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Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

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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