Tag: vedic methods of multiplication

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Multiple choice vedic methods of multiplication history of mathematics maths

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.

  1. $15\ minutes$
  2. $10\ minutes$
  3. $5\ minutes$
  4. $9\ minutes$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice vedic methods of multiplication history of mathematics maths

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

  1. $990\ km$
  2. $1200\ km$
  3. $1320\ km$
  4. $1440\ km$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $AB = 50\times 12 + 60 \times 12$
$= 600 + 720 = 1320\ km$.

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

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

  1. $\dfrac {1}{\sqrt {{A}}}$
  2. $\dfrac {2}{\sqrt {{A}}}$
  3. $\dfrac {3}{\sqrt {{A}}}$
  4. $None$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

  1. $109363$
  2. $100363$
  3. $109373$
  4. $109563$
Reveal answer Fill a bubble to check yourself
A Correct answer
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