Geethika uses 4 cups of water to cook 2 cups of rice every day. One day when some guests visited her house, she needed to cook 6 cups of rice. How many cups of water will she need to cook 6 cups of rice? We come across many such situations in our day-to-day life, where we observe a change in one quantity brings in a change in the other quantity. For example,

- What is the weight of 20 food packets, if the weight of 40 packets is 35.17 Kg? Clearly, the weight of 20 food packets is less.
- If we deposit more money in a bank, what can you say about the interest earned? Definitely, the interest earned also will be more.
- What happens to the total number of items sold, if there are more customers? Clearly, the total number of items sold will increase.
- What happens to the total time taken to accomplish a task, if there are more workers? Definitely, it takes less time with more number of workers.

In the above examples, we can observe that change in one quantity leads to change in another quantity. So to help Geethika, we need to study some types of variations.

**Direct Proportion**

On the occasion of the School Anniversary, the Head Master of the school decided to take up a plantation of saplings. The number of students in each class is given below in the form of a table.

Class |
VI |
VII |
VIII |
IX |
X |
---|---|---|---|---|---|

Number of Students |
7 |
10 |
11 |
14 |
17 |

Each student has to plant two saplings. Find the number of saplings needed for plantation for each class.

Class |
VI |
VII |
VIII |
IX |
X |
---|---|---|---|---|---|

Number of Students |
7 |
10 |
11 |
14 |
17 |

Number of Saplings required |
14 |
20 |
22 |
28 |
34 |

What can you say regarding the number of saplings required? What kind of change do you observe in the number of students and number of saplings required? Either both increase or both decrease.

number of saplings required 14 20 22 2

———————————— = —– = —– = —– = ………. = —– = 2

number of students 7 10 11 1

Here, 2 is a constant and is called the** ****constant of proportion.**** **As the ratio is the same, we call this variation as * Direct Proportion. *If x and y are any two quantities such that both of them increase or decrease together and

**x/y**remains constant (say k), then we say that x and y are in

**Direct Proportion.**This is written as

**x ∝ y**and read as x is directly proportional to y.

∴ (x/y) = k —-> x = ky where k is constant of proportion

∴ Similarly, if y1 and y2 are the values of y corresponding to the values of x1 and x2 of x respectively, then

x1 x2

—– = —–

y1 y2

Now, let’s help our friend Geethika in finding the total number of cups of water required to cook 6 cups of rice. As mentioned earlier, Geethika uses 4 cups of water to cook 2 cups of rice.

Number of cups of rice |
2 |
6 |
---|---|---|

Number of cups of water |
4 |
12 |

Here, it can be observed that the number of cups of water required increases with an increase in the number of cups of rice.

number of cups of water 4 12 2

——————————– = —– = —– = —–

number of cups of rice 2 6 1

In this case, 2 is the constant of proportion. Here, the two quantities number of cups of water and the number of cups of rice are said to be in direct proportion to each other.

number of cups of water ∝ number of cups of rice

Let us solve some example problems now,

### Examples

**Example 1: A vertical pole of 10 m height casts a 20 m long shadow. Find the height of another pole that casts an 80 m long shadow under similar conditions?**

**Solution: **

The length of the Shadow is directly proportional to the height of the pole.

Height of Pole |
10 |
? |
---|---|---|

Length of Shadow |
20 |
80 |

So, (x1 / y1) = (x2 / y2). Here, x1 = 10m y1 = 20m x2 = ? and y2 = 80m.

Upon substituting the values,

(10 / 20) = (x2 / 80)

x2 = (10 x 80) / 20

x2 = 40m

Therefore, the height of another pole is x2 = 40m.

**Example 2: If the cost of 50m of cloth is Rs. 1500, then what will be the cost of 10m of that cloth?**

**Solution:**

The cost of the cloth is directly proportional to the length of the cloth.

Length of cloth |
50m |
10m |
---|---|---|

Cost of cloth |
₹1500 |
? |

So, (x1 / y1) = (x2 / y2). Here, x1 = 50m y1 = Rs.1500 x2 = 10m y2 = ?

Upon substituting the values,

(50 / 1500) = (10 / y2)

y2 = (10 x 1500) / 50

y2 = 300

Therefore, the cost of 10m cloth is Rs.300

**Example 3**:** Following are the vehicle parking charges near a Bus Station.**

Number of Hours (x) |
Parking Charges (y) |
---|---|

up-to 4 hours |
Rs.40 |

up-to 8 hours |
Rs.80 |

up-to 12 hours |
Rs.120 |

up-to 24 hours |
Rs.240 |

**Check if the parking charges and parking hours are in direct proportion?**

**Solution:**

We can observe that the parking charges (y) increase with the increase in the number of hours (x). Let’s calculate the value of (x / y). If it is a constant, then they are in direct proportion. Otherwise, they are not in direct proportion.

x 4 8 12 24 1

—– = —– = —– = —– = —– = —–

y 40 80 120 240 10Here, (1 / 10) is constant and is called the constant of proportion. You can easily observe that all these ratios are equal. So they are in Direct Proportion.

