__Surds__**:- **What actually are surds? its actually irrational numbers in the form of the root (√). Yes, that’s true. You are quite known to any numbers in the form “n√a”, where a is a natural number, and n is any positive integer.

Let’s look into some examples to have a clear picture of what actually I’m trying to say.

We call root 2, root 5 or three root 17 as –

√2, √5, ∛17 which are some examples of surds.

To be more conceptually clear, they are the** “nth root of a number”.**

Now, you must be clear with what actually surds are. Let’s proceed to the laws of surds.

There are some laws related to surds, which I think most of you know. Once have a look.

__Laws of surds__

^{n}√a = a^{1/n}^{n}√ab =^{n}√a ×^{n}√b = a^{1/n}× b^{1/n}^{n}√a/b =^{n}√a /^{n}√b- (
^{n}√a )^{n}= a ^{m}√ (^{n}√a )=^{mn}√a- (
^{n}√a)^{m}=^{n}√a^{m}

This is what laws of surds actually are. There were no hardcore formulas for you to find difficult. So, it’s better to learn them by heart.

Before going to the next topic I will let you know certain comparisons between surds. These are quite important when it comes to the exam perspective.

__Comparison between surds:-__

- If two Surds are of the same order i.e. √𝑎
^{m}and √𝑏^{m}, Then the one whose base is larger, is the larger of two.

e.g. √29^{3} , √15^{3} , √69^{3} , √126^{3} ⇒ √15^{3} < √29^{3} < √69^{3} < √126^{3}.

**(2)** If two Surds are of distinct order, we change them into the Surds of the same order. This order is the L.C.M. of the orders of the given Surds.

To understand it better look at the example given below:-

** Ques.** Which is larger √6 &

__Solution__

As the orders of the two numbers are distinct, first we need to find out the L.C.M. of the orders. Since the orders here are 2 and 3, whose L.C.M. is 6. So, we multiply and divide our numerator and denominator of the order by 6.

Let a = √6 = (6)^{1/2}

What we have to do is now multiply and divide numerator and denominator of the order by 3

= (6)^{1×3/2×3} = (6)^{3/6 }, which can now be written as –

= ^{6}√6^{3} = ^{6}√216 (since, 6^{3} =216)

And now let b = ^{3}√4 = (4)^{1/3} = (4)^{1×2/3×2} = (4)^{2/6 }= ^{6}√4^{2}= ^{6}√16

We see that a and b both have the same orders and hence we can conclude that a> b {we learned that in (1)}.

Therefore, √6 >√4^{3}.

Thus, we see that it is quite easy and understandable. As it must be clear so now look some of the questions asked in __SSC-CGL EXAM.__

Here we go…

**Ques:- Which of the following is TRUE? (SSC-CGL,TIER II)**

- 1/
^{3}√12 > 1/^{4}√29> 1/√5 - 1/
^{3}√29 > 1/^{3}√12> 1/√5 - 1/ √5 > 1/
^{3}√12> 1/^{4}√29 - 1/ √5 > 1/
^{4}√29> 1/^{3}√12 - Only I (2) Only II (3)Only III (4)Only IV

Solution:- What we see, we have three numbers

Let a= 1/ ^{3}√12

b= 1/^{4}√29

c = 1/√5

There is nothing new to it. As we have learned from the comparison no. (2) same implies here as well. You can also look at the example we just did above.

So, since all the numbers have distinct orders we will find the L.C.M. of the orders. So here it goes.

L.C.M. of 3,4,2 is 12.

So we’ll do the same what we did previously.

a= 1/ ^{3}√12 = 1/ (12)^{1/3}= 1/(12)^{1×4/3×4 }= 1/ (12)^{4/12 }= 1/^{12}√20736

b= 1/ ^{4}√29 = 1/ (29)^{1/4}= 1/(29)^{1×3/4×3 }= 1/ ^{12}√24389

c=1/ √5 = 1/ (5)^{1/2}= 1/(5)^{1×6/2×6 }= 1/ ^{12}√15625

Now, as we have all our terms in the same order we can easily distinguish which is comes first.

Here, 1/ √5 > 1/^{3}√12> 1/^{4}√29

So, our answer is III.

This is the pattern how questions are actually asked in the __QUANT SECTION OF SSC-CGL EXAM.__

Also, you see questions asked from this section are generally of this form. Of course, yes you have to focus on your speed. Enhance it.

Let’s look into another example asked in __SSC-CGL.__

** Ques**:- Which of the following statements(s) is/are TRUE?

- 2/ 3√5 < 3/2√5 < 5/4√3
- 3/ 2√5 < 2/3√3 < 7/4√5

(i) Only I (ii) Only II

(iii) Both I and II (iv) Neither I nor II

**Solution: **The solution to this problem is the same as we have done previously, but what makes it unique is the denominator term.

