**13.1 INTRODUCTION**Mohan prepares tea for himself and his sister. He uses 300 ml of water, 2 spoons of sugar, 1 spoon of tea leaves and 50 ml of milk.

How much quantity of each item will he need, if he has to make tea for five persons?

If two students take 20 minutes to arrange chairs for an assembly, then how much time would five students take to do the same job ?

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. **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.

Observe that change in one quantity leads to change in the other quantity.

How do we find out the quantity of each item needed by Mohan? Or, the time five students take to complete the job ?

To answer such questions, we now study some concepts of variation.

**13.2 DIRECT PROPORTION**

Consider the following situation. Suppose you go to a grocery shop to buy eggs. If the shopkeeper is selling 2 eggs for Rs. 4, then what amount is required to buy 5 eggs ? We can solve this problem using the unitary method.

First, let us determine the cost of 1 egg.

It is given that cost of 2 eggs = Rs. 4

so, Cost of 1 egg = $\frac{Rs.4}{2}$ = Rs. 2

Hence, cost of 5 eggs = 5 × Rs. 2 = Rs. 10

What do we observe in the above example ?

One simple thing that we observe is that as the number of eggs increases, its cost also increases. Such situations are examples of direct proportion. In our day-to-day lives, we come across various such situations. For example, if a car moves with constant speed, then the distance covered by it is in direct proportion with the time taken to cover the distance.

Direct proportion can be defined as follows.

When two variable quantities increase or decrease simultaneously such that their ratio remains unchanged, they are said to be in direct proportion. It is said that one variable “varies directly” from the other. |

Suppose a car uses 4 litres of petrol to travel a distance of 60 km. How far will it travel using 12 litres? The answer is 180 km. How did we calculate it? Since petrol consumed in the second instance is 12 litres, i.e., three times of 4 litres, the distance travelled will also be three times of 60 km. In other words, when the petrol consumption becomes three-fold, the distance travelled is also three fold the previous one. Let the consumption of petrol be x litres and the corresponding distance travelled be y km.

Now, complete the following table :

Petrol in litres (x) | 4 | 8 | 12 | 15 | 20 | 25 |

Distance in km (y) | 60 | 120 | 180 | 225 | 300 | 37.5 |

We find that as the value of x increases, value of y also increases in such a way that the ratio $\frac{x}{y}$ does not change; it remains constant (say k). In this case, it is $\frac{1}{15}$ .

We say that x and y are in direct proportion, if $\frac{x}{y}=k$ or x = ky.

In this example, $\frac{4}{60}=\frac{12}{180}$, where 4 and 12 are the quantities of petrol consumed in litres (x) and 60 and 180 are the distances (y) in km. So when x and y are in direct proportion, we can write $\frac{{x}_{1}}{{y}_{1}}=\frac{{x}_{2}}{{y}_{2}}$ [y_{1}, y_{2} are values of y corresponding to the values x_{1}, x_{2} of x respectively]

The consumption of petrol and the distance travelled by a car is a case of direct proportion. Similarly, the total amount spent and the number of articles purchased is also an example of direct proportion.

Two variables, x and y, are in direct proportion, if $\frac{x}{y}=k$, or x = ky where k is a constant. |

Suppose variables x and y are in direct proportion. If y_{1} and y_{2} are the values of y corresponding to the respective values of x_{1} and x_{2}, then $\frac{{x}_{1}}{{y}_{1}}=k$ and $\frac{{x}_{2}}{{y}_{2}}=k$ .

Therefore, we can write the equation as follows:

$\frac{{x}_{1}}{{y}_{1}}=\frac{{x}_{2}}{{y}_{2}}$

Let us now discuss some examples based on this concept.**Example 1**. The scale of a map is given as 1 : 10000. The distance between two buildings in a city on the map is 5 cm. What is the actual distance between the two buildings? **Solution. **Here, the distance on the map and the actual distance between the buildings are in direct proportion. Let the actual distance between the buildings be x.