**Example 4: If the cost of 35 rice bags of the same size is Rs. 28,000. What is the cost of 100 rice bags of the same kind?**

**Solution:**

We know that if the number of rice bags purchased increases then the cost also increases. Therefore, the cost of rice bags varies directly with the number of rice bags purchased.

Number of rice bags (x) |
35 |
100 |
---|---|---|

Cost (y) |
Rs. 28,000 |
? |

So, (x1 / y1) = (x2 / y2) Here x1 = 35 y1 = Rs. 28000 x2 = 100 y2 = ?

Upon substituting the values,

(35 / 28000) = (100 / y2)

y2 = (100 * 28000) / 35

y2 = 80,000

Therefore, the cost of 100 rice bags of the same size is y2 = Rs. 80,000

**Inverse Proportion**

A Parcel company has a certain number of parcels to deliver. If the company engages 36 persons, it takes 12 days. If there are only 18 people is, it will take 24 days to finish the task. You see as the number of persons is halved time taken is doubled, if the company engages 72 people, will the time taken be half? Yes, it is. Let’s have a look at the table

Number of Persons |
36 |
18 |
9 |
72 |
108 |
---|---|---|---|---|---|

Time Taken |
12 |
24 |
48 |
6 |
4 |

How many persons shall a company engage if it wants to deliver the parcels within a day?

Two quantities change in such a manner that, if one quantity increases, the other quantity decreases in the same proportion and vice versa, which is called **Inverse Proportion. **In the above example, the number of persons engaged and the number of days are inversely proportional to each other. Symbolically, this is represented as

1

number of days required ∝ ———————————-

number of persons engaged

If x and y are in inverse proportion, then **x ∝ (1 / y)**

x = k / y —-> xy = k where k is constant of proportionality

If y1 and y2 are the values of y corresponding to the values of x1 and x2 of x respectively then

x1 x2

—- = —- or x1y1 = x2y2 ( = k )

y1 y2

### Examples

**Example 1: If 36 workers can build a wall in 12 days, how many days will 16 workers take to build the same wall? (assuming the number of working hours per day is constant)**

**Solution:**

If the number of workers decreases, the time to take built the wall increases in the same proportion. Clearly, the number of workers varies inversely to the number of days.

So here **x1y1 = x2y2** where x1 = 36 workers x2 = 16 workers and y1 = 12 days y2 = ( ? ) days

No. of Workers |
No. of days |
---|---|

36 |
12 |

16 |
y2 |

Since the number of workers are decreasing

36 ÷ x = 16 –> x = 36 / 16

So the number of days will increase in the same proportion i.e,

(36 / 16) * 12 = 27 days

Substitute, (36 / 16) = (y2 / 12) —> y2 = (12 * 36) / 16 = 27 days.

Therefore 16 workers will build the same wall in 27 days.

**Example 2: A car takes 4 hours to reach the destination by traveling at a speed of 60 km/h. How long will it take if the car travels at a speed of 80 Km/h? **

**Solution:**

As speed increases, time is taken decreases in the same proportion. So the time is taken and varies inversely to the speed of the vehicle, for the same distance.

**Method 1**

Speed |
Time |
---|---|

60 |
4 |

80 |
x |

(60 / 80) = (x / 4)

60 x 4 = (80 x x)

x = (60 x 4) / 80 = 3hrs.

**Method 2**

Speed |
Time |
---|---|

60 |
4 ÷ x |

80 |
y |

(60)(x) = 80 and 4 ÷ x = y

x = 80 / 60

4 ÷ (80 / 60) = y

y = (4 x 60) / 80 = 3hrs.

Therefore, the time taken to cover the distance at a speed of 80 Km/h is 3hrs.

**Example 3: 6 pumps are required to fill a tank in 1 hour 40 minutes. How long will it take if only 10 pumps of the same type are used?**

**Solution:**

Let the desired time to fill the tank be x minutes. Thus, we have the following table.

Number of pumps |
6 |
10 |
---|---|---|

Time (in minutes) |
100 |
x |

The lesser the number of pumps more will be the time required to fill the tank.

So, this is a case of inverse proportion.

Hence, (100)(6) = (x)(10) [x1 y1 = x2 y2]

or (100 x 6) / 10 = x

or x = 60 minutes

Thus, time taken to fill the tank by 10 pumps is 60 minutes or 1 hour.

**Example 4: A school has 7 periods a day each of 45 minutes duration. How long would each period become, if the school has 5 periods a day? (assuming the number of school hours to be the same)**

**Solution:**

Let the desired duration of each period be x minutes. Thus, we have the following table.

Number of periods |
7 |
5 |
---|---|---|

Time for each period (in minutes) |
45 |
x |

The lesser the number of periods a day, the more will be the duration of each period.

so, this is a case of inverse proportion.

Hence, (7)(45) = (x)(5) [x1 y1 = x2 y2]

or (7 x 45) / 5 = x

or x = 63 minutes

Thus, time duration of each period if the school has 5 periods a day is 63 minutes or 1 hour 3minutes.