Monika takes $\dfrac{1}{6}$ hour to walk to the playground.

In ancient India, squares and circular altars were used for household rituals.

The geometry of the Vedic period originated with the construction of altars (or vedis) and fireplaces for performing Vedic rites. Square and circular altars were used for household rituals, while altars, whose shapes were combinations of rectangles, triangles and trapeziums, were required for public worship.

Let the number be $x$.

Since,

$ \dfrac{x}{2}=12 $

$ x=24 $

Since,

$ \Rightarrow \dfrac{3}{4}\times 24 $

$ \Rightarrow 18 $

Hence, this is the answer.

Sonia talked to Shivani for $=\dfrac{1}{4}$ hour

Victor can throw a ball $=50\dfrac{3}{5}$ feet

Let the number be $x$. Then
$60$% of $\cfrac{3}{5}$ of $x=36$
$\Rightarrow$ $\cfrac{60}{100}\times \cfrac{3}{5}\times x=36$
$\Rightarrow$ $x=(36\times \cfrac{25}{9})=100$
$\therefore$ Required number $=100$

5/4 , 5/9
Difference of cross product = 45 - 20 = 25
If the difference of the cross product is positive then the first fraction is larger.
hence 5/4 is larger.

'Ekadhikena Purvena'   means  “by one more than the previous”
So, Ekadhikena Purvena of the number 37 is 38.

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

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.

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.

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

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

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

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

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

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

Let the length of the rod be $x$ $m$
$\therefore$ Length of the rod coloured be red$=\cfrac{1}{10}$ $x$ $m$
$\therefore$ Length of the rod coloured be orange$=\cfrac{1}{20}$ $x$ $m$
Length of the rod coloured be yellow$=\cfrac{1}{30}$ $x$ $m$
Length of the rod coloured be green$=\cfrac{1}{40}$ $x$ $m$
Length of the rod coloured be blue$=\cfrac{1}{50}$ $x$ $m$
Length of the rod coloured be black$=\cfrac{1}{60}$ $x$ $m$
Length of the rod coloured be violet$=12.08m$
$\therefore$ According to question
$\cfrac { 1 }{ 10 } x+\cfrac { 1 }{ 20 } x+\cfrac { 1 }{ 30 } x+\cfrac { 1 }{ 40 } x+\cfrac { 1 }{ 50 } x+12.08=x$
$\Rightarrow x-\cfrac { 49 }{ 200 } x=12.08\Rightarrow \cfrac { 151 }{ 200 } x=12.08\Rightarrow x=16$
$\therefore$ Length of the rod $=16m$

After interchanging the signs,
$2\times 15 + 15\div 3 - 8 = ?$
$\Rightarrow 2\times 15 + 5 - 8 = ?\Rightarrow 30 + 5 - 8 =?$
$? = 35 - 8 = 27$.

$ 101 \dfrac {3}{5} = \dfrac {101 \times 5 + 3}{5} = \dfrac {505+3}{5} = \dfrac {508}{5}$

Shorter candle ray ${L} _{1}$ burns in 3.5 hours.
So in 2 hours $\displaystyle\frac{{L} _{1}}{3.5} \times 2$ of the candle will burn
Remaining length $= {L} _{1} - \displaystyle\frac{2{L} _{1}}{3.5} = \displaystyle\frac{1.5}{3.5}{L} _{1}$
The longer candle, ${L} _{2}$ burns in 5 hours. So in 2 hours $\displaystyle\frac{{L} _{2}}{5} \times 2$ will burn
Remaining length $= {L} _{2} - \displaystyle\frac{2{L} _{2}}{5} = \displaystyle\frac{3}{5} {L} _{2}$
Equating $\displaystyle\frac{1.5}{3.5} {L} _{1} = \displaystyle\frac{3}{5}{L} _{2}$
$\Rightarrow   \displaystyle\frac{{L} _{1}}{{L} _{2}} = \displaystyle\frac{3}{5} \times \displaystyle\frac{3.5}{1.5} = \displaystyle\frac{7}{5}$
Note : There is a mistake in question. It should read the short and the long ones can burn respectively for 5 hours and 3.56 hours. The answer $\left(\displaystyle\frac{{L} _{1}}{{L} _{2}}\right)$ will come out to be $\displaystyle\frac{5}{7}$ in that case.

The given Fraction are $\displaystyle \frac { 2 }{ 5 }, \frac {3} {10} , \frac {9} {10}, \frac {16} {35} $
LCM of 5, 10, 14, 35 = $\displaystyle (5\times 2\times 7\times)= 70 $
Now change each of the following into an equivalent fraction having 70 as its denominator.
Now, $\displaystyle \frac { 2 }{ 5 } =\frac { 2\times 14 }{ 5\times 14 } =\frac { 28 }{ 70 } \ \frac { 3 }{ 10 } =\frac { 3\times 7 }{ 10\times 7 } =\frac { 21 }{ 70 } \ \frac { 9 }{ 14 } =\frac { 9\times 5 }{ 14\times 5 } =\frac { 45 }{ 70 } \ and\quad \frac { 16 }{ 35 } =\frac { 16\times 2 }{ 35\times 2 } =\frac { 32 }{ 70 } \ Clearly,\quad \frac { 28 }{ 70 } >\frac { 21 }{ 70 } <\frac { 45 }{ 70 } >\frac { 32 }{ 70 } \ Hence,\quad \frac { 3 }{ 10 } <\frac { 2 }{ 5 } <\frac { 16 }{ 35 } <\frac { 9 }{ 14 } $

$\dfrac { 1 }{ \sqrt { 11-2\sqrt { 30 }  }  } -\dfrac { 3 }{ \sqrt { 7-2\sqrt { 10 }  }  } -\dfrac { 4 }{ \sqrt { 8+4\sqrt { 3 }  }  } $

Statement-1: Total quantity of milk bought = 7$\dfrac{1}{2}$litres = $\dfrac{15}{2}$litres
Quantity of milk consumed = 5$\dfrac{3}{4}$litres
                                            = $\dfrac{23}{4}$ litres
$\therefore$ Quantity of milk left = $\big(\dfrac{15}{2}$ - $\dfrac{23}{4}$$\big)$ litres
= $\big(\dfrac{30-23}{4}\big)$ litres = $\dfrac{7}{4}$ = 1$\dfrac{3}{4}$ litres
Statement-2: Let total number of pages in the book be x.
According to question,
$\dfrac{3}{5}$x + 80 = x
$\Rightarrow$ x - $\dfrac{3}{5}$x = 80 $\Rightarrow$ x = $\dfrac{80\times5}{2}$
$\Rightarrow$ x = 200
$\therefore$ Total number of pages = 200
Hence, only statements-2 is true.

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.

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 $AB = 50\times 12 + 60 \times 12$
$= 600 + 720 = 1320\ km$.