Tag: maths

Questions Related to maths

Find Ekanyunena Purvena of the number $37$.

vedic methods of multiplication history of mathematics maths

  1. $35$

  2. $30$

  3. $36$

  4. None of these

Show answer
Correct Option: C
Explanation:

'Ekanyunena Purvena'   means  “by one less than the previous”
So, Ekadhikena Purvena of the number 37 is 37 - 1 = 36.

Identify the larger fraction between $\dfrac{2}{3}, \dfrac{5}{8}$ using Vedic mathematics.

vedic methods of multiplication history of mathematics maths

  1. $\dfrac{2}{3}$

  2. $\dfrac{5}{8}$

  3. Both are equal

  4. None of these

Show answer
Correct Option: A
Explanation:

2/3, 5/8
Difference of cross product = 16 - 15 = 1

If the difference of the cross product is positive then the first fraction is larger.
So, 2/3 is larger fraction.

Identify the larger fraction between $\dfrac{7}{12}, \dfrac{9}{20}$ using Vedic mathematics.

vedic methods of multiplication history of mathematics maths

  1. $\dfrac{7}{12}$

  2. $\dfrac{9}{20}$

  3. Both are equal

  4. None of these

Show answer
Correct Option: A
Explanation:

7/12, 9/20

Difference of cross product = 140 - 108 = 32

If the difference of the cross product is positive then the first fraction is larger.
So, 7/12 is larger fraction.

Find Ekadhikena Purvena of the number $200$.

vedic methods of multiplication history of mathematics maths

  1. $201$

  2. $199$

  3. $200$

  4. None of these

Show answer
Correct Option: A
Explanation:

'Ekadhikena Purvena'   means  “by one more than the previous”
So, Ekadhikena Purvena of the number 200 is 201.

Find Ekadhikena Purvena of the number $99$.

vedic methods of multiplication history of mathematics maths

  1. $101$

  2. $90$

  3. $95$

  4. None of these

Show answer
Correct Option: D
Explanation:

'Ekadhikena Purvena'   means  “by one more than the previous”
So, Ekadhikena Purvena of the number 99 is 100.

Find Ekanyunena Purvena of the number $99$.

vedic methods of multiplication history of mathematics maths

  1. $100$

  2. $98$

  3. $101$

  4. None of these

Show answer
Correct Option: B
Explanation:

'Ekanyunena Purvena'   means  “by one less than the previous”
So, Ekadhikena Purvena of the number 99 is 99 - 1 = 98.

Find the correct representation of Ekadhikena Purvena of the number $199$.

vedic methods of multiplication history of mathematics maths

  1. $199-1$

  2. $199+1$

  3. $199/100$

  4. None of these

Show answer
Correct Option: B
Explanation:

'Ekadhikena Purvena'   means  “by one more than the previous”
So, Ekadhikena Purvena of the number 199 is 199 + 1 = 200.

Identify the larger fraction between $\dfrac{11}{21}, \dfrac{12}{25}$ using Vedic mathematics.

vedic methods of multiplication history of mathematics maths

  1. $\dfrac{11}{21}$

  2. $\dfrac{12}{25}$

  3. Both are equal

  4. None of these

Show answer
Correct Option: A
Explanation:

11/21, 12/25
Difference of cross product = 275 - 252 = 23

If the difference of the cross product is positive then the first fraction is larger.
So, 11/21 is larger fraction.

$\displaystyle\frac{1}{5}$ of $\displaystyle\frac{2}{7}$ of $\displaystyle \frac{8}{3}$ of $4095=?$

vedic methods of multiplication history of mathematics maths

  1. $642$

  2. $598$

  3. $648$

  4. $475$

  5. None of these

Show answer
Correct Option: E
Explanation:

$\displaystyle \frac{1}{5}$ of $\displaystyle\frac{2}{7}$ of $\displaystyle\frac{8}{3}$ of $4095$
$=\displaystyle\frac{1}{5}\times \frac{2}{7}\times \frac{8}{3}\times 4095=624$.

Hence the correct answer is option E

The value of $3\dfrac{1}{12}$ - $\big[ 1\dfrac{3}{4} $+$\big[$ 2$\dfrac{1}{2}$ - $\big(1\dfrac{1}{2}$ - $\dfrac{1}{3}$ $\big)\big]\big]$

vedic methods of multiplication history of mathematics maths

  1. $\dfrac{1}{2}$

  2. $2$

  3. $1$

  4. $0$

Show answer
Correct Option: D
Explanation:

We have, 3$\dfrac{1}{12}$ - $\bigg[$ 1$\dfrac{3}{4}$ +$\big[2\dfrac{1}{2}$ - $\big(1\dfrac{1}{2}$ - $\dfrac{1}{3}\big)\big]\bigg]$
= $\dfrac{37}{12}$ - $\bigg[$ $ \dfrac{7}{4}$ +$\big[\dfrac{5}{2}$ - $\big(\dfrac{3}{2}$ - $\dfrac{1}{3}$ $\big)\big]\bigg]$
= $\dfrac{37}{12}$ - $\bigg[\dfrac{7}{4}$ + $\big[\dfrac{5}{2}$ - $\big(\dfrac{9-2}{6}\big)\big]\bigg]$   $\bigg[ \because$ LCM of 2,3 =6$\bigg]$
= $\dfrac{37}{12}$ - $\big[\dfrac{7}{4}$ + $\big(\dfrac{5}{2}$ - $\dfrac{7}{6}\big)\big]$
=$\dfrac{37}{12}$ - $\big(\dfrac{7}{4}$ + $\big(\dfrac{15-7}{6}\big)\big]$        $\bigg[\because$ LCM of 2, 6 = 6$\bigg]$
=$\dfrac{37}{12}$ - $\big[\dfrac{7}{4}$ + $\dfrac{8}{6}$ $\big]$
= $\dfrac{37}{12}$ - $\big[\dfrac{21+16}{12}$ $\big]$            $\big[\because$ LCM of 4, 6 = 12 $\big]$
= $\dfrac{37}{12}$ - $\dfrac{37}{12}$ = 0