Houses/workers/days problem (direct and inverse variation)

Today I want to feature a great video from a fellow blogger and Youtuber, Dave Marain. He solves this question, which at first can sound intimidating:

If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days?

It sounds like you'd need algebra, proportions, inverse or direct variation, etc. And true, you could use those. But his method completely avoids all that and is based on setting up a simple TABLE, AND using common sense!

|  HOUSES  |  WORKERS  |  DAYS  |
----------------------------------
|    3     |    10     |  60    |
----------------------------------
|    1     |    10     |        |
----------------------------------
|    2     |    10     |        |
----------------------------------
|    5     |           |        |
----------------------------------

See the video below:




And here's a link to his other Youtube videos: www.youtube.com/user/MathNotationsVids.

Houses/workers/days problem (direct and inverse variation)

Today I want to feature a great video from a fellow blogger and Youtuber, Dave Marain. He solves this question, which at first can sound intimidating:

If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days?

It sounds like you'd need algebra, proportions, inverse or direct variation, etc. And true, you could use those. But his method completely avoids all that and is based on setting up a simple TABLE, AND using common sense!

|  HOUSES  |  WORKERS  |  DAYS  |
----------------------------------
|    3     |    10     |  60    |
----------------------------------
|    1     |    10     |        |
----------------------------------
|    2     |    10     |        |
----------------------------------
|    5     |           |        |
----------------------------------

See the video below:




And here's a link to his other Youtube videos: www.youtube.com/user/MathNotationsVids.

More new worksheet generators

At HomeschoolMath.net, you can now make free worksheets for place value & scientific notation (such as write a number in expanded form or in scientific notation), and also for proportions, including simple proportion word problems.

The links are:
Place value & scientific notation worksheets
Proportion worksheets

More new worksheet generators

At HomeschoolMath.net, you can now make free worksheets for place value & scientific notation (such as write a number in expanded form or in scientific notation), and also for proportions, including simple proportion word problems.

The links are:
Place value & scientific notation worksheets
Proportion worksheets

New Math Mammoth books

There are now three new books available in the Blue Series. They are all for grades 5-6. The material for these books came from the Light Blue 5th and 6th grade curricula. Please click on the links to learn more and see samples.

1. 

   Math Mammoth The Four Operations
   (with a Touch of Algebra)

   The main topics studied in this book are simple equations, expressions that involve a variable, the order of operations, long multiplication, long division, and graphing simple linear functions.
   
   The idea is not to practice each of the four operations separately, but rather to see how they are used together in solving problems and in simple equations. We are trying to develop student's algebraic thinking. Many of the ideas in this chapter are preparing them for algebra in advance.
   
   

2. Math Mammoth Ratios & Proportions & Problem Solving

   This worktext concentrates, first of all, on two important concepts: ratios and proportions, and then on problem solving.
   
   First, we study thoroughly the concept of ratio, including how it connects with fractions and the aspect ratio in rectangles.
   
   Solving proportions is divided into three separate lessons. In the first one, we solve proportions by thinking through equivalent ratios. In the second one, the usual method of cross-multiplying is introduced. Then follows a lesson that explains just why cross-multiplying is allowed. Then there is more practice with solving proportions and word problems. We also study scaling geometric figures and floor plans.
   
   The last lessons concentrate on various kinds of word problems that can be solved with the help of bar or block diagrams (the same kind as in Singapore Math). These involve problems with fractional parts, and then problems involving ratios. The diagrams become a powerful tool to solve such problems without the use of algebra.
   
   
3. 

   Math Mammoth Place Value 5

   Math Mammoth Place Value 5 is a short worktext that covers value up to the trillions—that is, numbers up to 15 digits—rounding them, and using a calculator.
   
   The first lesson covers place value up to billions. Then, we study exponents, and right after that, place value up to trillions, writing the numbers in the expanded form using exponents. After working with addition a little, there are two lessons about calculator usage.
   
   

New Math Mammoth books

There are now three new books available in the Blue Series. They are all for grades 5-6. The material for these books came from the Light Blue 5th and 6th grade curricula. Please click on the links to learn more and see samples.

1. 

   Math Mammoth The Four Operations
   (with a Touch of Algebra)

   The main topics studied in this book are simple equations, expressions that involve a variable, the order of operations, long multiplication, long division, and graphing simple linear functions.
   
   The idea is not to practice each of the four operations separately, but rather to see how they are used together in solving problems and in simple equations. We are trying to develop student's algebraic thinking. Many of the ideas in this chapter are preparing them for algebra in advance.
   
   

2. Math Mammoth Ratios & Proportions & Problem Solving

   This worktext concentrates, first of all, on two important concepts: ratios and proportions, and then on problem solving.
   
   First, we study thoroughly the concept of ratio, including how it connects with fractions and the aspect ratio in rectangles.
   
   Solving proportions is divided into three separate lessons. In the first one, we solve proportions by thinking through equivalent ratios. In the second one, the usual method of cross-multiplying is introduced. Then follows a lesson that explains just why cross-multiplying is allowed. Then there is more practice with solving proportions and word problems. We also study scaling geometric figures and floor plans.
   
