have a + b + c = 18k and a + b = 12k 

⇒ (a + b + c) - (a + b) = 18k - 12k

∴ c = 6k

Where b + c = \(\frac{{23k}}{2}\)

⇒ b = \(\frac{{23k}}{2}\) - 6k

∴ b = ​\(\frac{{11k}}{2}\)​

We have a + b = 12k

⇒ a = 12k - \(\frac{{11k}}{2}\)

∴ a = \(\frac{{13k}}{2}\)

So we have a = \(\frac{{13k}}{2}\), b = \(\frac{{11k}}{2}\), c = 6k​​

⇒ \(\frac{{132\ :\ 156\ :\ 143}}{{13 \times 11 \times 6}}\)

∴ Ans :- 132 : 156 : 143

Methods are the ways/styles of the transaction of content in the classroom for effective learning. In the learning- centred approaches the focus is how students construct their own knowledge on the basis of their prior experience. 

Project Work in Mathematics-

• The project-based learning is a learner-centred method in which the students are challenged to do something by themselves outside the realm of normal classwork.
• Project-based learning is an individual or group activity that goes on over a period of time, resulting in a product, presentation, or performance.
• A project of mathematics contains rich activities, active participation, freedom to students and correlation with other subjects. 
• It promotes inquiry skills, enhances problem-solving skills and establishes interdisciplinary linkages.
• The project based learning followed the following steps- 
  • Providing situations
  • Choosing and purposing
  • Planning for the project
  • Executing the project
  • Judging the project
  • Recording the project

Thus from above-mentioned points, it is clear that Projects make scoring easy in Mathematics is not true.

Mathematics is the study of numbers, shape, quantity, and patterns. The nature of mathematics is logical and it relies on logic and connects learning with learners' day-to-day life.

• Teaching methods of mathematics include problem-solving, lecture, inductive (induction), deductive, analytic, synthetic, heuristic and discovery method. Teacher adopts any method according to the needs and interests of students.

Induction Method: 

Induction approach is based on the process of induction. It is a method of constructing a formula with the help of a sufficient number of concrete examples. 

• Induction means to provide a universal truth by showing, that if it is true for a particular case.
• It starts from examples and reach towards generalizations.
• Example: Square of an odd number is odd and the square of an even number is even.
• Inductive approach proceeds from-
  • Particular to general
  • Known to unknown
  • Simple to complex
  • Example to formula

Hence, it could be concluded that the appropriate method for the establishment of the formulae in mathematics is induction method.

Deductive Method:

• The deductive method is based on deduction. At first, the rules are given and then students are asked to apply these rules to solve more problems.
• It begins with general premises and through logical argument, comes to a specific conclusion.

Analysis Method:

• It is a generic process combining the power of the Scientific Method with the use of a formal process to solve any type of problem.
• Analysis means breaking up into components. All scientifically based problem-solving approaches use the analytical method.

Synthesis Method: 

• The word “synthetic’ is derived from the word ‘synthesis which means to combine. 
• In this method, we combine several facts, perform cer­tain mathematical operations, and arrive at the solution.

Giftedness refers to talent. The gifted students are essentially exceptional and generally show consistency in their performances and exhibit superiority in general intelligence. 

•  A gifted child may or may not have a very high I.Q. and giftedness can be in one area and more than one area. They have great intellectual curiosity, creativity, imagination, and willingness to create and investigate.

• All the gifted students have some special needs to fulfill their curiosity and to enhance their creativity.
• Enrichment provides more opportunities to gifted child to go deeper or to range more widely than the average child in his intellectual, social, and artistic experience.
• Enrichment programme can be done by:
  • emphasizing the creative and other experimental work that will help the gifted students to learn and discover on their own.
  • co-operative planning and activity that provides an opportunity for leadership training and experiences in social adjustment.
  • the development of self-study habits by giving them assignments that requires brainstorming and other cognitive skills.
  • providing rich and efficient and prompt library service that will help them to garnish their knowledge and thought process in the long run.
  • acceleration of class promotion that refers to moving higher-ability children to classes beyond their age group yet sharing similar interests and abilities.

Hence, it could be concluded that the provision of supplementary reading materials, reference books and general literature is not an enrichment program for gifted students in Mathematics.

Formula Used:

The n^th term of an AP is given by:

a_n = a + (n – 1) × d      ----(i)

where a_n = n^th term of the AP,

a = first term of the AP,

n = number of terms within the AP,

d = difference between successive terms in the AP.

Calculation:

The numbers between 1 and 99, which are divisible by 3, are:

3, 6, 9,......., 99.

Using the equation (i), we get:

99 = 3 + (n – 1) × 3

⇒ n = 33

So, the remaining terms, not divisible by 3 = 99 – 33 = 66

Among these 66 terms, 5 appears once each in the following terms:

5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 56, 57, 58, 59, 65, 75, 85, and 95.

But, 15, 45, 51, 54, 57 and 75 are divisible by 3, and hence excluded.

So, the total number of remaining terms with 5 appearing only once = 18 – 6 = 12.

And appears twice in the term 55.

So, the total number of times the digit 5 appears = 12 + (1 × 2) = 14

∴ The total number of times 5 appears between 1 to 99, when excluding the numbers divisible by 3, will be 14

Correct answer is: (c) 2³X3³

Correct answer is: (a) Terminating

Correct answer is: (c) 1

CosƟ = I-cos²Ɵ = sin²Ɵ 

Therefore 

Sin²Ɵ + sin⁴Ɵ = cosƟ + cos²Ɵ = 1