Compound Variation: Direct variation is the interrelation between two variables whose ratio equals a constant value. In other words, direct variation is a condition where an increase in one quantity sources a corresponding increase in the other quantity or a decrease in one quantity results in a decrease of the other quantity.

When the value of one quantity increases regarding the decrease in other or vice-versa, we call it inverse variation. It means that the two quantities act opposite.
If a variable quantity directly varies as the product of two or more variable quantities, then the first variable quantity jointly varies as the other variable quantities. It is called a joint variation or compound variation.

Direct Variation

If two variable quantities \(A\) and \(B\) are so related that when \(A\) increases (or decreases) in a certain ratio, \(B\) also increases (or decreases) in the same ratio, then we say that \(A\) varies directly as \(B\) or simply \(A\) varies as \(B\), or we may say, \(A\) is proportional to \(B\). We then write \(A ∝ B\).

Now, \(A ∝ B\) when \(\frac {A}{B} =\) constant or \(\frac {A}{B} = k\) or \(A = kB\), where \(k\) is a non-zero constant. This constant \(\left( {k \ne 0} \right)\) is called the constant of variation. Therefore, if \(A ∝ B\) then, \(A = kB\), where \(k\) is a constant of variation. Conversely, if \(A = kB\) then, \(A ∝ B\).

Inverse Variation

A variable quantity \(A\) is said to vary inversely as another variable quantity \(B\), if they are so related that while \(A\) increases (or decreases) in a certain ratio, \(B\) decrease (or increases) in the same ratio. In this case, we may also state that \(B\) varies inversely as \(A\). In symbol, we write \(A ∝ \frac {1}{B}\). Now, \(A ∝ \frac {1}{B}\) when \(AB =\) constant or, \(AB = m\) or, \(A = m.\frac{1}{B}\). Where \(m\) is a non-zero constant.

Therefore, if \(A ∝ \frac {1}{B}\), then \(A = m.\frac{1}{B}\) or, \(AB = m\), where \(m\) is a constant of variation. Conversely, if \(A = m.\frac{1}{B}\) or, \(AB = m\), then \(A ∝ \frac {1}{B}\).
If \(\frac {A}{B} =\) constant, then \(A ∝ B\)
If \(AB =\) constant, then \(A ∝ \frac {1}{B}\)
If \(A\) inversely varies as \(B\) then \(A\) directly varies as \(\frac {1}{B}\).

Examples of Direct and Inverse Variation

In our daily routine, we observe variations in the values of several quantities depending upon the proportion in values of some other quantities. For example, our income is directly varied to how many hours we work.
Work more hours to get more salary, which means the increase in the value of one quantity also increases the value of another quantity. A decrease in the value of one quantity also decreases the value of another quantity. In this case, as mentioned above, two quantities are named to exist in direct variation.
Some examples are:

1. Marks are directly varied to performance.
2. The cost of the groceries is directly varied to its weight
3. Pressure is inversely varied to the volume of a gas at a given temperature.
4. The number of workers is inversely varied to the time taken by them to complete the work.
5. Speed is inversely varied to the time taken to travel.

Constant of Variation

If two quantities \(a\) and \(b\) existing in direct variation can be expressed as:
\(a ∝ b\)
\(\frac {a}{b} = K\)
\(a = Kb\)
\(K\) is known as the non-zero constant of proportionality.
If \(a_1,\;b_1\) are the initial values and \(a_2,\;b_2\) are the final values of quantities existing in direct variation, then, they can be expressed as
\(\frac{a_1}{b_1} = \frac {a_2}{b_2} = K\)

Example: \(y\) is directly proportional to \(x\), and when \(x = 9\) then \(y = 81\). What is the constant of proportionality?
Solution: Given: \(y\) is directly proportional to \(x\):
\(⇒ y = Kx\)
Substituting \(x = 9\) and \(y = 81\), we get
\(81 = K × 9\)
\(⇒ K = \frac {81}{9}\)
\(⇒ K = 9\)
Therefore, the constant of proportionality is \(9\).

Compound Variation

The variation involving two or more quantities is called compound variation. In other words, a compound variation can be defined as two or more variations that take place in a proportion. It is also known as joint variation. If a variable quantity \(A\) jointly varies as the two variable quantities \(B\) and \(C\) then \(A ∝ BC\) or, \(A = nBC\), where \(n\) is a non-zero constant.

Let \(A,\,B,\) and \(C\) be three variable quantities. If \(A ∝ B\) when \(C\) remains constant and \(A ∝ C\) when \(B\) remains constant, then \(A ∝ BC\) when both \(B\) and \(C\) varies.

