 maths > commercial-arithmetics

Direct and Inverse Variation Pair

what you'll learn...

overview

The objective of this lesson is to show that direct and inverse variations are two sides of the same problem. A number of examples are provided to understand this fact.

number of days ×$\times$ rate (coins per day) =$=$ overall earnings

number of books ×$\times$ weight of a book =$=$ weight of the parcel

length of cloth ×$\times$ rate =$=$ overall cost

number of machines $\times$ number of hours $\Rightarrow$ number of machine-hours

number of pipes $\times$ number of hours $\Rightarrow$ amount filled

number of hens $\times$ number of days $\Rightarrow$ amount of food grain

example type 1

Consider the problem: A person earns $300$ coins in $10$ days. How much will she earn in $25$ days?

This problem is an example of direct variation.

As the number of days increases, the amount earned increases.

Consider the problem: A person earns $30$ coins per day for $10$ days. If the earning rate is reduced to $20$ coins per day, how many days the person has to work to earn the same amount?

This is an example of inverse variation.

As the earning rate decreases, the number of days increases.

Consider the problem: A person earns $300$ coins in 10 days. How much will she earn in $25$ days?

It is noted that the underlying mathematical operation is multiplication.

$10$ days $\times$ rate = $300$coins". Rate can be calculated as $30$ coins per day.

Consider the two problems. The underlying operation for both is
number of days $\times$ rate (coins per day) $=$ overall earnings (coins)

In the first problem,
•  number of days = $10$ days (multiplicand)
•  overall earnings = $300$ coins (product)
•  rate is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Earnings in $25$ days = $\frac{300}{10} \times 25$

In the second problem,
•  number of days = $10$ days (multiplicand)
•  rate of earnings = $30$ coins per day (multiplier)
•  overall earnings is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.

Number of days to earn the same amount at $20$ coins per day = $30 \times 10 / 20$

example type 2

Consider the problem: $3$ books are sent by postal-mail weighing $6$kg. How many books will be there in a parcel weighing $20$kg?

It is a direct variation problem.

As the weight of the parcel increases, the number of books increases.

Consider the problem: $3$ books, weighing $2$kg each, can be sent in a parcel. How many books of $3$kg each can be packed in a parcel of same weight?

It is an inverse variation problem.

As the weight of a book increases, the number of books decreases.

Consider the problem: $3$ books are sent by postal-mail weighing $6$kg. How many books will be there in a parcel weighing $20$kg?

The underlying mathematical operation in this is multiplication.

number of books $\times$ weight of a book $=$ weight of the parcel

Consider the two problems. The underlying operation for both is
number of books $\times$ weight of a book $=$ weight of the parcel

In the first problem,
•  number of books = $3$ (multiplicand)
•  weight of parcel = $6$kg (product)
•  weight of each book is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Number of books in a $20$kg parcel = $\frac{3}{6} \times 20 = 10$books

In the second problem,
•  number of books = $3$ (multiplicand)
•  weight of each book = $2$ kg (multiplier)
•  weight of the parcel is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Number of books of $3$kg weight = $3 \times 2 / 3 = 2$ books.

example type 3

Consider the problem: A person can buy $3$ meter cloth costing $600$ coins. What will be the cost of $7$m of the same cloth?

It is a direct variation problem.

As the length increases, the price of the cloth increases.

Consider the problem: A person can buy $3$ meter cloth which costs $200$ coins per cloth. If the person chooses a different cloth costing $150$ coins per cloth, what is the length of the cloth she can buy?

This is an inverse variation problem.

As the price decreases, the length of the cloth increases.

Consider the problem: A person can buy $3$ meter cloth costing $600$ coins. What will be the cost of $7$m of the same cloth?

The underlying mathematical operation is multiplication.

length of cloth $\times$ rate $=$ overall cost

Consider the two problems. The underlying operation for both is
length of cloth $\times$ rate $=$ overall cost

In the first problem,
•  length of cloth $=$ $3$ meter (multiplicand)
•  overall cost $= 600$ coins (product)
•  rate (multiplier) is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Cost of $7$ meter cloth $= \frac{600}{3} \times 7 = 1400$ coins

In the second problem,
•  length of cloth $=$ $3$ meter (multiplicand)
•  rate = $200$ coins (multiplier)
•  overall cost (product) is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Length of cloth at cost $150$ coins $= 3 \times 200 / 150 = 4$ meter

example type 4

Consider the problem: $5$ machines can process $90$ boxes of chemical in $15$ hours. How long would $3$ machines take?

This is an inverse variation problem.

As the number of machined decrease, the number of hours increases.

Consider the problem: $5$ machines can process $90$ boxes of chemical in 15 hours. If $3$ machines are used, how much chemical can be processed in the same number of hours?

This is a direct variation problem.

As the number of machines decrease, the amount produced decreases.

Consider the problem: $5$ machines can process $90$ boxes of chemical in $15$ hours. How long would $3$ machines take?

