15th National Maths Convention 2022 (VIRTUAL MODE)

All India Ramanujan Maths Club is organizing 15th National Maths Convention from 12 and 13 Feb, 2022 where you can show your talents and win certificates and prizes.

Read the instruction carefully and click the link given below to register yourself before 5th February 2022

 A student can participate in maximum 2 events.

The program is divided into two parts- Junior Level and Senior Level. 

 Junior Level - Class6 to 8 

Senior Level - Class 9 to  12 , Even 1st yr college student can participate 

LAST DATE OF REGISTRATION :- 5th February 2022

Email to selected students will be sent by 7th February 2022.

GET READY WITH YOUR PUZZLE/ MODEL/ POEM/ESSAY  BEFORE OPENING THE LINK. TO UPLOAD YOUR CONTENTS IN THE LINK.

                                              Exploring Mathematics Award

Students who are very creative and have written Mathematical blogs/ designed websites/ created you tube channels can only apply under this category.

Age Limit :- Below 16 years

                                                      Quiz Competition

Syllabus :-  General Mathematics

                          History of Mathematics

                         Know your Mathematicians

                          Identify Mathematical Shape/ Curve/ Film / Book

Question on the mathematician that you can check on the blog page itself. In the past we have conducted several quiz on Aryabhata and Ramanujan so get an idea of what type of question will be asked in quiz section.

Level 1 :- Written test through Google Form will be conducted in the first phase. 12 students with maximum correct answer will be selected for Final round.

1st level Exam will be held on 6th February. Only registered students can participate in the exam.

Level 2 : PPT Round with selected students

 Junior Exam will be held in the morning while Senior Exam will be held in the evening. 

                 Mathematical Model Topics (Junior/Senior)

 a) Make a mathematical model from kitchen accessories.

b)      Working model on geometry

c)       Model on the work of any of Indian mathematicians

d)      Concept of graph and its use in co-ordinate geometry

e)      Mind map of algebraic equation/ expression/ identities

f)        Pythagorean triplet multi model

g)       Model on Set Theory

h)      Multi-layer Car Parking Model

i)        Mathematics in Bio Technology, IT, Biology

j)        Mathematical Model of DNA

                      Topic for Junior and Senior are same.                                         

                           Mathematics Puzzle (Junior/ Senior) Topic

  Calendar

b)      Dice

c)       Puzzle from Indian/Western monuments, temples

d)      Puzzle that puzzled even mathematician

e)      Impressive puzzle around the world

                             Mathematical Poem Topic (Junior/Senior)

                

  Role of Mathematic

  Zero

Number game

Ramanujan

e     Mathematics in daily life

       Why I love Maths

                                      Word Limit for Poem :- 100 - 200 words

y                Mathematical Essay Topic (Junior/ Senior)

 a ) Relevance of Ramanujan in today’s world

b)  If I were Aryabhata/ Brahmagupta

c) Mathematics in Kitchen

 d) Importance of Numbers

 e)India the country of Numbers  

 f)Contribution of Indian Mathematicians in Freedom Struggle

g) Use of ICT in Mathematic

h)Math Phobia : How to address it

i) Journey of Pi in 100 years

j)Career in Maths

k) Had the shape of Ice-cream cone be Spherical?

                                                      Word Limit :- 200- 500

Font:- 12 Times New Roman

   REGISTRATION LINK FOR STUDENTS 

                          EVENTS FOR TEACHERS

  Extempore: - A teacher has to present his view on the topic given on the spot. The maximum time allocated is 7 minutes for presenter and 3 minutes for judges. Topic of Extempore are ---

Ramanujan

 Indian freedom struggle and contribution of Mathematicians

 Mathematicians you would like to be

Mathematics in War planning

 Cryptography and Maths

Use of Mathematics in preserving monuments

History of Indian Mathematics

Zero

Career in Maths

 Paper Presentation: - A researcher/ teacher can present his paper on the given topic in virtual mode if selected. A 200-300 words write up (Abstract) need to be sent well before the due date. The top 10 entries will get a chance to present their paper live on the designated day. 

Use of ICT in teaching Maths

Mathematics in Gaming

Maths in NEP, Your take

 Practical use of Imaginary Number 

Anecdotes :  Relevance in teaching Maths

   REGISTRATION LINK FOR TEACHERS

            Award for Teachers / RP/ 

All India Ramanujan Maths Club invites application from teachers / teaching professionals / Professors working in different parts of country who will be conferred National Award under different categories.

