IQOM 2025 Syllabus

the syllabus for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, here's a detailed breakdown of the topics typically covered:

The IOQM syllabus is comprehensive and covers advanced topics from five major areas of mathematics, generally aligning with Class 8 to 12 level, but with an Olympiad-level rigor:

1. Basic Mathematics:

 * Number Systems (integers, rational, real, complex numbers)

 * Basic Inequality Concepts (AM-GM, Cauchy-Schwarz, Rearrangement)

 * Modulus & Greatest Integer Functions

 * Logarithmic Functions

2. Number Theory:

 * Prime Numbers, Prime Factorization

 * Divisibility Rules, GCD, LCM, Euclidean Algorithm

 * Modular Arithmetic (Congruences, Chinese Remainder Theorem)

 * Diophantine Equations (Linear & Pell's)

 * Number Bases (Binary, Octal, Hexadecimal)

 * Arithmetic Functions (Euler's Totient \phi, Möbius \mu, \sigma Functions)

 * Fermat's Little Theorem and Euler's Theorem

3. Algebra:

 * Simplification & Factorization of Algebraic Expressions

 * Inequalities (AM-GM, Cauchy-Schwarz, Rearrangement)

 * Polynomials: Vieta's Formulas, Newton's Identities, Roots, Fundamental Theorem of Algebra, Factor Theorem, Rational Root Theorem

 * Complex Numbers: Operations, Roots of Unity, De Moivre's Theorem

 * Sequences & Series (AP, GP, Convergence, Infinite Series)

 * Functional Equations (Cauchy, Jensen, etc.)

 * Binomial Theorem and Combinatorics

 * Quadratic Equations and Expressions

 * System of Linear Equations

4. Combinatorics:

 * Counting Principles (Multiplication, Addition, Inclusion-Exclusion)

 * Permutations and Combinations

 * Pigeonhole Principle

 * Recurrence Relations (Homogeneous, Non-Homogeneous)

 * Principle of Inclusion and Exclusion (PIE)

 * Graph Theory: Basics, Trees, Eulerian & Hamiltonian Paths, Graph coloring, Connectivity

 * Combinatorial Geometry & Theorems

 * Generating Functions

 * Combinatorial Identities (Catalan Numbers, Hockey Stick, Vandermonde's)

5. Geometry:

 * Euclidean Geometry: Points, lines, planes, angles, triangles (congruence, similarity), quadrilaterals, circles (tangents, secants, angles, theorems), polygons

 * Geometric Transformations: Reflection, rotation, translation, dilation, isometries, similarities

 * Coordinate Geometry: Distance formula, slope and equations of lines, circles, conic sections

 * Trigonometry: Sine, cosine, tangent, and their properties; trigonometric identities and equations; applications in geometry

 * Locus problems

 * Symmetry and tessellations

The IOQM exam aims to test problem-solving skills, mathematical reasoning, and creativity. While the syllabus broadly aligns with the NCERT curriculum for Classes 8-12, the problems are often more challenging and require a deeper understanding and application of concepts.

2025 is perfect square year

2025 is a perfect square year and this rare and fascinating mathematical milestone makes the year 2025 remarkable. The total sum of its first two digits and last two digits (20 + 25) square is 2025 - a perfect square of 45 (45² = 2025) and it’s also the product of two squares (9² x 5² = 2025). Furthermore, it’s the sum of three squares (40² + 20² + 5² = 2025). On top of that, 2025 is the sum of the cubes of digits from 1 to 9 (1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ + 9³ = 2025) and 2025 is the square of the sum of the first nine positive integers (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)² = 2025. We thought it is a beautiful reminder of Nicomachus's theorem (the square of the nth triangular number is equal to the sum of the first n cubes). 1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ + 9³ = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)² = 2025 Perfect squares are always positive because a negative number multiplied by a negative number results in a positive number and 2025 marks the first perfect square year since 1936. Till now, there have been 45 perfect square years in total. The previous perfect square year was 1936 (44²), which occurred 89 years ago. Before that, the perfect square year was 1849 (43²), which happened 87 years earlier (44²) and it’s also the only perfect square year this generation will experience, as the next perfect square year will be in 2116 (46²), 91 years from now. The sequence of perfect square years began back in 1 AD. As the number grows larger, the gaps between perfect square years also get bigger, making each one feel even more significant. Let’s make 2025, a mathematical marvel, a year to remember!

Using a Clinometer to Measure Height

In this Instructable, We'll show how to use a clinometer to measure the height of a tall object 

What we need;

Clinometer

Tape measure

Paper

Pen or pencil

Assistant

Step 1: Pick a Spot

Pick a spot to measure your object (I measured a telephone pole). You should be far enough away from your object that you can see the top of it, and you need to be on level ground with the base of the object. I like to set something down by my feet once I've picked my spot, so that I can easily come back to it.

