Condition for direct proportionality is $\frac{x}{y}$= constant and condition for inverse proportionality is xy= constant

So, the given equation y = x + 3; y is neither directly nor inversely proportional to x.

Let us assume that her distance covered in cm in the map be x.

The scale and kilometer it represent are directly proportional
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ = $\frac{A_2}{B_2}$

18 x = 72

x = $\frac{72}{18}=4cm$

Let n be the number of boxes.
As there are 12 crayons in each box, 2400 crayons fits into
$\frac{2400}{12}=200boxes$
Now the number of boxes produced will be directly proportional to time taken.
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ = $\frac{A_2}{B_2}$

$\frac{No.ofcrayonsboxes}{time}=\frac{200}{4}=\frac{n}{15}$

So, n = 750

The number of bottles and the number of boxes are in inverse proportion.
AB = constant
$A_1\times B_1$ = $A_2\times B_2$
Suppose the required number of boxes is  $x$

Then, $\frac{25}{x}$ = $\frac{20}{12}$

Number of boxes, x = $\frac{(25\times 12)}{20}=15$

Let the required number of toys produced by 7 machines in 5 minutes be n.

As the time increases, the number of toys increases. Hence they are in direct proportion.

n = 490 x 5= 2450

So, 7 machines can make 2,450 toys in 5 minutes.

As the number of machines increases, number of toys increases. They are in direct proportion.

Let k is the number of toys produced by 10 machines in 5 minutes.

Then, k = $\frac{2450\times 10}{7}$

$⟹$ k = 3500

So, the number of toys produced = 3,500

Expanding the given equation, we get
 $xy+4x=1$
$\text{Here,}xy=1-4x,\text{which is not a constant}$. Similarly,
$\frac{x}{y}=(\frac{1}{4y}-\frac{x}{4}),\text{which is not a constant}$.
$\text{So,}y\text{is not proportional to}x.$

Given, two quantities x and y are said to vary in inverse proportion.
Then, $x\propto \frac{1}{y}$ or $y\propto \frac{1}{x}$.
In any case, we get $xy=\text{constant}$
Hence the given statemnet is true.

The ratio of the length of the pole to the length of the shadow remains constant.
The length of pole and the length of the shadow are directly proportional
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ = $\frac{A_2}{B_2}$

$\frac{Lengthofpole}{lengthofshadow}=\frac{560}{320}=\frac{n}{500}\text{(where n is length the second pole)}$ 

n = 875 cm

$\therefore$ the height of the pole  = 875 cm

Cost of painting the pole = 875 x 5 = ₹ 4375