Basics of percent of change - with videos

The two videos below have to do with basics of percent of change, where you are given the initial and final quantities, and you have to calculate the the percent of change (percent change, percentage change).

The main formula I use is to calculate the DIFFERENCE in the quantities divided by the original quantity. Then this fraction is written as a decimal, and as a percent.





In the second video, I solve two word problems that involve percentage change - one having to do with percentage increase in area, and another where the price of a washer is discounted by 10%, then by another 10%, and we are asked the total discount percentage--yet the PRICE of the washer is NOT given!



Teachers - feel free to use these problems in your teaching!

I have recently been making lots of videos, aiming to produce full playlists for certain topics such as percent, decimals, and fractions. I also have many older videos on various topics. Click the image on the left to see my Youtube channel and see what I have uploaded thus far.

What fraction of the elephants is blue?

Someone sent me a nice word problem to solve:

 Image courtesy of Wednesday Elf - Mountainside Crochet

Bob went to the zoo last week and was captivated by the lovely baby elephants. Some were pink and some were blue. When he counted the elephants, he found that the number of pink elephants was 2 1/2 times the number of blue elephants. What fraction of the elephants was blue?

First of all, think: which ones are less, blue or pink? Clearly, the blue ones are less. Make those kind represented by ONE BLOCK or one unit.

So, let's let the BLUE elephants to be |----|  (one block).

Then, the pink ones would be |----|----|--| (2 1/2 times as many)

Now, to get the fraction asked, I could use those little dashed lines in my blocks... See, I made each block have 4 little dashes ---- and the half-block has two.

So, the Blue elephants are "four dashes" and the pink ones are "ten dashes". I know that in reality we don't know how many blue or pink elephants there are, but this won't matter if we only want to know what FRACTION or what PART of the elephants are blue. So we might as well assign some quantities here that fit the problem.

From this, we can write the fraction. Blue elephants are 4, total elephants are 14, so 4/14 of the elephants are blue. This simplifies to 2/7.

You can also get the same answer this way: blue elephants are 1 block, and there are a total of 3 1/2 blocks of elephants. The fraction is 1 / (3 1/2) which is equivalent to 2/7 (multiply both numerator and denominator by 2).

Tiling word problem

Someone sent me this problem to solve, so here goes:

Photo courtesy of Brajeshwar

Square polystyrene tiles, 50 cm by 50 cm, are used to cover the ceiling of a classroom measuring 7.4 m by 4.5 m

1. Find the number of tiles that are needed.

2. Find the total cost if one tile costs $130.

Now, consider the room that is 7.4 m by 4.5 m. The tiles will not fit exactly along the 7.4-meter side, so we will need to cut some of the tiles. So we might as well consider it to be 7.5-meter side for the purpose of figuring how many tiles we need.

Along the 7.5 meter side we would need 15 tiles, each 0.5 m. Why? Because 0.5 fits into 7.5 exactly 15 times. It's essentially a division problem... 7.5 ÷ 0.5 but you can solve it mentally by thinking that each meter takes 2 tiles, so 7.5 meters takes 7.5 x 2 = 15 tiles.

Similarly, along the 4.5 meter side you would need 9 tiles, each 0.5 m.

Therefore in total we need 15 x 9 = 135 tiles. Some of those will need cut!

The cost will be 135 x $130 = $17,550. This would be expensive tiling... but I realize the person sending the question might have used some other kind of money units (not US dollars)!

Problem solving videos--bar/block model

These videos show you examples of how to use the bar or block model in solving math word problems. The examples are all about 5th grade level. Enjoy! I hope they are of help!

First I solve the following word problem using a bar model (Singapore math style), taken from Math Mammoth grade 5 curriculum: Brenda and Lily shared the cost of a $11.70 lunch so that Brenda paid two times as much as Lily. Find their shares.





Next problem is this: One rake is $5.60 more than the other, and together they cost $22.70. How much does the cheaper rake cost?




And lastly, John spent 3/10 of his money, and had $45.57 left. How much did he have initially?

