45 minutes

2 hours

12

700

Let the number of workers employed to build the wall in 30 hours be y.
We have the following table

Obviously more the number of workers, faster will they build the wall. So, the number of hours and number of workers vary in inverse proportion.
So 48 × 15 = 30 × y
Therefore,
\(\frac{48 \times 15}{30}=y \)  or y = 24
i.e., to finish the work in 30 hours, 24 workers are required.

Let the number of sheets which weigh \(2\frac { 1 }{ 2 } \) kg be x. We put the above information in the form of a table as shown below:

More the number of sheets, the more would their weight be. So, the number of sheets and their weights are directly proportional to each other.
So, \(\frac{12}{40}=\frac{x}{2500}\) or \(\frac{12 \times 2500}{40}=x\) 
or 750 = x 
Thus, the required number of sheets of paper = 750.
Two quantities x and y which vary in direct proportion have the relation x = \(=k y \text { or } \frac{x}{y}=k\) 
Here, \(k=\frac{\text { number of sheets }}{\text { weight of sheets in grams }}=\frac{12}{40}=\frac{3}{10}\) 
Now x is the number of sheets of the paper which weight \(2 \frac{1}{2} \mathrm{~kg}[2500 \mathrm{~g}]\).
Using the relation x = ky, \(x=\frac{3}{10} \times 2500=750\) 
Thus, 750 sheets of paper would weight \(2\frac { 1 }{ 2 } \) kg.

Let the height of the tree be x metres. We form a table as shown below:

Note that more the height of an object, the more would be the length of its shadow.
Hence, this is a case of direct proportion. That is, \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) 
We have \(\frac{14}{10}=\frac{x}{15} \text { (Why?) }\) 
or \(\frac{14}{10} \times 15=x\) 
or \(\frac{14 \times 3}{2}=x\) 
So 21 = x
Thus, height of the tree is 21 metres.
Alternately, we can write \(\mathrm{e} \frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}} \text { as } \frac{x_{1}}{x_{2}}=\frac{y_{1}}{y_{2}}\) 
so x₁ : x₂ = y₁ : y₂
or 14 : x = 10 : 15
Therefore, 10 × x = 15 × 14
or x = \(\frac{15 \times 14}{10}=21\) 

Suppose the length of cloth is x metres and its cost, in Rs, is y.

As the length of cloth increases, cost of the cloth also increases in the same ratio. It is a case of direct proportion.
We make use of the relation of type \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) 
(i) Here x₁ = 5, y₁ = 210 and x₂ = 2
Therefore, \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) gives \(\frac{5}{210}=\frac{2}{y_{2}}\) or 5y₂ = 2 x 210 or y2 = \(\frac{2 \times 210}{5}=84\) 
(ii) If x₃ = 4, then \(\frac{5}{210}=\frac{4}{y_{3}}\) or 5y₃ = 4 x 210 or y3 = \(=\frac{4 \times 210}{5}=168\) 
[Can we use \(\frac{x_{2}}{y_{2}}=\frac{x_{3}}{y_{3}}\) here? Try!]
(iii) If x₄ = 10, then \(\frac{5}{210}=\frac{10}{y_{4}}\) or y₄ = \(=\frac{10 \times 210}{5}=420\) 
(iv) If x₅ = 13, then \(\frac{5}{210}=\frac{13}{y_{5}} \text { or } y_{5}=\frac{13 \times 210}{5}=546\) 
[Note that here we can also use \(\frac{2}{84} \text { or } \frac{4}{168} \text { or } \frac{10}{420} \text { in the place of } \frac{5}{210}\)]

Here, x₁= 2, y₁= 9 x 10⁶, x₃ = 1.2 and y₃ =?
Now, by using the relation\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 3 } }{ { y }_{ 3 } } \)
\(\frac { 2 }{ 9\times { 10 }^{ 6 } } =\frac { 1.2 }{ { y }_{ 3 } } \Rightarrow 2\times { y }_{ 3 }=1.2\times 9\times { 10 }^{ 6 }\)
\(\Rightarrow { y }_{ 3 }=\frac { 10.8\times { 10 }^{ 6 } }{ 2 } =5.4\times { 10 }^{ 6 }\Rightarrow { y }_{ 3 }=5.4\times { 10 }^{ 6 }\)
Hence, there are 5.4 \(\times\) 10⁶ crystals of sugar in 1.2 kg of sugar.

x = 10

It is the case of direct proportion.
Let number of pencils be x and cost of pencil be Rs.y
\(x\propto y\)
x₁ =10, y₁ = Rs.90, x₂ =19,y₂ =?
Using formula,
\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \Rightarrow \frac { 10 }{ 90 } =\frac { 19 }{ { y }_{ 2 } } \)
y₂ x 10 = 90 \(\times\)19
\(\ { \therefore y }_{ 2 }=\frac { 90\times 19 }{ 10 } =171\)
So, cost of 19 pencils is Rs.171.

100 days

13 h

(i) If the number of workers increases, then time to complete the job would decrease. So, it is the case of inverse proportion.
(ii) For longer distance, more time would be required. So, it is not the case of inverse proportion.
(iii) For more area of cultivated land, more crops would-he harvested. So, area of cultivated land and the crop harvested are not in inverse proportion.
(iv) If speed of a vehicle is more, then time to cover a fixed journey would be less. So, it is a case of inverse proportion.
(v) For more population, less area per person would be there. So, it is a case of inverse proportion

 When x₁ = 100, y₁ = 60, then x₁y₁= 100 \(\times\) 60 = 6000
When x₂ = 200, y₂ = 30, then x₂y₂ = 200 \(\times\) 30 = 6000
When x₃ = 300, y₃ = 20, then x₃y₃ = 300 \(\times\) 20 = 6000
When x₄ = 400, y₄ = 15, then x₄y₄ = 400 \(\times\) 15 = 6000
Here, all the values of x and y are equal to 6000
x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 6000
x and y are in inverse proportion.