Mathematics · Quantitative Aptitude

Surds and Indices

362 Questions

Surds and indices questions focus on evaluating square roots, cube roots, and fractional exponents. These topics are a core part of the quantitative aptitude section in many competitive exams. Practicing these problems builds speed and accuracy for solving numerical equations.

Square roots evaluationCube roots calculationExponents and powersFractional exponentsSurds multiplication

Surds and Indices Questions

Multiple choice

Given √(224) r = 13r, the value of radix r is

  1. 10

  2. 8

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

sqrt(224 base r) = 13 base r implies 224 base r = (13 base r)^2. (2*r^2 + 2*r + 4) = (r + 3)^2 = r^2 + 6r + 9. Solving r^2 - 4r - 5 = 0 gives (r-5)(r+1)=0. Since r > 0, r=5.

Multiple choice

The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x2 – N = 0. If i denotes the iteration index, the correct iterative scheme will be
  1. $X_{i+1} = \frac{1}{2}\Big(X_i + \frac{N}{X_i}\Big)$
  2. $X_{i+1} = \frac{1}{2}\Big(X_i^2 + \frac{N}{X_i^2}\Big)$
  3. $X_{i+1} = \frac{1}{2}\Big(X_i + \frac{N^2}{X_i}\Big)$
  4. $X_{i+1} = \frac{1}{2}\Big(X_i - \frac{N}{X_i}\Big)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Newton-Raphson iteration for f(x) = 0 is x_{i+1} = x_i - f(x_i)/f'(x_i). For f(x) = x^2 - N, f'(x) = 2x. Thus, x_{i+1} = x_i - (x_i^2 - N)/(2x_i) = (2x_i^2 - x_i^2 + N)/(2x_i) = (x_i^2 + N)/(2x_i) = 1/2 * (x_i + N/x_i).

Multiple choice maths cube and cube root estimation of cube root estimation of cube roots cubes and cube roots

Find the cube root of the number $120.$

  1. $4.1$
  2. $4.2$
  3. $4.7$
  4. $4.9$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We use the Babylonian Algorithm for cube roots here
According to the algorithm, the cube root is given by the formula 
$x _{n+1}=\dfrac{\left (2x _n+\left (\dfrac N{x _{n^2}}\right )\right )}{3}$
where,
  • $N$ is the number for which cube root is to be found
  • $x _{n}$ is the initial approximation of the cube root
  • $x _{n+1}$ is the subsequent improvement on the cube root 

In this case,
$N = 120$
    $x _0 =4$ since $4^3<40 <5^3$

      $ \therefore$ $x _1 = \dfrac{\left ((2\times4)+\left (\dfrac {120} {4^2}\right )\right )}{3} = \dfrac{\left (8+\left (\dfrac {120}{16}\right )\right )}{3}=4.9$

      $\Rightarrow x _2 =\dfrac{ \left (2\times4.9+\left (\dfrac {120}{(4.9)^2}\right )\right )}{3} = \dfrac{\left (9.8+\left (\dfrac {120}{24.01} \right )\right )}{3}= \dfrac{(9.8+4.99)}{3} = 4.9$

      We can see the value stabilizes around $4.9$. 

      Hence the answer is $'D'.$
      Multiple choice maths cube and cube root estimation of cube root estimation of cube roots cubes and cube roots

      What is the approximate value of the cube root of the number $9?$

      1. $2.08$
      2. $2.19$
      3. $2.34$
      4. $2.51$
      Reveal answer Fill a bubble to check yourself
      A Correct answer
      Explanation

      First multiply and divide by $1,000,000,$ we get
      $\sqrt[3]{\dfrac{9\times 1000,000}{1000,000}}$
      $\sqrt[3]{9,000,000} \div 100$
      Now find the closest cube root of $9,000,000.$
      $208^3 = 8,998,912,$ therefore we can say that $\sqrt[3]{9}$ $\sim$ $\dfrac{208}{100} \sim 2.08$