How to help a student with a fraction word problem

Today, my daughter had to tackle this word problem in her 5th grade Math Mammoth (lesson Multiply Mixed Numbers):

The square on the right measures 10 in. × 10 in. and
the rectangle inside it measures 6 7/8 in. × 3 1/2 in.
How many square inches is the colored area?

This requires the student to

• multiply mixed numbers, 
• subtract mixed numbers, 
• understand about area, and 
• understand how to find the "colored" area by subtraction of areas. 

So it is a multi-step word problem.

She didn't understand it, she said. My first "help" was this:
"Let's say we change those fractions to whole numbers 6 and 3. Can you mark those in the image? Would you be able to solve the problem now?"

The strategy I used is:
If you can't solve the problem at hand, change it and make it easier. Then try to solve the easier problem.

She was able to mark 6 and 3 on the sides of the rectangle (that is inside the square). But she said she couldn't solve it. She said it's not possible to find the area of the colored area!


Then I asked her, "Is there anything you CAN find? Is there anything you CAN solve using this information?"

And this is another great strategy in solving any problem (whether math or not): If you can't find the answer to the question in the problem, solve what you CAN solve. That might lead you to find the answer to your question somewhere along the way!

I said, "Well, we CAN find the area of the square. It is 100 square inches. We CAN find the area of the rectangle inside it."

THEN immediately after I said that, she saw it: "OHH! SUBTRACT!" And on she went to multiply the mixed numbers in the problem. So the story had a happy ending!

---

Here's the complete solution:

First multiply the mixed numbers 6 7/8 and 3 1/2 to find the area of the rectangle. Keep in mind they need to be changed into fractions before multiplying.

6 7/8 × 3 1/2 = 55/8 × 7/2
= 385 / 16 square inches

Change this into a mixed number. Here one needs to use long division to divide 385 ÷ 16 = 24 R1. This tells us the whole number part is 24. The remainder,1, tells us how many 16th parts are "left over."

So, 385 / 16 = 24 1/16 square inches. So this is the area of the rectangle.

Now, the area of the surrounding square is simply 10 in x 10 in. = 100 square inches.

And lastly, the area of the colored area is found by subtracting 100 − 24 1/16 = 75 15/16 square inches.

Green tea word problem

Photo credit: leojmelsrub

Ahmed sent me in this kind of word problem:


A tea producer want to market mixed green tea leaves at $14 per pound. how many pounds of high mountain green tea leaves worth $20 per pound must be mixed with 90 pounds of regular green tea leaved worth $10 per pound?

I have solve many problems like this one on my blog, but it never hurts to solve some more. This can be solved with algebra, using a chart. I've done that before for similar problems... so if you are reading this, and you feel a bit "rusty" in this area, try to make the chart yourself first, before you read further!

For the chart, we also need to choose a variable or several. In this case it is easy:  the unknown is obviously what is asked, or the amount of high mountain green tea. Note also that the cost is always the price per pound times the amount.

mountain green   regular green  the mixture
 tea leaves  tea leaves
--------------------------------------------------------------------------
amount |   x   90   90 + x
--------------------------------------------------------------------------
cost   |  20x   $900   14(90 + x)
--------------------------------------------------------------------------

The chart is ready. Its purpose is to help us write an EQUATION of some sort which will solve x.

So where can we find something equals something? It comes from the cost. The COST of mountain green tea + the COST of regular green tea = COST of the mixture.

20x + 900 = 14(90 + x)

20x + 900 = 1260 + 14x

6x = 360

x = 60 

Now, let's check. That's always the last step in solving equations.

We need 60 pounds of mountain green tea leaves mixed with 90 pounds of regular green tea leaves. The mixture will weigh 150 pounds. The cost of mountain green tea leaves will be $1200, the cost of regular green tea will be $900, and the total cost will be $2100. Calculating cost per pound: $2100 / 150 lb = $14 per pound, so it checks.

Two problems about fractional parts

I have 2 questions on fractions which I can't solve.

There were 3/5 as many adults as children on a bus. At the next bus stop, 6 adults and 6 children boarded the bus. As a result, there were 2/3 as many adults as children on the bus. How many people were on the bus at first?

