Strategies to do Mental Division Fast

Out of the 4 most basic arithmetic operations – addition, subtraction, multiplication and division, generally the most feared one is division. However, you can learn few simple division tricks and shortcuts to become comfortable with divisions.

How do you mentally divide?

You generally find multiplication more effortless as compared to division, right? So to mentally solve division, simply convert the division into multiplication. Yes, you heard me right. Hold on to see how this makes things relatively easier for you.

In this article, I will share with you the trick to divide any number by 5, 25 and 125 by converting these into single digit multiplications. You can refer to an earlier article where certain Vedic Mathematics trick are used for division of more difficult numbers.

Believe me, you will be amazed at the simplicity of this method. So stay tuned and read ahead.

How do you divide by 5?

To divide any number by 5, you don’t actually need to perform any division. You will get your answer in 2 simple steps.  

First step, you simply need to multiply the given number by 2.

Let us take some examples to understand this further.

2323/5 = 2323×2 = 4646

Second step, put place a decimal just before one digit from the right. So in this case the decimal will come before 6. Hence the answer would be 464.6

Thus, 2323/5 = 464.6

Isn’t that simple? You just need to know multiplication or table of 2.

Another example,

Step 1: 23487/5 = 23487×2 = 46974

Step 2: placing the decimal before one digit from right: 4697.4    

Did you ever think it could be this easy?                                                                                                                                      

How do you divide by 25?

To divide any number by 25, you simply need to multiply the given number by 4 and then place a decimal before 2 digits from the right.

Let’s try with an example, 5830/25

Step 1: multiply the given number by 4

5830×4 = 23320

Step 2: placing a decimal point before 2 digits from the right

Another example, 719835/25

Step 1: Multiplying by 4:

719835*4 = 2879340

Adjusting the decimal by 2 places

28793.40, that’s the answer.

This is fun.

How do you divide by 125?

To divide any number by 125, multiply the number by 8 and then put a decimal point before 3 digits from the right. Let’s take an example, 2650/125

Step 1: Multiplying by 8:

2650*8 = 21200

Adjusting the decimal by 3 places

21.200 = 21.2, that’s the answer.

Another example, 348847/125

Step 1: Multiplying by 8:

348847*8 = 2790776

Adjusting the decimal by 3 places

2790.776, that’s the answer.

Don’t you love this?

Explanation of the shortcut trick for division

For those who’re interested in understanding what’s going on here, you can read further.

Let us first understand for 5.

x/5 = x*2/10 = (x/10) * 2 = 0.x * 2

thus, to divide any number by 5, we can multiply the same number by 2 and adjust the decimal point by one place.

Similarly, we can understand shortcut we adopted above for multiplication by 25.

x/25 = x * 4/100 = (x/100) * 4 = 0.0x * 4

Thus, to divide any number by 25, we can multiply the same number by 4 and adjust the decimal point by 2 places.

Stretching the method further, to divide anything by 125 we multiply the number by 8 because –

x/125 = x*8/1000 = (x/1000) * 8 = 0.00x * 8

Once you understand the logic explained above, you don’t have to memorize the steps in these tricks. In fact, by using the same logic you can do quick division by numbers like 50, 250, 500 and so on. So please don’t memorize, please understand and then apply.

If you’ve slightest doubt, ask your questions in the comments below.

Best books for JEE Mains Physics/Chemistry/Maths

JEE Main is one of the competitive exams which is the gateway for different engineering institutes of India including NITs and IIITs. It is also one of the hurdles that an IIT aspirant needs to cross, to be eligible for one of the toughest known exam “IIT JEE ADVANCE”. Roughly 20% of the registered students clear IIT JEE Main every year.

A set of 30 questions in each subject, Mathematics, Chemistry and Physics in a time period of 180 minutes (3 hours) is not a task that can be counted easy. One has to be fast and accurate while solving questions, since any mistake can lead to negative marking.

The marking scheme is quite simple for the exam, as every correct answer awards 4 marks and every incorrect deducts 1 from his aggregate total. So it is clear that one is being evaluated in a maximum score of 360.

JEE Main is one of the exam that is based on basic concepts and understanding of different principles/theorems that are being taught in 11th and 12th standard. The syllabus of the exam includes all the topics that are included in NCERT books.

The exam has all types of questions, some of them are based on principles (like Archimedes, Bernoulli, etc.), theorems (Pythagoras, Theorems for triangles, etc.) and applications of theory (Projectile, Equations of motion, etc.). It has both theoretical and numerical type of problems, with 4 multiple choices. Among the 4 choices, only one choice is most appropriate or correct.

One can follow multiple books to practice and learn the concepts that are required to crack the exam. The most recommended books for understanding of basic concepts for JEE Main is NCERT books (for all the three subjects). It is counted as a baseline material for JEE Main and learning all the principles, formulas and theorems. It is one of the best book for theoretical questions as some of the problems in JEE Main are manipulated from NCERT textbooks.

