Tag: vedic methods of multiplication

Questions Related to vedic methods of multiplication

Multiply $12, 18$ using NIkhilam formula (sub-base) of vedic mathematics.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $246$

  2. $236$

  3. $216$

  4. None of these

Show answer
Correct Option: C

Find the cubage of $65$ by Nikhilam formula of Vedic Mathematics.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $27005$

  2. $235625$

  3. $274625$

  4. None of these

Show answer
Correct Option: C

Find the deviation in the cubage of $96$ by Nikhilam formula of Vedic Mathematics.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $04$

  2. $-04$

  3. $6$

  4. None of these

Show answer
Correct Option: B

Find the representation of the rightmost two digits in the cubage of $96$ by Nikhilam formula of Vedic Mathematics.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $100-64$

  2. $-64$

  3. $64$

  4. None of these

Show answer
Correct Option: A

Perform division of $422\div 11$ using Nikhilam Sutra Method on base $10$. Also, find the quotient$(Q)$ and remainder$(R)$.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $Q=40$ and $R=3$

  2. $Q=39$ and $R=5$

  3. $Q=33$ and $R=3$

  4. $Q=38$ and $R=4$

Show answer
Correct Option: D

Multiply $9, 8, 15$ by Nikhilam formula of vedic maths.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $1060$

  2. $1070$

  3. $1080$

  4. None of these

Show answer
Correct Option: C

Multiply $101, 102, 103 $ by Nikhilam formula of vedic maths.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $1061166$

  2. $1061106$

  3. $1661106$

  4. None of these

Show answer
Correct Option: B

Find the cubage of $80$ by Nikhilam formula of Vedic Mathematics.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $512000$

  2. $512200$

  3. $514000$

  4. $516000$

Show answer
Correct Option: A

Multiply $32, 38$ using Nikhilam formula (sub-base) of vedic mathematics____________.

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $1216$

  2. $1261$

  3. $1621$

  4. $1612$

Show answer
Correct Option: A

Which of the following statements is CORRECT?

multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

  1. The product of $\dfrac{231}{119}$ and $\dfrac{117}{118}$ is greater than $\dfrac{231}{119}$

  2. The product of $\dfrac{17}{25}$ and $\dfrac{117}{225}$ is greater than $\dfrac{17}{25}$

  3. The product of $\dfrac{1735}{2001}$ and $\dfrac{2734}{2724}$ is greater than $\dfrac{1735}{2001}$

  4. $\dfrac{1}{3}$ of $\dfrac{4}{5}$ is greater than $\dfrac{3}{4}$ of $\dfrac{8}{7}$

Show answer
Correct Option: C
Explanation:

Option $[A]$  :  $\dfrac{117}{118}<1$.  
The product of any number with a number less than $1$ is less than that number.
So,  $\dfrac{231}{119}\times\dfrac{117}{118}<\dfrac{231}{119}$.

$\Rightarrow[A]$ is not correct.

Option $[B]$  :  $\dfrac{117}{225}<1$
The product of any number with a number less than $1$ is less than that number.
So,  $\dfrac{17}{25}\times\dfrac{117}{225}<\dfrac{17}{25}$.
$\Rightarrow[B]$ is not correct.

Option $[C]$  :  $\dfrac{2734}{2724}>1$
The product of any number with a number greater than $1$ is greater than that number.
So,  $\dfrac{1735}{2001}\times\dfrac{2734}{2724}>\dfrac{1735}{2001}$.
$\Rightarrow[C]$ is correct.

Option $[D]$  :
$\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{15}\approx0.267$   and   $\dfrac{3}{4}\times\dfrac{8}{7}=\dfrac{6}{7}\approx0.857$.
But  $0.857>>0.267$.
So, $[D]$ is not correct.


$\therefore$  The correct answer is  $[C]$.