 Aryabhata Award :-

 a)15 years of Teaching Experience in College/ University/ IIT / NIT ..                                                      b) Minimum 10 research paper publication                                                                                                c) Published 5 books with ISBN                                                                                                                  d) M.Sc /M . Phil or PhD ( NET/ GATE  qualified if Not PhD) is a must

Ramanujan Award :-     

a) 10 years of teaching experience                                                                                                                b) Publication in Newspaper, Magazine is a must                                                                                        c) Associated with NGO or working for the welfare of social backward                                                          community                                                                                                                                                d)M.Sc /M . Phil or PhD 

Best Teacher Award :- We select 6 teachers for this prestigious award. Minimum 5 years of teaching experience is a must. Working for society to uplift the level of mathematics will be an added advantage.  This year we have 3 additional award for our RP's associated with any of our sister organization and have studied Mathematics up to Graduation.

CLICK HERE TO APPLY FOR TEACHER AWARD

THANK YOU

Geometric Progression - K Bhanumoorthy

 

 Geometric Progression. (GP)

 

If the terms in a sequence are such that, each term is a constant multiple of preceding term, then we say terms are said to be in a GP. The multiple is called as common ratio and denoted by “r”.

Eg :    4,8,16,32, 64       r is 2.

           8,4,2, 1, ½           r is ½.

           7, -14, 28, -56    r is -2.

 In general, we take terms in GP as a, ar, ar²,ar³ ar⁴- - -        r is “r”.

In this GP also we can find n ^th, term and also sum to n terms of GP.

To note:     n ^th , term of a GP  is       a. r^r-1,      - - - - - - - - -        I

Let S denote sum to n terms

      S = a + ar+ ar² + ar³+ar⁴+ - - - - - - - - - - - - + ar^n-1   . . . . . . .  II

       Multiply both sides by ‘ r’ ,

    rS = ar + ar²+ ar³+ ar⁴+  - - - - - - - - - - - - - - +arⁿ  . . . . . . . ..  III

 

subtracting equation III from II , we get ,

S – rS = a – a rⁿ   ;   S ( 1 – r ) = a ( 1 – rⁿ ); so ,  S = a ( 1 – r ⁿ ) /( 1 – r).

Here condition for r is r < 1.

Note 1:  . If r > 1, then S = a (r ⁿ – 1) / (r – 1).

Note 2:   a, b, c are 3 consecutive terms of a GP, then b is said to be geometric mean of   a and c.  Further b/ c = c / b,  ie  b ²  = ac.

 

Let us discuss some problems for practice.

 

1.     Find the 40^th term of GP ,2, 14,28,56 - - - - - - - --   

Ans: a = 2, r =7   r > 1, hence to use the formula   

 40 th term is   2. (7 )^39. Is the answer.

 

2.Find n, given that 1024 is the n ^th term of GP 4,8,16,32, - - - - - -

Ans : a = 4 and r = 2 ,using the formula ,we get  4x 2^n-1  = 1024,

 2² x 2^n-1 = 1024; 2 ^{2+n -1} = 1024; 2 ^{n +1} = 1024; 2 ^{n +1} = 2 ¹⁰ ,

  Since the base are the same powers are equal, n + 1 = 10,

 Hence n = 9. Is the answer.

 

3.find the sum to 18 terms of the GP  3,6,12, 24, - -- - - - --    - - - - - - -

Ans : a = 3, r = 2, r > 1, to use the formula ,

S = a (r ⁿ – 1) / (r – 1); 3. (2¹⁸ – 1) / (2 – 1)   = 3 (2 ¹⁸ – 1) is the answer.

 

4. If 5, x, y, 40 are in GP, find x and y.

    From the given data, x ² = 5y ; and  y ² = 40x  ; on substituting for y ,

Y = x ² / 5, we get x ⁴ = 25 x 40x, x ³ = 1000, hence x =10 and y = 20 

(For 5y = 100).

 

5. If the first, second and fifth terms of an AP are three consecutive terms of a GP, find the common ratio.

Ans: Let a, a+ d, a+2d, a+3d, a+4d be in AP, with ‘a’ as the first term, common difference as ‘d’.

As per given data in the question, a, a+ d, a+4d are three consecutive terms in GP.   - - - - - - - - I

 Ie using the principle of GP, we get (a+ d)² = a(a+4d),

                                  a²+2ad+d² = a² + 4d², we get d = 2a, by cancelling common terms.