Once an aspirant has completed the theoretical study from NCERT books, he/she can refer below-mentioned books for practice and improving on their concepts.


There is one well known book by a professor in IIT Kanpur, naming “Concepts of physics by H.C Verma”. The book comes in 2 volumes(Concepts of Physics Vol.1 and Vol.2) and are the most recommended books for JEE aspirants. This book includes short answer type questions to sharpen one’s basic understanding. Also this book includes, single choice and multiple choice questions that can be very helpful to build up your concepts and apply whatever you have learnt in your theory. At the end of every topic, you can find exercise problems, which can be really fruitful to score good marks in JEE Main.

That single book can surely make you crack JEE Main with a very good score along with NCERT Physics book. If you feel you still have time, you can check DC Pandey (Arihant Publications) and Resnic Halliday.



Concentrate on NCERT for theoretical questions and learning of concepts including chemical reactions and formulas. NCERT chemistry can be referred for solving theoretical problems, but it doesn’t have good number of problems. For practice of problems and extending your knowledge towards different topics, you can refer these books-

For Organic chemistry-

  • Organic chemistry by O.P Tandon
  • Concept of Organic Chemistry by M.S Chauhan

For Inorganic chemistry

  • Inorganic chemistry by J.D Lee
  • Inorganic chemistry by O.P Tandon

For Physical chemistry-

  • Numerical Chemistry by R.C Mukherjee
  • Physical Chemistry by O.P Tandon

Practice as many problems (numerical as well as theoretical) from the mentioned books when you are through with NCERT books and you surely can score really well in Chemistry.


Mathematics is considered as one of the subjects that play a vital role in cracking IIT JEE Main and scoring a good rank. For learning concepts and theorems, one can refer Mathematics NCERT book, but it doesn’t have very good level problems for practice.

For practicing Mathematical problems and to get idea of implementation of theorems/principles, you can refer these books-

  • Objective Mathematics by R.D Sharma.
  • IIT Mathematics by M.L Khanna
  • Trigonometry by S.L Loney
  • Geometry by S.L Loney
  • Higher Algebra by Hall and Knight

Once you are through with the above mentioned books, you must practice mock test papers and solve previous year exam papers. Some of the mock test papers are released by JEE organising committee (CBSE earlier, NTA from 2019 onwards) and some are available from different publications. Some of the publications are-

  • Arihant publications(Most recommended)
  • Disha publications
  • Balaji publications

Mock tests can help you to get an idea of where you stand and a rough estimate of how much you can score. The last two to three weeks before the exam must be dedicated to solving previous year’s problems. You should take last 15 year papers of IIT JEE Main and solve them. That will get you ready for your IIT JEE Main exam and you will be surely score really good.

Along with theory, numerical questions will help you boost your marks and get you a very good rank. As it is said for competitive exams-

“More you practice, closer you move towards scoring good marks”.

Don’t forget to download JEE Main preparation app on Google Play Store. It provides free notes, topic-wise study material, quizzes, mock test, question papers, and all the updates related JEE Mains/Advanced exam.

Quick & Easy Calculations to Save Money while Shopping

On my recent visit to a supermarket, I realized the struggle and decision making fatigue which shoppers go through while making their household purchases.

I decided to write this post to ease this struggle to the best of my capacity.

For ease of understanding I’ve made categories based on the type of decisions we need to make.

Comparison Shopping

All the examples given below are from the items I picked up from the grocery section of the supermarket.

1st comparison 

  • 24 mangoes for Rs.699
  • 72 mangoes for Rs.1899

To compare these options, first thing you do is round off the numbers.

  • 24 mangoes for Rs.699 700
  • 72 mangoes for Rs.1899 1900

After that, you need to compare the price by making the quantity same. Ideal way is to scale up the lower quantity.

In this case if you multiply smaller quantity, i.e. 24 by 3 you get 72.

Hence 24×3 mangoes in 1st case will be Rs.700×3 = Rs.2100

Where in 2nd case, 72 mangoes are available for Rs.1900

So you know, if you can consume 72 mangoes, it’s available at a bargain. However, we need to consider the perishable nature of mangoes and the quantity we can consume in a reasonable time till which mangoes can remain fresh. 


2nd Comparison

  • 85 gm of honey for Rs.149
  • 200 gm of honey for Rs.299

Again, 1st thing is to round off the prices.

  • 85 gm of honey for Rs.149 150
  • 200 gm of honey for Rs.299 300

Next, easier thing to do in this case will be to scale it up and see.

