Directly proportional: If an increase/decrease in one quantity leads to an increase/decrease in  the other quantity  or vice-versa, the two quantities are said to be directly proportional to each other.  x and y are directly proportional if the ratio y/x is constant i.e.  y can be written as a constant multiple of x. i.e. y = kx. It is also denoted as: y ∝ x.
Thus, Speed is directly proportional to distance
here as speed decreases, distance also decreases or vice -versa.
Example 1: If 15 toys cost Rs. 234, what do 35 toys cost?
Solution:

Let the required cost be Rs. x. then,
More toys, More cost (Direct Proportion)
∴ 15 : 35 :: 234 : [latex]x[/latex] ⇔ (15 x [latex]x[/latex]) = (35 x 234) ⇔ [latex]x[/latex] = ([latex]\frac{35 \times 234}{15}[/latex]) = 546.

Example 2: If the wages of 6 men for 15 days-be Rs. 2100, then find the wages of 9 men for 12 days.
Solution:

Let the required wages be Rs. [latex]x[/latex].
More men, More wages (Direct Proportion)
Less days, Less wages (Direct Proportion)
\begin{equation} \left. \begin{array}{@{}r@{\quad}ccrr@{}} \textrm{Men} & & 6 : 9 \\ \textrm{Days} & & 15 : 12 \\ \end{array} \mkern18mu\right\} \textrm{:: 2100 : x} \end{equation} ∴ (6 x 15 x [latex]x[/latex]) = (9 x 12 x 2100) ⇔ [latex]x[/latex] = ([latex]\frac{9 \times 12 \times 2100}{6 \times 15}[/latex]) = 2520.
Hence, the required wages are Rs. 2520.

Example 3: If 20 men can build a wall 56 meters long in 6 days, what length of a similar wall can be built by 35 men in 3 days?
Solution:

Let the required length be [latex]x[/latex] meters.
More men, More length built (Direct Proportion)
Less days, Less length built (Direct Proportion)
\begin{equation} \left. \begin{array}{@{}r@{\quad}ccrr@{}} \textrm{Men} & & 20 : 35 \\ \textrm{Days} & & 6 : 3 \\ \end{array} \mkern18mu\right\} \textrm{:: 56 : x} \end{equation} ∴ (20 x 6 x [latex]x[/latex]) = (35 x 3 x 56) ⇔ [latex]x[/latex] = [latex]\frac{(35 \times 3 \times 56)}{120}[/latex] = 49.
Hence, the required length is 49 m.

Inversely proportional: If one quantity increases then the other quantity decreases or vice - versa, then those quantities are said to be indirectly proportional to each other.  x and y are inversely proportional to each other, if the product yx is a constant i.e.   y = k/x. It is also denoted as: y ∝ [latex]\frac{1}{x}[/latex]
Thus, Time taken to finish a work is inversely proportional to the number of persons working at it.
Here as the number of persons increases, the time taken to complete a work decreases or vice- versa.

Example 1: If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do it?
Solution:

Let the required number of hours be [latex]x[/latex]. Then,
Less men, More hours (Indirect Proportion)
∴ 15 : 36 :: 25 : [latex]x[/latex] ⇔ (15 x [latex]x[/latex]) = (36 x 25) ⇔ [latex]x[/latex] = [latex]\frac{36 \times 25}{15}[/latex] = 60.
Hence, 15 men can do it in 60 hours.

Example 2: If 15 men, working 9 hours a day, can reap a filed in 16 days, in how many days will 18 men reap the field, working 8 hours a day?
Solution:

Let the required number of days be [latex]x[/latex].
More men, Less days (Indirect Proportion)
Less hours per day, More days (Indirect Proportion)
\begin{equation} \left. \begin{array}{@{}r@{\quad}ccrr@{}} \textrm{Men} & & 18 : 15 \\ \textrm{Days} & & 8 : 9 \\ \end{array} \mkern18mu\right\} \textrm{:: 16 : x} \end{equation} ∴ (18 x 8 x [latex]x[/latex]) = (15 x 9 x 16) ⇔ [latex]x[/latex] = ([latex]\frac{15 \times 144}{144}[/latex]) = 15.
Hence, required number of days = 15.