$\therefore \frac{1}{10000}=\frac{5cm}{x}$

$\Rightarrow $ x = (10000 × 5) cm

x = 50000 cm

x = $\frac{50000}{100}m$

x = 500 m

Thus, the actual distance between the buildings is 500 m. **Example 2**. If 1 kg 600 g of rice is sufficient for 20 people, then what quantity of rice will be sufficient for 27 people ? Also calculate for how many people 2 kg 400 g of rice will be sufficient. **Solution. **Let x kg of rice be sufficient for 27 people and 2 kg 400 g of rice be sufficient for y number of people.

1 kg 600 g = 1.6 kg

2 kg 400 g = 2.4 kg

The given information can be represented by the following table :

Quantity of rice (kg) : x | 1.6 | x | 2.4 |

Number of people : y | 20 | 27 | Y |

Here, the quantity of rice and the number of people are in direct proportion.

$\therefore \frac{1.6}{20}=\frac{x}{27}$ and $\frac{1.6}{20}=\frac{2.4}{y}$

$x=\frac{27\times 1.6}{20}andy=\frac{2.4\times 20}{1.6}$

x = 2.16 kg and y = 30 people

x = 2 kg 160 g and y = 30 people

Hence, 2 kg 160 g of rice will be sufficient for 27 people and 2 kg 400 g of rice will be sufficient for 30 people. **Example 3**. A car travels at a constant speed of 35 km/h. How far can it travel in 15 minutes?**Solution. **Speed of car = 35 km/hThis means that the car travels 35 km in 60 minutes.

Let the car travel x km in 15 minutes.

The given information can be represented by the following table :

Distance covered by the car (km) : x | 35 | x |

Time taken by the car (min) : y | 60 | 15 |

Here, the distance covered by the car and the time taken by it are in direct proportion.

$\therefore \frac{35km}{60min}=\frac{x}{15min}$

$\Rightarrow x=\left(\frac{35}{60}\times 15\right)km$

**$\Rightarrow x=\left(\frac{35}{4}\right)km=8.75km$**Thus, the car travels 8.75 km in 15 minutes.

**Example 4**. If 12 machines can be assembled in 4 hours, then how many machines can be assembled in 8 hours ?

**Solution.**Let x machines be assembled in 8 hours.

**The given information can be represented by the following table :**

Number of machines : x | 12 | x |

Time taken (hours) : y | 4 | 8 |

Here, the number of machines and the time taken to assemble them are in direct proportion.

$\therefore \frac{12}{4}=\frac{x}{8}$

$x=\frac{12\times 8}{4}$

$x=24$

Thus, 24 machines can be assembled in 8 hours. **Example 5**. At a particular time of the day, the length of the shadow of a 28 feet high tree is 1.4 feet. Determine the height of a tree that has a shadow of length 2.3 feet at the same time of the day ?**Solution.** Let the height of the tree having shadow length 2.3 feet at the given time be x feet. The given information can be represented by the following table :

Height of the tree (ft) : x | 28 | x |

Length of the shadow (ft) : y | 1.4 | 2.3 |

At a particular time of the day, the height of the tree and length of its shadow are in direct proportion.

$\therefore \frac{28ft}{1.4ft}=\frac{x}{2.3ft}$

$x=\left(\frac{28\times 2.3}{1.4}\right)ft$

$x=46ft$

Hence, the height of the second tree is 46 feet. **Example 6**. State whether the given situations involve two variables in direct proportion.

1. Distance covered by a car and the quantity of petrol consumed by it

2. Number of workers and the time taken by them to complete a work

3. Speed of a person and the time taken by him/her to cover a fixed distance

4. Speed of a person and the distance covered by him/her in a fixed time

5. Time period and simple interest if the rate of interest is fixed

6. Time period and rate of interest if the simple interest is fixed

7. Area of cultivated land and the crop harvested**Solution.** (1) The quantity of petrol consumed by a car increases if the distance covered by it also increases. Thus, the distance covered by a car is in direct proportion with the quantity of petrol consumed by it.

(2) More number of workers will take less time to complete a work. Thus, the number of workers and the time taken by them to complete the work are not in direct proportion.

(3) If a person travels at a higher speed, then he/she takes less time to cover a fixed distance. Thus, the speed of a person and the time taken by him/her to cover a fixed distance are not in direct proportion.

(4) If a person increases his/her speed, then he/she will cover more distance in a fixed time. Hence, the speed of a person and the distance covered by him/her in a fixed time are in direct proportion.