If we multiply the smaller quantity by 2, we get

85×2 = 170 gm for Rs.150 x 2 = Rs.300

Comparing once again,

In 1st case, we get 170 gm for Rs.300

In 2nd case, we get 200 gm of Rs.300

Thus clearly 2nd case is more beneficial.


3rd comparison

  • 5kg of detergent powder for Rs.153
  • 4kg of detergent powder for Rs.418 plus a bucket free.

We need to scale up the smaller quantity to match the larger quantity. Then the price comparison will be apple to apple. That too of the same size.

In the above case you saw that just simple multiplication was good enough to do that. However, here you need to do some quick thinking.

You remember HCF taught in school? Don’t panic. No worries even if you don’t.

HCF or highest common factor means finding a number which divides both the numbers.

Let me explain this for the above example. If we have smaller packets of 500 gm each, for buying 1.5 kg you need 3 packets and for buying 4 kg you need 8 packets.

So now, if 1.5kg is at Rs.153, 1 packet of 0.5kg is for Rs.153/3 = Rs.51

Therefore, 8 packets (i.e. 4kgs) will be for Rs.51*8 = Rs.408.

So, if you ignore the free bucket available you’re better off buying the smaller packet i.e. of 1.5 kg compared to 4 kg pack.

However, getting a bucket by paying additional Rs.10 is not a bad deal. So you should go for 4 kg packet.

In an earlier article I’ve touched upon calculations of gross margin, cost and sale price. That article will also help you in calculations of discounts. Have a look.

I hope I helped you to overcome your dilemma during shopping when you are spoilt for choices and one offer seems to be better than the other.

Did you all like my tips? If yes, write in the comments below. Stay tuned for another set of situations on Discount Shopping.

All you wanted to know about Polygons and Interior Angles

Geometric figures like pentagon, hexagon, octagon, etc. are very intriguing geometric two dimensional figures with their own peculiar properties. In general, we call these figures polygons. A little later you will find the proper definition of polygons.

At times, it might look little scary or daunting to understand the properties of these polygons. If you know polygons in and out, this article may not be for you. It is for learners who would like to understand polygons in a simple common sensical manner.

So let us delve deep into the world of polygons. So, here we go with the most obvious question first –

What is a polygon?

Polygon is any shape made up of 3 or more connecting lines on a flat sheet surface. Hence, polygons are 2-dimensional closed figures made up of straight lines. Thus the intersecting lines have to terminate at the point of intersection.

If you would like to dive deeper into the formal definition and terminology related to polygons, I recommend you check out Chegg’s page on polygons

What is an interior angle of a polygon?

The angles formed in the interior or inside of a polygon where two pair of sides intersect are called interior angles.

interior angle in a polygon

What is a regular polygon?

A polygon in which all sides are equal (equilateral) and all angles are equal (equiangular). Otherwise, it’s an irregular polygon.

How do you find the sum of the interior angles of a polygon?

I can directly give you the formula to calculate the sum of all interior angles of a polygon. But I want you to dive a little deeper and understand how that formula is arrived.

Our goal for today is to figure out how the interior angles of a polygon change as the number of sides of the figure increases. So let us start with the polygon which has smallest number of sides, i.e. triangles.


To understand interior angles of a polygon, we have to keep triangles (type of polygon) as the starting point. Triangle has 3 interior angles and 3 sides. It’s a known fact that sum of these 3 interior angles of a triangle is always 180°.

Little out of context, but the moment we think of triangles Pythagoras Theorem is generally the first things which comes to our mind. In a previous post, I have tried exploring, was Pythagoras Theorem actually proved by Pythagoras?

3 sided polygon, sum of interior angles = 180°

Moving forward and looking at polygons with higher number of sides –


Quadrilateral is polygon with 4 sides and obviously 4 interior angles.

If we connect the opposite corners of the quadrilateral (i.e. the diagonal) divides the quadrilateral into 2 triangles. If we add the sum of interior angles of theses 2 triangles, we get 180°+180°=360°. Thus the sum of interior angles in a 4 sided polygon is always 360°.

Another way of looking at it: square is a polygon with all sides equal, i.e. it’s a regular polygon. We know in a square each of the 4 angles is equal to 90°. Thus sum of all the interior angles is 90°x4 = 360°

4 sided polygon, sum of interior angles = 360°


Pentagon Polygon

In a 5 sided polygon, also called pentagon, you can join opposite 2 vertices from any one vertex and you will get 3 triangles. Again, we know that sum of interior angles of a triangle is 180°.

Thus, sum of interior angles of 3 triangles = sum of all interior angles of a pentagon = 180°x3 = 540°

5 sided polygon, sum of interior angles = 540°


Have you noticed that, as we keep increasing the number of sides in a polygon the interior angles keep increasing.

As we move further, i.e. for each increase in number of sides of a polygon, 180° gets added to the sum of interior angles.