Example 3: A garrison of 3300 men had provisions for 32 days, when given at the rate of 850 gms per head. At the end of 7 days, a reinforcement arrives and it was found that the provisions will last 17 days more, when given at the rate of 825 gms per head. What is the strength of the reinforcement?
Solution:

The problem becomes:
3300 men taking 850 gms per head have provisions for (32 - 7) or 25 days. How many men taking 825 gms each have provisions for 17 days?
Less ration per head, more men (Indirect Proportion)
Less days, More men (Indirect Proportion)
\begin{equation} \left. \begin{array}{@{}r@{\quad}ccrr@{}} \textrm{Ration} & & 825 : 850 \\ \textrm{Days} & & 17 : 25 \\ \end{array} \mkern18mu\right\} \textrm{:: 3300 : x} \end{equation} ∴ 825 x 17 x [latex]x[/latex] = 850 x 25 x 3300 or [latex]x[/latex] = [latex]\frac{850 \times 25 \times 3300}{825 \times 17}[/latex] = 5000.
∴ Strength of reinforcement = (5500 - 3300) = 1700.

1. If 35 bottles cost Rs. 800, what is the cost of 45 bottles?
Solution:

Given that,
Cost of 35 bottles = Rs. 800
Let the required cost be [latex]x[/latex]
Here it is clear that as bottles increased cost also increases, so direct proportion is applied.
Now, Ratio is 35 : 45 :: 800 : [latex]x[/latex]
⇒ 35 x [latex]x[/latex] = 45 x 800
⇒ 35 x [latex]x[/latex] = 36000
⇒ [latex]x[/latex] = [latex]\frac{36000}{35}[/latex]
⇒ [latex]x[/latex] = 1028.5 ≅ 1029
Therefore, the cost of 45 bottles = Rs. 1029.

2. Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate how many bottles could 10 such machines produce in 4 minutes?
Solution:

Let the required number of bottles be [latex]x[/latex]
From the data, it is very clear that more machines more bottles and more minutes more bottles
So, direct proportion is applied
Ratios are:
For machines ⇒ 6 : 10 :: 270 : [latex]x[/latex]
For minutes ⇒ 1 : 4 :: 270 : [latex]x[/latex]
Now, by combining them
⇒ 6 x 1 x [latex]x[/latex] = 10 x 4 x 270
⇒ 6 x [latex]x[/latex] = 10800
⇒ [latex]x[/latex] = [latex]\frac{10800}{6}[/latex]
⇒ [latex]x[/latex] = 1800
Therefore, such 10 machines in 4 minutes can produce 1800 bottles.

3. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?
Solution:

Let the required number of days be [latex]x[/latex]
It is clear that, less men, more days. So, inverse proportion is applied
Ratio is 27 : 36 :: 18 : [latex]x[/latex]
⇒ 27 x [latex]x[/latex] = 36 x 18
⇒ [latex]x[/latex] = [latex]\frac{36 * 18}{27}[/latex]
⇒ [latex]x[/latex] = 24
Therefore, required number of days = 24 days.

4. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs.When the smaller wheel has made 21 revolutions, then find the number of revolutions made by the larger wheel?
Solution:

Let the required number of revolutions made by larger wheel be [latex]x[/latex]
here more cogs, less revolutions. So, indirect proportion is applied.
Ratio is 14 : 6 :: 21 : [latex]x[/latex]
⇒ 14 x [latex]x[/latex] = 21 x 6
⇒ [latex]x[/latex] = [latex]\frac{126}{14}[/latex]
⇒ [latex]x[/latex] = 9
Therefore, number of revolutions made by larger wheel = 9.

5. 3 pumps, working 8 hours a day, can empty a tank in two days. how many hours a day must 4 pumps work to empty the tank in 1 day?
Solution:

Let the required number of working hours per day = [latex]x[/latex]
here more pumps, less working hours per day and less days, more working hours per day. So they are inversely proportional to each other.
for pumps ⇒ 4 : 3 :: 8 : [latex]x[/latex]
for days ⇒ 1 : 2 :: 8 : [latex]x[/latex]
By combining them,
4 x 1 x [latex]x[/latex] = 3 x 2 x 8
⇒ 4 x [latex]x[/latex] = 48
⇒ [latex]x[/latex] = [latex]\frac{48}{4}[/latex]
⇒ [latex]x[/latex] = 12
Therefore, number of working hours per day = 12 hours.