Few more examples for direct proportion are as follows:
(i) Length of the cloth purchased and its total cost.
(ii) Number of months and total salary.
(iii) Number of hours of production and the amount of the commodity produced

Present duration each period = (50 - 5) minutes = 45 minutes.
Let the present number of period be 'x'.
Since, more the number of periods, less is the duration of a period.
\(\therefore\) It is a case of inverse variation.
We have:

\(\therefore \ 9\times 50=x\times 45\Rightarrow x=\frac { 9\times 50 }{ 45 } =10\)

Thus the required number of periods = 10.

More is the number of persons, less is the time to complete the task.
\(\therefore\) It is a case of inverse variation,
Now, we have:

\(\therefore \ 5\times 4=4\times x\Rightarrow x=\frac { 5\times 4 }{ 4 } =5\)
Thus, the required number of days = 5.

Number of persons added = 5
\(\therefore\) Present number of persons = 20 + 5 = 25
Since, for more persons, the food will last less number of days
\(\therefore\) It is a case of inverse variation.
\(\therefore \ 25\times x=20\times 10\Rightarrow x=\frac { 20\times 10 }{ 25 } =8\)
\(\therefore\) The food will now last for 8 days.

5km/h

44 more cows

Rs.7200

180 mm of mercury

It is given that
\(mass\ of\ the\ rod\propto Length\ of\ therod\)
Let mass of the rod = m
and Length of the rod = l
\(m\propto l\)
\(\frac { { m }_{ 1 } }{ { l }_{ 1 } } =\frac { m_{ 2 } }{ { l }_{ 2 } } \)
Here, m₁ = 192 g,l₁ = 16 cm,m₂ = 105 g and l₂ = ?
It is given that
\(\frac { { m }_{ 1 } }{ { l }_{ 1 } } =\frac { m_{ 2 } }{ { l }_{ 2 } } \quad \Rightarrow \frac { 192 }{ 16 } =\frac { 105 }{ { l }_{ 2 } } \)
\({ l }_{ 2 }=\frac { 16\times 105 }{ 192 } =8.75cm\)

If the number of boxes increases, the space required to keep them also increases.
This is a case of direct proportion.
Let number of boxes = n The space required to keep them = v
\(\frac { { n }_{ 1 } }{ { v }_{ 1 } } =\frac { n_{ 2 } }{ { v }_{ 2 } } \)
Here, n₁ = 2, v₁ = 1000 cm³ and n₂ = 200, v₂ =?
\(\Rightarrow \frac { 2 }{ 1000 } =\frac { 200 }{ { v }_{ 2 } } \Rightarrow { v }_{ 2 }=\frac { 1000\times 200 }{ 2 } =100000\)

Given, l varies directly as m, i.e.\(l\propto m\)
l₁ = 5,l₂ = ? \(m_1=\frac{2}{3}\),\(m_2=\frac{16}{3}\)
\(\frac { { l }_{ 1 } }{ { m }_{ 1 } } =\frac { l_{ 2 } }{ { m }_{ 2 } } \)
\(\frac { 5 }{ \frac { 2 }{ 3 } } =\frac { { l }_{ 2 } }{ \frac { 16 }{ 3 } } \Rightarrow \frac { 5\times 3 }{ 2 } =\frac { { 3\times l }_{ 2 } }{ 16 } \)
\({ l }_{ 2 }=\frac { 5\times 3\times 16 }{ 2\times 3 } =40\)

We will find \(\frac{p}{q}\)in each case.
\(\frac { p }{ q } =\frac { { 1 }^{ 2 } }{ { 1 }^{ 3 } } =1,\frac { p }{ q } =\frac { { 2 }^{ 2 } }{ { 2 }^{ 3 } } =\frac { 1 }{ 2 } ,\frac { p }{ q } =\frac { { 3 }^{ 2 } }{ { 3 }^{ 3 } } =\frac { 1 }{ 3 } ,\frac { p }{ q } =\frac { { 4 }^{ 2 } }{ { 4 }^{ 3 } } =\frac { 1 }{ 4 } \)
Here, \(\frac{p}{q}\)is not constant.
So, in this case, the quantities does not vary directly with each other.

120 men

Let the food will last for x days.
Then, we have the following table:

Here, the number of animals increases, so the number of days will decreases. Therefore, this is a case of inverse proportion.
\(20\times 6=30\times x\Rightarrow x=\frac { 20\times 6 }{ 30 } =4\)
Hence, the food would last for 4 days, if there were 10 more animals in his cattle.

Let each children gets X number of sweets.
Then, we have the following table:

Here, less the number of students, more would be the number of sweets each get, i.e. if the number of children decreases, then the number of sweets each get will increases.
This is a case of inverse proportion.
24 x 5 = 20 X x    [∴ x₁y₁ = x₂y₂]
\(x=\frac { 24\times 5 }{ 20 } =6\)
Hence, each children will get 6 sweets, if the number of the children is reduced by 4.