 

Hence by substituting d = 2a in I, we get, a, a+2a, a+8a, ie  a, 3a,9a,

 

Hence the common ratio is 3.

 

Progression - K. Bhanumoorthy

 

Progressions:

 

                                  “A teacher affects eternity

                       He can never tell when his influence stops.”

-         - - Henry Adams

-          

There are 3 progressions namely, Arithmetic Progression, (AP), Geometric Progression (GP), Harmonic progression (HP).

 Let us discuss one by one. Before we define the AP, let me say what a sequence is? what a series is, this would facilitate to understand better.

A sequence is an arrangement of numbers in a particular order. A series is defined as the sum of the elements of a sequence.

Its applications are there in our daily life. It helps us to analyse a pattern. Building a ladder with sloping sides – the length/distance between two sticks /landing remains the same. Multiples of a number say 5, are 5, 10, 15 ,20 - - - -, house numbers in a street, monthly rent, increase of salary for every year, page number of a book, annual measurement of the diameter and circumference of a tree trunk form an AP, foot prints of animals / people along a track the distance covered be in AP, paying an interest for every month if anyone has availed a loan. Travelling in a taxi one has to pay initial rate plus rate per kilometer shall be added. It forms an AP; waiting for a bus predicting when the next bus would come provided the traffic moves with constant speed. We know in a clock; time progresses in perfect AP second by second. Sum of AP series, its application, we do use in research in science subject, nuclear science, space science.

In a sequence if the terms are such that the difference between two consecutive terms is constant, (to be specific difference between the second term and the first term, difference between the third term and the second term, so on), we say that the terms are said to be in AP. The difference is called as the common difference (cd) and is denoted by “d’.

Eg:    4,7,10,13, cd - - - - - - - - - - cd is 3;

         2, -1, -4, -7 - - - - - - - - cd is -3 ( -1 - 2)

        -3,1, 5, 9, - - - - - - - - - .cd is 1 –(-3) = 1 + 3 = 4,cd is 4.

In general, we form a sequence as, a, a+ d, a+2d, a+3d ,…., cd is  d.

To find the n^th term of an AP:  we can proceed in this way.

Given the  first term as 6 and common difference is 3, fine the n^th   term.

The sequence is 6, 9,12, 15, -------------.

1 ^st term is 6,

2 ^nd term is    6 + 1(3) = 6 + 3 = 9

3 ^rd term is 6 + 2(3) = 6 + 6 = 12.

4^th term           6 + 3(3) = 6 + 9 = 15., proceeding in this way, to find

n^th term         6 + (n-1) 3     6 + 3n -3 = 3n +3.

We can generalize as n^th term is a +( n-1) d - - - - - - - - - - I

There is a formula to find the sum to n terms of an AP:

   Let us take AP as   a, a+ d, a+2d, a+3d, ----------------------------------(1)

    n^th term is a + (n-1) d, let us denote by ‘l’

ie   we can write sum of sequence 1 as           S= a+a+d+a+2d+a+3d -      - -, l      (2)         

Rewriting the sum of sequence 2 from end   S = l+l-d+l-2d+l-3d, l-4d ------, a      (3)

Adding (2) and (3) we get n times of (a+ l) = n (a+ l). If S denotes the sum of series, then we get 2 S   = n (a + l), hence sum to n terms is given as,

                     S = n/2 (a + l), substituting for l. we get S = n/2.{ 2a + (n-1) d}

Note: If a, b, c are three consecutive terms in AP, then b = (a + c) / 2; b is called as arithmetic mean of a and c.

                                           Problems for practice:

 1. Find 40 ^th term of the AP  8, 5, 2, - -,- - - - - - -  -

Cd is 5 – 8 = - 3, first term is 8, using the formula we get,

 40 ^th term    = 8 + (40 -1) ( -3) = 8 + 39 x – 3 = 8 - 117 = - 109 is the answer.

 2 Find the value of n, given that - 49 is the n ^th term. AP is 11, 8, 5, - - - - - -   

  Using the formula again n ^th term, a + (n -1) d   = - 49, here a = 11 and d = - 3.

  On substituting we get, 11 + (n -1) x (- 3) = - 49, 11 – 3n +3 = -49,

- 3n + 14 = - 49, on simplification we get n = 21 is the answer (for - 3n = - 63)

3. If 3, x, y, 18 are in AP find the values of x and y.

We can use the arithmetic mean formula. x = (3 + y) / 2, y = (x + 18) / 2

 On simplification we get 2 x = 3 + y   ( i) , 2y = x + 18, substituting for y in equation (i) we get 4 x = 6 + x + 18 , the value of x is  8 , hence value of y  is 11  .

