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How to tackle Math Olympiad questions – 2

January 7th, 2012 in Uncategorized

In the last article we talked about some general practices that will stand you in good stead while taking part in the Maths Olympiad (or any exam/test in general). In this edition, we’ll talk about various additional techniques that could help you perform better in Math Olympiads.

Most of the time the questions in such competition are of the objective type question. This can make things simpler for you (on the other hand, depending on the ingenuity of the question setter, it can also make it harder for you). The way it makes it easier for you is in two ways

One is that you can easily check if the answer you compute is part of the options provided to you
The second is that sometimes, just sometimes, you can do some amount of guessing to eitherget the right answer, or narrow down your possibilities

For example, take the following question taken from http://www.edugain.com

Question: The number 11449 can represented by a 107 x 107 square grid. Out of the following numbers,

which number can not be represented on the square grid?

a. 14641 b. 90601

c. 9216 d. 16122

Here, you could work out the square root of all the numbers and try to figure out the answer, but a quicker way may be to notice that a square number cannot end in 2. In addition, the question reads ”which number”, which implies there is only one choice that is true.

So the answer here is obviously d) – 16122

Similarly, in a lot of these cases, you can eliminate at least one of the choices – there can be many variants that you could identify by just looking at the question. Ask yourself some mental questions like “Can the answer be an odd number?” etc. Of course, each mental question you ask yourself will be specific to that question.

Sometimes, a question can seem rather hard at a quick glance, but if you read it once more slowly, you can probably see past the apparent complexity into a simple solution.

Consider the question (also taken from http://www.edugain.com)

Question: Archana’s roll number is a two digit number. Her friend Balvinder’s roll number has the same digits as Archana’s roll number, but with digits interchanged. If they add their roll number, and divide the sum by 11. Find the remainder of this division.

a. 1 b. 2

c. 0 d. Can not be determined without knowing the roll number

At first glance, it may seem really hard – how can you know without really trying? Here you can use one of two methods. One is not really what I would call mathematical, but if you are pressed for time, you could easily use it for such question. And what’s that technique?

It’s simple – all you do is try it out on a couple of example numbers. Imagine that Archana’s roll number is 61. Then Balvinder’s roll number would be 16. Add 61 and 16, and you get 77. The remainder when 77 is divided by 11 is 0.

However, you also need to know how to solve this directly (after all, you are here because you want to learn mathematics).

The way to do that is to think of it as follows

Archana’s number has two digits – say they are “ab” where a and b could be any of 0,1,2,3,4,5,6,7,8 or 9.

You can represent it mathematically as

10a + b

Then Balvinder’s number would be “ba” – mathematically you would say

10b + a

Add them up

10a + b + 10b + a = 11a + 11b = 11(a+b)

This is of course divisible by 11 leaving a remainder of 0.

Simple, and more satisfying to do it this way, isn’t it?

So you should always keep in mind that no matter how hard the question looks, it is very likely that you have already been taught the tricks to solving it. There may be a few questions that may use concepts you haven’t been taught yet (to identify the really advanced students), but there’s not much you can do about that. For a majority of the questions though, you already know all the techniques you need to solve them. It’s just a question of identifying them and applying them.

You should also know your strengths and use them appropriately. For instance, you may be very good with spatial questions and at visual data – then go for the geometry or figure related questions first. If numbers dance in your head, go for the arithmetic questions. Get them out of the way, and then move on to the other sections.

One final note, that I mentioned in the previous article. Once you’ve found a solution, put it back in the question and double check that your solution is indeed right.

How to tackle Math Olympiad questions – Part 1

January 7th, 2012 in Uncategorized

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Math Olympiad papers can seem rather daunting, isn’t it? And rightly so – a well set Olympiad question will test your fundamentals and analytical skills, and challenge your way of thinking.

Irrespective of the fact that most of the so-called “Math Olympiads” conducted by various businesses in our country are nowhere near international standards, the questions in even those are generally “tougher” than the ones you face generally in school tests or exams.

In solving these types of questions, you generally need to be able to think quickly on your feet – one factor that makes these questions harder is that the time available to you is limited. A second factor is that you may have negative marking for wrong answer, so sometimes it would seem better to leave the question rather than putting in an answer you are not sure about. To add to the confusion, sometimes it is possible that more than one answer is correct.