(5) For a fixed rate of interest, if the time period is more, then the simple interest will also be more. Hence, the time period and the simple interest are in direct proportion if the rate of interest is fixed.

(6) For a fixed simple interest, if the time period is more, then the rate of interest will be less. Hence, the time period and the rate of interest are not in direct proportion if the rate of interest is fixed.

(7) The more the area of land cultivated, the more will be the amount of crop harvested. Hence, the area of cultivated land and the crop harvested are in direct proportion.**13.3. VERIFICATION OF DIRECT PROPORTION**

Consider the following situation :

You go to a stationery shop to buy some pens. If each pen costs Rs. 5, then what amount is required to buy 4 such pens? Also, determine the amount you need to pay to buy 10 such pens?

It is given that cost of 1 pen = Rs. 5

Hence, cost of 4 pens = 4 × Rs. 5 = Rs. 20

Similarly, cost of 10 pens = 10 × Rs. 5 = Rs. 50

This information can be represented in the table form, where the number of pens is denoted by variable x and their corresponding cost is denoted by variable y, as

Number of pens : x | 1 | 4 | 10 |

Cost (Rs.) : y | 5 | 20 | 50 |

If we observe the ratio of the corresponding values of x and y, then we see that

$\frac{1}{5}=\frac{1}{5};\frac{4}{20}=\frac{1}{5};\frac{10}{50}=\frac{1}{5}$

We, thus, observe that as the number of pens (x) increases, their cost (y) also increases in such a manner that their ratio $\left(\frac{x}{y}\right)$ remains constant, say k. Thus, in this case, the value of k is $\frac{1}{5}$.

Hence, we say that x and y are in direct proportion, if $\frac{x}{y}=k$ or x = ky

Thus, to check whether the variables x and y are in direct proportion, we need to find the ratio $\frac{x}{y}$ for their corresponding values. If this ratio remains constant, then the variables are in direct proportion, otherwise, they are not.

Let us now discuss some more examples based on this concept. **Example 7**. The following table lists the distance covered by a person and the corresponding time taken by him to cover this distance. Check whether the distance covered changes in direct proportion with the time taken.

Distance covered |
Time taken |

4 km | 16 minutes |

6 km | 24 minutes |

12 km | 48 minutes |

20 km | 72 minutes |

**Solution. **The above information can be represented in a table by taking the distance covered (km) as variable x and the corresponding time taken (minutes) as y.

Distance covered (km) : x | 4 | 6 | 12 | 20 |

Time taken (minutes) : y | 16 | 24 | 48 | 72 |

Thus, we can find the ratio $\frac{x}{y}$ for the corresponding values of x and y as

$\frac{4}{16}=\frac{1}{4};\frac{6}{24}=\frac{1}{4};\frac{12}{48}=\frac{1}{4};\frac{20}{72}=\frac{5}{18}$

Since the ratio $\frac{x}{y}$ does not remain constant, the distance covered by the person does not change in direct proportion with the time taken.**Example 8**. Observe the following tables and find whether x and y are directly proportional.

(a) | x | 2 | 9 | 14 | 15 | 17 | 19 |

Y | 6 | 27 | 42 | 45 | 51 | 57 |

(b) | x | 50 | 45 | 37.5 | 30 | 25 | 5 |

y | 40 | 36 | 30 | 24 | 20 | 4 |

(c) | x | 50 | 48 | 42 | 39 | 30 | 17 |

y | 40 | 38 | 32 | 29 | 20 | 7 |

**Solution. **

(a) | x | 2 | 9 | 14 | 15 | 17 | 19 |

Y | 6 | 27 | 42 | 45 | 51 | 57 |

Thus, $\frac{2}{6}=\frac{1}{3}$

$\frac{9}{27}=\frac{1}{3}$

$\frac{14}{42}=\frac{1}{3}$

$\frac{15}{45}=\frac{1}{3}$

$\frac{17}{51}=\frac{1}{3}$

$\frac{19}{57}=\frac{1}{3}$

Since the ratio $\frac{x}{y}$ remains constant for the corresponding values of x and y, the variables x and y are directly proportional.