Types of Compound Variation

The four different types of compound variation are given below.

1. Direct – Direct variation
2. Direct – Inverse variation
3. Inverse – Direct variation
4. Inverse – Inverse variation

Methods of Solving Compound Variation

The methods to solve the problems based on compound variation are given below:

1. Proportion method
2. Multiplicative factor method
3. Formula method

(i) Proportion method: From the given data, determine whether they are in direct or inverse variation. Then, the value of the unknown can be determined using:
The product of the extremes \(=\) The product of the means

(ii) Multiplicative factor method: Let us consider the table to know better how to solve the problem using the multiplicative factor method.

Quantity 1	Quantity 2	Quantity 3
\(a\)	\(b\)	\(c\)
\(x\)	\(d\)	\(e\)

In the above table, the unknown quantity is \(x\).

1. Compare the unknown quantity (Quantity 1) with the known values (Quantity 2 and quantity 3)
2. If quantity 1 and quantity 2 are in direct variation, then take the multiplying factor as \(\frac {d}{b}\) (take the reciprocal)
3. If quantity 1 and quantity 3 are in inverse proportion, then take the multiplying factors as \(\frac {c}{e}\) (no change)
4. The value of the unknown quantity \(x\) can be determined by using \(x = a \times \frac{d}{b} \times \frac{c}{e}\)

(iii) Formula method: From the given data, identify the person \((P)\), days \((D)\), hours \((H)\) and work \((W)\) and use the below formula
\(\frac{{{P_1} \times {D_1} \times {H_1}}}{{{W_1}}} = \frac{{{P_2} \times {D_2} \times {H_2}}}{{{W_2}}}\)
Here, the suffix \(1\) and \(2\) denotes the value from statements \(1\) and \(2\) respectively.

Solved Examples – Compound Variation

Q.1. A farmer needs \(30\,\rm{kgs}\) of food to feed \(5\) cows for \(2\) weeks. Find how many kilograms of food is required to feed \(10\) cows for \(5\) weeks? Apply the three methods and find the solution.
Ans: To determine the type of variation, let us tabulate the data.

\(\rm{Kgs}\)	Cows	Weeks
\(30\)	\(5\)	\(2\)
\(x\)	\(10\)	\(5\)

Compare the unknown value \(x\) with the known values(Cows and Weeks).
Proportion method:
Compare \(\rm{Kgs}\) with cows. If the number of cows increases then, the quantity of food also increases. So, it is a direct proportion.
Therefore, the proportion is \(30 : x :: 5 : 10\) …….(i)
Compare \(\rm{Kgs}\) with Weeks. If the number of weeks increases, then the quantity of food also increases. So, it is a direct proportion.
Therefore, the proportion is \(30 : x :: 2 : 5\) …….(ii)
Hence, the given problem is in Direct – Direct variation.
We know that, The product of the extremes \(=\) The product of the means
Therefore, from (i) and (ii)
\(30 : x :: 5 : 10\) and \(30 : x :: 2 : 5\)

Extremes	Means
\(30 : 10\)	\(x : 5\)
\(5\)	\(2\)

\( \Rightarrow 30 \times 10 \times 5 = x \times 5 \times 2\)
\( \Rightarrow x = \frac{{30 \times 10 \times 5}}{{5 \times 2}}\)
\( \Rightarrow x = \frac{{1500}}{{10}}\)
\( \Rightarrow x = 150\)
Therefore, the farmer needs \(150\;\rm{kgs}\) of food to feed \(10\) cows for \(5\) weeks.
Multiplicative factor method:

\(\rm{kgs}\)	Cows	Weeks
\(30\)	\(5\)	\(2\)
\(x\)	\(10\)	\(5\)

Compare \(\rm{kgs}\) with Cows. If the number of cows increases then, the quantity of food also increases. So, it is a direct proportion.
Therefore, the multiplying factor is \(\frac {10}{5}\)
Compare \(\rm{kgs}\) with Weeks. If the number of weeks increases, then the quantity of food also increases. So, it is a direct proportion.
Therefore, the multiplying factor is \(\frac {5}{2}\)
Hence, the given problem is in Direct- Direct variation.
\(x = 30 \times \frac{{10}}{5} \times \frac{5}{2}\)
\(x = 150\)
Therefore, the farmer needs \(150\;\rm{kgs}\) of food to feed \(10\) cows for \(5\) weeks.
Formula method:
Here, \(P_1 = 5,\,D_1 = 2,\,W_1 = 30\) and \(P_2 = 10,\,D_2 = 5,\,W_2 = x\)
Substituting these values in the formula \(\frac{{{P_1} \times {D_1}}}{{{W_1}}} = \frac{{{P_2} \times {D_2}}}{{{W_2}}}\), we get
\(\frac{{5 \times 2}}{{30}} = \frac{{10 \times 5}}{x}\)
\(x = \frac{{30 \times 50}}{{10}}\)
\(x = 150\)
Therefore, the farmer needs \(150\;\rm{kgs}\) of food to feed \(10\) cows for \(5\) weeks.