So the general rule or formula is,

Sum of Interior Angles = (n-2) x 180°


How to find the interior angles of a polygon?

In a ‘n’ sided polygon, the sum of interior angles of a polygon = (n-2) x 180°

A regular polygon is equi-angular; thus each interior angle will be equal. In a ‘n’ sided or ‘n’ angled polygon,

Each Interior Angle = Sum of Interior Angles / no. of sides = [(n-2) x 180°]/n

For example, in a dodecagon (12 sided figure shown below),

dodecagon 12 sided polygon

Sum of interior angles

= (n-2) x 180°

= (12-2) x 180°

= 10 x 180°

= 1800°

And in a regular dodecagon, each interior angle

= 1800°/12

= 150°

What is an exterior angle? How to find an exterior angle?

An exterior angle in any polygon is an angle formed by one side of the polygon and the extension of an adjacent side of the polygon.

Interior and exterior angles are supplementary angles.

Exterior angle = 180° – Interior angle

Thus, if interior angle is , exterior angle = 180° – X°

Another important point to keep in mind is that,

Sum of all exterior angles = 360°

So, in a regular polygon of n sides, each

Exterior angle = 360°/number of sides in a polygon

Example, in a hexagon, each exterior angle will be 360°/6 = 60°

How to find the number of sides of a polygon, if one interior angle is given?

The easiest way to finding the number of sides, is to first find out the exterior angle.

So let’s say the interior angle given is 144°, how to find the number of sides in the polygon?

First step is to calculate the exterior angle, which in this case = 180° – 144° = 36°

Now since it’s known that sum of all exterior angles is 360°, thus the formula is

Number of sides in a polygon = 360°/exterior angle

In above case, number of sides = 360°/36° = 10 sides. Thus it’s a decagon.

For your further reading, you can check this previous post where I’ve explained the ratio of area and volume derived from ratio of sides.

Objective of QuickerMaths is to make mathematics fun, quick and simple. I would love to hear from you in comments below.

Eggs-cellent Riddles & Brain Teasers

You always knew that eggs are full of nutrition but here on QuickerMaths, it will help you strengthen your brain cells too. Here are some riddles & puzzles around eggs for you to solve.

First Egg Riddle 



If one and a half hens lay one and a half eggs in one and a half days, how many eggs does one hen lay in one day?

Leave your answers below.

Second Egg Riddle

2 fathers and 2 sons sat on the table to eat eggs for breakfast. They ate exactly three eggs, each person had an egg. Now you need to explain how that’s possible?

Third Egg Riddle

Let say you started selling a basket of eggs.

First customer buys one-half of your eggs plus one-half of an egg. Second customer buys one-half of your eggs plus one-half of an egg.

Third customer buys one-half of your eggs plus one half an egg.

At this point you have sold all of your eggs, and you never broke an egg. How many eggs did you start with?

7 Best Ways to Master GMAT Quantitative Ability Section

Muddled up as how to prepare for the Quantitative ability section of GMAT? The Quantitative ability section of the exam is one of the most feared sections of any entrance. Although the difficulty level isn’t too much, the test is built to put students in a pinch. It is a computer adaptive test, i.e., the difficulty of every next question changes according to your rate of correct answer.

Get all information for GMAT in French here

Correct answers will result in points being awarded and difficulty being raised for the next question on-screen, while wrong answers will result in similar difficulty level and no points awarded for that question. So, those planning to appear for the exam must have a look below to get an idea of the best ways to master the section.

The test evaluates four key skills of a student, i.e. analytical writing, quantitative analysis, verbal skills, and reading skills. The language of the exam is English, with emphasis on grammar, algebra, geometry, and arithmetic. The exam also assesses analytical writing and problem-solving abilities of the students. GMAC believes that data sufficiency, logic, and critical reasoning skills are extremely vital to businesses in the real world.

GMAT 2018 Paper Pattern

To prepare for any exam, you must first be familiar with the structure of that exam. There are four sections in a GMAT exam. These are analytical writing assessment, integrated reasoning section, quantitative section, and verbal section. The exam is 3 hours and 7 minutes long, with time divided unequally amongst all the three sections.

GMAT Test Section Questions Question Types Timing
Analytical Writing Assessment 1 Topic Analysis of Argument 30 Minutes
Integrated Reasoning 12 Questions Multi-Source Reasoning
Graphics Interpretation
Two-Part Analysis
Table Analysis
30 Minutes
Quantitative 31 Questions Data Sufficiency
Problem Solving
62 Minutes
Verbal 36 Questions Reading Comprehension
Critical Reasoning
Sentence Correction
65 Minutes

As you can see in the table above, there are two types of questions in the Quantitative section:

  • Problem Solving (PS)
  • Data Sufficiency (DS)

The Problem Solving section has the same multiple choice format that is popular for every standardized test in today day and age. There are five options, out of which only one can be the correct answer.

