Let the distance covered on the map be 'x' km.
\(\therefore\) We have:

It is a case of direct variation.
\(\therefore \ \frac { 20 }{ 72 } =\frac { 1 }{ x } \Rightarrow 20\times x=1\times 72\)
\(\Rightarrow x=\frac { 72 }{ 20 } =\frac { 18 }{ 5 } =3.6\)
Thus, the required distance on the map is represented as 3.6 cm.

Let the required number of bottles to be filled in 5 hours be x
Since, more number of bottles, more number of hours would be required.
\(\therefore\) The given quantities very directly
\(\therefore \ \frac { 540 }{ x } =\frac { 6 }{ 5 } \Rightarrow 6\times x=5\times 540\)
\(\Rightarrow \ x=\frac { 5\times 540 }{ 6 } =5\times 90=450\)
Thus, the required number of bottles = 450

Here, as the base increased, the required number of green pigments will also increase.
\(\therefore\) The quantity vary directly:
i.e.,\(\frac { 1 }{ 6 } =\frac { 4 }{ { x }_{ 1 } } =\frac { 5 }{ { x }_{ 2 } } =\frac { 6 }{ { x }_{ 3 } } \) 
\(\therefore\) \(\frac { 4 }{ { x }_{ 1 } } =\frac { 1 }{ 6 } \Rightarrow 1\times { x }_{ 1 }=4\times 6\Rightarrow { x }_{ 1 }=24\)
\(\frac { 1 }{ 6 } =\frac { 5 }{ { x }_{ 2 } } \Rightarrow 1\times { x }_{ 2 }=5\times 6\)
\(\Rightarrow { x }_{ 2 }=\frac { 5\times 6 }{ 1 } =30\)
\(\frac { 1 }{ 6 } =\frac { 6 }{ { x }_{ 3 } } \Rightarrow 1\times { x }_{ 3 }=6\times 6\)
\(\Rightarrow \ { x }_{ 3 }=\frac { 6\times 6 }{ 1 } =36\)
Thus, the required unknown quantities are: x₁ = 24, x₂ = 30, and x₃ = 36.

It is given that, a water tank casts a shadow 21 m long and a tree of height 9.5 m casts a shadow 8 m long at the same time. Here the lengths of the shadows are directly proportional to their heights.
Let height = h and the length of shadow = l
Here, h1 = 9.5 m, l₁= 8 m, h2 = X, l₂ = 21 m
For direct proportion,
\(\frac{h_1}{l_1}=\frac{h_2}{l_2} \Rightarrow \frac{9.5}{8}=\frac{x}{21}\)
∴ x = \(\frac{9.5\times 21}{8}=24.9\)m
∴ The height of the tank is 24.9 m.

It is clear that more the number of spokes, less the measure of angle between a pair of consecutive spokes. So, it is the case of inverse proportion.
Here, x₂ = 6,y₂ = 60 and x₃ = 8,y₃ = ?
We know that, x₂ y₂ = x₃y₃
\(\therefore \ 6\times 60=8\times { y }_{ 3 }\Rightarrow { y }_{ 3 }=\frac { 6\times 60 }{ 8 } =\frac { 6\times 15 }{ 2 } \)
[numerator and denominator by 4]
y₃ = 45°
x₃ =8,y₃ = 45 and x₄ =10,y₄ = ?
x₃y₃ = x₄y₄ = 8 x 45 =10 x y₄
\(\ { y }_{ 4 }=\frac { 8\times 45 }{ 10 } \frac { 8\times 9 }{ 2 } \)
[dividing numerator and denominator by 5]
y₄ = 36°
x₄ = 10,y₄ = 36 and x₅ = 12,y₅ = ?
x₄y₄ = x₅y₅ = 10\(\times\)36 = 12\(\times\)y₅
\({ y }_{ 5 }=\frac { 10\times 36 }{ 12 } =\frac { 10\times 36 }{ 12 } \)
[dividing numerator and denominator by 6]
y₅ = 30°
Now, the complete table is given below:

(i) Given,
4 x 90° = 6 x 60° = 8 x 45° = 10 x 36° = 12 x 30° = 360°
i.e.x₁y₁= x₂y₂ = x₃y₃ = x₄y₄ = x₅y₅ = 360°
So, it is clear that it is the case of inverse proportion. Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.
(ii) Let the measure of angle be x⁰.
Lesser the number of spokes, more will be the angle between a pair of consecutive spokes. So, it is the case of inverse proportion.
Here, x₁= 4, y₁ = 90° and x₂ = 15, y₂= x
We know that, x₁y₁=x₂y₂
4 x 90 = 15 x x
\(x=\frac { 4\times 90 }{ 15 } =4\times 6=24^{ 0 }\)
Hence, the angle between a pair of consecutive spokes on a wheel with 15 spokes is 24°.
(iii) Let number of spokes be n.
Lesser the number of spokes, more will be the angle between a pair of consecutive spokes. So, it is the case of inverse proportion.
Here, x₁ = 4, y₁ = 90° and x₂ = n; y₂ = 400
We know that, x₁y₁ = x₂y₂
\(\therefore 4\times 90=n\times 40\Rightarrow n=\frac { 4\times 90 }{ 40 } =9\)
Hence, 9 spokes would be needed, if the angle between a pair of consecutive spokes is 40°.