So how do you tackle this. The first step is of course (to reiterate a point made in an earlier article we wrote) is “Don’t Panic”. Take a deep breath, relax, read the questions and do your best.

In this article we talk about some general practices you should follow. In the next part of this article, we’ll talk about some specific mathematical techniques that could help you tackle such exams better.

The first step is make sure all your resources are there with you – pencils, erasers, spare paper for computations (if allowed). It seems trivial, but these are things that you don’t want to worry about once the test starts. Now, start reading the paper. There’s no standards here – different people do it in

different ways. Some people start from a fixed location, typically the beginning of the paper, but some folks even like to start at the end (the theory being that the examiners would have gotten tired of finding tough questions, and would start putting easier questions towards the end – of course, no sensible examiner would set papers that way). Some other folks like to look for questions in topics they are familiar in, and try those first. Whatever approach you are comfortable with is fine.

The most important thing, and this is where a lot of people make their mistake – read the question completely and carefully. Spend as much time you need on this part. Any mistake you make here will ensure you get the wrong answer.

Let me repeat this – read the question carefully and completely. Find out what the question is really all about. Take careful note of all the values and data provided in the quesion.

Then, and only then, should you start trying to find the answer.

And once you have solved it, or think you have – check that your answer indeed is what the question is asking for (If, for example, the question asked for speed of a car, and you have 121 seconds as your answer – something has gone wrong)

Second, keep track of the time. Keep a general idea of how much time you have per question. It need not be exact – for example, if you have 2 hours and there are 50 questions, just think that you’ll have about 2 minutes per question. Of course, you may do some questions faster, but keep this general timeframe in mind per question. If you are completely confident of a question, you could take a little more time (say 3 minutes) on it, but generally try not to spend more than the computed time per question. In fact, ideally you should target leaving the last 10 minutes for a quick revision.

Now, if you haven’t attempted all the questions when you find there are 10 or so minutes left then it’s time to start picking and choosing from the remaining questions. Run through them quickly to see which ones look easiest for you to solve. Then try those. The next tip is on using your spare paper effectively. It can get very messy quickly, so you should mark off specific areas for questions for which you need to use the spare paper for computation. You could use a grid kind of structure if that helps – but the aim is to use

it effectively and neatly. And make sure that for questions where you use it, copy all the values correctly to the spare sheet. Double check it, and triple check. Do the same thing while copying the

answer back.

In the next part of this series, we’ll look at techniques specific to mathematics to help you do better in Olympiads

Solving Mathematical Problems – Part 1

November 10th, 2011 in Uncategorized

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When I was younger, and still at school – there used to be a day I used to dread. Well, not the whole day, but some hours. OK – less than that. A minute or two, actually.

The first time I saw the math question paper in front of me.

All the problems, the numbers, and the geometric figures began swimming in front of my eyes. I’m sure some of you have also had the feeling.

It’s only when I grew older and actually began to understand the subject – I realized that there was nothing to worry about. Mathematics is not really all that hard – as long as you knew the basics of the subject. If you really, really know the basic principles, and the techniques that you can apply, then it’s just a matter of using them in the right fashion to solve any problem.

More importantly, no problem is too small, and you can always learn something new from any problem. Even failures in solving a problem can teach you something new. All you need is a curious spirit and a willingness to try. As the great 20^th century mathematician George Polya said “Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery”. He goes on to add “Such experiences at a susceptible age may create a taste for mental work and leave their imprint on the mind and character for a lifetime”.

And you, my young friends are of such a age – when you are learning new things, and are full of curiosity about the world around you. If you were to instill this “taste for mental work” at this age, then it will be a lifelong companion for you, and you will have succeeded in effectively using the only organ that separates us from the other animals that inhabit our world – our brains and our capacity for thinking.

Isn’t that something worth striving for?

And you know, even at your age, that the only way to learn how to solve problems is, well, by actually solving them. Try and solve as many of them as you can. Practice, practice, practice. Sites like www.edugain.com are good sources of problems that challenge your mind. And while solving each problem, try and think about the various ways in which you could solve it (where possible, of course).

With this series of articles, we’ll try and look at various problems and techniques on solving problems in mathematics.