The other format, Data Sufficiency, is unique to the GMAT exam, with unique rules too, which require different strategies as to all other tests. This section, thus, requires the most amount of practice and work.

7 Best Tips for GMAT Quantitative Ability preparation

If you are preparing for the GMAT exam, you will need all the help you can get to clear the exam. Provided below are a few tips you can follow in order to better prepare for the Quantitative section of the exam.

Strengthen your Basics

The mathematical concepts tested in GMAT are extremely simple, consisting of basic arithmetic, algebra, and geometry. The only problem is that students tend to forget the basics as time moves on. Your GMAT preparation should first and foremost cover the basics, and only after completing those should you think about going ahead with further preparations. The best way to remember formulae is to create flashcards and stick them around in your room. That way, every time you walk by a formula, your eyes will tend to hover over the flashcard.

Practise tests and mock tests

It should be obvious that practise will make you better eventually. The more you practise, the easier you will find the test to be. Thankfully, practise tests don’t have to be very expensive. Several online resources provide free practice content to use.

After every iteration of a practice or a mock test, you would also need to analyse your performance. Review the results and note the questions that you have answered incorrectly. Improve upon these particular areas identify your area of weakness. Majority of the GMAT exam questions revolve around students’ familiarity with different types of questions and avoiding common mistakes. With ample practice, you will be able to realise which questions are trick questions, thus also saving you a lot of time. One practice every week should be a comfortable place to start.

Pay special attention to Data Sufficiency questions

The Quantitative section is the most difficult section of GMAT primarily due to the Data Sufficiency portion. They would require you to think a little differently, but the more you practice them, the easier they become. There are several key points to remember when working on Data Sufficiency questions. Read the provided statements individually, and very carefully. Only after carefully evaluating the statements, make your answer choice.

Data sufficiency requires only sufficiency, not the actual answer, which means that if a problem states if the value of a variable can be determined, you only have to see whether it can be or cannot be determined, without actually solving for the value. You are just trying to find out if there’s enough information to answer the question, but you don’t actually have to find the answer.

Memorize the Five Answer Choices

There are always the same five answer choices for every Data Sufficiency Question. These answer choices are:

  • Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
  • Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
  • Both statements together are sufficient, but neither statement alone is sufficient.
  • Each statement alone is sufficient.
  • Statements (1) and (2) together are not sufficient.

If you were to memorise these statements, you could save precious few seconds for every problem you attempt. You would only need to read the statement provided alongside the question and place judgement based on them, by clicking on the answer choices.

Be careful with Graphs, Charts, and Tables

A lot of questions in GMAT quantitative section will require you to read and interpret information provided on charts, graphs, and tables. It is extremely important that you read the axes, the key, units of measurement, etc. correctly so that you don’t misinterpret the data.

Use the rough paper in exam

Even if you feel like the GMAT quantitative section is too easy for you, it would only benefit you to use a paper for calculations as much as possible. Writing down your calculations will help you notice any mistake you might have made before you press the answer and move on to the next question. Writing also forces you to make sure you’re thinking through progress in steps instead of leaps, which can help reduce mistakes further. Remember, use of a calculator is forbidden in the GMAT exam.

Read the Questions carefully

Last but not the least, this is the most crucial yet the most ignored piece of advice that one can offer to any student aspiring to crack any national level exam. It is simply because students tend to make more mistakes when they are fatigued.

One of the most common mistakes on the GMAT exam is to misinterpret or read the question incorrectly. The GMAT exam purposefully throws in questions with difficult language, or questions that can mistakenly be read differently.

Instead of asking “Which of the following may be false?” GMAT will present the question as “Which of the following may not be true?” which might be misunderstood as “Which of the following may be true?” While this may sound a bit far-fetched right now, the atmosphere of pressure inside the hall and the fatigue during the exam itself could easily lead to such mistakes.

Thus, make sure that you read every question carefully so you can save yourself from these easily avoidable mistakes.


Top colleges offering Bachelors in Mathematics in India

There are various colleges which offer Bachelors in Mathematics course for the aspiring students. It is the responsibility of the students to find out the best colleges for the courses they wish to opt for. Below is the list of some of the top colleges offering Bachelors in Mathematics course:

Indian Statistical Institute

It is an institute which is of national importance that has been established and recognized under the act of Indian parliament in the year 1959. It has been established in the year 1931 which is a public university and is considered one of the oldest and prestigious universities which focus on the study of stastics and mathematics.
It is one of the leading training and research institute in the fields of computer science, quantitative aptitude, statics, and related sciences and so on. The headquarters of the university is located in West Bengal, Kolkata.

