Joint variation - a quantity dependent on two variableS

Today we want to look into joint variation. In previous lessons, we have considered direct variation with solved examples. In the last lesson, we dealt with inverse variation and also solved easy and hard examples. Now its time to understand joint variation and we will definitely solve two or more examples. What is joint variation though?

In joint variation, the quantity in question has a direct relationship with more than one variable. In most cases, the main object is changed or affected by changes in the other two quantities with whom it is jointly related. Joint variation is always expressed in terms of values and figures between the quantity and the variables it is related to.

Examples of Joint variation in Real life

• Savings: How much money an individual saves could be directly proportional to their income and inversely proportional to their expenditures. This means that the more money a person makes, and the less spending they do, the more money they could save. In this example, Savings is jointly proportional to expenditure and income level.

• Soccer: The chances of of a soccer team winning a match is directly proportional to the number of goals scored and inversely proportional to number of goals considered. If a team scores more goals and concedes less, they are more likely to win matches.

With the two examples above, it is clear that joint variation happens in real life too. But then, we can always express these relationships in form of mathematical expressions. So lets see how to solve joint variation in numbers.

Solving joint variation examples.

Example 1: To fuel his car, a driver went to a petrol station where the amount of time he spends there is directly proportional to the number of vehicles and inversely proportional to the number of pumps that service customers. The station fuels 20 vehicles in 10 minutes using 5 petrol pumps. Solve for the following:

(a) If the station has 2 pumps, how long does it take to fuel 50 vehicles?
(b) If 40 vehicles are to be fueled in 20 minutes, how many pumps will be required.

Solution

Let time = T
Number of vehicles = N
Number of pumps = P
Constant = k

(First Step: We need to obtain the mathematical relationship between the 3 quantities and then find k )

T ∝ N/P
(Next Step: Introducing the constant k, we have)
T = Nk/P
(Next Step: Use the initial values given to solve for k)

T = 10, N = 20, P = 5.
T = Nk/P
(Next Step: Substituting the values)
10 = 20k/5
(Next Step: Cross multiply to make k the subject)
20k = 10x5
20k = 50
k = 50/20
k = 2.5

Having found the value of the constant k, we can easily solve for the other unknowns.

(a) If the station has 2 pumps, how long does it take to fuel 50 vehicles?

Solution

T = Nk/P
T = 50 x 2.5/2
T = 125/2
T = 62.5 minutes

(b) If 40 vehicles are to be fueled in 20 minutes, how many pumps will be required.

Solution

T = Nk/P
20 = 40 x 2.5/P
20 = 100/P
(Next Step: Cross multiply to make P the subject)
20P = 100
(Next Step: Collect like terms. 20 turns to division sign when it crosses the equals sign)
P = 100/20
P = 5

Example 2: A varies directly as the square of B and inversely as C. When A = 18, B = 3 and C = 4. Solve for A when B = 6 and C = 9.

Solution

(First Step: We need to obtain the mathematical relationship between the 3 quantities and then find k )
A ∝ B²/C
(Next Step: Introducing the constant k, we have)
A = B²k/C
(Next Step: Use the initial values given to solve for k)
18 = 3²k/4
(Next Step: Cross multiply to make k the subject)
9k = 18 x 4
9k = 72
(Next Step: Collect like terms)
k = 72/9
k = 8

Now that k is know, Lets solve for A. We are asked to find A when B = 6 and C = 9.

A = B²k/C
A = 6²x8/9
A = 36 x 8/9
A = 288/9
A=32

Conclusion

We have seen examples of joint variation and how to solve it. Basically, one quantity is directly related to two. Or directly related to one of the variables and inversely related to the other. In all cases, once you identify the nature of their relationship and establish the constant, its easy to quickly find the values of the unknown.

Related

• Solving real world examples of Inverse variation