So what’s the very first thing you need to keep in mind when you see a problem in front of you? I’m not sure if many of you have heard of a set of books called “The Hitchhiker’s guide to the galaxy”. You’re all probably a little too young to have read them (but I recommend you do try them when you are a teenager and can appreciate them), but there’s this catchphrase in that book that applies to this situation. It’s just two words

DON’T PANIC

Yes – Don’t Panic. It’s just a problem, and no matter how complex it looks like, it is solvable using stuff you already know. Just take a deep breath, and relax.

Now, step 2. Read the question carefully. Sift out the relevant facts and numbers. The reason it probably looks complex is because it’s a long problem (not always true – some of the toughest problems are very easy to state – we’ll talk about Goldbach’s conjecture in some later articles when talking about unsolved problems).

Anyway, back to solving our problem. Pull out the relevant information and relations. See what the problem statement actually relates to among the various subjects you have learnt. In most cases, this will be straightforward, but in some cases this may be tricky. This will especially be true when the questions are of the Higher-Order-Thinking variety.

Let’s start with a very simple problem (this problem can also be found on http://www.edugain.com website).

Q: Sulekha has 8 pair of red socks and 9 pair of brown socks. All the socks are in a bag, if Sulekha picks the socks without looking at the bag, how many socks does she have to remove from the bag before she can be sure that she has a pair of a single colour?

Now, this may seem simple to some of you, but there are a lot of students who get a little confused when they see this question. So how should they approach this problem?

First, though the question tells a story about Sulekha and her rather odd collection of socks, remember that those are not important points.

The key point is that there are 8 pairs of type A, and 9 pairs of type B in a set/group.

But let’s look at the technique to solve it in terms of the stated problem.

Sulekha’s requirement is to get one pair of single colour, so she would stop once she gets either two red socks, or two brown socks.

Let’s use some simple logic to see what the possibilities are. For ease of simplicity, consider the symbols R = Red and B = Brown,

Now if she takes only one sock from the bag, she could get either (R) or (B). In this case, she definitely cannot have two of the same colour, as she took only one sock (Simple logic – you need two for a pair)
If she takes two socks out of the bag, she could get any of the following combination after she picks up both the socks - (RR) or (BB) or (BR). Note the third combination. This is the possibility of one RED and one BROWN. Since we cannot predict what she will get, so this is also not enough. Getting two socks does not guarantee there are one Black and one Red socks.
However, if she takes three socks, these are the four possibilities after all 3 socks are removed, (RRR), (BBB), (RRB) or (BBR). Note that the order of the socks does not matter, i.e. RRB is same as RBR or BRR. Here you can see that in all possibilities she is getting one pair of same colour socks. Even if one of the socks in the triplet is a different colour, atleast two of them will have the same colour. Problem solved. She just need to pick three socks to be sure that she has got at least one pair single colour.

In subsequent articles we will look at different techniques of solving problems.

In the meantime, maybe you can think about this related problem:

Sanjana has 4 pair of white socks and 9 pair of black socks. All the socks are in a bag, if Sanjana picks the socks without looking at them, she has to remove ____ socks from the bag before she can be sure that she has a pair of white color.

LCM of Fractions

June 26th, 2011 in Uncategorized

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The Least Common Multiple (LCM) of two integers x and y, is the smallest positive integer that is a multiple of both x and y. Generally LCM is used for adding fractions where denominators are not same. Most of the kids know how to calculate LCM, but I was surprised to learn that most of the kids are not aware of the physical significance of LCM.

I was interacting with some of the grade 6 students and I asked them about LCM, and all kids in class said that they know LCM very well. So I asked them to find LCM of 1/2 and 1/3 and surprisingly no one could answer. They were trying to apply regular method of finding LCM of integers, and of-course that did not help them in finding LCM of fractions.

Problem was that kids just learn the method of solving questions in textbooks, and do not pay much attention to theory. Had any student used the definition of LCM (and not the regular method of finding LCM) they could have easily solved this question.

Anyway lets forget the regular method of finding LCM and try to solve this using definition of LCM.

So as per the definition of LCM, we have to find a number which can be fully divided by 1/2 and 1/3. If you think about it you will find that answer is 1. Since one is fully divisible by both 1/2 and 1/3.

1/ (1/2) = 2
1/ (1/3) = 3

Now lets take another example. Find LCM of 1/6 and 1/9.
Answer for this one is 1/3, since,
(1/3) / (1/6) = 2
(1/3) / (1/9) = 3

Ok, these were simple cases so we could do just by thinking about it, but we might have to do this for more complex fractions. Now lets formalize a method to find LCM of any two fractions.

