Denise has made a good post on the concept of division, which I heartily recommend. She deals with a study where Finnish researchers gave this problem about division and remainders to high school students and pre-service teachers:
We know that:
498 ÷ 6 = 83.
How could you use this relationship (without using long-division) to discover the answer to:
491 ÷ 6 = ?
[No calculators allowed!]
I really like the question. To solve it, you need to TRULY understand what DIVISION and remainders are all about!
Now, let's think about it. Have you ever seen a pattern in division and remainders, like the one below?
20 ÷ 4 = 5
21 ÷ 4 = 5 R1, or 5 1/4
22 ÷ 4 = 5 R2, or 5 2/4
23 ÷ 4 = 5 R3, or 5 3/4
24 ÷ 4 = 6
25 ÷ 4 = 6 R1, or 6 1/4
26 ÷ 4 = 6 R2, or 6 2/4
27 ÷ 4 = 6 R3, or 6 3/4
28 ÷ 4 = 7
29 ÷ 4 = 7 R1, or 7 1/4
30 ÷ 4 = 7 R2, or 7 2/4
31 ÷ 4 = 7 R3, or 7 3/4
Students need to see and do such patterns when they are first learning basic division.
The pattern shows that every fourth number is evenly divisible by 4, and the ones in between have remainders 1, 2, or 3 in order. If the answer is given as a mixed number, the remainder is the numerator.
Back to 498 ÷ 6 = 83. Since 498 is divisible by 6, so is the number just 6 less than 498, or 492. In fact, 492 ÷ 6 = 82, or in other words, the quotient is one less than 83.
This makes sense when thinking of division as, "How many times does it fit?" If 6 fits into 498 exactly 83 times, then it fits into 492 one less time, or 82 times.
Continuing, also 492 − 6 = 486 is divisible by 6, and this time 486 ÷ 6 = 81.
We can now build the pattern from 486 onward until we have 491 on our list:
486 ÷ 6 = 81
487 ÷ 6 = 81 R1 or 81 1/6
488 ÷ 6 = 81 R2 or 81 2/6
489 ÷ 6 = 81 R3 or 81 3/6
490 ÷ 6 = 81 R4 or 81 4/6
491 ÷ 6 = 81 R5 or 81 5/6
492 ÷ 6 = 82
So, 491 ÷ 6 = 81 R5 or 81 5/6. Problem solved.
Sunday, November 22, 2009
Spread of H1N1 (swine) flu and mathematics
I came across an interesting blog post by Murray Borne titled H1N1 and the Logistic Equation. It explains how a logistic function can be used to model the spread of a virus or a disease in a given population.
Now, maybe you don't know what is logistic function or equation. It is shown in the blogpost; it uses the exponential function as a part of it. Basically, it is like an exponential growth function but it is limited after a certain point so that the growth tapers off, and approaches a certain (upper) limit.
Murray shows the graph, and then shows a real-life example about the spread of swine flu in Mexico last spring. It's a great, yet fairly simple, example of how mathematics is used for modeling real-life situations.
You could definitely use it as such an example with your students, even if you don't understand a THING about logistic equations. You see, seeing how math is used is definitely inspiring and motivating to a lot of students - especially when it ties in with some current "hot topic" such as the H1N1 flu.
Now, maybe you don't know what is logistic function or equation. It is shown in the blogpost; it uses the exponential function as a part of it. Basically, it is like an exponential growth function but it is limited after a certain point so that the growth tapers off, and approaches a certain (upper) limit.
Murray shows the graph, and then shows a real-life example about the spread of swine flu in Mexico last spring. It's a great, yet fairly simple, example of how mathematics is used for modeling real-life situations.
You could definitely use it as such an example with your students, even if you don't understand a THING about logistic equations. You see, seeing how math is used is definitely inspiring and motivating to a lot of students - especially when it ties in with some current "hot topic" such as the H1N1 flu.
10/10 and the Metric Week
Today is 10th of October or 10/10. As we know, the metric system is based on number 10. Thus, the week ending today has been designated as the metric week..
The National Council of Teachers of Mathematics (NCTM) started the National Metric Week tradition in 1976. Please read a little bit more about the history of the Metric Week here.
You could celebrate the Metric Week doing some metric units puzzles and quizzes. Here's a link to postal stamps and cards celebrating the metric system.
NCTM has tons of related resources so I'll mainly point you there.
I grew up using the metric system, and so I've actually had to learn the imperial system while authoring math materials. These days, it seems, I know the U.S. system better than the people around me... But the metric is sure easier, as far as calculations and conversions go, since you just have to remember it goes by 10s.