Step 2: Measure Angle

Here's where we bust out our handy clinometer. Look through the straw of your clinometer at the top of the light pole (or whatever object you're measuring). The weighted string should hang down freely, crossing the protractor portion of the clinometer. Read the angle shown, and subtract from 90° to find your angle of vision from your eye to the top of the pole (it can be helpful here to have an assistant to read the measurement while you look through the straw). Record your results on your paper.

From my spot, my clinometer (read by my assistant) showed 55°. Subtracting from 90°, that indicated that I looked at an angle of 35° to the top of the telephone pole.

Step 3: Measure Distance

Once you have your angle of vision, use your tape measure to find the distance from the spot you're standing to the base of the object you're measuring (an assistant comes in handy here, too). We must know how far away you are to accurately calculate the height.

My spot was 15.6 meters from the base of the telephone pole I measured.

Step 4: Find Your Eye-height

The last piece of data you need to calculate the height of your object is the height from the ground to your eye (your eye-height). Have your assistant help you measure this using your tape measure.

My eye height was recorded for this example as 1.64 meters.

Step 5: Draw a Picture

Time to move inside. In calculating the height of the object you just measured, I find it helpful to begin by drawing a picture and labeling it with all of the information I have.

Step 6: Model As a Triangle

The next step is to simplify your drawing to model your system as a right triangle. Label your triangle with the angle you read on your clinometer as well as the distance you were standing from the object (we don't need the eye-height just yet).

Step 7: Solve for X

We can find x in this triangle (which represents the portion of the height from eye-level up) by using some basic trigonometry, specifically the tangent ratio of the triangle:

tan(angle) = x / distance

Multiply by the distance on both sides and you get:

x = tan(angle) * distance

Use a calculator to multiply these together and get a decimal value (be sure your calculator is in 'degrees' mode, rather than 'radians'!).

In my example:

tan(35°) = x / 15.6

x = tan(35°) * 15.6

x = 10.92 meters

Step 8: Combine With Eye Height

To find the height of your object, bring this x value back to the original drawing. By labeling it, we can see that the height of the object, h, is equal to the x value we just found plus the eye-height we measured earlier:

h = x + (eye-height)

In my example:

h = 10.92m + 1.64m

h = 12.56m

There you have it! A few basic classroom materials and a little bit of trigonometry and you can measure the height of anything around you!

Deleted portion of class 12 maths exercise-wise

Chapter 1 – Relations and Functions

Deleted topics – Composite functions, Inverse of a function.

Exercise 1.3 & 1.4 (Deleted)

Miscellaneous – Q1, Q3, Q9, Q11, Q12, Q13, Q14, Q15, Q18, Q19 (Deleted)

Chapter 2 – Inverse Trigonometric Functions

Exercise 2.2 – Q1 to Q4, and Q12 to Q15 (Deleted)

Miscellaneous – Q3 to Q8, Q12, 13, 14, 16 (Deleted)

Chapter 3 – Matrices

Exercise – 3.4 (Deleted)

Chapter 4 – Determinants

Exercise – 4.2 (Deleted)

Miscellaneous – Q2, 4, 6, 11, 12, 13, 15, 17 (Deleted)

Chapter 5 – Continuity and Differentiability

Exercise – 5.8 (Deleted)

Chapter 6 – Applications of Derivatives

Exercise – 6.3, 6.4 (Deleted)

Miscellaneous – Q1, 4, 5, 20, 21, 22, 23, 24 (Deleted)

Chapter 7 – Integrals

Exercise – 7.8 (Deleted)

Miscellaneous – Q40 (Deleted)

Chapter 8 – Applications of the Integrals

Exercise – 8.2 (Deleted)

Miscellaneous – Q2, 6, 7, 10, 12, 13, 14, 15, 18, 19

Chapter 9 – Differential Equations

Exercise – 9.3 (Deleted)

Miscellaneous – Q3, 5

Chapter 10 – Vectors (No Deletion)

Chapter 11 – Three-dimensional Geometry

Exercise – 11.3 (Deleted)

Miscellaneous – Q8 to Q 23 (Except Q9 and Q 20) except 2 questions, Q8 to 23 are deleted.

Chapter 12 – Linear Programming

Exercise – 12.2 (Deleted)

Miscellaneous Exercise (Deleted)

Chapter 13 – Probability

Exercise – 13.4 (Questions are mixed) We don’t have to calculate variance.

Exercise – 13.5 (Deleted)

Miscellaneous – Q4, 5, 6,9, 10

Nishtha 2.0 Diksha Portal Module 1 Quiz 2021 Answer

Which of the following Statement is not Valid for Curriculum.

Answer – It Provides the Vision for education at various Levels in India.

2. Ram is Student with Visual Impairment which of the following would be most useful for orientation and mobility training from him ?

Answer – Tactile maps of the School Building

3. Which of the following is not important for Promoting Inclusive Education

Answer – Socio – Economic Status of Teacher

4. Inclusive Education is based on the Principal of —

Answer – World Brotherhood

5. Major Concern of School Curriculum is __

Answer – Holistic Development of Children

6. Right of Children to Free and Compulsory Education (RTI) Act provides that every

Answer – Between the age of 6 – 14 years and till completion of Elementary

7. If a Student ask a question in the classroom, then it would be beneficial to

Answer – Answer the Question & Encourage asking more questions.