(b) |
x | 50 | 45 | 37.5 | 30 | 25 | 5 |

y | 40 | 36 | 30 | 24 | 20 | 4 |

Thus, $\frac{50}{40}=\frac{5}{4}$

$\frac{45}{36}=\frac{5}{4}$

$\frac{37.5}{30}=\frac{5}{4}$

$\frac{30}{24}=\frac{5}{4}$

$\frac{25}{20}=\frac{5}{4}$

$\frac{5}{4}=\frac{5}{4}$

Since the ratio $\frac{x}{y}$ remains constant for the corresponding values of x and y, the variables x and y are directly proportional.

(c) | x | 50 | 48 | 42 | 39 | 30 | 17 |

y | 40 | 38 | 32 | 29 | 20 | 7 |

Thus, $\frac{50}{40}=\frac{5}{4}$

$\frac{48}{38}=\frac{24}{19}$

$\frac{42}{32}=\frac{21}{16}$

$\frac{39}{29}=\frac{39}{29}$

$\frac{30}{20}=\frac{3}{2}$

$\frac{17}{7}=\frac{17}{7}$

Since the ratio $\frac{x}{y}$ does not remain constant for the corresponding values of x and y, the variables x and y are not directly proportional.**13.4. INVERSE PROPORTION**

Two quantities may change in such a manner that if one quantity increases, the other quantity decreases and vice versa. For example, as the number of workers increases, time taken to finish the job decreases. Similarly, if we increase the speed, the time taken to cover a given distance decreases.

To understand this, let us look into the following situation.

Zaheeda can go to her school in four different ways. She can walk, run, cycle or go by car. Study the following table.

Walking | Running | Cycling | By car | |

Speed in km/hour | 3 | 6 | 9 | 45 |

Time taken (in minutes) | 30 | 15 | 10 | 2 |

Observe that as the speed increases, time taken to cover the same distance decreases.

As Zaheeda doubles her speed by running, time reduces to half. As she increases her speed to three times by cycling, time decreases to one third. Similarly, as she increases her speed to 15 times, time decreases to one fifteenth. (Or, in other words the ratio by which time decreases is inverse of the ratio by which the corresponding speed increases).

Let us consider another example. A school wants to spend Rs. 6000 on mathematics textbooks. How many books could be bought at Rs. 40 each? Clearly 150 books can be bought. If the price of a textbook is more than Rs. 40, then the number of books which could be purchased with the same amount of money would be less than 150. Observe the following table.

Price of each book (in Rs.) | 40 | 50 | 60 | 75 | 80 | 100 |

Number of books that can be bought | 150 | 120 | 100 | 80 | 75 | 60 |

Observe that as the price of the books increases, the number of books that can be bought, keeping the fund constant, will decrease. Ratio by which the price of books increases when going from 40 to 50 is 4 : 5, and the ratio by which the corresponding number of books decreases from 150 to 120 is 5 : 4. This means that the two ratios are inverses of each other.

Notice that the product of the corresponding values of the two quantities is constant; that is, 40 × 150 = 50 × 120 = 6000.

If we represent the price of one book as x and the number of books bought as y, then as x increases y decreases and vice-versa. It is important to note that the product xy remains constant. We say that x varies inversely with y and y varies inversely with x. Thus two quantities x and y are said to vary in inverse proportion if there exists a relation of the type xy = k between them, k being a constant. If y_{1}, y_{2} are the values of y corresponding to the values x_{1}, x_{2} of x respectively then x_{1}y_{1} = x_{2}y_{2} (= k), or $\frac{{x}_{1}}{{x}_{2}}=\frac{{y}_{1}}{{y}_{2}}$. We say that x and y are in inverse proportion.

“Two variables x and y will be in inverse proportion, if xy = k, where the constant k is known as the constant of proportionality of the inverse proportion”. |

**Example 9**. In each of the following statements, find the situation where two variables are in inverse proportion.

1. Distance covered by a car and the amount of petrol required

2. Number of workers and time taken by them to complete the work

3. Speed of a person and distance covered by him in a fixed time

4. Length and breadth of a rectangle to keep its area constant**Solution. **(1) If we want to cover more distance, then we will require more amount of petrol. Hence, distance covered by a car and amount of petrol is in direct proportion, not in inverse proportion.