In fact, to fully operate in this world, it seems it's best to know both.
Converting between metric and U.S. measuring systems
Here are a few helpful guidelines if you find yourself having to switch between one or the other. I have these conversion factors memorized from much of use. Not that everyone else would actually use these all as much as I do, being in the business of authoring math materials, but anyway:
1 quart ≈ 1 liter, but 1 quart is slightly less.
1 liquid ounce ≈ 30 ml. From this, one can figure out that 1 cup ≈ 240 ml and 4 C ≈ 960 ml.
1 inch = 2.54 cm
It's an awkward number but I need to use this conversion factor constantly, when working with images on my computer, which has a resolution of 96 pixels per inch... but I need the image to print out as 5 cm long or whatever.
You could use 1 inch ≈ 2.5 cm.
Another way is to think about those typical student rulers which are 12 inches = 30 cm. So... 4 inches is 10 cm.
Also... 1 inch might very well be the length of your thumb's last bone (the bone that contains the nail). Check! And 1 cm might very well be the width of any of the other fingernails. Check!
1 yard ≈ 1 meter; better yet 1 yard ≈ 90 cm and 1 meter = 100 cm. You see, that student ruler was 12 inches = 1 foot ≈ 30 cm.
1 mile ≈ 1.6 km. Or, 5,000 meters or 5 K is a popular running distance... about 3 miles.
1 pound is about 450 g, but actually it's easier to remember 1 kg = 2.2 lb.
It's fairly easy to multiply by 2.2. I have a metric scale; I might weigh about 55 kg. 55 x 2 = 110 and 55 x 0.2 = 11. So I weigh about 121 lb.
And 1 ounce ≈ 30 g. Just like 1 liquid ounce was about 30 ml.
To convert anything, go to Google and type
25 lb to kg
8 m to inches
22 km to miles
etc.
The National Council of Teachers of Mathematics (NCTM) started the National Metric Week tradition in 1976. Please read a little bit more about the history of the Metric Week here.
You could celebrate the Metric Week doing some metric units puzzles and quizzes. Here's a link to postal stamps and cards celebrating the metric system.
NCTM has tons of related resources so I'll mainly point you there.
I grew up using the metric system, and so I've actually had to learn the imperial system while authoring math materials. These days, it seems, I know the U.S. system better than the people around me... But the metric is sure easier, as far as calculations and conversions go, since you just have to remember it goes by 10s.
In fact, to fully operate in this world, it seems it's best to know both.
Converting between metric and U.S. measuring systems
Here are a few helpful guidelines if you find yourself having to switch between one or the other. I have these conversion factors memorized from much of use. Not that everyone else would actually use these all as much as I do, being in the business of authoring math materials, but anyway:
1 quart ≈ 1 liter, but 1 quart is slightly less.
1 liquid ounce ≈ 30 ml. From this, one can figure out that 1 cup ≈ 240 ml and 4 C ≈ 960 ml.
1 inch = 2.54 cm
It's an awkward number but I need to use this conversion factor constantly, when working with images on my computer, which has a resolution of 96 pixels per inch... but I need the image to print out as 5 cm long or whatever.
You could use 1 inch ≈ 2.5 cm.
Another way is to think about those typical student rulers which are 12 inches = 30 cm. So... 4 inches is 10 cm.
Also... 1 inch might very well be the length of your thumb's last bone (the bone that contains the nail). Check! And 1 cm might very well be the width of any of the other fingernails. Check!
1 yard ≈ 1 meter; better yet 1 yard ≈ 90 cm and 1 meter = 100 cm. You see, that student ruler was 12 inches = 1 foot ≈ 30 cm.
1 mile ≈ 1.6 km. Or, 5,000 meters or 5 K is a popular running distance... about 3 miles.
1 pound is about 450 g, but actually it's easier to remember 1 kg = 2.2 lb.
It's fairly easy to multiply by 2.2. I have a metric scale; I might weigh about 55 kg. 55 x 2 = 110 and 55 x 0.2 = 11. So I weigh about 121 lb.
And 1 ounce ≈ 30 g. Just like 1 liquid ounce was about 30 ml.
To convert anything, go to Google and type
25 lb to kg
8 m to inches
22 km to miles
etc.
Cell size and scale
Just a neat link my hubby found this morning...
http://learn.genetics.utah.edu/content/begin/cells/scale/
You can zoom in to see these various things starting from a coffee bean and down to a skin cell, human egg, red blood cell, bacteria, viruses, hemoglobin, glucose and molecules, etc., all the way "down" to a carbon atom.