Courses offered

  • B Stat (Hons)
  • B Math (Hons)
  • M Stat
  • M Math
  • MS in Quantitative Economics
  • MS in Library and Information Science
  • M Tech in Computer Science
  • M Tech in Quality, Reliability and Operations Research
  • JRF in Statistics
  • JRF in Computer Science
  • JRF in Mathematics

Maharishi Markandeshwar University, Ambala

It is a deemed to be university in Mulana in the state of Haryana. It was established in the year 1993 in the name of Maharishi Markandeshwar and was founded by Tarsem Garg. The institute is the leading symbol of education in terms of technical, medical and other professional streams. It has been accredited by the NAAC team with an A grade.

The university is committed to excel in research and innovation and the skill development of the students. It has developed an industry oriented education system which helps the students in order to make the students leaders in the professional world.

Courses offered

  • Electrical Engineering
  • Biotechnology
  • Civil Engineering
  • Computer Science & Engineering
  • Mechanical Engineering which includes Specialization in Automobile Engineering and Mechatronics
  • Electronics & Communication Engineering
  • Computer Science with Specialization in Software Development Program (American Pattern)

Chennai Mathematical Institute

It is an institute of central excellence which has been formed for the teaching and research in the mathematical sciences which is established in the year 1989 with the motto of bridging the gap between the teaching and the research in the fields of mathematics and various other allied subjects.

It has been recognized by the government of India under the section 3 of the UGC act, 1956. The objective of the institute is to provide the excellent quality education to the students in the best possible way.

Courses offered

  • B.Sc. (Hons.) in Mathematics and Computer Science (3 year integrated course).
  • B.Sc. (Hons.) in Mathematics and Physics (3 year integrated course).
  • M.Sc. in Data Science
  • M.Sc. in Mathematics
  • M.Sc. in Applications of Mathematics
  • M.Sc. in Computer Science
  • Ph.D. in Mathematics
  • Ph.D. in Computer Science
  • Ph.D. in Physics

Indian Institute of Technology (IIT)

It is one of the autonomous public institutes of higher education situated in India. It is governed by the Institute of technology Act, 1961 which has been declared as the institution of national importance in India. There are various branches of the institute which has been established in various other cities of India to impart excellent education to the students in the fields of technology, science and mathematics. Every year, a large number of students gear up to take admission in this university.

Courses offered


  • Aerospace Engineering
  • Biological Sciences and Bio-Engineering
  • Chemical Engineering
  • Civil Engineering
  • Computer Science and Engineering
  • Electrical Engineering
  • Materials Science and Engineering
  • Mechanical Engineering

Lovely Professional University, Jalandhar

It is a private university situated in Jalandhar, Punjab. It was established in the year 2005 by the lovely international trust under the lovely professional university act, 2005 which started its operation in the year 2006. It has the largest single campus in terms of the private university in India. The campus spreads over 600 acres of land on the outskirts of Jalandhar and accommodates more than 24,000 students.

It has been recognized by the UGC, NCTE and AIU. It was also ranked 18th as India’s best college in the year 2017. It offers various courses in Mathematics which are spread over 3 years across six semesters. In order to get admissions in such courses, the candidates should have scored 60% and above in the secondary examination or any other degree equivalent to it. The candidates also need to clear the LPUNEST Exam 2018-2019 in order to take admission in the examination.


  • Master of Science in Mathematics
  • M. Sc (Mathematics) – M. Tech (Computer Science Engineering) (Dual Degree)
  • Doctor of Philosophy in Mathematics (Full-time & part-time)


  • Master of Science in Physics
  • M. Sc (Physics) – M. Tech (Computer Science Engineering) (Dual Degree)
  • Master of Science (hons.) in Physics
  • M. Sc (Hons.) (Physics) – M. Tech (Computer Science Engineering) (Dual Degree)
  • Master of Philosophy in Physics
  • Doctor of Philosophy in Physics (Full-time & part-time)

How to prepare for mathematics in UPSC examination?

Preparing for UPSC exam is a challenging task but choosing a good optional subject increases your chances for qualifying. Mathematics is one such key player. If you have a mathematics background and have interest in that subject, then prefer maths as your optional subject. Maths has always been known as a good scoring subject. For the people who hate mugging, Mathematics is a gift. There is no hit and try involved in it. If you know the formula, you know the answer.

Choosing mathematics has its own advantages such as it requires more of application part and zero mugging. Secondly not many UPSC aspirants take up maths as the optional subjects, so there is less competition. Thirdly, the study material is easily available. And lastly, this subject has a limited syllabus unlike other subjects in which current affairs are also involved. Here are few easy tips to get high marks in the exam.

  1. Number of attempts

    During the exam, if you are unable to solve the question in two attempts, leave it. You will not be able to solve that. Focus on other questions. If a problem is taking too much time, then hold on, you may be following a wrong approach to solve it.