A solution using bar diagrams (Singapore style):

children |----|----|----|----|----|

adults   |----|----|----|

Then we have:

children |----|----|----|----|----| +6

adults   |----|----|----| +6

Here look at the difference of children and adults. There are two "blocks" more children than adults. We know the number of adults is 2/3 of the number of children... therefore the DIFFERENCE of two blocks must be 1/3 of the children.

Now look:

children |----|----|----|----|----| +6

Two of those blocks is 1/3 of the total... the other two blocks is another 1/3 of the total... so |----| +6 or one block and 6 must be 1/3 of the total.
This means the +6 must be one block. Or, one block = 6. This now solves the problem, because originally we had 8 "blocks" of people, or 48 people.


Solution with algebra:
Let c be the number of children at first. There were (3/5)c adults at first.

Then 6 more adults and 6 more children come in, so now we have c + 6 children, and (3/5)c + 6 adults.

Now, the amount of adults = (2/3) of the amount of children.

(3/5)c + 6 = (2/3)(c + 6)

As this equation has fractions, let's multiply both sides by 15 to start.

9c + 90 = 10(c + 6)

9c + 90 = 10c + 60

30 = c

There were 30 children and 18 adults at first, or 48 total people.

---

2/5 of the counters in a box were red and the rest were blue. After putting another 48 blue counters into a box, 3/4 of the counters were blue. How many counters were in the box at first?

A solution using bar diagrams (Singapore style):

blue |----|----|----|

red  |----|----|

Then:

blue |----|----|----| + 48

red  |----|----| 

...and now 3/4 of the counters are blue. So, |----|----| (the red ones) or two blocks is 1/4 of the counters. Therefore there are a total of 8 blocks of counters. but I drew five blocks and 48. So, 48 must be worth three "blocks", or one block is 16.

At first we had five "blocks" of counters, or 5 x 16 = 80 counters.

CHECK:
80 counters... 32 red and 48 blue.
Add 48 blue. Now we have 32 red and 96 blue, a total of 128.
Reds are 32/128 = 8/32 = 1/4. That checks.

A solution using algebra:
At first, we have (2/5)c counters red and (3/5)c counters blue. ONce we hadd 48 blue counters, we have (2/5)c red counters and (3/5)c + 48 blue counters, and c + 48 total counters. It says now 3/4 of the counters are blue. We write that as an equation:

(3/4) of all counters = number of blue counters

(3/4)(c + 48) = (3/5)c + 48 Multiply this by 20.

15(c + 48) = 12c + 960

15c + 720 = 12c + 960

3c = 240

c = 80

There were originally 80 counters.

A problem with a chord: find the radius

Today I had the opportunity to solve a real math problem involving a circle and a chord of known length in it. I had to find the radius. It wasn't a textbook problem or a puzzle on some website, but a math problem I needed to solve for my own needs.

For a tiny while I thought I could find the answer online, but I didn't, so I'm writing it out in case someone else needs it -- they should be able to find this solution by searching the Internet.


I wanted to make a kind of "moon-sliver shapes" in CorelDraw, to use as watermarks in my new books. I have the height and the width of the "sliver".

Here's the problem mathematically:
I have a chord of a circle, 17 mm in length in my example, and the other distance marked in the image is 5 mm. I need to find the radius of the circle, AND the angle measure of the arc of the circle that makes the sliver's rounded part.





At first, like I said, I searched around if there was some theorem or formula that would tell me what I needed directly. I didn't find any, but I did realize that I can use this theorem to solve my problem:

If two chords intersect, the product of the segments of one chord equals the product of the segments of the other chord (see proof).


My problem looks like this:




My chord intersects the diameter of the circle, which is a chord too. The two parts of the first cord are 8.5 and 8.5, and the two parts of the other are 5 and d − 5. Thus I get the equation

5(d − 5) = 8.5²

From this it is quick to solve that d = 19.45. Then, the radius is of course half that, or 9.725.

On to the second part of my problem: to find the angle measure.

