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

Find the value of cube root of the number $45$. (Round off your number to the nearest whole number)

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We need to find value of $\sqrt[3]{45}$
Take, $n = 45$, choose any starting value of $x$.
So, $3^3$ is $27 < 45$
So, $x = 3$
$x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next} = $ $\dfrac{2}{3}3+\dfrac{45}{3\times 3^2}$
$x _\text{next} = 3.6666$
So, the nearest whole number for the cube root $45$ is $4$.

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

Estimate the value of cube root of the number $1333$.

  1. $10.99$
  2. $20.10$
  3. $12.45$
  4. $10.56$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We need to find $\sqrt[3]{1333}$
Take, $n = 1333$, choose any starting value of $x$.
So, $11^3$ is $1331 < 1333$
So, $x = 11$
$x _\text{next}$ $=$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next}$ $=$ $\dfrac{2}{3}11+\dfrac{1331}{3\times 11^2}$
$x _\text{next}$ $= 10.999 $    ....(1)
Assume $x = 10.99$
$x _\text{next} =$ $\dfrac{2}{3}10.99+\dfrac{1331}{3\times 10.99^2}$
$x _\text{next} = 10.99$     ....(2)
Since we are getting $10.99$ in (1) and (2)
So, the approximate value of $\sqrt[3]{1333}$ $= 10.99$

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 514.

  1. 8.0104

  2. 8.1104

  3. 8.2104

  4. 8.3104
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We use the Babylonian Algorithm for cube roots here
According to the algorithm, the cube root is given by the formula 
$x _{n+1}=\dfrac{(2x _n+(N/x _{n^2}))}{3}$
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,
$N = 514$
    $x _0 =8$ since, $8^3<514 <9^3$

      $ \therefore$ $x _1 = \dfrac{((2\times8)+(514 /8^2))}{3} = \dfrac{(16+(514/64))}{3}=8.0104$

      $\Rightarrow x _2 = \dfrac{((2\times8.0104+(514/(8.0104)^2))}{3} = \dfrac{(16.02.08+((514/64.1666))}{3}=8.0104$

      We can see the value stabilizes around $8.0104$. Hence the answer is A.

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

      Estimate the cube root of the number $40.$

      1. $3.1$
      2. $3.2$
      3. $3.4$
      4. $3.9$
      Reveal answer Fill a bubble to check yourself
      C Correct answer
      Explanation

      We use the Babylonian Algorithm for cube roots here

      According to the algorithm, the cube root is given by the formula 
        $x _{n+1}=\dfrac{(2x _n+(N/x _{n^2}))}{3}$
        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 
        $N = 40$
          $x _0 =3$ since $3^3<40 <4^3$


            $ \therefore$ $x _1 = \dfrac{((2\times3)+(40 /3^2))}{3} = \dfrac{(6+(40/9))}{3}=3.4$

            $\Rightarrow x _2 = \dfrac{(2\times3.4+40/(3.4)^2)}{3} = \dfrac{(6.8+(40/11.56))}{3}= \dfrac{(6.8+3.46)}{3} = 3.4$

            We can see the value stabilizes around 3.4. Hence, the answer is C.

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

          Find the value of cube root of the number $1290$. (Round off your number to the nearest whole number)

          1. $9.2$
          2. $10.2$
          3. $10.96$
          4. $11$
          Reveal answer Fill a bubble to check yourself
          D Correct answer
          Explanation

          We need to find value of $\sqrt[3]{1290}$
          Take, $n = 1290$, choose any starting value of $x$.
          So, $10^3$ is $1000 < 1290$
          So, $x = 10$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} =$ $\dfrac{2}{3}10+\dfrac{1290}{3\times 10^2}$
          $x _\text{next} =  10.966$
          So, the nearest whole number for the cube root $1290$ is $11$.

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

          Find the value of cube root of the number $6860$. (Round off your number to the nearest hundredth)

          1. $19.003$
          2. $19.00$
          3. $19.32$
          4. $19.008$
          Reveal answer Fill a bubble to check yourself
          B Correct answer
          Explanation

          We need to find value of $\sqrt[3]{6860}$
          Take, $n = 6860$, choose any starting value of $x$.
          So, $19^3$ is $6859 < 6860$
          So, $x = 19$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} = $ $\dfrac{2}{3}19+\dfrac{6860}{3\times 19^2}$
          $x _\text{next} = 19.00085 $    ....(1)
          So, the approximate value of $\sqrt[3]{6860}$ nearest hundredth place is $19.00$.

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

          Find the value of cube root of the number $823$. (Round off your number to the nearest whole number)

          1. $8$
          2. $7$
          3. $9$
          4. $10$
          Reveal answer Fill a bubble to check yourself
          C Correct answer
          Explanation

          We need to find value of $\sqrt[3]{823}$
          Take, $n = 823$, choose any starting value of $x$.
          So, $9^3$ is $729 < 823$
          So, $x = 9$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} =$ $\dfrac{2}{3}9+\dfrac{823}{3\times 9^2}$
          $x _\text{next} = 9.386$
          So, the nearest whole number for the cube root $823$ is $9$.

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

          Find the value of cube root of the number $2486$. (Round off your number to the nearest whole number)

          1. $11$
          2. $12$
          3. $13$
          4. $14$
          Reveal answer Fill a bubble to check yourself
          D Correct answer
          Explanation

          We need to find value of $\sqrt[3]{2486}$
          Take, $n = 2486$, choose any starting value of $x$.
          So, $13^3$ is $2197 < 2486$
          So, $x = 13$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} =$ $\dfrac{2}{3}13+\dfrac{2486}{3\times 13^2}$
          $x _\text{next} = 13.56$
          So, the nearest whole number for the cube root $2486$ is $14$.