 

4.Is 280 a term of the AP ?  11+15+19+23+27 + - - - - - - - - - - - 

   Ans: a = 11 and cd = 4. Let us find the which term of the sequence we get 280?

  To use n ^th term formula.  280 = 11 + (n-1) x (4); 280 = 11 + 4n -4= 4n +7;

       280 – 7 = 4n, 273 = 4n , n = 273 / 4 = 68 .25, n cannot be  a fraction ,hence we can

   Conclude 280 is not term in the given sequence. 

5.Find the sum of all integers between 50 and 100.

  The given series is 51+52+53+54+55+      - - - - - - - +- 99  ------(i)

  Here a = 51 d = 1, Let S denote the sum. using the formula sum to n terms of an AP

         S = (49/ 2) x {51+ 49} = 49 x 50 = 3675 is the answer.

6.      Find the sum of the first 50 odd numbers.

Ans:  The given sequence is 1 +3+5+7+9+11+13+  - - - - - - - - - +99.

  There are 50 odd numbers, we can use the formula (n/2) (a +l) l denoting last term.

     S   = 50/ 2 (1 +99) = 25 x 100 = 2500 is the sum.

 

-                                                                             - - - - - - - - -  to  be continued GP.

 

 

 

 

 

 

 

The Collins Perfect Maths League 2021

                    The Collins Perfect Maths League 2021

Collins Learning India is the education publishing division of HarperCollins Publishers, India. It publishes textbooks and digital resources for all core and supplementary subjects mapped to the CBSE, ICSE, State Board and International curricula.

All India Ramanujan Maths Club is a National level Maths Club working since 1993 with a mission to remove math phobia from the mind of young kids. AIRMC has so far organized 14 National level Maths Convention in different parts of India where students from 23 states do participate in different mathematical activities without any registration charge.

Rules for Quiz

1. 2 students from a school (forming a team) can register for the quiz, using the registration links, for each category (of class 3-5 and classes 6-8). 

2. The first round will be an online quiz round. It would open to the registered teams on 2 November 2021 from 5:00 to 5:30 pm, for each category (of classes 3-5 and classes 6-8).  

3. The average of the combined marks (of the 2 students) will be considered for preparing the result.

4. The first round is a selection round. The best 6 teams (consisting of 2 students) will make to Round 2 (zonal level). 

5. Round 2 and the Grand finale are live rounds. 8 teams (2 top teams from each zone) will be selected for the Grand Finale to be held on 15 November 2021.

Syllabus of Exam

1. Group A - For students of Class 3 to 5

No specific syllabus is prescribed for the test but you can follow NCERT text book. Besides that some Mental Ability questions, Reasoning, Crypt Arithmetic,  Mathematical Poems, Mathematical Shapes and Mathematician identification will be included in the test.

2. Group B - For Student of Class 6 to 8

For syllabus Refer text book prescribed by NCERT. The test will include 25-30 questions from different topics on Mathematics, Mathematical reasoning, Crypt Arithmetic, Mathematical Shapes, Identify the Mathematicians, Mathematical films etc.

Mock Test

To have an idea of test you can take the test by clicking the following link. The mock test is a practice test and has nothing to do with the pattern of the FIRST ROUND test. Though this test has been designed to make you familiar with the pattern of the quiz.

MOCK TEST LINK FOR GROUP A (3-5)

MOCK TEST LINK FOR GROUP B (6-8)

KEEP CHECKING THE FACEBOOK PAGE OF COLLINS PUBLISHING HOUSE FOR UPDATED INFORMATION.

Registration link for Class 3-5: https://forms.gle/Yw2k1ZdtwaRHVMzs9

Registration link for Class 6-8: https://forms.gle/Xhpg4DZsh8sgx4we6

COMBINATION -- A BRIEF INTRODUCTION --- BY K BHANUMOORTHY

                                                                  COMBINATION

 Combination:  Combination is nothing but selection of objects/things/numbers from a set of objects/things/numbers. Here the order is not important. Some formulae to remember for the combination. 

The number of combinations of ‘n; distinct objects taken ‘r’ at a time (0< r < n) is given by 

C (n, r) = n! / r! (n-r)!

 Note: C (n, n) = 1 = C (n,0); C (n, r) = C (n, r-1); C (n, r-1) + C (n, r) = C (n+1, r).                                   

