Detailed Solutions: Surds and Indices
Surds and Indices is one of the most important chapters in Mathematics for SSC, Banking, Railway, NDA, and other competitive exams. This topic helps students understand square roots, cube roots, powers, and exponential expressions, which are frequently used in algebra and arithmetic.
In this post, you will get the Top 10 Important Questions on Surds and Indices with Solutions to boost your exam preparation.
What is Surds and Indices in Mathematics?
Surds are irrational roots such as √2, √3, or ∛5.
Indices (or exponents) show the power of a number, such as 2³, 5², or 10⁻¹.
These topics are used to simplify expressions, solve equations, and handle powers in mathematics.
Question: 1
सरल कीजिए : \(\sqrt{-\sqrt{5} + \sqrt{5 + 4\sqrt{6 + 2\sqrt{5}}}} \)
[A] 2 [B] 5
[C] \(\sqrt{2}\) [D] 0
⚡Solution:
Answer (c) : \(\sqrt{2}\)
Step – 1: \(6 + 2\sqrt{5}\) को \( a^2 + b^2 + 2ab \) में बदलें जिससे \((a + b )^2\) प्राप्त होगा ।
\(\Rightarrow \) यहाँ \(2ab = 2\sqrt{5}\) अर्थात् \(a = 1 = \sqrt{1} \) तथा \(b = \sqrt{5} \)
\(\Rightarrow \) \(6 + 2\sqrt{5}\) = \(1 + 5 + 2\sqrt{5}\)
\(\Rightarrow \)\( (\sqrt{1})^2 + (\sqrt{5})^2 + 2\sqrt{5}\) = \( ( 1 + \sqrt{5} ) ² \)
Step – 2: \(\sqrt{-\sqrt{5} + \sqrt{5 + 4\sqrt{( 1 + \sqrt{5} ) ²}}} \)
\(\Rightarrow \)\(\sqrt{-\sqrt{5} + \sqrt{5 + 4( 1 + \sqrt{5} )}} \)
\(\Rightarrow \)\(\sqrt{-\sqrt{5} + \sqrt{5 + 4 + 4\sqrt{5} )}} \)
\(\Rightarrow \)\(\sqrt{-\sqrt{5} + \sqrt{9 + 4 \sqrt{5} }} \)
Step – 3: पुनः \(9 + 4\sqrt{5}\) को \( a^2 + b^2 + 2ab \) में बदलें जिससे \((a + b )^2\) प्राप्त होगा ।
\(\Rightarrow \) यहाँ \(2ab = 4\sqrt{5}\) अर्थात् \(a = 2\) तथा \( b = \sqrt{5} \)
\(\Rightarrow \) \(9 + 4\sqrt{5}\) = \(4 + 5 + 2 × 2 × \sqrt{5}\)
\(\Rightarrow \)\( (2)^2 + (\sqrt{5})^2 + 4\sqrt{5}\) = \( ( 2 + \sqrt{5} ) ² \)
Step-4: \(\sqrt{-\sqrt{5} + \sqrt{( 2 + \sqrt{5} ) ² }} \)
\(\Rightarrow \)\(\sqrt{-\sqrt{5} + ( 2 + \sqrt{5} ) } \)
\(\Rightarrow \)\(\sqrt{-\sqrt{5} + 2 + \sqrt{5} } \)
\(\Rightarrow \)\(\sqrt{2}\)
Question: 2
सरल कीजिए : \(\sqrt{8 – 2\sqrt{15}}\)
[A] \(\sqrt{5}\) + \(\sqrt{3}\) [B] \(3\) – \(\sqrt{5}\)
[C] \(5\) – \(\sqrt{3}\) [D] \(\sqrt{5}\) – \(\sqrt{3}\)
⚡Solution:
Answer (d) : \(\sqrt{5}\) – \(\sqrt{3}\)
Step – 1: \(8 – 2\sqrt{15}\) को \( a^2 + b^2 – 2ab \) में बदलें जिसके बाद \((a – b )^2\) प्राप्त होगा ।
\(\Rightarrow \) यहाँ \(2ab = 2\sqrt{15}\) अर्थात् \(a = \sqrt{3} \) तथा \(b = \sqrt{5} \)
\(\Rightarrow \) \(8 – 2\sqrt{15}\)
= \(3 + 5 – 2\sqrt{15}\)
\(\Rightarrow \)\( (\sqrt{3})^2 + (\sqrt{5})^2 – 2\sqrt{3}\sqrt{5}\)
= \( ( \sqrt{3} – \sqrt{5} ) ² \)
Step – 2: \(\sqrt{( \sqrt{3} – \sqrt{5} ) ²} \)
\(\Rightarrow \)\( (\sqrt{3} – \sqrt{5} ) \)
Question: 3
सरल कीजिए : ( \(2^5\) × \(2^2\)) ÷ \(2^4\)
[A] 4 [B] 3
[C] 2 [D] 8
⚡Solution:
Answer (d) : 8
Step-1: Product Rule: When multiplying with the same base, add the powers: \(2^5\) × \(2^2\) = \(2^7\).
Step-2: Quotient Rule: When dividing with the same base, subtract the powers: \(2^7\) ÷ \(2^4\) = \((2)^{7-4}\)
Step-3: Answer : \(2^3\) = 8
Question: 4
\(x\) का मान ज्ञात करें : \((9)^{x-1}\) = 27
[A] 4 [B] 3.5
[C] 2.5 [D] 3
⚡Solution:
Answer (c) :
Step-1 : Express both sides with the same base (Base 3).
\((3^2)^{x-1}\) = \(3^3\)
Step-2: Multiply the powers: \((3)^{2x-2}\) = \(3^3\)
Since the bases are the same, the exponents must be equal: \(2x – 2 = 3\)
Solve for \(x: 2x\) = 5 \(\rightarrow \) \(x = 2.5\)
Why Surds and Indices Questions Are Important for Exams
- Forms the foundation of Algebra
- Used in Quadratic equations
- Frequently asked in SSC, Banking, and Railway exams
- Helps in solving roots and powers quickly
Conclusion
Mastering Surds and Indices will help you score high in mathematics exams. Practice these important questions regularly to improve speed and accuracy.