∵ Time taken by 2 students = 20 minutes
∴ Time taken by 1 student = 20 x 2 minutes
∴ Time taken by 5 students \(={20\times2\over5}\) minutes = 8 minutes
We come across many such situations in our day-to-day life, where we need to see variation in one quantity bringing in variation in the other quantity. Quantities so related are called variables and the relation is called variation
For example:
(i) If the number of articles purchased increases, the total cost also increases.
(ii) More the money deposited in a bank, more is the interest earned.
(iii) As the speed of a vehicle increases, the time taken to cover the same distance decreases.
(iv) For a given job, more the number of workers, less will be the time taken to complete the work

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.

Rs.120

29 days

(a) 9 bags
(b) 440 kg

Let the distance covered on the map be x cm. Then, we have the following table:

It is the case of direct propotion.
∴ \(\frac{24}{168}=\frac{1}{x} \Rightarrow x=\frac{168}{24}\)=7
So, distance covered on the map is 7 cm.
(b) Let the distance covered on the road be x km. Then, we have the following table:

It is case of direct proportion.
So, \(\frac{24}{x}=\frac{1}{11} \Rightarrow x=24\times 11 \Rightarrow x=264\)
Hence, distance covered on the road is 264 km.
(c) Here, concept of direct proportion is used. We use road map to show the long distances in smaller units. 

Given, l∝ m ⇒ l = km
where, k is any constant.
⇒ \(k=\frac{l}{m}\)
(i) \(k=\frac{l}{m}\)
(ii) Given, l = 6 and m = 18 
∵ \(k=\frac{l}{m}\)=\(k=\frac{6}{18}\)
∴ \(k=\frac{1}{3}\)
(iii) Given, m = 33 and l =?
∵ \(k=\frac{l}{m}\)⇒\(\frac{1}{3}=\frac{l}{33}\)    [∵ \(k=\frac{1}{3}\) from part (ii)]
∴ l = 11
(iv) Given, l = 8 and m =? 
∵ \(k=\frac{l}{m}\)
⇒ \(\frac{1}{3}=\frac{8}{m} \Rightarrow m=24\)

7.2

Let the height of vertical pole and length of shadow be x m and y m, respectively
Now, we can make a table as shown below:

As the height of the vertical pole increases, the length of the shadow also increases. So, it is the case of direct proportion.
Here, x₁ = 5 m 60 crn = 5 m + 60 cm
\(=5m+\frac { 60 }{ 100 } m\)
x₁= 5 m+ 0.6 rn= 5.6 m
x₂ = 10 m 50 cm= 10 m + 50 cm = 10 m+\(\frac { 50 }{ 100 } m\)
= 10 m+ 0.50 m= 10.5 m
y₁=3m 20cm = 3m + 20cm = 3m + \(\frac { 20 }{ 100 } m\)
= 3 m+ 0.2 m= 3.2 m
Now,by using the relation \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \) 
\(\frac { 5.6m }{ 3.2m } =\frac { 10.5m }{ ym } \Rightarrow y\times 5.6=3.2\times 10.5\)
\(y=\frac { 3.2\times 10.5 }{ 5.6 } =\frac { 33.6 }{ 5.6 } =\frac { 336\times 10 }{ 56\times 10 } \Rightarrow y=6\)
Hence, the length of the shadow cast by another pole is 6 m.
(ii) Here, we can make a table as shown below:
Now,by using the relation \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)
\(\frac { 5.6m }{ 3.2m } =\frac { x }{ 5m } \Rightarrow x\times 3.2m=5.6\times 5m\)
\(x=\frac { 5.6\times 5 }{ 3.2 } =\frac { 56\times 5\times 10 }{ 32\times 10 } =\frac { 35 }{ 4 } m\Rightarrow x=8.75m\)
Hence, the height of the pole is 8.75 m or 8 m 75 cm.
\(\left[ \because 8.75m=8m+0.75m=8m+\frac { 75 }{ 100 } m=8m+75cm \right] \)

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.

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.