Topics to be learn : Direct Variation Inverse Variation Time, Work, Speed

Direct Variation :

Direct variation occurs when two quantities increase or decrease together in the same proportion, such as the relationship between the number of items purchased and their total cost. 

• If x and y are two quantities such that \(\frac{x}{y}\) = constant, then there is direct variation between x and y.
• There is direct variation between two quantities x and y. This statement can be written as x ∝ y, using mathematical symbol. ∝. (∝ Greek letter alpha) is used to denote variation.
• x ∝ y is written in the form of an equation as x = ky, where k is a constant of variation.
• x = ky is the equation of direct variation.
• Direct variation is also known as directly proportion.

Practical Examples :

The following table illustrates the direct variation between the number of notebooks (x) and their cost (y), where k = \(\frac{1}{20}\)

Number of notebooks (x)	12	3	9	24	50	1
Cost in Rupees (y)	240	60	180	480	1000	20

Additional Applications:

• Geometry: The circumference (c) of a circle is directly proportional to its radius (r): c ∝ r.
• Physics: The area of a circle (A) is directly proportional to the square of its radius (r): A ∝ r².
• Fluid Dynamics: The pressure of a liquid (p) varies directly as the depth (d) of the liquid: p ∝ d.
• Labor: Total remuneration paid to laborers is in direct variation with the number of laborers employed.

Solved Examples :

Q.1. x varies directly as y, when x = 5, y = 30. Find the constant of variation and equation of variation.

Solution:

x varies directly as y, that is as x ∝ y

∴ x = ky             ......... k is constant of variation.

when x = 5, y = 30, is given

∴ 5 = k × 30

∴ k = \(\frac{5}{30}\)

∴ k = \(\frac{1}{6}\) …(constant of variation)

∴ equation of variation is x = ky, that is x = \(\frac{y}{6}\)  or y = 6x

 

Q.2. Cost of groundnuts is directly proportional to its weight. If cost of 5 kg groundnuts is ₹ 450 then find the cost of 1 quintal groundnuts. (1 quintal = 100 kg)

Solution:

Let the cost of groundnuts be x and weight of groundnuts be y.

It is given that x varies directly as y

 ∴ x ∝ y or x = ky

It is given that when x = 450 then y = 5, hence we will find k.

x = ky

∴ 450 = 5k

∴ k = \(\frac{450}{5}\) = 90 (constant of variation)

∴ equation of variation is x = 90y.

∴ if y = 100,

x = 90 × 100 = 9000

∴ cost of 1 quintal groundnut is ₹ 9000.

Inverse variation :

Inverse variation describes a relationship where one value increases as the other decreases, commonly seen in time and speed or workforce size and project duration.

• If x and y are two quantities such that x × y = constant, then there is inverse variation between x and y.
• x varies inversely as y is written as x ∝ \(\frac{1}{y}\)
• If x ∝ \(\frac{1}{y}\), x = \(\frac{k}{y}\)   ... (k is a constant)
• x = \(\frac{k}{y}\) or xy = k is the equation of variation.

Practical Examples :

The following table demonstrates the inverse relationship between the number of students in a row and the total number of rows required for a drill (k = 240):

Number of students per row	40	10	24	12	8
Number of Rows	6	24	10	20	30

Additional Applications:

• Acoustics: The wavelength of sound (l) and its frequency (f) are in inverse variation (l ∝ \(\frac{1}{f}\) ).
• Optics: The intensity of light (I) varies inversely with the square of the distance (d) from the source: I ∝ \(\frac{1}{d^2}\) .
• Logistics: The number of pipes of the same size required to fill a tank is inversely proportional to the time taken.

Solved Examples :

Q.1. f ∝ \(\frac{1}{d^2}\), when d = 5, f = 18

Hence, (i) if d = 10 find f. (ii) when f = 50 find d.

Solution:

 f ∝ \(\frac{1}{d^2}\)

∴ f × d² = k, when d = 5 and f = 18.

∴ 18 × 5² = k

∴ k = 18 × 25 = 450 (constant of variation)

(i) if d = 10 then f = ? f × d2 = 450 ∴ f × 102 = 450 ∴ f × 100 = 450 ∴ f = 4.5

(ii) if f = 50, then d = ? f × d2 = 450 ∴ 50 × d2 = 450 ∴ d2 = 9 ∴ d = 3 or d = -3

Time, Work, Speed :

Time and Work :

The number of workers employed on a job and the time taken to complete that job are inversely proportional.

• Formula: d.n = k, where d is days (or hours) and n is the number of workers.
• Example : If 15 women can harvest a crop in 8 days (k = 120), reducing the time to 6 days requires increasing the workforce to 20 women ( \(\frac{120}{6}\) = 20).

Speed and Time :

The time taken to cover a certain distance by a vehicle and its speed is related to variation.

The time taken to cover a fixed distance is inversely proportional to the uniform speed of the vehicle.

• Formula: s.t = k, where s is speed and t is time.
• Example : A vehicle traveling at 48 km/hr takes 6 hours to complete a journey (k = 288). If the speed is increased to 72 km/hr, the time taken decreases to 4 hours (\(\frac{288}{72}\) = 4).

The examples related to time, work and speed are the examples on variation.

Steps to solve variation problems :

• Identify the type of variation: Determine if the relationship is direct (x/y = k) or inverse (xy =k ).
• Find the constant of variation (k):Use a provided pair of values to calculate k.
• Establish the equation of variation: Write the relationship using the calculated k.
• Solve for the unknown: Substitute the remaining given value into the equation to find the final result.

Table of Variation Types :

Feature	Direct Variation	Inverse Variation
Basic Definition	Variables move in the same direction.	Variables move in opposite directions.
Symbolic Form	x ∝ y	x ∝ \(\frac{1}{y}\)
Equation	x = ky	x.y = k
Constant (k)	Ratio ()	Product (x × y)
Key Example	Consumption of petrol vs. Distance.	Number of workers vs. Time to finish.