Q.2. If \(y\) varies jointly as \(x\) and \(z\), and \(y = 12\). When \(x = 2\) and \(z = 3\), find \(y\) when \(x = 7\) and \(z = 4\).
Ans: Applying the formula method, we get
\(\frac{{{x_1}{z_1}}}{{{y_1}}} = \frac{{{x_2}{z_2}}}{{{y_2}}}\)
\( \Rightarrow \frac{{2 \times 3}}{{12}} = \frac{{7 \times 4}}{y}\)
\( \Rightarrow 6y = 336\)
\( \Rightarrow y = 56.\)

Q.3. A soap factory makes \(600\) units in \(9\) days with the help of \(20\) machines. How many units can be made in \(12\) days with the help of \(18\) machines?
Ans:

Machines	Days	Units
\(20\)	\(9\)	\(600\)
\(18\)	\(12\)	\(x\)

\( \Rightarrow \frac{{20 \times 9}}{{600}} = \frac{{18 \times 12}}{x}\)
\( \Rightarrow 20 \times 9 \times x = 600 \times 18 \times 12\)
\( \Rightarrow x = \frac{{600 \times 18 \times 12}}{{20 \times 9}}\)
\( \Rightarrow x = 720\;\rm{units}\).

Q.4. \(195\) men working \(10\) hours a day can finish a job in \(20\) days. How many men are employed to finish the job in \(15\) days if they work \(13\) hours a day?
Ans: Let \(x\) be the number of men required

Days	Hours	Men
\(20\)	\(10\)	\(195\)
\(15\)	\(13\)	\(x\)

\( \Rightarrow 20 \times 10 \times 195 = 15 \times 13 \times x\)
\( \Rightarrow x = \frac{{20 \times 10 \times 195}}{{15 \times 13}}\)
\( \Rightarrow x = 200\) men.

Q.5. In a construction site, if \(20\) women can complete a work in \(7\) days, how long will it take to complete the same work if \(6\) women left the work?
Ans: Given that \(20\) women can complete work in \(7\) days.
Person days \(=\) Number of persons \(\times\) Number of days
Person days \(= 20 \times 7 = 140\)
If \(6\) women left the work, then the number of persons becomes \(20\, -\, 6 = 14\).
Now, substituting the number of persons and person-days in the formula, we have:
\(140 = 14 \times \rm{days}\)
Days \(= \frac {140}{14}\)
Days \(= 10\)
Therefore, it takes \(10\) days to complete the same work.

Summary

In this article, we have discussed the definitions of direct and inverse variation and examples of direct variation and inverse variation, constant of variation. Also, we have studied the definition of compound variation. The variation involving two or more quantities is called compound variation.

It can be defined as two or more variations that happen in a proportion, also known as joint variation. We saw the types of compound variation and the methods to solve the problems based on compound variation and did some example problems on the same.

Study About Coefficient of Variation

Frequently Asked Questions (FAQs)

Q.1. What is the compound variation?
Ans: The variation involving two or more quantities is called compound variation. In other words, a compound variation can be defined as two or more variations that occur in a proportion.

Q.2. What are the types of compound variation?
Ans: The four different types of compound variation are given below.
1. Direct – Direct variation
2. Direct – Inverse variation
3. Inverse – Direct variation
4. Inverse – Inverse variation

Q.3. Name the methods to solve the compound variation.
Ans: The methods to solve the problems based on compound variation are given below:
1. Proportion method
2. Multiplicative factor method
3. Formula method

Q.4. What is direct proportion examples?
Ans: Some examples of direct proportions are:
1. The amount of heat transferred is directly varying from the temperature change.
2. Marks are directly varied to performance.
3. The cost of the groceries is directly varied to its weight.

Q.5. How do you find the ratio of a compound?
Ans: The compounded ratio of the two ratios \(a : b\) and \(c : d\) is the ratio \(ac : bd\), and that of \(a : b,\,c : d\) and \(e : f\) is the ratio \(ace : bdf\).

Now you are provided with all the necessary information on compound variation and we hope this detailed article is helpful to you. If you have any queries regarding this article, please ping us through the comment section below and we will get back to you as soon as possible.