The underlying mathematical operation in this is multiplication.

number of machines $\times$ number of hours = number of machine-hours equivalent of $90$ boxes

Consider the two problems. The underlying operation for both is
number of machines $\times$ number of hours $\Rightarrow$ number of machine-hours equivalent of $90$ boxes

In the first problem,
•  number of machines = $5$ (multiplicand)
•  number of hours = $15$ (multiplier)
•  number of machine-hours (product) remains unchanged.
Multiplicand and multiplier are in inverse proportion.

$3$ machines would take $= 5 \times 15 / 3 = 25$ hours.

In the second problem,
•  number of machines = $5$ (multiplicand)
•  number of machine-hours is equivalently $90$ boxes of chemical (product)
•  number of hours (multiplier) remains unchanged.
Multiplicand and product are in direct proportion.
Amount of chemical processed by $3$ machines $= \frac{90}{5} \times 3 = 54$ boxes

example type 5

Consider the problem: $4$ pipes can fill a tank in $60$ mins. How long does it take to fill the tank with $9$ pipes?

This is an inverse variation problem.

As the number of pipes increase the time to fill the tank decreases.

Consider the problem: $4$ pipes can fill a tank in $60$ mins. How much of the tank will be filled by $3$ pipes in the same time duration?

This is a direct variation problem.

As the number of pipes decrease the amount filled decreases.

Consider the problem: $4$ pipes can fill a tank in $60$ mins. How long does it take to fill the tank with $9$ pipes?

The underlying mathematical operation is multiplication.

number of pipes $\times$ number of hours $\Rightarrow$ amount filled

Consider the two problems. The underlying operation for both is
number of pipes $\times$ number of hours $\Rightarrow$ amount filled

In the first problem,
•  number of pipes = $4$(multiplicand)
•  number of hours = $60$ mins (multiplier)
•  amount filled (product) is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Time to fill the tank with $9$ pipes = $4 \times 60 / 9$

In the second problem,
•  number of pipes = $4$(multiplicand)
•  amount filled = $1$ tank (product)
•  time (multiplier) remains unchanged.
Multiplicand and product are in direct proportion.
The amount filled by $3$ pipes $= \frac{1}{4} \times 3$

example type 6

Consider the problem: A farmer has enough grains to feed $300$ hens for 18 days. If he buys $100$ more hens, how long would the food last?

This is a problem of inverse variation.

Consider the problem: A farmer has enough grains to feed $300$ hens for $18$ days. If he buys $100$ more hens, how much more grains he has to buy to feed the entire $400$ hens?

This is a problem of direct variation.

As the number of hens increase, the amount of grains required increases.

Consider the problem: A farmer has enough grains to feed $300$ hens for $18$ days. If he buys $100$ more hens, how long would the food last?

The underlying mathematical operation in this is multiplication.

number of hens $\times$ number of days $\Rightarrow$ amount of food grain

Consider the two problems. The underlying operation for both is
number of hens $\times$ number of days $\Rightarrow$ amount of food grain

In the first problem,
•  number of hens $= 300$ (multiplicand)
•  number of days $= 18$(multiplier)
•  amount of food grain (product) is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
The time the food lasts $= 300 \times 18 / 400$ days

In the second problem,
•  number of hens $= 300$ (multiplicand)
•  Amount of food grain $= 1$ part(product)
•  number of days (multiplier) remains unchanged.
Multiplicand and product are in direct proportion.
Amount of food required for $400$ hens $= \frac{1}{300} \times 400$,
This includes the food-grain already available.

example type 7

Consider the problem: A person makes a car in $20$ days. If $4$ persons work together, how long does it take to complete a car?

This is a problem of inverse variation.

As the number persons increase, the number of days to complete a work decreases.

Consider the problem: A person makes a car in $20$ days. If $4$ persons work together, how many cars can they make in $20$ days?

This is a problem of direct variation.

As the number persons increase, the number of cars produced increases.

Consider the problem: A person makes a car in $20$ days. If $4$ persons work together, how long does it take to complete a car?

The underlying mathematical operation is multiplication.

number of persons $\times$ number of days $\Rightarrow$ number of car

Consider the two problems. The underlying operation for both is
number of persons $\times$ number of days $\Rightarrow$ number of cars

In the first problem,
•  number of persons = $1$ (multiplicand)
•  number of days = $20$ days(multiplier)
•  number of cars (product) remains unchanged.
Multiplicand and multiplier are in inverse proportion.
number of days to complete the car for $4$ persons $= 1 \times 20 / 4$

In the second problem,
•  number of persons = $1$ (multiplicand)
•  number of cars = $1$ (product)
•  number of days (multiplier) is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Number of cars $4$ persons can make = $\frac{1}{1} \times 4$ cars

summary

Direct and Inverse Variation Pair : Multiplicand and product are in direct variation.
Multiplier and product are in direct variation.
Multiplicand and Multiplier are in inverse variation.

Outline