          Multiple choice maths multiply and divide division trick division division of numbers

          Solve it 
          $\dfrac {\left( {{{\left( {245 + 232} \right)}^2} - {{\left( {245 - 232} \right)}^2}} \right)}{\left( {245 + 232} \right)}$

          1. $4$
          2. $2$
          3. $232$

          4. none of these
          Reveal answer Fill a bubble to check yourself
          D Correct answer
          Explanation
          $=\dfrac{{\left(245+232\right)}^{2}-{\left(245-232\right)}^{2}}{\left(245+232\right)}$
          $=\dfrac{\left(245+232-245+232\right)\left(245+232+245-232\right)}{\left(245+232\right)}$
          $=\dfrac{2\left(232\right)2\left(245\right)}{\left(245+232\right)}$
          $=\dfrac{2,27,360‬}{477}=476.65$
          Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

          The greater number between $\sqrt{17}-\sqrt{12}$ and $\sqrt{11}-\sqrt{6}$ is ____.

          1. $\sqrt{17}-\sqrt{12}$
          2. $\sqrt{11}-\sqrt{6}$

          3. Both are equal

          4. Cannot comare
          Reveal answer Fill a bubble to check yourself
          B Correct answer
          Explanation

          $\sqrt{17}=4.12\ \sqrt {12}=3.46\ \therefore\sqrt{17}-\sqrt{12}=0.66$

          $\sqrt{11}=3.32\ \sqrt6=2.45\ \therefore\sqrt{11}-\sqrt6=0.87$

          $\sqrt{11}-\sqrt6>\sqrt{17}-\sqrt{12}$

          Multiple choice maths indices negative indices law of indices laws of indices

          $\left { \left (\dfrac {3}{4}\right )^{-1} - \left (\dfrac {1}{4}\right )^{-1}\right }^{-1} = ?$

          1. $\dfrac {3}{8}$
          2. $\dfrac {-3}{8}$
          3. $\dfrac {8}{3}$
          4. $\dfrac {-8}{3}$
          Reveal answer Fill a bubble to check yourself
          B Correct answer
          Explanation

          We need to find value of $\left { \left (\dfrac {3}{4}\right )^{-1} - \left (\dfrac {1}{4}\right )^{-1}\right }^{-1} $
          It can be written as $\left (\dfrac {4}{3} - 4\right)^{-1}$ $=$ $\left (\dfrac {-8}{3}\right)^{-1}$ $=$ $-\dfrac {3}{8}$

          Multiple choice maths indices negative indices law of indices laws of indices

          $\left {\left (\dfrac {1}{3}\right )^{-3} -\left (\dfrac {1}{2}\right )^{-3}\right } \div \left (\dfrac {1}{4}\right )^{-3} = ?$

          1. $\dfrac {19}{64}$
          2. $\dfrac {64}{19}$
          3. $\dfrac {27}{16}$
          4. $\dfrac {16}{27}$
          Reveal answer Fill a bubble to check yourself
          A Correct answer
          Explanation
          We need to find value of $\left \{\left (\dfrac {1}{3}\right )^{-3} -\left (\dfrac {1}{2}\right )^{-3}\right \} \div \left (\dfrac {1}{4}\right )^{-3} = ?$
          $\left (\dfrac{1}{3}\right)^{-3}=3^{3}$
          $\left (\dfrac{1}{2}\right)^{-3}=2^{3}$
          $\left (\dfrac{1}{4}\right)^{-3}=4^{3}$
          So, given expression becomes,
          $\dfrac {3^{3}-2^{3}}{4^{3}}$ $=\dfrac {19}{64}$. 
          Hence, A is the right option.
          Multiple choice maths indices negative indices law of indices laws of indices

          If $\sqrt [ 3 ]{ a+\sqrt { b }  } =7+4\sqrt { 3 } $, then $\sqrt [ 3 ]{ { a }^{ 2 }-b } =$

          1. $0$
          2. $1$
          3. $-1$
          4. $7$
          Reveal answer Fill a bubble to check yourself
          B Correct answer
          Explanation

          We have.

          $ \sqrt[3]{a+\sqrt{b}}=7+4\sqrt{3} $

          $ {{(a+\sqrt{b})}^{1/3}}=(7+4\sqrt{3}) $

          $ a+\sqrt{b}={{(7+4\sqrt{3})}^{3}}\ \ ......\ \ (1) $


          $ \text{Similarly,} $

          $ a-\sqrt{b}={{(7-4\sqrt{3})}^{3}}\ \ ......\ \ (2) $


          On multiplying (1) and (2) to. We get,

          $ (a+\sqrt{b})(a-\sqrt{b})={{(7+4\sqrt{3})}^{3}}{{(7-4\sqrt{3})}^{3}} $

          $ {{a}^{2}}-b={{\left[ (7+4\sqrt{3})(7-4\sqrt{3}) \right]}^{3}} $

          $ {{a}^{2}}-b={{\left[ 49-16\times 3 \right]}^{3}} $

          $ {{({{a}^{2}}-b)}^{1/3}}=(1) $

          $ \sqrt[3]{{{a}^{2}}-b}=1 $


          Hence, this is the answer