(2) If number of workers will be more, then they will take less time to complete the work. Hence, the number of workers and time taken by them to complete the work are in inverse proportion.

(3) If a person will increase his speed, then he will cover more distance in a fixed time. Hence, the speed of a person and distance covered by him in a fixed time are not in inverse proportion.

(4) If we will increase the length of rectangle, then we will have to decrease its breadth in order to keep its area constant. Hence, to keep the area of a rectangle constant, its length and breadth should be in inverse proportion. **Example 10**. A packet of chocolates is to be distributed among 25 children such that each of them will get 4 chocolates. How many more chocolates would each of them get if the number of children is reduced by 5 ?**Solution. **Let each child get x more chocolates. Hence, each child will get (4 + x) chocolates. There were 25 children. If number of children is reduced by 5, then there are 25 − 5 = 20 children We can represent the given situation by constructing a table between the number of children and number of chocolates as shown below.

Number of children | 25 | 20 |

Number of chocolates | 4 | 4 + x |

The number of children and the number of chocolates that each child got are in inverse proportion.

$\therefore $ 25 × 4 = 20 (4 + x)

4 + x = $\frac{25\times 4}{20}$

4 + x = 5

x = 5 − 4 = 1

Therefore, each child will get 1 more chocolate if 5 children are reduced.**Example 11**. In one full day in a school, there are 8 periods of 40 min each. The school management decided to increase the number of periods in a day so that more classes can take place in a single day. However, they want to keep the school duration the same.

What will be the duration of new periods if the number of periods is increased to 10 ?**Solution. **Let the duration of new periods be x.The given situation can be represented with the help of the following table.

Number of periods | 8 | 20 |

Time duration (minutes) | 40 | x |

Here, the number of periods and time duration of each period are in inverse proportion.

$\therefore $ 40 × 8 = x × 10

x = $\frac{40\times 8}{10}$ = 32

Therefore, the duration of new periods will be 32 minutes.**13.5. VERIFICATION OF INVERSE VARIATION**

The following table shows the speed of a train and time taken to cover the same distance.

Speed (in km/hr) |
Time taken (in hr) |

20 | 6 |

30 | 4 |

40 | 3 |

50 | 2.4 |

Are the speed of the train and the time taken to cover the distance in inverse proportion? We know that

“Two variables x and y will be in inverse proportion if xy = k, where the constant k is known as the constant of proportionality of the inverse proportion”. |

Therefore, to check whether the two variables x and y of a given situation are in inverse proportion or not, we have to calculate the product of the value of variable x with its corresponding value of the variable y. If all these products are equal, then we can say that the variables x and y are in inverse proportion, otherwise not.

Using this concept, let us check whether the speed of train and time taken to cover the distance is in inverse proportion or not.

In the given table, there are two variables − speed of the train and time taken to cover the distance. For the first observation, the product of these variables is 20 × 6 = 120.

Similarly, for the other observations, the products of corresponding values of the variables are

30 × 4 = 120

40 × 3 = 120

50 × 2.4 = 120

Here, the products of the values of the variables for all observations are the same.

Thus, the speed and time taken to cover the same distance are in inverse proportion.

Using this concept, we can check whether the variables of a given situation, given in a tabular form, are in inverse proportion or not.

Let us discuss one more example based on verification of inverse proportion. **Example 12**. Observe the following tables and check whether x and y are related inversely or not.

(i) |
x |
1 |
2 |
5 |
10 |

y |
20 |
10 |
4 |
2 |

(ii) |
x |
7 |
8 |
9 |
10 |

y |
27 |
20 |
12 |
7 |

**Solution. **We know that two quantities x and y are in inverse proportion, if x × y = constant. (i) Here, 1 × 20 = 20

2 × 10 = 20

5 × 4 = 20

10 × 2 = 20

It can be seen that x × y = 20, which is constant for each observation.

Therefore, x and y are in inverse proportion.

(ii) Here, 7 × 27 = 189

8 × 20 = 160

9 × 12 = 108

10 × 7 = 70

It can be seen that x × y is not constant for each observation.

Therefore, x and y are not in inverse proportion.