                             

1.Find the number of diagonals in a polygon having 15 sides.

    Ans: (15 x 14) / (2) - 15 = 90 diagonals. {(n x n-1) ÷2}- n = n(n-3)/2

  2.There are 12 persons in a party and if each, two of them shake hands with        each other, how many handshakes happen in the party.?

Ans: Two persons hand shake is counted as one. Hence there are C (12 ,2) ways ; we get the result as 66 after simplification using the formula.

3.A committee of 3 persons is to be constituted from a group of 2 men and 3 women in how many ways can this be done if the committee to have 1 man and 2 women.?

Ans: 1 man out of 2 men be selected in C (2,1) ways; 2 women be selected in C (3,2) ways. Answer is 2 x 3 = 6 Ways.

4. 4 cards are to be chosen from a pack of 52 playing cards. If 2 cards of Red and 2 cards of Black are to be chosen, find the number of ways.

Ans: There are 26 Red cards and 26 Black cards. 2 red cards from 26 red cards C (26, 2) ,2 black cards from 26black cards C (26, 2). So

C (26, 2) x C (26, 2) = 2 x (26! / 22! .4!) = 2 x 23 x 25 x 26 = 29900 Answer. Formula used.

5.Find the number of ways in which a committee of 6 members can be formed from 8 men and 4 Women, to have at least 3 women in the committee.

Ans: Given 8 Men and 4 Women. The committee to have 6 members, with the condition to have at least 3 women.  So, there are 2 possibilities.

                          I   3 women, 3 men, II.  4 women, 2 men 

Required Answer: C (4,3) x C (8, 3) + C (4, 4) x C (8, 2)

On simplification we get 4 x 56 + 28 = 252.  Note; (C (4,4) =1, C (4,3) = C (4,1)

6.Find the number of straight lines that can be formed by joining 20 points of which 4 points are collinear.

Ans:  4 points are collinear, it forms a straight line, we get a straight line by joining 2 points. Hence solution is C (20, 2) --- C (4,2)   + 1 = 196 – 6 +1 = 185.

Note Since given 4 points are collinear to be subtracted as they form C4,2) lines in case of not collinearity.

7. In an examination a student has to answer 10 out of 13 questions, it is compulsory to choose at least 4 questions from the first 5 questions. Find the number of choices.

Ans:   Possibilities   are   I:   4 + 6 = 10; II   5 + 5 = 10 

Hence, we get C (5, 4) x C (8 ,6)   + C (5, 5) x C (8, 5),

After simplification   we get 5 x 28 + 1x 56 = 140 + 56 = 196.

8.Let Tn be the number of all possible triangles formed by joining vertices of an n –sided regular polygon. If Tn+1 -- Tn = 10, then, find the value of n.?

Ans:  We use the formula C (n, r) + C (n, r-1) = C (n+1, r) -----I

 Given Tn+1 – T n = 10    ie. C (n+1,3) –C (n, 3) = 10; using the formula I, we get 

  C (n ,2) = 10, by solving we get n= 5 (we get Q equation where factors are 

   (n-5) and (n+4) so the answer. Is n = 5

9. Find the number of ways in which one can select three distinct integers between 1 and 30 both inclusive, whose sum is even.

Answer: there are two possibilities 1. All are even, 2. Two odd and one even.

There are 15 odd integers 15 even integers. Hence the number of ways is,

C (15,3) + C (15, 2) x C (15 ,1); C (15,3) = 455, C (15, 2) = 105, C (15 ,1) ,

455 + 105x 15 = 455 + 1575 =   2030 is the ANSWER.

10.Find the number of ways distributing 10 distinct coins among 2 (two) people such that each will get at least one coin.

Answer: What are the possibilities let us see. 

Select one coin and give it one person, the remaining to the second person. This can be done in C (n,1) ways; in the same way select two coins give them to one person, and the remaining to the second person. This can be done in n(n,2) ways. Proceeding in these ways, the number of ways is 

C (n,1) + C (n,2) + C(n;3) + Cn.4) + …………. + C (n, n-1), ----I

We know C(n,0) + C (n,1) + C(n,2) + C(n,3) + ……………C (n, n-1) = 2n, (sum of binomial coefficients C0+ C1+C2+ --------+ Cn)

Now adding and subtracting C (n.0) and C(n,n) , in equation I,we get ,

The answer as 2 n - 2 since C (n.0) + C (n, n) = 1 + 1 = 2.