Tag: maths

Questions Related to maths

What is the approximate value of the cube root of the number $12?$

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

  1. $2.6$

  2. $2.1$

  3. $2.8$

  4. $2.2$

Show answer
Correct Option: D
Explanation:

$\sqrt[3] {\dfrac{12\times 1000}{1000}}$
$\sqrt[3]{12000} \div 10$
Now find the closest cube root of $12000.$
$22^3 = 10648,$ therefore we can say that $\sqrt[3]{12}$ $\sim$ $\dfrac{22}{10} = 2.2$

If $\displaystyle \sqrt[3]{3\left ( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right )}=2$ then $\displaystyle \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}=$_________

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

  1. $\displaystyle \frac{8}{3}$

  2. 0

  3. 1

  4. -1

Show answer
Correct Option: A
Explanation:

$\displaystyle \sqrt[3]{3\left ( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right )}=2$ 
Let us assume $\left( \sqrt [ 3 ]{ x } -\frac { 1 }{ \sqrt [ 3 ]{ x }  }  \right) $=a
So, $\sqrt [ 3 ]{ 3a } =2$
Taking  cube both sides,
$3a=8$
$a=\frac { 8 }{ 3 } $
So, $\left( \sqrt [ 3 ]{ x } -\frac { 1 }{ \sqrt [ 3 ]{ x }  }  \right) =\frac { 8 }{ 3 } $

The simplest form of $\sqrt[3]{768}$ is

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

  1. $2\sqrt[3]{12}$

  2. $4\sqrt[3]{12}$

  3. $3\sqrt{12}$

  4. $3\sqrt[3]{6}$

Show answer
Correct Option: B
Explanation:

$\sqrt[3]{768}=\sqrt[3]{2^3\times2^3\times2\times2\times3}=2\times2\sqrt[3]{12}=4\sqrt[3]{12}$

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

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: D
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 $4$.

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

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

  1. $10.99$

  2. $20.10$

  3. $12.45$

  4. $10.56$

Show answer
Correct Option: A
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 $\sqrt[3]{1333}$ $= 10.99$

Find the cube root of the number 514.

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

  1. 8.0104

  2. 8.1104

  3. 8.2104

  4. 8.3104

Show answer
Correct Option: A
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.

      Estimate the cube root of the number $40.$

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

      1. $3.1$

      2. $3.2$

      3. $3.4$

      4. $3.9$

      Show answer
      Correct Option: C
      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.

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

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

          1. $9.2$

          2. $10.2$

          3. $10.96$

          4. $11$

          Show answer
          Correct Option: D
          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 $11$.

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

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

          1. $19.003$

          2. $19.00$

          3. $19.32$

          4. $19.008$

          Show answer
          Correct Option: B
          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 $19.00$.

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

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

          1. $8$

          2. $7$

          3. $9$

          4. $10$

          Show answer
          Correct Option: C
          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 $9$.