In measuring scale, you go from millimeters (0.001 or 10-3m) to micrometers (0.000001 or 10-6m) to nanometers (0.000000001 or 10-9m) to picometers (0.000000000001 or 10-12m).
http://learn.genetics.utah.edu/content/begin/cells/scale/
You can zoom in to see these various things starting from a coffee bean and down to a skin cell, human egg, red blood cell, bacteria, viruses, hemoglobin, glucose and molecules, etc., all the way "down" to a carbon atom.
In measuring scale, you go from millimeters (0.001 or 10-3m) to micrometers (0.000001 or 10-6m) to nanometers (0.000000001 or 10-9m) to picometers (0.000000000001 or 10-12m).
Percentages with mental math
Find 10% of some example numbers (by dividing by 10).
Find 1% of some example numbers (by dividing by 100).
Find 20%, 30%, 40% etc. of these numbers.
FIRST find 10% of the number, then multiply by 2, 3, 4, etc.
For example, find 20% of 18. Find 40% of $44. Find 80% of 120.
I know you can teach the student to go 0.2 × 18, 0.4 × 0.44, and 0.8 × 120 - however when using mental math, the above method seems to me to be more natural.
Find 3%, 4%, 6% etc. of these numbers.
FIRST find 1% of the number, then multiply.
Find 15% of some numbers.
First find 10%, halve that to find 5%, and add the two results.
Find 25% and then 75% of some numbers. 25% of a number is 1/4 of it, so you find it by dividing by 4. For example, 25% of 16 is 4. To find 75%, first find 25% and multiply that by 3.
Calculate some simple discounts. If an item is discounted 20%, 15%, 25%, 75% etc., then find the new price.
For example, a book costs $40 and is discounted by 15%. What is the new price?
First find 15% of $40 (10% of $40 is $4 and 5% of $40 is $2... so 15% of it is $6). Then subtract $40 - $6. So the new price is $34.
"40% of a number is 56. What is the number?" - types of problems.
You CAN do this mentally: First FIND 10% of the number, and then multiply that result by 10, and you'll get 100% of the number - which is the number itself.
So if 40% is 56, then 10% is 14 (divide by 4). Then, 100% of the number is 140. This result is reasonable, because 40% of this number was 56, so the actual number (140) needs to be more than double that.
"34% of a number is 129. What is the number?" (A calculator will help here.)
You don't need to write an equation. You can just first find 1% of this number, and then find 100% of the number.
If 34% of a number is 129, then 1% of that number is 129/34. Find that, and multiply the result by 100.
Hope you enjoyed these little mental math ideas! They also help students understand the concept of percent where they don't end up relying too much on mechanical calculations or equations.
Find 1% of some example numbers (by dividing by 100).
Find 20%, 30%, 40% etc. of these numbers.
FIRST find 10% of the number, then multiply by 2, 3, 4, etc.
For example, find 20% of 18. Find 40% of $44. Find 80% of 120.
I know you can teach the student to go 0.2 × 18, 0.4 × 0.44, and 0.8 × 120 - however when using mental math, the above method seems to me to be more natural.
Find 3%, 4%, 6% etc. of these numbers.
FIRST find 1% of the number, then multiply.
Find 15% of some numbers.
First find 10%, halve that to find 5%, and add the two results.
Find 25% and then 75% of some numbers. 25% of a number is 1/4 of it, so you find it by dividing by 4. For example, 25% of 16 is 4. To find 75%, first find 25% and multiply that by 3.
Calculate some simple discounts. If an item is discounted 20%, 15%, 25%, 75% etc., then find the new price.
For example, a book costs $40 and is discounted by 15%. What is the new price?
First find 15% of $40 (10% of $40 is $4 and 5% of $40 is $2... so 15% of it is $6). Then subtract $40 - $6. So the new price is $34.
"40% of a number is 56. What is the number?" - types of problems.
You CAN do this mentally: First FIND 10% of the number, and then multiply that result by 10, and you'll get 100% of the number - which is the number itself.
So if 40% is 56, then 10% is 14 (divide by 4). Then, 100% of the number is 140. This result is reasonable, because 40% of this number was 56, so the actual number (140) needs to be more than double that.
"34% of a number is 129. What is the number?" (A calculator will help here.)
You don't need to write an equation. You can just first find 1% of this number, and then find 100% of the number.
If 34% of a number is 129, then 1% of that number is 129/34. Find that, and multiply the result by 100.
Hope you enjoyed these little mental math ideas! They also help students understand the concept of percent where they don't end up relying too much on mechanical calculations or equations.
