Surds & Indices Questions and Answers

Q: Which among \(3\sqrt{8}\), 3, \(\sqrt[3]{1000}\), 5 is greatest ?

The greatest is \(\sqrt[3]{1000}\).Evaluate each number:\[ \sqrt[3]{1000} = 10, \quad \sqrt[3]{8} = 2 \]So the values are 10, 2, 3 and 5, and \(\sqrt[3]{1000} = 10\) is the greatest. The stem prints the first number as \(3\sqrt{8}\), while option C has \(\sqrt[3]{8}\); even read as \(3\sqrt{8} \approx 8.49\), it is still less than 10, so the answer does not change.

Q: The value of \(\left(\frac{1}{6^{-2}}\right) \times (81)^{-3/4}\) is equal to :

The value is \(\frac{4}{3}\).Use \(\frac{1}{6^{-2}} = 6^{2}\) and \(81 = 3^{4}\).\[ \begin{aligned} 6^{2} \times (3^{4})^{-3/4} &= 36 \times 3^{-3} \\ &= \frac{36}{27} = \frac{4}{3} \end{aligned} \]Reading \(6^{-2}\) in the denominator as \(\frac{1}{36}\) in the numerator gives \(\frac{1}{972}\), and \(\frac{3}{4}\) is the reciprocal of the answer.

Q: If \( m \) and \( n \) are whole numbers such that \( m^{n} = 121 \), the value of \( (m-1)^{n+1} \) is:

\( 121 = 11^{2} \), so \( m = 11 \) and \( n = 2 \).\[ (m-1)^{n+1} = (11-1)^{3} = 10^{3} = 1000 \]The whole-number condition rules out other roots such as \( 121^{1} \) with \( n = 1 \), which would need \( m = 121 \) - and the standard convention takes the prime-power form.

Important Instructions

21.

Which among \(3\sqrt{8}\), 3, \(\sqrt[3]{1000}\), 5 is greatest ?

A.5
B.3
C.\(\sqrt[3]{8}\)
D.\(\sqrt[3]{1000}\)

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: D

The greatest is \(\sqrt[3]{1000}\).

Evaluate each number:

\[ \sqrt[3]{1000} = 10, \quad \sqrt[3]{8} = 2 \]

So the values are 10, 2, 3 and 5, and \(\sqrt[3]{1000} = 10\) is the greatest. The stem prints the first number as \(3\sqrt{8}\), while option C has \(\sqrt[3]{8}\); even read as \(3\sqrt{8} \approx 8.49\), it is still less than 10, so the answer does not change.

Discuss in Forum

22.

The value of \(\left(\frac{1}{6^{-2}}\right) \times (81)^{-3/4}\) is equal to :

A.\(\frac{2}{3}\)
B.\(\frac{3}{4}\)
C.\(\frac{4}{3}\)
D.\(\frac{1}{2}\)

Exam Year(s)

UKPSC Lower PCS - 2025,

Answer: C

The value is \(\frac{4}{3}\).

Use \(\frac{1}{6^{-2}} = 6^{2}\) and \(81 = 3^{4}\).

\[ \begin{aligned} 6^{2} \times (3^{4})^{-3/4} &= 36 \times 3^{-3} \\ &= \frac{36}{27} = \frac{4}{3} \end{aligned} \]

Reading \(6^{-2}\) in the denominator as \(\frac{1}{36}\) in the numerator gives \(\frac{1}{972}\), and \(\frac{3}{4}\) is the reciprocal of the answer.

Discuss in Forum

23.

\[ \frac{1}{1+a^{(n-m)}} + \frac{1}{1+a^{(m-n)}} = \ ? \]

A.0
B.1/2
C.1
D.\( a^{(m+n)} \)

Answer: C

\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}} \)

\(= \frac{1}{(1+\frac{an}{am})}+\frac{1}{(1+\frac{am}{an})} \)

\(= \frac{am}{(am+an)}+\frac{an}{(am+an)} \)

\(= \frac{(am+an)}{(am+an)} \)

\(= 1\)

Discuss in Forum

24.

If \( m \) and \( n \) are whole numbers such that \( m^{n} = 121 \), the value of \( (m-1)^{n+1} \) is:

A.1
B.10
C.121
D.1000

Answer: D

\( 121 = 11^{2} \), so \( m = 11 \) and \( n = 2 \).

\[ (m-1)^{n+1} = (11-1)^{3} = 10^{3} = 1000 \]

The whole-number condition rules out other roots such as \( 121^{1} \) with \( n = 1 \), which would need \( m = 121 \) - and the standard convention takes the prime-power form.

Discuss in Forum

25.

\((0.04)^{-1.5}\) = ?