  2. Tips and tricks

    Practice some good tips and tricks to solve the problems. The more problems you solve, you efficiency will automatically be increased. Try to adopt Vedic Maths techniques. This will help you to solve even the biggest of equations within fractions of seconds.

    A few tips and tricks using Vedic Maths :

    Step 1:
    Let us consider multiplication of three digit numbers 208 x 206.

    Step 2:
    Now, deduct the last digit from the respective numerals.
    208 – 8 = 200
    206 – 6 = 200

    Step 3:
    Pick any one number and add it with the unit digit of another number.
    208 + 6 = 214

    Step 4:
    Now, multiply the result obtained in step 1 and step 2.
    214 * 200 = 42800

    Step 5:
    Then, multiply the unit digits of the given numbers.
    8 * 6 = 48

    Step 6:
    Add the values obtained in step 4 and step 5.
    42800 + 48 = 42848
    208 x 206 = 42848

  3. Avoid silly mistakes

    Maths is one such subject where a single silly mistake can deviate you from the actual answer. Be careful of the sign the number is carrying. Practice, practice and practice till you become perfect.

  4. Make a formula chart

    Make a chart (or two) of all the formulas and paste them in your study room. Don’t be hesitant in looking through the formula in case you forget. After some practice, the formulas will automatically be memorized by you.

  5. Try solving by hand

    Rather than just reading any solution, better take a pen and paper and try solving it. There is a huge difference between solving in mind and solving on paper.

    It can be illustrated through an example :

  6. New Topics

    Some new topics have been introduced such as Mechanics and Fluid dynamics. Try to understand the concepts in the beginning of your preparation period. After reading all the solved examples, then start off with previous year question papers. This will give you an idea what and how much to study for the prelims.

  7. Writing answers

    If you are writing short answers then give one or two lines for introduction and then move straight to the point. In case of long answers, give introduction and explain wherever necessary and make a perfect conclusion at the end.

  8. Join Test Series

    Joining test series is an effective way to improve your time management and self evaluation skills. If not daily, then give a mock test at least twice a week for a sound preparation.

  9. Keep yourself relaxed

    You cannot solve any mathematics problem if you lack focus or concentration. Keep all the distractions away from you and be cool & relaxed. This way you will not only be able to understand the problem but will also hit the right answer too.

    Mathematics is the first love of science. If studied and applied properly, mathematics will fetch you good marks and ultimately a good UPSC score.

4 Tips for Building a Good Relationship With Your Child’s Math Tutor

Today, getting a tutor for a child having difficulties with math is a normal course of action to take. If your child needs help with understanding certain concepts in geometry, algebra, calculus, or with applying the unitary method in grade 2, there is nothing wrong with asking other parents or going online to find a tutor for your little one.

Hiring a tutor or enrolling a child in a tutorial program, in general, is a popular option among parents. It is something that many parents decide to do to help their kids excel at or meet the demands of a difficult subject (or more than one). It is one of the best ways parents can show their love and support for the children who are having a hard time keeping up with their schoolwork.

Working with Your Child’s Tutor

As a parent, your role in your child’s academic progress won’t stop or be diminished if you decide to enroll your child in an after-school math tutorial program. Your youngster will experience more benefits from a tutorial program if you have a good working relationship with his tutor.

To make sure you and your child’s tutor are on the same page regarding his improvements and continuous progress in math, follow these helpful tips:

  1. Practice constant and open communication

    Communication is one of the vital pillars of a tutoring process. From the start, ask questions if you have any, share your feedback and raise any concerns you might have about the program. Make sure the tutor has all the necessary information regarding the key needs and preferences of your child as well.
    Before scheduling a talk with the tutor or learning center, come up with a list of questions regarding their procedures and policies. Get details about their assessment process and their teaching methodologies as well.

  2. Set clear goals for the tutorial program

    Before the formal tutorial program starts, create a list of goals that you want your child to achieve. Discuss these goals with the tutor. Ensure you highlight the particular concepts or areas your kid needs help with.
    When coming up with your list of goals, talk to your child. Also, keep in mind that a tutor isn’t just a person you hire to help your child do his math assignments or review for an upcoming test. If your child needs help with this subject, work with the tutor to ensure that the whole problem will be tackled and not just one area will be solved or managed.
    One goal that you should never forget to list is to see a change in your child’s attitude towards math. Aside from getting higher grades, your tutor should also play an important role in changing your youngster’s negative perceptions about math, and in the process inspire the student to actively participate in studying it.

