Algebra is the easiest topic under SSC CGL 2017 because one is not expected to learn any formulas and all the questions can be solved within 10 seconds with a jugaad, which Team SSCHacks likes to call a SUPERHACK.

We will begin with the concept of symmetrical expressions(as we call it). **A symmetrical expression is the ****one in which the weight of all the variables (a, b, c, etc.)** is equal. Examples will make things clear.

Examples of symmetrical expressions

a^{3}+ b^{3} + c^{3}

3a + 3b + 3c

a^{2} + b^{2} + c^{2}

a + b + c

ab + bc + ca

Examples of non symmetrical expressions

a+b- c

2a + 3b + 3c

a^{3} + b^{2} + c^{3}

a + b + c^{2}

**Hack 1: “Whenever you encounter a symmetrical equation in any question, you can safely ****assume : a = b = c (even if it is not given in the question)”**

Let’s solve previous year questions:

Q. 1.Here you can see that the LHS as well as the RHS of the

equation is symmetrical, hence a = b = c

Answer : (A)

Q. 2.

We put a = b = c, hence (a+c)/b becomes (a+a)/a, or 2

Answer : (B)

Q. 3.

In this question we have to find the value of x.

Here the equation is completely symmetrical, hence we assume a = b = c

Put b=a, c=a (so that the whole equation is in terms of ‘a’)

Now LHS becomes 3(x-a^{2})/2a

RHS = 12a

Solving this, you will get, x = 9a^{2}

From here we get that the value of x is 9a^{2} Now put a = b = c in all the 4 options and check which

option gives you the value 9a^{2}

A) 9a^{2}

B) 3a^{2}

C) 3a^{2}

D) 0

Answer : (A)

Q. 4.bc + ab + ca = abc is symmetrical and hence we can assume

a=b=c

Now put b=a and c=a in this equation. We will get 3a^{2}= a^{3}

So a = 3

Now put a=b=c=3 in the expression whose value we have to

find. You will get the answer as 1.

Answer : (B)

**Hack 2****: “When only a single equation is given and based on that you have to find the value ****of an expression, you can assume the value of variables yourself. But make sure to assume ****only such values that will not make the denominator zero”**

Examples :

Q . 5

In this question, only a single equation is given, i.e., x + y + z = 0, and based on this equation we have to find the value of an expression We can assume x = 1,

y = 1 and z = 0 (such that x + y + z = 0) Now on putting these values in the expression, we get the answer as 2.

Answer : (D)

Q . 6

a + b = 1

Let’s assume a = 1 and b = 0

Put the values in the expression, and you will get 0.

Answer : (A)

Q . 7

In this question the values of x and y both depend on a constant ‘a’. We can assume any value for ‘a’ and this will give the values of x and y. Let us assume a=1

This will give x = 2 and y = 0,

Put these values in the expression and you will get the answer as 16.

Answer : (A)

Q . 8

Pick values for a, b and c, such that their sum is 2s. Let us assume a = 2s, b = s and c = -s

(here you should not assume a,b or c to be zero because that will make the elimination of options difficult)

Put these values in the expression and you will get 2s^{2}

Now check all the four options to see which of them will give the value 2s^{2} on putting a=2s, b=s and c=-s

Answer : (C)

Algebra should be a cakewalk for you from now on, we guess !!

Isn’t it?

If not, we will be updating more posts and mock tests to help you achieve what you have planned for. Till then, work hard and work smart.

Happy Learning!

All the Best !

**-Team**

Perfectly awesome work guys!

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Thank you Ayush 🙂

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In the beginning you have mentioned a+b+c under both ‘symmetrical’ and ‘non symmetrical’ expressions.

Doubt: bc + ab + ca = abc is symmetrical and hence we can assume

a=b=c

How? Lhs is symmetrical in itself, rhs is individually symmetrical. So, do we see both Lhs and rhs for checking symmetrical expressions?

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Error rectified, it is ‘a+b-c’ under unsymmetrical category. Also second part of your question, the RHS is not symmetrical, as we can consider this Symmetry only in questions where we can compare the expressions or there is a ‘plus’ sign between expressions.

To check for symmetry search for the types that are mentioned in the post. Considering SSC CGL these methods are enough to identify the symmetry.

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Also, how does one actually check symmetrical expressions just in case one fails to conclude from normal visualization in a particular question or case? Is there also a proper method for it?

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Always remember the powers will always be the same of all the expressions in symmetry.

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Sir i am not able to understand the difference ..

SYMMETRIC EXPRESSIONS are those expressions which have SAME COEFFICIENT and SAME POWERS ??

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Sir what exactly are SYMMETRIC EXPRESSIONS ??

Those expressions in which POWER of each expression is same and COEFFICIENT of each expression is same ????

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Mink, do not get confused, it is very simple. See if you are comparing an equation in which both sides you have variables. Then, the equation is symmetric if the number of operators are same and their powers on both sides. Even if powers are not some then also it works most of the times but there may be some exceptions, but they won’t be applicable for SSC.

Like in example 4 above, ab+bc+ca= axbxc. Now you can see here that in LHS the degree is one, and both side three operators are there. Actually there are many ways with which you can identify the symmetry of expressions. Just solve the examples yourself and some similar questions, you will definitely get the idea. Happy Learning

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