Lets try to find LCM of (a/b) and (c/d). If b and d were same it was easy to find LCM since if denominators are same, we just need to find LCM of numerators, hence LCM of (a/b) and (c/b) would be LCM(a,c)/b. So we have to first make denominators of both the fractions same.

So here are the steps to find LCM of a/b and c/d

Find the LCM of b and d = LCM(b,d)
Multiply numerator and denominator of first fraction by LCM(b,d)/b.
Multiply numerator and denominator of first fraction by LCM(b,d)/d.
After this multiplication, denominator of both fractions are same.
Find LCM of new numerators.
The answer is LCM(numerators)/LCM(b,d)

Lets see this using example of 2/9 and 8/21.

LCM of 9 and 21 = 63,
Now multiply first fraction by 63/9 = 7
Multiply second fraction by 63/21 = 3
So now first fractions is (2 x 7)/(9 x 7) = 14/63
and second fraction is (8 x 3)/(21 x 3) = 24/63
Now since both denominators are same, LCM of numerators 14 and 24 = 168.
Hence LCM of 2/9 and 6/21 is (168/63)
After simplification 168/63 = 8/3

Now lets check if our answer (8/3) is correct or not by dividing this by 2/9 and 8/21.
(8/3) / (2/9) = 12
(8/3) / (8/21) = 7

You can see that this is fully divisible by both fractions.

Now we have understood the concept, lets try to find a formula for quickly solving this.
If you observe the new numerators after multiplication, they are (a/b)*LCM(b,d) and (c/d)*LCM(b,d).
So answer is
LCM ( (a/b)*LCM(b,d) , (c/d)*LCM(b,d) ) / LCM(b,d)

After simplification this will come down to a simple formula,

LCM( (a/b) , (c/d) ) = LCM(a,c)/HCF(b,d)

The Magic of “Pi”

January 29th, 2011 in Uncategorized

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Let’s start with the circle – can there be a more simple shape. Just take any point, and map out all the other points on a plane that are at some fixed distance from it. And you get a circle.

The word itself is from the Greek for circle – kirkos (Oddly enough, the word “circus” comes from the same root). Of course, our ancestors were familiar with the circle much before they had any language. In nature, they could see the moon, and the sun. When they cut a tree, they could see that the trunk had a circular cross-section. They invented the wheel – so they knew about the shape. And they believed it to be a symbol of perfection.

So there you have it – a simple shape, nothing seemingly complicated about it. A point. Another set of points that are equidistant from it. What could be simpler?

Ah, but in this seeming simplicity lie some wonderful things. Let’s just take one of them – the number “pi”.
What is pi, you ask. Well, it’s just the ratio of the circle’s circumference to it’s diameter. This ratio is the same for all circles – no matter how large or how small.

So what is the value of this number?

This is where the story starts to get strange. When you were in 3rd grade, you were probably told that the value of this number, this “pi” is 3 (which makes sense, you really didn’t know much about decimal numbers at that point). Then in grade 4 or so, when you got used to fractions and decimals, you were probably told to use the value 22/7 or 3.14 for this value to solve problems.
Here’s the funny part – although we can define what this number is (remember – ratio of the circumference to diameter of any circle) we can’t tell the exact value of the number. Sure, we know that goes something like “3.1415926535,,,” but there is no precise value. The part after the decimal point never ends. It goes on and on into infinity, and the numbers don’t repeat. What this means is, unlike a fraction like 1/3, which is also infinite (0.333333…) we can’t get a repeating pattern that lets us predict what, for example, the billionth digit after the decimal is (for 1/3, the billionth, or trillionth digit after the decimal point is always going to be “3″). It must be said that mathematicians recently have come up with some clever tricks that let you compute the value of any digit of the value of pi very fast (and without knowing any of the previous digits), but that involves rather advanced mathematics, and we won’t go into that here.

Another fascinating fact about pi is that it can be proven (some very clever German mathematicians showed this in the 19th century) that you can never get an equation with a finite number of operations on integers to give you the value of “pi”.

Mathematicians have been trying to find the value of pi for over 4000 years. By 1900 BC, Egyptian and Babylonian mathematicians knew the value to within 1%. Indian mathematicians also knew a very good approximation – the Shatapatha Brahmana , a 6th century BC work from India, gave the value as 339/108. Of course, today we know the value to trillions of digits.