Mixture problems - algebra 1
A merchant made a mixture of 150lb. of tea worth $109.50 by mixing tea worth $1.25 a pound with tea worth $.65 a pound. How many pounds of each kind did he use?
Organizing the information in a table or chart is usually very helpful in dealing with mixture problems. Other than that, it helps to study several examples and practice solving them yourself. After a while, it gets easier and patterns begin to emerge.
The first problem has two unknowns. Let x be the amount of more expensive tea, and y the amount of the cheaper tea (in pounds).
In our table, we will look at the amounts of tea (in pounds), price per pound, AND the amount the tea is worth, which is (the amount) times (the price).
amount | price per lb | worth------------------------------------- x | $1.25 | 1.25x------------------------------------- y | $0.65 | 0.65y-------------------------------------
Then we add one more row to the table that has to do with the MIXTURE, or the total.
amount | price per lb | worth------------------------------------- x | $1.25 | 1.25x------------------------------------- y | $0.65 | 0.65y-------------------------------------150 lb | ?? | $109.50
Now we get our equations. First of all, x + y = 150. And secondly, 1.25x + 0.65y = 109.50.
This gives you a system of two linear equations to solve, using any standard technique. For example, you can solve from the first that y = 150 − x and substitute that into the second.
The solution is: x = 20, y = 130.
Check: we have 20 lbs of tea costing $1.25 per pound, so it is worth $25.
We have 130 lbs of tea costing $0.65 per pound, so it is worth $84.50. Total worth is $109.50. It checks.
--------------------------------------------------------------------------------
A pharmacist has 10 oz. of salt and water of which 4 oz. are salt. How may ounces of water must he add so that 5% of the new solution is salt.
This is a very typical (and routine) problem from algebra 1 textbooks. Here, our table will have one row for the original situation, and another for the final situation. We are checking the amounts of salt and water, and then the total amount.
| salt | water | total-------------------------------------1st situation | 4 | 6 | 10-------------------------------------2nd situation | 4 | ? | ?-------------------------------------
The KEY is that there is no salt added, only water. Our unknown is the amount of water added.
| salt | water | total---------------------------------------1st situation | 4 | 6 | 10---------------------------------------2nd situation | 4 | 6 + x | 10 + x---------------------------------------
The equation is gotten from the statement that 5% of the new solution is salt. This means that 5% of the total (which is 10 + x) is salt (which we know to be 4 oz).
0.05(10 + x) = 4.
0.5 + 0.05x = 4
0.05x = 3.5
x = 3.5 / 0.05 = 70.
He needs to add 70 oz of water.
--------------------------------------------------------------------------------
In 110 lb. of an alloy of tin and copper, the amount of tin was 5lb. less than 1/3 that of the copper. How may pounds of tin were there?
Again, we organize this into a table:
| tin | copper | total--------------------------------------| t | c | 110---------------------------------------
We have TWO unknowns: the amount of tin and the amount of copper. Right there we get one equation: t + c = 110. The statement "the amount of tin was 5lb. less than 1/3 that of the copper" allows us to build another equation relating t and c.
t = (1/3)c - 5
Again, a system of equations. Since t is expressed in terms of c in the equation above, I use that to substitute to the first equation:
(1/3)c - 5 + c = 110
(4/3)c = 115
c = 115 * 3 / 4
c = 86.25.
But it asked for t, so t is 110 - 86.25 = 23.75 lb.
I hope these examples were helpful in dealing with "mixture" type problems in algebra.
Organizing the information in a table or chart is usually very helpful in dealing with mixture problems. Other than that, it helps to study several examples and practice solving them yourself. After a while, it gets easier and patterns begin to emerge.
The first problem has two unknowns. Let x be the amount of more expensive tea, and y the amount of the cheaper tea (in pounds).
In our table, we will look at the amounts of tea (in pounds), price per pound, AND the amount the tea is worth, which is (the amount) times (the price).
amount | price per lb | worth------------------------------------- x | $1.25 | 1.25x------------------------------------- y | $0.65 | 0.65y-------------------------------------
Then we add one more row to the table that has to do with the MIXTURE, or the total.
amount | price per lb | worth------------------------------------- x | $1.25 | 1.25x------------------------------------- y | $0.65 | 0.65y-------------------------------------150 lb | ?? | $109.50
Now we get our equations. First of all, x + y = 150. And secondly, 1.25x + 0.65y = 109.50.
This gives you a system of two linear equations to solve, using any standard technique. For example, you can solve from the first that y = 150 − x and substitute that into the second.
The solution is: x = 20, y = 130.
Check: we have 20 lbs of tea costing $1.25 per pound, so it is worth $25.