  3. Involve your child’s teacher

    For your child to benefit more from the tutorial program, inform his teacher about this after-school activity. It is important that the teacher and tutor work together so that they are on the same page in terms of the lessons and goals.
    If possible, schedule parent-teacher conferences and tutor meetings to coincide with the time your child gets his progress report. It is essential that the tutor communicates with your child’s teacher to get feedback on his performance in the classroom. They don’t even have to meet personally; they can correspond by email or private messages. Just make sure your kid’s tutor and teacher coordinate regularly.

  4. Be involved

    Lastly, aside from finding a good tutor and bringing your child to the tutorial center during his sessions, you also need to take a proactive role in the whole process. At the very least, at the end of each session, ask the tutor what topics they covered today and what difficulties your child encountered. Ask the tutor how you can help reinforce what your child learned from the session.
    Although this may mean relearning continuity and different ability and other mathematical concepts so that you can give some exercises for your child to work on at home, your efforts will be rewarded in the end.
    Your role in helping your child overcome his fear or dislike of math and getting higher grades doesn’t stop at finding a good tutorial center and bringing him here every session. For your child to get more from each tutorial lesson, you must be totally involved in the process and have a good working relationship with the tutor.

Maloy Burman is the Chief Executive Officer and Managing Director of Premier Genie FZ LLC. He is responsible for driving Premier Genie into a leadership position in STEM (Science, Technology, Engineering and Mathematics) Education space in Asia, Middle East and Africa and building a solid brand value. Premier Genie is currently running 5 centers in Dubai and 5 centers in India with a goal to multiply that over the next 5 years.

Mentally Solve Problem of Averages in Seconds

In all competitive examinations and otherwise also in our day to day calculations we encounter issues related to averages. Especially when the composition of the group changes it becomes difficult to do computations.  Let us discuss quick and easy solutions to such problems.

When a person leaves a group and another person replaces him, then there can be 2 scenarios:

Scenario I: When the average age increases then,

Age of the new comer = age of person who left + (no. of persons in the group * increase in average age)

Scenario II: When the average decreases then,

Age of the new comer = age of the person who left – (no. of persons in the group * decrease in average age)

Let us try to understand these with a simple example –

Question: The average age of 45 persons is decreased by 1/9 year when one of them whose age is 60 years is replaced by a new comer. What is the age of the new comer.


Age of the new comer = age of the person who left – (no. of persons in the group * decrease in average age)

Age of the new comer = 60 – (45 * 1/9) = 55 years

Isn’t that easy to calculate using the formula I just gave you. However, to intuitively answer such questions you need to keep following things in mind

Firstly, the average age is reducing, which means an older man is replaced by relatively younger person. Hence the answer has to be lesser than 60. Hence we subtract from 60 and not add to it.

Secondly, average reduction of 1/9 kg for a group of 45 people means, in total there is a wait reduction of 1/9 per person x 45 persons = 5 kg for the entire group. This is result of the replacement of an old person by a relatively younger person in the group. That’s the reason we’re subtracting 5 from 60, to get the age of the new comer.

B. When a person joins a group without any replacement, then there can be 2 scenarios:

Scenario I: When the average age increases then,

Age of the incoming person = previous average age of the group + no. of persons including the person who joined * increase in the average age value

Scenario II: When the average age decreases then,

Age of the incoming person = previous average age of the group – no. of present persons including the person who joined * decrease in the average age value

I will explain this further with an example,

Question: The average age of 20 teachers is 45 years which is decreased by 6/7 years when a student joins the group. Then what is the age of the student?


Age of the outgoing person = previous average age of the group – new no. of persons including the person who joined * decrease in the average age value

Therefore, age of the student = 45 – 21 * 6/7 = 27 years

Here also you need to observe that since the average age is going down when the new person is joining, the age of new comer will be lesser than the average, hence subtraction.

Moreover, the decrease is equal to 6/7 per person for each person in the new group. That is 21 times 6/7 = 18 kgs, which needs to be subtracted from the overall average to get the age of the new comer.

C. When a person leaves a group without any replacement, then there can be 2 scenarios:

Scenario I: When the average age increases then,

Age of the outgoing person = previous average age of the group – no. of persons excluding the person who left * increase in the average age value

Scenario II: When the average age decreases then,

Age of the outgoing person = previous average age of the group + no. of present persons excluding the person who left * decrease in the average age value

Hope we’ve covered all scenarios and with the help of the above logical formulas we will be able to do our calculations.

Practice questions for you to solve (try to solve mentally)

Question 1: In a boat there are 8 men whose average weight is increased by 1 kg when a man of 60 kg is replaced by a new man. What is the weight of the new comer?

Question 2: In a class there are 30 boys whose average weight is decreased by 200 grams when one boy whose weight was 25 kgs leaves the class and a new comer is admitted. Find the weight of the new comer?

Question 3: A cricketer has a certain average for 9 innings. In the 10th innings, he scores 100 runs, thereby increasing his average by 8 runs. His new average is?