Math Teachers at Play carnival

Denise has done a good job putting together Math Teachers at Play carnival #20. Lots of stuff to read, head on over!

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.

Mixture problems - algebra 1

I am hoping you can help me. I can not remember how to solve mixture problems and how to set them up. Examples are as follows:

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.

Fact families on a whiteboard

I just found this picture that I took of the fact families my 4-year old wrote on the whiteboard - totally on her own.



There was a time she loved writing fact families like this every day. Being able to choose different color markers plus it being on the whiteboard seemed to be the main motivating factors, because she didn't want to write them on blank paper... Kids are funny.

Then again, it allows us teachers to use colorful markers as a "motivational tool" : )

Anyway, I was really happy that she had grasped the concept.

Review of Math Apprentice

Math Apprentice is a new free website, meant to show students how math is used in real world. In the game, you are like an apprentice at various companies, applying your math skills to challenges similar to those encountered in the real world and real companies.

Main Street - Click to enlarge

To begin, you click the button on the home page of the site that says "Explore the Math". Then choose your character, and you'll be on the main street (see screenshot above) . Then use arrow keys to move right or left, and click to select a company to visit.

The companies you can visit are:

• Sweet Treat Cafe - baking pies

Sweet Treat Cafe - Click to enlarge

• Wheelworks - constructing bicycles and exploring gear ratios
• Game Pro! - keep track of the distance between superhero and the villain in a computer game, using Pythagorean Theorem
• Spacelogic - study speed of a spacecraft & slope, and then angle & distance commands to get the space rover where it needs go.
• Trigon Studios - Explore the usage of sine and cosine functions to create rhytmic or repeating motion of animated objects.
• Doodles - explore polar curves created with sine and cosine. These can be like stars, flowers, or spirals.
  

Doodles - Click to enlarge

• Builders, Inc. - calculate areas and perimeters of shapes

Builders, Inc. Click to enlarge

• Adventure Rides - study the angle of elevation and height of a roller coaster

While the site is targeted to grades 4-7, many of the mathematical ideas are actually far more advanced than that. In the simulations involving sine and cosine, for example, all you have to do is change the values in the equations using sliders and observe. Also, in some activities there are instructions given how to calculate things.

It says the about page: "Some of the mathematics in Math Apprentice may seem advanced for its targeted age group, grades 4-7. That's ok. It's important for students to interact with math concepts beyond the standards. This is where the joy of math can often be found."

Nevertheless, I feel some activities are definitely best reserved for students who have studied the concept (such as Pythagorean Theorem in Game Pro! company).

In general, I think Math Apprentice has well-made and interesting activities, and kids are sure to enjoy it!

Percentages with mental math

(This is an older post that I have revised plus added a video to it.)
In this article I want to explore some ideas for using MENTAL math in calculating percents or percentages.

I have also made a video about this topic:



And here are the ideas:

1. Find 10% of some example numbers (by dividing by 10).
2. Find 1% of some example numbers (by dividing by 100).
3. 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.
4. Find 3%, 4%, 6% etc. of these numbers.
   FIRST find 1% of the number, then multiply.
5. Find 15% of some numbers.
   First find 10%, halve that to find 5%, and add the two results.
6. 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.
7. 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.
   
8. "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.
9. "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.