__maths__>__Commercial Arithmetics__>__Unitary 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

The answer is "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?

Is it direct variation or inverse variation?

- direct variation
- inverse variation
- inverse variation

The answer is "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 = `300`coins
- `10` days `xx` rate = `300`coins
- `10` days `xx` `300` coins = rate

The answer is "`10` days `xx` rate = `300`coins". 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 `6`kg. How many books will be there in a parcel weighing `20`kg?

Is it direct variation or inverse variation?

- direct variation
- direct variation
- inverse variation

The answer is "direct variation".

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?

Is it direct variation or inverse variation?

- direct variation
- inverse variation
- inverse variation

The answer is "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 `6`kg. How many books will be there in a parcel weighing `20`kg?

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 `6`kg. How many books will be there in a parcel weighing `20`kg?

• There are `3` books each 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? 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 = `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 = `3/6 xx 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 xx 2 // 3 = 2` books.

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?

Is it direct variation or inverse variation?

- direct variation
- direct variation
- inverse variation

The answer is "direct 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

The answer is "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 `7`m 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 `7`m 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

The answer is "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

The answer is "direct 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

The answer is "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

The answer is "direct 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

The answer is "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

The answer is "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?

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

The answer is "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

The answer is "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?

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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