 ## How to cover important topics of Algebra for JEE Main?

To command on mathematics in IIT JEE examination, Algebra is one of the key to get through. This portion of mathematics covers various topics, such as, Set Theory, Quadratic Equations, Logarithms, Permutations, Matrices and Determinants and Probability. In order to attain success in this field it involves the understanding of the topic versus marks ratio. There are direct questions being asked on Determinants and Matrices. Permutations and Combinations as well as Probability are most important sections. In IIT JEE Main exam mostly questions are fetched on Complex numbers, Probability and progressions & series.

There are many portions in Algebra section and it has a weightage which is most often equal to the weightage of Calculus and together they cover around 65 percent of entire Maths. So let’s find out the important topics of Algebra for JEE Main:

1: Complex Numbers and Quadratic Equations

Algebra of complex numbers, modulus and argument (or amplitude) of a complex number, square root of a complex number, triangle inequality, Quadratic equations in real and complex number system and their solutions, Relation between roots and coefficients etc.

2: Matrices and Determinants

Types and algebra of matrices, properties of determinants order two and three, area of triangles using determinants. Adjoint and inverse of a square matrix, Test of consistency and solution of simultaneous linear equations in two or three variables.

3: Permutations and Combinations

Fundamental principle of counting, permutation as an arrangement and combination as selection

4: Mathematical Induction and Reasoning

Statements, logical operations. ‘and’, ‘or’, ‘implies’, ‘implied by’, ‘if and only if’. Understanding of tautology, contradiction, converse and contrapositive.

5: Binomial Theorem

Binomial theorem for a positive integral index, general term and middle term, properties of Binomial coefficients and simple applications.

6: Sequences and Series

Arithmetic and Geometric progressions, Relation between A.M. and G.M. Sum up to n terms of special series: Sn, Sn2, Sn3. Arithmetico – Geometric progression.

7: Statistics and Probability

Measures of Dispersion

Calculation of mean, median, mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data.

Probability

Probability of an event, addition and multiplication theorems of probability, Baye’s theorem, probability distribution of a random variable, Bernoulli trials and Binomial distribution.
The above topics are bifurcated into different subtopics which must be emphasised on. A few points to study and cover Algebra are:

1. Priority wise: Progressions & Series, Permutations & Combinations, Binomial, probability, Complex numbers, Matrices & determinants (these are arranged in the order of ease and weightage combined)
2. Algebra requires huge practice with the blend of understanding the concepts. Start any topic with NCERT and solve ALL the questions even if you find them extremely easy.
3. Once you are done with that move on to R D Sharma objective and then finally any one of Cengage Algebra G Tejwani or Sk Goyal Arihant publication can be considered the reference book.
4. Always go through solved examples first and then start solving the unsolved questions
5. At the end of any chapter, go for previous year questions on that chapter and this time focus on speed and accuracy.

If you wish to be updated regarding all the updates of JEE Main 2018, download Gradeup App. It connects all IIT-JEE aspirant on one platform where you can share find good study material resources, topic-wise notes, video lectures, practice quizzes & tests, and ask your doubts and discuss with experts and aspirants.

All the best!!

Please follow and like us: ## Do you struggle with solving simultaneous equations?

By simultaneous I mean equations with multiple unknown variables. Generally the number of equations given will be equal to the number of equations.

Let’s take an example,
3x + 4y = 18
5x + 7y = 31

### Methods Taught at Schools

In our schools, we are taught to solve for x by equating the co-efficient of y by multiplying both the equations by some constants in such a manner that you get the same resultant value and then subtracting one equation from the other.

For instance in this case, to find the value of x, we will multiply first equation by 7 and second equation by 4 and get 28 in both the cases. This is done so that we get a zero on subtracting.
3x + 4y = 18 …………………(i) x 7
5x + 7y = 31 …………………(ii) x 4

We get 2 equations, where co-efficient of y is same
21x + 28y = 126
20x + 28y = 124

Subtracting second equation obtained above from the first one we get
(21x – 20x) + (28y – 28y) = 2
Hence, x = 2

Plugging the value of x in equation (i) we get y = 3

### Problems with the Above Methods

• This method can become quite laborious, especially when the co-efficients of the unknowns are such that they have to be multiplied by large numbers to make them equal to eliminate one of the unknown by adding or subtracting as the case may be.
• The above calculations become cumbersome, when the co-efficient(s) are large prime numbers.
• This method involves multiple steps where we need to do multiplication and addition/subtraction.
• Also, there is no chance of using this method to solve the problem mentally as one has to keep track of the equations and various computations.

### Solving Simultaneous Equations the Smarter Way

Our new method will give us the final answer in fractions, i.e. in Numerator and Denominator for both the variables: x and y.

