Tag: vedic methods of multiplication

Questions Related to vedic methods of multiplication

Two pipes $A$ and $B$ can fill a cistern in $37\dfrac {1}{2}$ minutes and $45$ minutes respectively. Both pipes are opened, the cistern will be filled just in half an hour, if the pipe $B$ is turned off after.

vedic methods of multiplication history of mathematics maths

  1. $15\ minutes$

  2. $10\ minutes$

  3. $5\ minutes$

  4. $9\ minutes$

Show answer
Correct Option: D
Explanation:

Let the capacity of cistern be $225$ units $\left (LCM\ of \dfrac {75}{2}\ and\ 45\right )$
$A$ does $= \dfrac {225}{75}\times 2 = 6\ units/ min$
$B$ does $= \dfrac {225}{45} = 5\ units/ min$.
Let pipe is turned off after $x$ minutes.
According to the question,
$6\times 30 + 5\times x = 225$
$5x = 225 - 180 = 45$
$x = 9$
After $9$ minutes, pipe $B$ is turned off.

Two trains starts from stations $A$ and $B$ and travel towards each other at speed of $50\ km/hr$ and $60\ km/hr$ respectively. At the time of their meeting, the second train has travelled $120\ km$ more than the first. The distance between $A$ and $B$ is

vedic methods of multiplication history of mathematics maths

  1. $990\ km$

  2. $1200\ km$

  3. $1320\ km$

  4. $1440\ km$

Show answer
Correct Option: C
Explanation:

Speed of train $A = 50\ kmph$
Speed of train $B = 60\ kmph$
Since, time is constant
$Speed \propto$ Distance covered
$\dfrac {S _{A}}{S _{B}} = \dfrac {D _{A}}{D _{B}}$
$\dfrac {50}{60} = \dfrac {D _{A}}{D _{B}}\Rightarrow \dfrac {5}{6} = \dfrac {D _{A}}{D _{B}}$
Given that train $B$ has travelled $120\ km$ extra.
$6x - 5x = 120$
$x = 120$
The distance between $A$ and $B = 6x + 5x = 11x$
$= 11\times 120 = 1320$
Alternate Method:
Let train $A$ start form station $A$ and $B$ from station $B$.
Let the trains $A$ and $B$ meet after/ hours.
$\therefore$ Distance covered by train $A$ in $t$ hours $= 50t$
Distance covered by train $B$ in $t$ hours $= 60t\ km$.
According to the question,
$60t - 50t = 120$
$\Rightarrow t = \dfrac {120}{10} = 12\ hours$.
$\therefore$ Distance $AB = 50\times 12 + 60 \times 12$
$= 600 + 720 = 1320\ km$.

If $A, {A} _{1}, {A} _{2}, {A} _{3}$ be the area of the in-circle and ex-circles, then $\dfrac {1}{\sqrt {{A} _{1}}}+\dfrac {1}{\sqrt {{A} _{2}}}+\dfrac {1}{\sqrt {{A} _{3}}}$ is equal to

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $\dfrac {1}{\sqrt {{A}}}$

  2. $\dfrac {2}{\sqrt {{A}}}$

  3. $\dfrac {3}{\sqrt {{A}}}$

  4. $None$

Show answer
Correct Option: A
Explanation:

$A _1={\pi}{r _1}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-a)^{2}}$

$A _2={\pi}{r _2}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-b)^{2}}$
$A _3={\pi}{r _3}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-c)^{2}}$
$A={\pi}{r}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s)^{2}}$
$\dfrac{1}{\sqrt{A _1}}+\dfrac{1}{\sqrt{A _2}}+\dfrac{1}{\sqrt{A _3}}=\dfrac{1}{\sqrt{\pi}}\bigg[\dfrac{s-a}{\Delta}+\dfrac{s-b}{\Delta}+\dfrac{s-c}{\Delta}\bigg]=\dfrac{1}{\sqrt{\pi}\Delta}[3{s}-(a+b+c)]=\dfrac{s}{\sqrt{\pi}\Delta}=\dfrac{1}{\sqrt{A}}$

(?)$-19657-33994=9999$

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $63650$

  2. $53760$

  3. $59640$

  4. $61560$

  5. None of these

Show answer
Correct Option: A
Explanation:

Let $x-53651=9999$
Then, $x=9999+53651=63650$

Perform the indicated operations:
$+7(-2)+(-8)+(+3)=$

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. -17

  2. -18

  3. -19

  4. -20

Show answer
Correct Option: C
Explanation:

The value of $+7(-2)+(-8)+(+3) $

$= -14-8+3 $
$= -19$

The algebraic expression for the statement: 'Product of x and a are subtracted from the product of b and y'.

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. ax - by

  2. x + a - by

  3. by - ax

  4. xa - b - y

Show answer
Correct Option: C
Explanation:

Product of x and a = xa

product of b and y = by
On subtraction: by-ax

(?)$+3699+1985-2047=31111$

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $34748$

  2. $27474$

  3. $30154$

  4. $27574$

Show answer
Correct Option: B
Explanation:

$x+3699+1985-2047=31111$
$\Rightarrow$ $x+3699+1985=31111+2047$
$\Rightarrow$ $x+5684=33158$
$\Rightarrow$ $x=33158-5684=27474$

$(4300731)-$? $=2535618$

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $1865113$

  2. $1775123$

  3. $1765113$

  4. $1675123$

  5. None of these

Show answer
Correct Option: C
Explanation:

Let $4300731-x=2535618$
Then, $x=4300731-2535618=1765113$

Subtract $347657$ by $238294$ using vinculum numbers.

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $109363$

  2. $100363$

  3. $109373$

  4. $109563$

Show answer
Correct Option: A
Explanation:

1) The first thing we do is write these numbers one on top of the other.

347657
238294

2) Now we just start subtracting vertically, and whenever our number is negative we represent it as a vinculum number.

347657
238294

3) This next subtraction is 7 – 8 = -1…so we just write this as 1 and continue on.

347657
238294
1 1 1 443

4) The ‘1‘ and ‘4‘ here are considered as two separate groups (since theres a non-vinculum number in-between them), so each is subtracted from 10. This has the effect of reducing the “previous one” by one.

109363

In $32$ Ekanyunena Purvena of the digit $2$ is?

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $13$

  2. $12$

  3. $22$

  4. $10$

Show answer
Correct Option: C