Theory & Concepts

SSC CGL Surds and Indices Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Surds & Indices specifically designed for the SSC CGL 2025-26 pattern.

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25 min readDifficulty: Easy-Intermediate

Surds and Indices are the backbone of algebraic simplification in SSC exams. Mastering these standard rules and rationalization techniques will save you critical minutes during calculation-heavy questions.

Learning path

Laws of Indices
Surds Rationalization
Infinite Series Hacks
10 Exam-Level Examples

1. Basic Laws of Indices

Memorize these fundamental rules of exponents to simplify powers quickly:

Multiplication & Division

a^m \times a^n = a^{m+n}

\frac{a^m}{a^n} = a^{m-n}

Power of Power

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

Note:

(a^m)^n \neq a^{m^n}

2. Rationalization of Surds

Never leave a root in the denominator. Multiply the numerator and denominator by the conjugate.

Conjugate Rule

To rationalize $\frac{1}{\sqrt{a} + \sqrt{b}}$ , multiply by its conjugate $(\sqrt{a} - \sqrt{b})$ .

\frac{1}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{a} - \sqrt{b}}{a - b}

If $a - b = 1$ (e.g. $\sqrt{3} + \sqrt{2}$ ), the denominator becomes 1, so the result is just the conjugate!

3. Infinite Series Hacks

SSC loves asking these specific infinite root patterns. Use these instant shortcuts:

Plus/Minus Series

\sqrt{n + \sqrt{n + \dots}} = \text{Larger factor}

Find two consecutive factors of $n$ . E.g., for $12$ (factors 4,3), answer is 4.

Product Series

\sqrt{a\sqrt{a\sqrt{a\dots}}} = a

If multiplied infinitely, the answer is the number itself.

4. 10 Solved examples

Question 01Exam Pattern

Find the value of $ \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots \infty}}} $.

Correct answer: b) 4

Solution

Step 1: The given expression is an infinite addition series of the form $ \sqrt{n + \sqrt{n + \dots}} $.

Step 2: Factorize $ n = 12 $ into two consecutive integers.

Step 3: Factors are $ 4 \times 3 = 12 $.

Final calculation: Since the sign is '+', the answer is the larger factor. Result = 4.

Question 02Exam Pattern

Find the value of $ \sqrt{7\sqrt{7\sqrt{7\dots \infty}}} $.

$ 7^2 $

Correct answer: a) 7

Solution

Step 1: The given expression is an infinite multiplication series of the form $ \sqrt{a\sqrt{a\sqrt{a\dots}}} $.

Final calculation: For an infinite product series, the value is always $ a $. Therefore, result = 7.

Question 03Exam Pattern

Find the largest number among: $ \sqrt{2}, \sqrt[3]{3}, \sqrt[4]{4}, \sqrt[6]{6} $.

$ \sqrt{2} $

$ \sqrt[3]{3} $

$ \sqrt[4]{4} $

$ \sqrt[6]{6} $

Correct answer: b) $ \sqrt[3]{3} $

Solution

Step 1: Express the roots as fractional powers: $ 2^{1/2}, 3^{1/3}, 4^{1/4}, 6^{1/6} $.

Step 2: Take the LCM of the denominators (2, 3, 4, 6), which is 12.

Step 3: Multiply each power by 12 to bring them to a common base.

$ 2^{12/2} = 2^6 = 64 $

$ 3^{12/3} = 3^4 = 81 $

$ 4^{12/4} = 4^3 = 64 $

$ 6^{12/6} = 6^2 = 36 $

Final calculation: The largest value is 81, which corresponds to $ \sqrt[3]{3} $.

Question 04Exam Pattern

Find the square root of $ 8 - 2\sqrt{15} $.

$ \sqrt{5} + \sqrt{3} $

$ \sqrt{5} - \sqrt{3} $

$ \sqrt{8} - \sqrt{15} $

$ \sqrt{3} - \sqrt{5} $

Correct answer: b) $ \sqrt{5} - \sqrt{3} $

Solution

Step 1: Use the standard square root identity: $ \sqrt{a+b - 2\sqrt{ab}} = \sqrt{a} - \sqrt{b} $.

Step 2: Find two numbers whose sum is 8 and whose product is 15.

Step 3: The numbers are 5 and 3. (Since $ 5+3=8 $ and $ 5 \times 3 = 15 $).

Final calculation: The square root is $ \sqrt{5} - \sqrt{3} $ (always write the larger number first to keep it positive).

Question 05Exam Pattern

Simplify: $ \frac{1}{\sqrt{3} + \sqrt{2}} $.

$ \sqrt{3} + \sqrt{2} $

$ \sqrt{3} - \sqrt{2} $

$ \sqrt{2} - \sqrt{3} $

Correct answer: b) $ \sqrt{3} - \sqrt{2} $

Solution

Step 1: Rationalize the denominator by multiplying the numerator and denominator by its conjugate, $ \sqrt{3} - \sqrt{2} $.

Step 2: Denominator becomes $ (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1 $.

Final calculation: The numerator is $ \sqrt{3} - \sqrt{2} $. Since the denominator is 1, the result is $ \sqrt{3} - \sqrt{2} $.

Question 06Exam Pattern

Find the value of $ \left(\frac{1}{64}\right)^{-2/3} $.