8. UDL works best with

Answer – Diverse group of students

9. A Syllabus ______

Answer – Provides class wise and subject wise list of themes and topics.

10. Two Children Having the same Disability ______________

Answer – May require different interventions by the teacher in the classroom

11. The National Education Policy 2020 (NEO 2020) Proposes Inclusion Fund for children

Answer – Gender Identities

12. Hidden Curriculum

Answer – May not be part of Formal course of Study.

13. Right of Children to Free and compulsory Education (RTE) Act was enacted and into force in the year.

Answer – Enacted in the year 2009 and came into force from April 2010

14. All of these are UDL principals Except

Answer – Multiple means of Behavior management.

15. The NEP 2020 Proposes to change existing design of School education from 10 + 2 to :-

Answer – 5+3+3+4

16. The First and second National Policies on Education in India were formulated in the

Answer – 1968 and 1986

17. Which of the following is not importance for promoting inclusive education

Answer – Socio – Economic Status of Teacher

18. Which of the following is not important for creation of inclusive classroom

Answer – Location of the school

19. Which of the following statement is not valid for Curriculumn

Answer – It provides the Vision for Education at Various levels in India.

20. Which of the following state area true for National Curriculum are

Answer – All

Principal of mathematical induction

Q1. n(n+1)(n+5) is a multiple of 3.

Proof:-

 We will prove it by using the formula of mathematical induction for all n ϵ N

Let P(n)=n(n+1)(n+5)=3d where d ϵ N

For n=1

P(1)=1(2)(6)=12 which is divisible by 3

Let P(k) is true 

P(k)=k(k+1)(k+5)=3m where m ϵ N

Since P(k+1) is true whenever P(k) is true.

So, by the principle of induction, P(n) is divisible by 3 for all n ϵ N

National Mathematics Day

Srinivasa Ramanujan, (born December 22, 1887, Erode, India—died April 26, 1920, Kumbakonam), Indian mathematician whose contributions to the theory of numbers include pioneering discoveries of the properties of the partition function.

When he was 15 years old, he obtained a copy of George Shoobridge Carr’s Synopsis of Elementary Results in Pure and Applied Mathematics, 2 vol. (1880–86). This collection of thousands of theorems, many presented with only the briefest of proofs and with no material newer than 1860, aroused his genius. Having verified the results in Carr’s book, Ramanujan went beyond it, developing his own theorems and ideas. In 1903 he secured a scholarship to the University of Madras but lost it the following year because he neglected all other studies in pursuit of mathematics.

Ramanujan continued his work, without employment and living in the poorest circumstances. After marrying in 1909 he began a search for permanent employment that culminated in an interview with a government official, Ramachandra Rao. Impressed by Ramanujan’s mathematical prowess, Rao supported his research for a time, but Ramanujan, unwilling to exist on charity, obtained a clerical post with the Madras Port Trust.

Ramanujan’s knowledge of mathematics (most of which he had worked out for himself) was startling. Although he was almost completely unaware of modern developments in mathematics, his mastery of continued fractions was unequaled by any living mathematician. He worked out the Riemann series, the elliptic integrals, hypergeometric series, the functional equations of the zeta function, and his own theory of divergent series, in which he found a value for the sum of such series using a technique he invented that came to be called Ramanujan summation. On the other hand, he knew nothing of doubly periodic functions, the classical theory of quadratic forms, or Cauchy’s theorem, and he had only the most nebulous idea of what constitutes a mathematical proof. Though brilliant, many of his theorems on the theory of prime numbers were wrong.

RAMANUJAN'S NUMBER 

1729 is the natural number following 1728 and preceding 1730. It is a taxicab number, and is variously known as Ramanujan's number and the Ramanujan-Hardy number, after an anecdote of the British mathematician G. H. Hardy when he visited Indian mathematician Srinivasa Ramanujan in hospital. He related their conversation.

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."

The two different ways are:

1729 = (1)^3 + (12)3 = (9)^3 + (10)^3

67 BPSC Math solution(Answer Key)

Q1. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day? 

Q9. The perimeter of a rhombus is 52 m and its shorter diagonal is 10 m. The length of the longer diagonal is

Solution

We have,

Parameter of a rhombus$=52m$Its diagonal$=10m$

Let the length of the other diagonal$=Xm$

 $Now,$

$Parameter\; of\; rhombus=4\times side$

52=4×side ⇒side=13m

Therefore, each side of rhombus $=13m$

Then, the length of other diagonal$=2\sqrt{{13}^2-{\left(\frac{10}{2}\right)}^2}$

x=2×144​

x=2×12

$x=24m$

We know that,

Area of rhombus$=\frac{1}{2}\times$ product of diagonals$=\frac{1}{2}\times 10\times 24$=5×24=120m.2     

Q10. In a 100 m race, A runs at 8 kmph. If A gives Ba start of 4 m and still beats him by 15 seconds, what is the speed of B?

Solution