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Thought-Process to Discover Knowledge

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mathsCommercial ArithmeticsUnitary Method; Direct & Inverse Variations

### Direct and Inverse Variation Pair

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.

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

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse 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?

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• 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?

What is the underlying mathematical operation in this?

• 10 days xx rate = 300coins
• 10 days xx rate = 300coins
• 10 days xx 300 coins = rate

The answer is "10 days xx rate = 300coins". Rate can be calculated as 30 coins per day.

Consider the two problems:

•  A person earns 300 coins in 10 days. How much will she earn in 25 days?

•  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?

The underlying operation for both is
number of days xx 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 = 300/10 xx 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 = 30xx10 // 20

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

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse variation

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

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

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• inverse variation

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

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

What is the underlying mathematical operation in this?

• number of books xx weight of parcel = weight of a book
• number of books xx weight of a book = weight of the parcel
• number of books xx weight of a book = weight of the parcel

The answer is "number of books xx weight of a book = weight of the parcel".

Consider the two problems:

•  3 books are sent by postal-mail weighing 6kg. How many books will be there in a parcel weighing 20kg?

•  There are 3 books each weighing 2kg each can be sent in a parcel. How many books of 3kg each can be packed in a parcel of same weight? The underlying operation for both is
number of books xx weight of a book = weight of the parcel

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

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 3kg weight = 3 xx 2 // 3 = 2 books.

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

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse variation

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?

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• inverse variation

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 7m of the same cloth?

What is the underlying mathematical operation in this?

• length of cloth xx rate = overall cost
• length of cloth xx rate = overall cost
• length of cloth xx overall cost = rate

The answer is "length of cloth xx rate = overall cost".

Consider the two problems:

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

•  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?

The underlying operation for both is
length of cloth xx 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 = 600/3 xx 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 xx 200 // 150 = 4 meter

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

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• inverse variation

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?

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse variation

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?

What is the underlying mathematical operation in this?

• number of machines xx number of hours = number of boxes
• number of machines xx number of hours rArr number of machine-hours equivalent of 90 boxes
• number of machines xx number of hours rArr number of machine-hours equivalent of 90 boxes

The answer is "number of machines xx number of hours = number of machine-hours equivalent of 90 boxes".

Consider the two problems:

•  5 machines can process 90 boxes of chemical in 15 hours. How long would 3 machines take?

•  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?

The underlying operation for both is
number of machines xx number of hours rArr 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 xx 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  = 90/5 xx 3 = 54 boxes

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

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• inverse variation

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?

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse variation

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?

What is the underlying mathematical operation in this?

• number of pipes xx number of hours rArr amount filled
• number of pipes xx number of hours rArr amount filled
• number of pipes xx amount filled rArr number of hours

The answer is "number of pipes xx number of hours rArr amount filled ".

Consider the two problems:

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

•  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?

The underlying operation for both is
number of pipes xx number of hours rArr 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 xx 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 = 1/4 xx 3

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?

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• 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?

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse 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?

What is the underlying mathematical operation in this?

• number of hens xx number of days rArr amount of food grain
• number of hens xx number of days rArr amount of food grain
• number of hens xx amount of food grain rArr number of days

The answer is "number of hens xx number of days rArr amount of food grain".

Consider the two problems:

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

•  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?

The underlying operation for both is
number of hens xx number of days rArr 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 xx 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 = 1/300 xx 400,
This includes the food-grain already available.

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?

Is it direct variation or inverse variation?

• direct variation
• inverse variation
• 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?

Is it direct variation or inverse variation?

• direct variation
• direct variation
• inverse 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?

What is the underlying mathematical operation in this?

• number of persons xx number of days rArr number of car
• number of persons xx number of days rArr number of car
• number of persons xx number of cars rArr number of days

The answer is "number of persons xx number of days rArr number of car".

Consider the two problems:

•  A person makes a car in 20 days. If 4 persons work together, how long does it take to complete a car?

•  A person makes a car in 20 days. If 4 persons work together, how many cars can they make in 20 days?

The underlying operation for both is
number of persons xx number of days rArr 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 xx 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 = 1/1 xx 4 cars

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.

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