   The last lessons concentrate on various kinds of word problems that can be solved with the help of bar or block diagrams (the same kind as in Singapore Math). These involve problems with fractional parts, and then problems involving ratios. The diagrams become a powerful tool to solve such problems without the use of algebra.
   
   
3. 

   Math Mammoth Place Value 5

   Math Mammoth Place Value 5 is a short worktext that covers value up to the trillions—that is, numbers up to 15 digits—rounding them, and using a calculator.
   
   The first lesson covers place value up to billions. Then, we study exponents, and right after that, place value up to trillions, writing the numbers in the expanded form using exponents. After working with addition a little, there are two lessons about calculator usage.
   
   

Making the fractions in a proportion

"How do you know how to make the fractions in a proportion?
When making them, how do you know where each number goes making the fraction, like which ones go on top of the fraction?"

Well, actually you can choose which quantity will go on top; the proportion WILL work either way!

But, sometimes people are used to always putting certain quantity on top and certain on the bottom. For example, if the question is about speed and the unit is "miles per hour", that tells you that miles go on top, and hours on bottom, because "per" means division (the fraction line).

However, you could still solve the proportion by putting hours in the numerator of the fractions and miles in the denominator, and the calculation will turn out alright.

Or, if the question is about "dollars per pound", then dollars go to the numerator and pounds in the denominator.

Let's look at this problem for example:

A car drives on constant speed. It can go 80 miles in 90 minutes. How long will it take for it to travel 100 miles?

You can make both fractions to be

miles
-----------
minutes

OR

minutes
------------
miles

Let's try the first way:

80 mi           100 mi
--------  =   ---------
90 min            x

To solve, cross multiply and you get 80x = 100 * 90, and then x = 900/8 = 112.5 minutes.

The other way it will be

90 min         x
--------  =  --------
80 mi         100 mi

To solve, cross multiply and you get 80x = 90 * 100

You see, the final equation ends up being the same, no matter which
quantities were on top of the fractions.

HOWEVER, one way is wrong: that is if you but "miles" on top in one
fraction, and "minutes" on top in the other... then you'll get it wrong:

90 min        100 mi
--------  =  -------
80 mi         x min

=> 90x = 100* 80 (WRONG)

Making the fractions in a proportion

"How do you know how to make the fractions in a proportion?
When making them, how do you know where each number goes making the fraction, like which ones go on top of the fraction?"

Well, actually you can choose which quantity will go on top; the proportion WILL work either way!

But, sometimes people are used to always putting certain quantity on top and certain on the bottom. For example, if the question is about speed and the unit is "miles per hour", that tells you that miles go on top, and hours on bottom, because "per" means division (the fraction line).

However, you could still solve the proportion by putting hours in the numerator of the fractions and miles in the denominator, and the calculation will turn out alright.

Or, if the question is about "dollars per pound", then dollars go to the numerator and pounds in the denominator.

Let's look at this problem for example:

A car drives on constant speed. It can go 80 miles in 90 minutes. How long will it take for it to travel 100 miles?

You can make both fractions to be

miles
-----------
minutes

OR

minutes
------------
miles

Let's try the first way:

80 mi           100 mi
--------  =   ---------
90 min            x

To solve, cross multiply and you get 80x = 100 * 90, and then x = 900/8 = 112.5 minutes.

The other way it will be

90 min         x
--------  =  --------
80 mi         100 mi

To solve, cross multiply and you get 80x = 90 * 100

You see, the final equation ends up being the same, no matter which
quantities were on top of the fractions.

HOWEVER, one way is wrong: that is if you but "miles" on top in one
fraction, and "minutes" on top in the other... then you'll get it wrong:

90 min        100 mi
--------  =  -------
80 mi         x min

=> 90x = 100* 80 (WRONG)

A certain proportion

I need help explaining how to solve:

 12      36
---- =  ----
 3w      63

Could you solve this if instead of 3w it was x?

Well you can do just that:

 12      36
---- =  ----
 x       63

Now cross multiply and go about it as usual. The answer is x = 21.

But, x is actually 3w. So 3w = 21. So w = 7.

This changing of variables is a very often used "trick" in mathematics.


Alternate solution.

Upon examining the proportion, we notice both the numbers in the first fraction are divisible by 3, and both numbers in the second fraction are divisible by 9.

 12      36
---- =  ----
 3w      63

So, we can write it as follows and simplify both fractions (the right and left sides):

3 * 4        9 * 4
------  =   -------
 3w          9 * 7


 4     4
--- = ---
 w     7

You can't find a simpler proportion than that, obviously w must equal 7.

A certain proportion

I need help explaining how to solve:

 12      36
---- =  ----
 3w      63

Could you solve this if instead of 3w it was x?

Well you can do just that:

 12      36
---- =  ----
 x       63

Now cross multiply and go about it as usual. The answer is x = 21.

But, x is actually 3w. So 3w = 21. So w = 7.

This changing of variables is a very often used "trick" in mathematics.


Alternate solution.

Upon examining the proportion, we notice both the numbers in the first fraction are divisible by 3, and both numbers in the second fraction are divisible by 9.