In this picture, I have a right triangle. I can therefore find the unknown angle α by using simple trigonometry, in this case the tangent.



The length 4.725 comes from the fact that the radius is 9.725, and then I subtract 5 from that.

The equation is tan α = 8.5/4.725, from which α ≈ 60.931°. The actual angle I want is double that, or about 121.862°.

I had to repeat this calculation several times for slivers of different "heights." (Or, actually I let Excel calculate the rest. Oh, how I love Excel!)

Here's one of the final pictures I made for my book:

Seniors & juniors algebra word problem

Here's a word problem that someone sent me recently:

The total number of girls in the combined junior and senior classes is equal to the number of boys in those two classes. If the senior class has 400 students and the junior class has 300 students, and if the ratio of boys to girls in the senior class is 5:3, what is the ratio of boys to girls in junior class?

This problem gives a lot of information, and it sounds like it can be solved many different ways. But the first task is to notice what we are given and what we are asked.

• We are asked about a ratio. 
• We're given one ratio, and all kinds of totals. There are boys and girls, senior and junior classes. In other words, there are four groups: senior boys, senior girls, junior boys and junior boys.

This sounds like it can be solved by setting up some equations and using algebra.

To start, you could for example notice the two facts given about the senior class: The senior class has 400 students and the ratio of boys to girls is 5:3.

From this it is easy to solve the number of boys and girls in the senior class.
The ratio 5:3 means 5/8 of them are boys and 3/8 of them are girls. Now, 5/8 of 400 is 250, and 3/8 of it is 150.

So seniors are solved... now to juniors. Let

B1 = boys in junior class
G1 = girls in junior class.

The first sentence of the problem gives us an equation:
G1 + 150 = B1 + 250

We also now that G1 + B1 = 300.

So let's solve this system of two equations:
G1 + 150 = B1 + 250
G1 + B1 = 300

I can subtract the bottom one from the top one to get:

150 - B1 = B1 - 50

200 = 2B1
B1 = 100

Since the total was 300, then G1 is 200.
And now the ratio of boys to girls, it is 100:200 or 1:2. All done!

---

You could also use straightforward algebra and set up four equations originally, using B1, G1, B2, and G2 for the numbers of boys and girls in junior and senior classes. You would get these four equations:

G1 + G2 = B1 + B2
B2 + G2 = 400
B1 + G1 = 300
B2/G2 = 5/3

And from these four you can solve all four unknowns, and then get the ratio of boys to girls in the junior class.

Which operation should you use in word problems?

I was asked just recently about word problems and WHAT operation to use in them.

Can you put a better explanation of when to use addition, subtraction, multiplication, and division, because my son is having a hard time understanding.

Many children have the same problem, and I have written about it in the past as well. In a nutshell, I feel the MAIN reason students have trouble is the way word problems are presented (or not presented) in math books.

Typically (and I still see this approach), when a lesson in a math book is on some operation, then the word problems in that lesson are typically solved using that particular operation. So, any intelligent kid who notices this pattern won't take time to decipher what the problem says, but will just take the two numbers that appear in the problem, and apply the operation that the lesson is about!

(For example, if the lesson is on subtraction, then the word problems are solved using subtraction.)

I have tried to AVOID this in my books from the start, and I think all other math curricula should follow suit, that whenever there are word problems to solve, that the operation or operations used to solve it SHOULD vary within the problem set.

Even in first grade addition, you can make two kinds of problems:

"Ann had seven marbles, and Mary had two. How many marbles do they have together?" SOLUTION: 7 + 2 = 9

OR

"Ann has seven marbles, but today she only found two. How many are missing?" SOLUTION: Think 2 + ___ = 7, or two and how many more makes seven?


So, starting in 1st grade, the word problems should never be such that they always involve the same operation. Then in 2nd grade, the word problems should proceed to easy 2-step problems.

When multiplication is introduced in 3rd grade, the word problems should be in two different ways so the child cannot automatically multiply the numbers in the problem, but has to think. For example:

Mary saw 12 duck legs. How many ducks were there?