More strangeness – “pi” appears everywhere in physics, maths and engineering. Even in places that seemingly have no connection to the circle. For example, if you were to try to compute the average height of all the people in the country, the formula for that would have a “pi” in it. In advanced physics, some of Einsteins equations trying to describe the nature of the universe have “pi”, as does Heisenberg’s equation governing the behavior of really small particles. Strange isn’t it? And all the more reason to learn maths. The secrets of the world around can only be understood through mathematics.

And yes, dont forget to celebrate National PI day (March 14, at 1:59. (3/14 1:59))

Algebra – A fascinating subject

January 11th, 2011 in Uncategorized

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Algebra is when most children get introduced to the concepts of “abstract” mathematics. Suddenly their world gets involved in quantities like “x” and “y”s. And if not introduced properly it can lead to fear of the subject. But if taught properly they can see that the “x” and “y” are like words that form a beautiful poem and they’ll delight themselves when they learn this secret language and understand the beauty of algebra.

Understanding and practice can go a long way in helping children here. Sites like EduGain (www.edugain.com) can help children in doing this.

Let’s look at a simple problem that can illustrate the beauty of this language (adults may like it too). Note that this article is meant only for children studying in grade 6 and above.

Suppose you take a long, really long string. You then cut out enough of it so that it can go all the way around the earth (I told you it was really long).

string around earth

So now you have this string that’s on the ground and encircling the earth along the equator.Now suppose you add another 30 centimeters to that string (that’s about the length of your long ruler). We then smoothen it out by pulling it up from the ground all over the equator so that it forms a proper circle again.

The question is “How high do you think the string will be above the ground now?”

Sounds very hard to solve, right. But let’s see…

We know that the circumference of a circle is C = 2 x π x R, where R is the radius.
(don’t worry about π – let’s just use the value of “3″ for now)

So, C = 2 x 3 x R = 6 x R

Now we don’t know the exact radius of the earth here, but we don’t need to. Let’s leave it as a symbol R. So the length of the original string = 6 x R. We’ll just call it 6R (and know it means the value of R, whatever it is, multiplied by 6).

So C = 6R

Now we added 30 cm to C. The length of the new string is now C+30.

What we need to find out is how much R increases.

Let’s say that R increased by some length Z. The value of Z is what we want to find

So for the string, we can say it has a new circumference (Cnew) a new radius (Rnew)

We also know that since it’s a circle

Cnew = 6Rnew

And we know that the new circumference Cnew = C+30,
and that
Rnew = R + Z (remember we already assumed R increased by a length Z which we are trying to find out).

So we get

C + 30 = 6(R + Z).

Expanding this, we get,

C + 30 = 6R + 6Z

But we already know that 6R = C

Let’s replace the 6R by C on the right side

C + 30 = C + 6Z

The two “C”s cancel out, leaving us with

30 = 6Z, or

Z = 30/6 = 5

Amazing isn’t it. The new string would be 5 cm above the ground all everywhere. Just by adding 30 cm extra.

And what’s more amazing – we solved it through Algebra

Tessellations

November 12th, 2010 in Uncategorized

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A tessellation is simply is a set of figures that can cover a flat surface leaving no gaps. To explain it in simpler terms – consider the floor of your house. That is a flat surface – called a “plane” in mathematical terms. And you’ll notice that the floor is covered with some tiles or marbles of different shapes. That is a good example of a “tessellation”. The one difference here is that technically a plane is infinite in length and width so it’s like a floor that goes on forever.

Of course, when we are talking about floors, the shapes used to cover it are mostly rectangles or squares (in fact, the word “tessellation” comes from the Latin word tessella – which means “small square”). The word “Tiling” is also commonly used to refer to “tessellations”.

There are different kinds of tessellations – the ones of most interest are tessellations created using polygons. If you use only one kind of polygon to tile the entire plane – that’s called a “Regular Tessellation”

Regular Tessellations

As it turns out, there are only three possible polygons that can be used here. There are only three rules to be followed when doing a “regular tessellation” of a plane

The tessellation must cover a plane (or an infinite floor) without any gaps or any overlaps.
All the tiles must be the same shape and size and must be regular polygons (that means all sides are the same length)
Each vertex (the points where the corners of the tiles meet) should look the same

Of course, you would have guessed that one is a square. What are the other two? They are triangles and hexagons. Let me show you examples of these two here.