7.2

Let the distance covered in the map = x cm
Now, we can make a table as shown below:

Here, more the distance covered by Rashmi, more wouldbe the length of distance on the map. So, this is a case of direr proportion.
Here, x₁ = 18,x₂ = 72,y₁=1 and y₂ = x?
Now,by using relation \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)
\(\frac { 18 }{ 1 } =\frac { 72 }{ x } \Rightarrow 18\times x=1\times 72\Rightarrow x=\frac { 1\times 72 }{ 18 } \Rightarrow x=4cm\)
Hence, the distance covered by her on the map is 4 cm.

Let the amount of sugar and number of crystals be x and y. As the amount of sugar increases, the number of crystals also increases in the same ratio. So, it is a case of direct proportion.
Here, x₁ = 2 y₁ = 9\(\times\)10⁶ and x₂= 5,y₂ =?
Now by using the relation\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)
\(\frac { 2 }{ 9\times { 10 }^{ 6 } } =\frac { 5 }{ { y }_{ 2 } } \Rightarrow 2\times { y }_{ 2 }=5\times 9\times { 10 }^{ 6 }\)
\(\Rightarrow { y }_{ 2 }=\frac { 45\times { 10 }^{ 6 } }{ 2 } =22.5\times { 10 }^{ 6 }\Rightarrow { y }_{ 2 }=2.25\times { 10 }^{ 7 }\quad \)
Hence, there are 2.25 x 10⁷ crystals of sugar in 5 kg of So sugar.

Given, P = Rs.1000, R = 8%
Simple interest for different time periods
(i) For T = 1yr,
\(Simple\ interest=\frac { PRT }{ 100 } =\frac { 1000\times 8\times 1 }{ 100 } =rs.80\)
(ii) For T = 2 yr,
\(Simple\ interest=\frac { PRT }{ 100 } =\frac { 1000\times 8\times 2 }{ 100 } =rs.160\)
(iii) For T = 3yr,
\(Simple\ interest=\frac { PRT }{ 100 } =\frac { 1000\times 8\times 3 }{ 100 } =rs.240\)
Compound interest for different time periods
(i) For T = 1yr
\(Compound\ interest=p\left( 1+\frac { R }{ 100 } \right) ^{ T }-p\)
\(=1000\left( 1+\frac { 8 }{ 100 } \right) ^{ 1 }-1000=1000\times \frac { 108 }{ 100 } -1000\)
= 1080 - 1000 = Rs.80
(ii) Compound interest = Rs.166.40
(iii)Compound interest = Rs. 259.712
Hence, the complete table is as follows:

Now, ratio of simple interest with time period are as follows:
\(For\ 1\ yr=,\frac { 80 }{ 1 } =80\)
\(For\ 2\ yr=,\frac { 160 }{ 2 } =80\)
\(For\ 3\ yr=,\frac { 240 }{ 3 } =80\)
Here, ratio of simple interest with time period is same for every year. Hence, simple interest changes in direct proportion with time period. Ratio of compound interest with time period are as follow:
\(For\ 1\ yr=,\frac { 80 }{ 1 } =80\)
\(For\ 2\ yr=,\frac { 166.40 }{ 2 } =83.20\)
\(For\ 3\ yr=,\frac { 259.712 }{ 3 } =86.5706\)
Here, ratio of compound interest with time period is not same for every year. Hence, compound interest does not change in direct proportion with time period.

When x = 6, y = 4, then
\(\frac { x }{ y } =\frac { 6 }{ 4 } =\frac { 6\div 2 }{ 4\div 2 } \ \Rightarrow \frac { x }{ y } =\frac { 3 }{ 2 } \) [HCF of 6 and 4= 2]
When x = 10,y = 8, then
\(\frac { x }{ y } =\frac { 10 }{ 8 } =\frac { 10\div 2 }{ 8\div 2 } =\frac { 5 }{ 4 } \quad \) [HCF of 10 and 8 = 2]
When x = 14,y= 12, then
\(\frac { x }{ y } =\frac { 14 }{ 12 } =\frac { 14\div 2 }{ 12\div 2 } =\frac { 7 }{ 6 } \) [HCF of 14 and 12 = 2]
When x = 18,y = 16, then
\(\frac { x }{ y } =\frac { 18 }{ 16 } =\frac { 18\div 2 }{ 16\div 2 } =\frac { 9 }{ 8 } \) [HCF of 18 and 16= 2]
When x = 22, y = 20, then
\(\ \frac { x }{ y } =\frac { 22 }{ 20 } =\frac { 22\div 2 }{ 20\div 2 } =\frac { 11 }{ 10 } \) [HCF of 22 and 20 = 2]
When x = 26, y = 24, then
\(\frac { x }{ y } =\frac { 26 }{ 24 } =\frac { 26\div 2 }{ 24\div 2 } =\frac { 13 }{ 12 } \) [HCF of 26 and 24= 2]
When x = 30, y = 28, then
\(\frac { x }{ y } =\frac { 30 }{ 28 } =\frac { 30\div 2 }{ 28\div 2 } =\frac { 15 }{ 14 } \) [HCF of 30 and 28 = 2]
From above, it is clear that the values of  \(\frac { x }{ y } \) is different for different valuesof x and y respectively .
So, these values of x and y are not directly proportional.