SSC CGL (Prelims) General: Surds and Indices questions with solutions

Q1. Simplify: \(\sqrt{20} + \sqrt{125}\)

1. 6 \(\sqrt{5}\)
2. 7 \(\sqrt{5}\)
3. 8 \(\sqrt{5}\)
4. 9 \(\sqrt{5}\)

Answer: 7 \(\sqrt{5}\)

\(\sqrt{20} = \sqrt{4\cdot 5} = 2\sqrt{5}\) and \(\sqrt{125} = \sqrt{25\cdot 5} = 5\sqrt{5}\). Adding them gives \(2\sqrt{5}+5\sqrt{5}=7\sqrt{5}\).

Q2. What is the value of \(\sqrt{8281} - \sqrt{2401}\)?

1. 32
2. 42
3. 52
4. 62

Answer: 42

\(8281 = 91^2\), so \(\sqrt{8281}=91\). Also, \(2401 = 49^2\), so \(\sqrt{2401}=49\). Therefore, the value is \(91-49=42\).

Q3. Simplify: \(\sqrt{7+4\sqrt{3}} + \sqrt{7-4\sqrt{3}}\).

1. 2
2. 2 \(\sqrt{3}\)
3. 4
4. 4 \(\sqrt{3}\)

Answer: 4

We can write \(7+4\sqrt{3} = (\sqrt{3}+2)^2\) and \(7-4\sqrt{3} = (2-\sqrt{3})^2\). Therefore, the expression becomes \((\sqrt{3}+2) + (2-\sqrt{3}) = 4\).

Q4. Simplify: \(\sqrt{3}\,(\sqrt{12}-\sqrt{27})\).

1. -3
2. 4
3. -5
4. 6

Answer: -3

We have \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\). So the expression becomes \(\sqrt{3}(2\sqrt{3}-3\sqrt{3})=\sqrt{3}(-\sqrt{3})=-3\).