As we did before we find the L.C.M. of the order, here we find out the L.C.M. of the whole denominator term.

So, the L.C.M. of 3√5, 2√5, 4√3 is 24√15.

I. 2/ 3√5 ×24√15 , 3/2√5 ×24√15 , 5/4√3 × 24√15

=> 16√3, 36√3 , 30√5

Clearly, we see that 16√3< 36√3 < 30√5

Therefore, 2/ 3√5 < 3/2√5 < 5/4√3.

II. 3/ 2√5, 2/3√3,7/4√5

We have to do the same here as well.

L.C.M. here of the denominator terms are 12√15.

3/ 2√5 × 12√15, 2/3√3 × 12√15, 7/4√5 × 12√15.

- 18√3, 8√5, 21√3

Here we see that 8√5 < 18√3 < 21√3

Then, 2/3√3 <3/ 2√5< 7/4√5.

So, our answer is option (i).

Simple? Isn’t it.

As now you must have been clear regarding Surds and its applications. So, let’s now move to __INDICES__

__INDICES__

Let’s begin with the definition of Indices.

Any number in the form a^{n}, a is any real number and n is a positive integer, is known as Indices. Here a is called the base and n is the index or exponent of a.

Like, surds there too are laws regarded to indices.

__Laws of Indices__**:-**

1. a^{m }× a^{n} = a^{m+n}

2. a^{x }/a^{y}= a^{x-y}

3. (a^{m})^{ n} = a^{mn} = (a^{n})^{ m}

4. a^{m^n} = a^{m×m×m……….n times}

5. (ab)^{ n} = 𝑎^{n}𝑏^{n}

6. (a/b ) ^{n}= a^{n }/ b^{n}, b ≠ 0

7. a^{0}= 1

You must have come across such type of indices in your textbook during your school days.

Let look at some questions:-

**Ques:-** 81× 81× 81× 81= 9^{?}

**Sol:-** As we know that 9^{2}= 81

So, above have four 81 they can be written as-

9^{2 }×9^{2}×9^{2}×9^{2} = 9^{2+2+2+2 }= 9^{8 } {Using the first law of indices]

So, we have 81× 81× 81× 81= 9^{8}, is our required answer.

**Ques:** Solve (1000)^{5}÷(10)^{3}

**Solution:-** As in the above problem 1000 can be written as 10^{3}.

We have,

(1000)^{5}= (10^{3})^{5}= 10^{15}

=> 10^{15}÷10^{3} = 10^{12} is the solution.

Since our objective is to solve questions regarding __SSC-CGL EXAM. __ So, let’s see what has been asked and proceed to those questions.

**Ques:** Which of the following statement (s) is/are TRUE?

I. (0.7)^{2}+ (0.07)^{ 2} + (11.1)^{ 2}> 123.8

II. (1.12)^{ 2}+ (10.3)^{ 2} + (1.05)^{ 2} > 108.3

(1) Only I (2) Only II

(3) Both I and II (4) Neither I nor II

**Solution : –**

I. Taking the L.H.S.

=> (0.7)^{2}+ (0.07)^{ 2} + (11.1)^{ 2}

=> 0.49 + 0.0049 + 123.21 = 123.7049 < 123. 8

So, I is wrong statement.

II. Taking the L.H.S.

- (1.12)
^{ 2}+ (10.3)^{ 2}+ (1.05)^{ 2} - (1+0.12)
^{ 2}+ (10+0.3)^{ 2}+ (1+0.05)^{ 2} - (1 + 0.0144 + 0.24) + (100 + 0.09 + 6) + (1 + 0.0025 + 0.1)
- 1.2544 + 106.09 + 1.1025 = 108.4469 > 108.3

We see that our second statement is true.

So, option (2) is correct.

**Ques:-** Which of the following statement (s) is/are TRUE?

I. (0.03/0.2) + (0.003/0.02) +( 0.0003/0.002) +(0.00003/0.0002) =0.6

II. (0.01) + (0.01)^{ 2} + (0.001)^{2}= 0.010101

(1) Only I (2) Only II

(3) Both I and II (4) Neither I nor II

**Solution:-**

I. (0.03/0.2) + (0.003/0.02) + (0.0003/0.002) +

(0.00003/0.0002)

= 3/20 + 3/20 + 3/20 + 3/20

=3/20×4 = 3/5 = 0.6

So, our first statement is true.

II. (0.01)+ (0.01)^{ 2} + (0.001)^{2}

= 0.01 + 0.0001 + 0.000001 = 0.010101.

So, our second statement Is also true.

So, the correct option is (3).