First, we need to find the numerator of the value of x in the above case, take the simple following steps:
Step #1: Cross-multilply the coefficient of y in the first equation by the constant term (RHS) of the second equation
Step #2: Subtract from it the cross-product of the y coefficient in the second equation and the constant term (RHS) of the first equation. So the numerator is 4×31 – 18×7 = 124-126 = -2.

Second, we need to find the denominator of the value of x:
Step #1: Cross-multiply the coefficient of y in the first equation by the coefficient of x in the second equation
Step #2: Subtract from it the cross-product of the y coefficient in the second equation and x coefficient in the first equation. Hence the denominator is 4×5 – 7×3 = -1
Hence, the value of x = -2/-1 = 2

Now, let’s try with a simpler example,
x+2y = 8
3x + y = 9

Using the above method, in a single line calculation you can say,
x = (2×9 – 1×8)/(2×3 – 1×1)
x = 10/5 = 2
Therefore, y = 3

Isn’t this amazingly simpler? With some practice you can comfortably apply this technique to solve simultaneous equation mentally.

If you liked this method, you must explore another Vedic Mathematics trick of solving a special class of simultaneous equations in seconds.

Do you find simultaneous linear equations difficult to solve? Do you think you can start using above method in solving equations?

Please follow and like us: ## Solving Simultaneous Equation using Vedic Mathematics

Let’s learn a very simple trick using which we can quickly solve linear algebraic equations with 2 variables using a Vedic Maths sutra named as Anurupye Shunyamanyat.  This is the 6th Sutra of Vedic Mathematics (flipkart affiliate link).

ANURUPYE SHUNYAMANYAT – which means, if one is in ratio, then the other one is zero. This is used in solving simple simultaneous equations in which one of the variables are in the same ratio to each other as the independent terms. This is useful in certain circumstances only but saves time when applicable.

Let us take a simple example

3x + 8y = 42

6x + 18y = 84

Co-efficient of “x” (which is a variable or unknown) are in the ratio 3:6 = 1:2 and the independent terms are also in the ratio 42:84 = 1:2

Wherever you see this happening, simply use the above sutra which says in such cases

y = 0

Therefore, x = 42/3 = 14

Hence problem solved. Simple isn’t it.

Let’s take another example –

11x + 7y = 28

23x + 21y = 84

Here, co-efficient of “y” are in the ratio 7:21 = 1:3 and the independent terms are also in the ratio 28:84 = 1:3

That means the above mentioned vedic maths sutra is applicable in this situation too.

Therefore,

y = 0

y = 28/7 = 4

If you’ve any question related to simultaneous equations, please post them as comment below. Share your views / feedback also.

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## Fun with Algebra

This is a guest post by Danielle Brooksis, a regular contributor on QuickerMaths.com. If you want to write a guest post, get in touch at vineetpatawari[at]gmail[dot]com.

Algebraic Equations for Fun!

There was a lot of positive feedback about the alphametic cryptatrithms, I posted a few months ago: so here is a little game that is based on the same idea, but with a different execution. This version is fun for those with a propensity towards algebra, and geared more toward middle school or high-school ages; however, if you are an adult, please enjoy them as well. There are a few less steps here than in the cryptarithms, but I’m sure y’all will get a kick out of them all the same!

1. (DD)^E  = DEED Read More
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## What’s your favorite equation?

It has been a while since we had a survey on any mathematics topic, so today I’d like to kick one off. What’s your favorite equation, formula, identity or inequality?

Try to keep it to your top 5 so that things don’t get out of hand – but please share which equation, formula, identity or inequality amuse you the most.

So – What’s your favorite equation, formula, identity or inequality in mathematics and if possible tell us why? Over to you!

Take into consideration all the branches of mathematics

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## Quick method to evaluate polynomials – Horner’s method

This is a guest post by Nandeesh H.N. of Kolkata

## How to find the value of a Polynomial Function?

Horner’s method is commonly used to find the roots of a polynomial function. However it can also be used to evaluate the polynomial function for a given value of x.

Suppose, we want to evaluate the polynomial

p(x) = 4x^5 – 3x^4 + 7x^3 + 6x^2 + 3x + 9 at x = 2.41.

The usual method of evaluation is to evaluate each product (such as 4*2.41^5 or 7*2.41^3) separately and then add. The drawback is that to evaluate any power of x, we go through all of the previous powers.

A slightly better method is to make a table of powers of 2.41 and put them in the given polynomial. Read More

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## Find the remainder – Vedic Algebra

### Vedic Algebra

I have got a mail from some QuickerMaths follower, to illustrate usage of Vedic Mathematics in branches of mathematics other than arithmetic. This post is for that purpose only. Here I am highlighting the usage of Vedic Mathematics in finding out the remainder when an algebraic expression is divided by another.

Finding out the remainder becomes extremely easy using Vedic Maths.

So lets begin with a simple example –

Find the remainder when

x3 + 4x2 + 6x – 7 is divided by (x + 5)

Solution: Read More

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