You may wonder why other shapes won’t work. Let’s try with pentagons and see what shape we come up with. You can see that there is a gap and that’s not allowed.

So what’s unique to those 3 shapes (triangle, square and hexagon)? As it turns out, the key here is that the internal angles of each of these three is an exact divisor of 360 (internal angle of triangle is 60, that of square is 90, and for a hexagon is 120). The mathematics to explain this is a little complicated, so we won’t look at it here

Semi-Regular Tessellations

If you use a combination of more than one regular polygon to tile the plane, then it’s called a “semi-regular” tessellation. If you look at the rules above, only rule 2 changes slightly for semi-regular tessellations. All the other rules are still the same.

For example, you can use a combination of triangles and hexagons as follows to create a semi-regular tessellation. There are eight such tessellations possible

Tesselations3

There are many other types of tessellations, like edge-to-edge tessellation (where the only condition is that adjacent tiles should share sides fully, not partially), and Penrose tilings. Each of these has many fascinating properties which mathematicians are continuing to study even today. Tessellations are also used in computer graphics where objects to be shown on screen are broken up like tessellations so that the computer can easily draw it on the monitor screen.

The magic of Primes

November 12th, 2010 in Uncategorized

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We all know what a prime number is. It is a number that is divisible only by 1 and itself. But do you know how fascinating these numbers are?
You could call them the building block of all numbers.

Why do I say that?
We all know numbers that are not prime numbers are called composite numbers. But you may not know that all composite numbers can be built from prime numbers in a unique way by multiplication. In fact, this fact is called the “Fundamental Theorem of Mathematics” – that any number can be written as a factor of two or more primes in a unique way (by unique I mean there is only one way to write it as a factor of primes).

For example, take 96. If you factor it in primes, you will get 96 = 2 * 48 = 2 * 2 * 24 = 2 * 2 * 2 * 12 = 2 * 2 * 2 * 2 * 6 = 2 * 2 * 2 *2 * 2 * 3

So 96 can be represented as 2 * 2 * 2 * 2 * 2 * 3. There is no other way to factorize it using primes.
And this is true for all prime numbers. Isn’t it fascinating?

There are many fascinating things about prime numbers and their role as the building block for all numbers.
But let me tell you about one strange fact – there is no largest prime number. It is just not possible to have something called the “largest prime number”. A Greek Mathematician called Euclid proved it more than 2000 years ago.

There is an active search going on for finding larger and larger prime numbers. The largest one found so far is 2^43,112,609 − 1. What is that? It is 2 multiplied by itself 43 million, one lakh, twelve thousand, six hundred and nine times, and one subtracted from that number. That number is 12 million digits long.
How big is that? Let me tell you – If you could write 80 numbers in a line, and your notebook page had 40 lines, and your notebook had 50 pages, it would take you 80 notebooks to write down this number. Of course, the search for still larger prime numbers continues.

Mathematics – Opening Children’s mind to a beautiful subject

October 27th, 2010 in Uncategorized

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We often see that children are scared of mathematics. This manifests itself as a reluctance to understand the subject, and ultimately in them pushing the subject aside.

However, we believe that we at EduGain can help them in overcoming this fear and to excel in a beautiful subject that can grow their minds and expand their horizons.

The truth is that Mathematics is a fascinating and unique subject. This is the sense that we need to instil in them.

Mathematics is unique in the sense that all other areas we study came from our curiosity about the universe around us. We looked up at the night sky, and from our curiosity about what we saw, came up with astronomy and physics. We looked at the proliferation of life around us, and developed the science of biology. And so on.

But mathematics is something that originated in the human mind. It is true that the start of mathematics came from daily activities, where we distilled the concepts of numbers and counting, but soon it surpassed those humble origins and became something we create from our minds – pure abstract thought. No wonder it is often called “The queen of the sciences”.

The reason why some children fear it is that it is projected in our education system as a very difficult area. However, this is not really true. We’ll go into the specifics of these in later articles, but one thing that really helps is practise. It’s true in all areas of human endeavour, but especially true for mathematics. Continous practise with various types of questions helps crystallize the concepts and that is when the fear of the subject goes away.

Sites like EduGain aim to aid the students in this aspect – by generating questions using an advanced engine, students get exposed to a variety of questions in all areas that helps them understand the concept thoroughly and overcome their fears of this beautiful subject.