OR

Grandma has 12 ducks, and each wore two blue ribbons. How many ribbons could you see when looking at all the ducks?

---

Now, recently I had an idea for help for all these elementary students that have trouble figuring out which operation to use. I THINK it might work fairly well, but I'd like to have your input on it.

Suppose we have a student in 3rd or 4th grade who can't do word problems. What if we used the word problems in Ray's Arithmetic with him/her, starting from the beginning?

Ray's is known to be full of word problems, starting from the simplest. It sort of uses word problems to teach arithmetic. It is an old book, and therefore it is available FREE online. See the links on the left for viewing it online or downloading.

See for example the word problems on page 26.

You can also purchase Ray's as printed books.

What do you think? Could Ray's get such students up to speed with word problems?

Fraction of a fraction word problem

A problem about fraction of a fraction...

The sixth-graders have a fundraieser. They raise enough money to reach 7/8 of their goal. Nikki raises 3/4 of this money. What fraction of the goal does Nikki raise?

The picture below shows first of all 7/8. Nikki raises 3/4 of this goal. We need to find 3/4 of 7/8.

It's not easy to directly see what is 3/4 of 7/8. So to do that, I divide each 1/8 piece into four pieces, and then color three of the four. That way I color 3/4 of each of the seven eighths.

Of course, those tiny pieces are now 1/32 parts. I have colored 3 x 7 = 21 of them. So, the colored part represents the fraction 21/32.

This problem is also simple to solve without a picture, if you understand what is asked. To find 3/4 of 7/8, you simply multiply those two fractions. The word "of" translates into MULTIPLICATION in fraction math!

 3       7      21
---  x  --- =  ----
 4       8      32

Two algebra 1 word problems (systems of linear equations)

Here are two problems for you to solve... OR to learn from me when I solve them. Both problems are for algebra 1, and use a system of 2 linear equations.

By the way, the comments have some wonderful ideas for solving these mentally, without using algebra. So please read them too!

---

Problem: John bought red pens for $4 apiece and blue pens for $2.80 apiece. If John purchased a total of 24 pens for $84, how many red pens did he purchase?

---

Solution: This is a typical problem that will have two variables and two equations.

Let r be the amount of red pens he buys, and b be the amount of blue pens he buys.

We get our first equation from this sentence: "He bought a total of 24 pens." So, r + b = 24.

We get the second equation from the fact that his total purchases were worth $84, and red pens cost $4, and blue pens cost $2.80

4r + 2.8b = 84

Now, just solve this system of two equations using your preferred method.

r + b = 24
4r + 2.8b = 84

I will multiply the top equation by -4, then add the two equations.

-4r - 4b = -96
4r + 2.8b = 84
-------------------
-1.2b = -12
From this, b = 10.

Then since r + b = 24, r must be 14.

---

Problem: In a group of 60 workers, the average salary is $80 a day per worker.  If some of the workers earn $75 a day and all the rest earn $100 a day, how many workers earn $75 a day?

---

Solution: To get started, first find out what are the unknowns. In this case, there are two. Some workers earn $75 a day, and some earn $100 (two quantities).

Let A = workers who earn $75 a day.
Let B = workers who earn $100 a day.

We need to have both variables when building the equations, even though the problem only asks for A.

THEN we need to somehow use the information given to build two equations. That is because to solve for two unknowns, you need two equations.

Well, there are 60 workers, so our first equation is pretty easy: A + B = 60.

Now, the first sentence may throw you off... but actually, the concept of average is not coming into the picture very much. The info in the first sentence is ONLY used to find out one useful fact: that the group of workers earns 60 x $80 = $4,800 in TOTAL each day. We use that to build our second equation, which has to do with the total earnings:

75A + 100B = 4,800

Now we have our two linear equations, and all that remains is to solve the system using any standard technique.

A + B = 60
75A + 100B = 4,800


Multiply the top equation by -100, and then add the two equations together:

-100A - 100B = -6000
75A + 100B = 4,800
-----------------------------
-25A = -1,200
A = 48

Then, B must be 12.