 12      36
---- =  ----
 3w      63

So, we can write it as follows and simplify both fractions (the right and left sides):

3 * 4        9 * 4
------  =   -------
 3w          9 * 7


 4     4
--- = ---
 w     7

You can't find a simpler proportion than that, obviously w must equal 7.

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

Problem on proportional reasoning

I'm just going to solve here on the blog another math problem that was sent in to me... Hopefully it helps some of you to learn how to solve problems (you can let me know!)

A man travelled 8 miles in the second hour. This is 1/7 times more than during the first hour, and 1/4 times more than he travelled during the third hour. What is the total miles he covered in three hours?

In this problem, it is easy to get "deceived" and think that you'd just go 1/7 × 8 miles or something like that.

But think first; did the man cover MORE miles during the first hour than during the second hour?

Yes; it says plainly that the 8 miles was 1/7 times MORE than what he covered during the first hour.

Is that 8 miles a LOT more, or a LITTLE BIT more than what he traveled during the first hour?

It's 1/7 times more, so it's a little bit more.

So simply mark as x the miles covered during the first hour. Then,

8 miles = 1 1/7 x

It's not 1/7 x, but 1 1/7 x, or 8/7x. If you put down 1/7 x, you're finding the seventh part, but it was seventh part MORE.

So 8 miles = 8/7x

Multiply both sides by 7, and divide by 8:

x = 7 miles.

Then the last hour. Again the miles traveled during the second hour are more than the miles traveled during the third hour. If miles traveled during the third hour are y, then we get,

8 miles = 1 1/4 y

8 miles = 5/4 y

y = 32/5 miles, or 6 2/5 miles.

Total he covered 7 mi + 8 mi + 6 2/5 mi = 21 2/5 miles or 21.4 miles.

Problem on proportional reasoning

I'm just going to solve here on the blog another math problem that was sent in to me... Hopefully it helps some of you to learn how to solve problems (you can let me know!)

A man travelled 8 miles in the second hour. This is 1/7 times more than during the first hour, and 1/4 times more than he travelled during the third hour. What is the total miles he covered in three hours?

In this problem, it is easy to get "deceived" and think that you'd just go 1/7 × 8 miles or something like that.

But think first; did the man cover MORE miles during the first hour than during the second hour?

Yes; it says plainly that the 8 miles was 1/7 times MORE than what he covered during the first hour.

Is that 8 miles a LOT more, or a LITTLE BIT more than what he traveled during the first hour?

It's 1/7 times more, so it's a little bit more.

So simply mark as x the miles covered during the first hour. Then,

8 miles = 1 1/7 x

It's not 1/7 x, but 1 1/7 x, or 8/7x. If you put down 1/7 x, you're finding the seventh part, but it was seventh part MORE.

So 8 miles = 8/7x

Multiply both sides by 7, and divide by 8:

x = 7 miles.

Then the last hour. Again the miles traveled during the second hour are more than the miles traveled during the third hour. If miles traveled during the third hour are y, then we get,

8 miles = 1 1/4 y

8 miles = 5/4 y

y = 32/5 miles, or 6 2/5 miles.

Total he covered 7 mi + 8 mi + 6 2/5 mi = 21 2/5 miles or 21.4 miles.

Writing some proportions

Got a question in today,

How can I solve the question, "write as many true proportions as you can with the numbers 3,6,9,12 and 18. Use a number only once".

This is a simple simple problem IF you know what the term "PROPORTION" means.

Proportion is simply an equation stating that one ratio is equal to another.

So to solve this, we play around and make ratios, and see if we can by happenstance come up with some equal ratios fitting the rule of using each number only once.

For example, 3:6 and 9:18 are two such ratios. So the PROPORTION then is (it needs an '=' sign)

3:6 = 9:18.

You can write the ratios using the fraction line, too.

Can you make others?

See also An idea of how to teach proportions.

Writing some proportions

Got a question in today,

How can I solve the question, "write as many true proportions as you can with the numbers 3,6,9,12 and 18. Use a number only once".

This is a simple simple problem IF you know what the term "PROPORTION" means.

Proportion is simply an equation stating that one ratio is equal to another.

So to solve this, we play around and make ratios, and see if we can by happenstance come up with some equal ratios fitting the rule of using each number only once.

For example, 3:6 and 9:18 are two such ratios. So the PROPORTION then is (it needs an '=' sign)

3:6 = 9:18.

You can write the ratios using the fraction line, too.

Can you make others?

See also An idea of how to teach proportions.

Teaching proportions

Many times kids do learn how to solve proportion problems in school (they manage to memorize the steps), but that seems to get forgotten in a flash after school is over. Maybe they only remember faintly something about cross multiplying but that's as far as it goes. How can we educators help them learn and retain? I finally posted the article about how to teach proportions.

Teaching proportions

Many times kids do learn how to solve proportion problems in school (they manage to memorize the steps), but that seems to get forgotten in a flash after school is over. Maybe they only remember faintly something about cross multiplying but that's as far as it goes. How can we educators help them learn and retain? I finally posted the article about how to teach proportions.