1/16

Correct answer: a) 16

Solution

Step 1: A negative exponent inverts the fraction. $ (1/64)^{-2/3} = (64)^{2/3} $.

Step 2: Express 64 as a power of 4. $ 64 = 4^3 $.

Step 3: Substitute: $ (4^3)^{2/3} $.

Final calculation: Multiply the powers. $ 4^{3 \times (2/3)} = 4^2 = 16 $.

Question 07Exam Pattern

If $ 2^x = 3^y = 6^{-z} $, find the value of $ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $.

-1

Correct answer: b) 0

Solution

Step 1: Let $ 2^x = 3^y = 6^{-z} = k $.

Step 2: Write numbers in terms of k: $ 2 = k^{1/x} $, $ 3 = k^{1/y} $, and $ 6 = k^{-1/z} $.

Step 3: We know that $ 2 \times 3 = 6 $. Substitute the k terms:

$ k^{1/x} \times k^{1/y} = k^{-1/z} $

Step 4: Using the multiplication rule of indices, add the powers: $ k^{1/x + 1/y} = k^{-1/z} $.

Final calculation: Equate powers: $ 1/x + 1/y = -1/z $. Transpose to get $ 1/x + 1/y + 1/z = 0 $.

Question 08Exam Pattern

Find the value of $ \frac{2^{n+4} - 2 \times 2^n}{2 \times 2^{n+3}} + 2^{-3} $.

7/8

1/8

Correct answer: a) 1

Solution

Step 1: Simplify the first fraction's numerator: $ 2^{n+4} - 2^{n+1} = 2^n(2^4 - 2^1) = 2^n(16 - 2) = 2^n \times 14 $.

Step 2: Simplify the denominator: $ 2 \times 2^{n+3} = 2^{n+4} = 2^n \times 2^4 = 2^n \times 16 $.

Step 3: The fraction becomes $ \frac{2^n \times 14}{2^n \times 16} = 14/16 = 7/8 $.

Final calculation: Add $ 2^{-3} $ (which is 1/8). Result = $ 7/8 + 1/8 = 8/8 = 1 $.

Question 09Exam Pattern

Simplify: $ \sqrt{5 + \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{49}}}}} $.

Correct answer: a) 3

Solution

Step 1: Start solving from the innermost root: $ \sqrt{49} = 7 $.

Step 2: Add to the next term: $ 29 + 7 = 36 $, and $ \sqrt{36} = 6 $.

Step 3: Next term: $ 19 + 6 = 25 $, and $ \sqrt{25} = 5 $.

Step 4: Next term: $ 11 + 5 = 16 $, and $ \sqrt{16} = 4 $.

Final calculation: Final term: $ 5 + 4 = 9 $, and $ \sqrt{9} = 3 $.

Question 010Exam Pattern

If $ a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} $ and $ b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} $, find $ a^2 + ab + b^2 $.

Correct answer: a) 8

Solution

Step 1: Recognize that $ a $ and $ b $ are reciprocals, so $ ab = 1 $.

Step 2: Find $ a+b $. $ a+b = \frac{\sqrt{5}+1}{\sqrt{5}-1} + \frac{\sqrt{5}-1}{\sqrt{5}+1} $.

Step 3: Using formula $ \frac{x+y}{x-y} + \frac{x-y}{x+y} = \frac{2(x^2+y^2)}{x^2-y^2} $. Here, $ x=\sqrt{5}, y=1 $.

Step 4: $ a+b = \frac{2(5 + 1)}{5 - 1} = \frac{12}{4} = 3 $.

Final calculation: We need $ a^2+ab+b^2 = (a+b)^2 - ab = (3)^2 - 1 = 9 - 1 = 8 $.

5. Strategy errors to avoid

Negative Exponent Myth

Students often think $ a^{-n} $ is a negative number. It's not! It's just a reciprocal: $ 1/a^n $. A negative power never changes the sign of the base.

Illegal Addition

You CANNOT add surds directly: $ \sqrt{a} + \sqrt{b} \neq \sqrt{a+b} $. Only like-surds (same base) can be added or subtracted.

Master the next modules

Simplification (BODMAS)Algebraic Identities Linear Equations

SSC CGL Surds and Indices Questions, Formulas & Tricks

Learning path

1. Basic Laws of Indices

Multiplication & Division

Power of Power

2. Rationalization of Surds

3. Infinite Series Hacks

Plus/Minus Series

Product Series

4. 10 Solved examples

Find the value of \( \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots \infty}}} \).

Find the value of \( \sqrt{7\sqrt{7\sqrt{7\dots \infty}}} \).

Find the largest number among: \( \sqrt{2}, \sqrt[3]{3}, \sqrt[4]{4}, \sqrt[6]{6} \).

Find the square root of \( 8 - 2\sqrt{15} \).

Simplify: \( \frac{1}{\sqrt{3} + \sqrt{2}} \).

Find the value of \( \left(\frac{1}{64}\right)^{-2/3} \).

If \( 2^x = 3^y = 6^{-z} \), find the value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \).

Find the value of \( \frac{2^{n+4} - 2 \times 2^n}{2 \times 2^{n+3}} + 2^{-3} \).

Simplify: \( \sqrt{5 + \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{49}}}}} \).

If \( a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \) and \( b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \), find \( a^2 + ab + b^2 \).

5. Strategy errors to avoid

Negative Exponent Myth

Illegal Addition

Master the next modules