We have 130 lbs of tea costing $0.65 per pound, so it is worth $84.50. Total worth is $109.50. It checks.
--------------------------------------------------------------------------------
A pharmacist has 10 oz. of salt and water of which 4 oz. are salt. How may ounces of water must he add so that 5% of the new solution is salt.
This is a very typical (and routine) problem from algebra 1 textbooks. Here, our table will have one row for the original situation, and another for the final situation. We are checking the amounts of salt and water, and then the total amount.
| salt | water | total-------------------------------------1st situation | 4 | 6 | 10-------------------------------------2nd situation | 4 | ? | ?-------------------------------------
The KEY is that there is no salt added, only water. Our unknown is the amount of water added.
| salt | water | total---------------------------------------1st situation | 4 | 6 | 10---------------------------------------2nd situation | 4 | 6 + x | 10 + x---------------------------------------
The equation is gotten from the statement that 5% of the new solution is salt. This means that 5% of the total (which is 10 + x) is salt (which we know to be 4 oz).
0.05(10 + x) = 4.
0.5 + 0.05x = 4
0.05x = 3.5
x = 3.5 / 0.05 = 70.
He needs to add 70 oz of water.
--------------------------------------------------------------------------------
In 110 lb. of an alloy of tin and copper, the amount of tin was 5lb. less than 1/3 that of the copper. How may pounds of tin were there?
Again, we organize this into a table:
| tin | copper | total--------------------------------------| t | c | 110---------------------------------------
We have TWO unknowns: the amount of tin and the amount of copper. Right there we get one equation: t + c = 110. The statement "the amount of tin was 5lb. less than 1/3 that of the copper" allows us to build another equation relating t and c.
t = (1/3)c - 5
Again, a system of equations. Since t is expressed in terms of c in the equation above, I use that to substitute to the first equation:
(1/3)c - 5 + c = 110
(4/3)c = 115
c = 115 * 3 / 4
c = 86.25.
But it asked for t, so t is 110 - 86.25 = 23.75 lb.
I hope these examples were helpful in dealing with "mixture" type problems in algebra.
Ratio word problem solved with block model and algebra
I guess it is time for some more problem solving, since someone sent this question in.
Two numbers are in the ratio of 1:2. If 7 be added to both, their ratio changes to 3:5. What is the greater number?
We can model the two original numbers with blocks. 1 block and 2 blocks makes the ratio to be 1:2.
|-------||-------|-------|
Now add the same thing to both (the 7):
7|-------|---||-------|-------|---| 7 The way I just happened to draw these suggests that I could just split the original block in two, and the problem is solved:
7|---|---|---||---|---|---|---|---| 7 Here, each little block is 7. The original larger blocks are 14 each.
So the original bigger number, which had two larger blocks, is 28, and the smaller number is 14.
Check:
Their ratio is 28:14 = 2:1. If you add 7 to both, you have 35 and 21, and their ratio is 35:21 = 5:3.
Solving the same problem using algebra
The two numbers in the ratio of 1:2 are x and 2x.
Once 7 is added to both, we have x + 7 and 2x + 7. Their ratio is 3:5, and we can write a proportion using fractions:
x + 7 3------- = ----2x + 7 5
Cross-multiply to get
5(x + 7) = 3(2x + 7)
5x + 35 = 6x + 21
35 - 21 = x
x = 14
The larger number was 2x or 28. We already checked this earlier.
Two numbers are in the ratio of 1:2. If 7 be added to both, their ratio changes to 3:5. What is the greater number?
We can model the two original numbers with blocks. 1 block and 2 blocks makes the ratio to be 1:2.
|-------||-------|-------|
Now add the same thing to both (the 7):
7|-------|---||-------|-------|---| 7 The way I just happened to draw these suggests that I could just split the original block in two, and the problem is solved:
7|---|---|---||---|---|---|---|---| 7 Here, each little block is 7. The original larger blocks are 14 each.
So the original bigger number, which had two larger blocks, is 28, and the smaller number is 14.
Check:
Their ratio is 28:14 = 2:1. If you add 7 to both, you have 35 and 21, and their ratio is 35:21 = 5:3.
Solving the same problem using algebra
The two numbers in the ratio of 1:2 are x and 2x.
Once 7 is added to both, we have x + 7 and 2x + 7. Their ratio is 3:5, and we can write a proportion using fractions:
x + 7 3------- = ----2x + 7 5
Cross-multiply to get
5(x + 7) = 3(2x + 7)
5x + 35 = 6x + 21
35 - 21 = x
x = 14
The larger number was 2x or 28. We already checked this earlier.
Subscribe to:
Posts (Atom)