Check: 48 x $75 + 12 x $100 = $4,800. It checks.

Algebra problem: two speeds

Photo courtesy of Ben Cooper

The speed of a freight train is 14km/h slower than the speed of a passenger train. The freight train travels 330 km, in the same time that it takes a passenger train to travel 360 km. Find the speed of each train.

Again, an algebra problem about speeds. Again, we will make a simple table about the two trains. The table will have columns for speed, distance, and time.

    
               | distance | speed  | time
-----------------------------------------------
Freight train  |  330     |   v    |
-------------------------------------------------
Passenger train|  360     | v + 14 |

Notice the problem says "in the same time". Let's call that t.

    
               | distance | speed  | time
-----------------------------------------------
Freight train  |  330     |   v    |   t
-------------------------------------------------
Passenger train|  360     | v + 14 |  t

Of course, the goal is to have an equation in a single variable, not in two variables (t and v).

Since the time is the same, we can build an equation of the time t = ... for both trains, and then set those expressions to be equal.

For the freight train, t = distance/speed = 330/v
For the passenger train, t = 360/(v + 14)

Now let's make those equal:

330/v = 360/(v + 14)

Cross multiply:

360v = 330(v + 14)

360v = 330v + 4620

30v = 4620

v = 154

So, the speed of the freight train is 154 km/h and the speed of the passenger train is then 168 km/h.

Check: How long will it take for the freight train to travel 330 km? Well, 330/154 hours, or 2.142857143 hours.

How long will it take for the passenger train to travel 360 km? Well, 360/168 hours, or 2.142857143 hours. Seems to match.

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.
   
   

Algebra problem: airplane's speed in still air

Photo by Caribb

Someone sent me this algebra problem:
An airplane flew for 6 hours with a tail wind of 60km/hr. The return flight against the same wind took 8 hours. Find the speed of the boat in still water.

Initially, this sounds like a trick problem, because you can't know the speed of the boat when all the information given is about an airplane!

But, let's assume they meant to ask the speed of the AIRPLANE in still air.

This problem has to do with constant speed. Constant speed ALWAYS involves TIME & DISTANCE.

And here we have two situations with two different speeds: First the airplane flies over there, with the wind helping. Then it flies back, and the wind is contrary of course.

It helps to organize our information in a table, once again. We'll need time, distance, and speed. And remember, speed = distance/time, or distance = speed * time. This time, time is known for both situations. Distance is not, but it is the same distance that way and back (call it d). Traveling there, the speed will be the sum of the airplane's speed (v) plus the wind speed (60). Traveling back, the wind is contrary so its speed is subtracted from the airplane's speed.

 |  Time  | Distance  | Speed
------------------------------------------
There |   6    |    d      | v + 60           
-----------------------------------------
Back |   8    |    d      | v − 60           

I'll copy the same table here, but instead of using d for distance, I'll calculate it using distance = speed × time.

 |  Time  | Distance  | Speed
------------------------------------------
There |   6    |  6(v+60)  | v + 60           
-----------------------------------------
Back |   8    |  8(v-60)  | v − 60      

Now, one just needs to find a way to make an equation... it will be a bit different in different word problems, but this time we get it from the fact that the distance over there is the same as the distance back:

6(v + 60) = 8(v − 60)

6v + 360 = 8v − 480

840 = 2v

v = 420

The speed of the airplane in still air is 420 km/h.

Check: For checking, we will calculate the distance it traveled. Traveling 6 hours with the speed (420 + 60) km/h or 480 km/h means the distance was 2,880 km. Traveling back took 8 hours, so the speed must have been 2,880 / 8 = 360 km/h, which is (420 − 60) km/h. So, the story "matches".

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.

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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.

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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.

Word problem Singapore way

Laura had 24 clips more than Holly. After she gave 5 clips to Holly, Laura had twice as many clips as Holly. How many clips did Laura have left?

This is from Singapore Challenging Word Problems book 3 (which they are now discontinuing, I heard).

I found two ways to solve this using the bar diagrams.

Solution 1. Notice that this is showing how I solved it initially, and at one point I had to adjust the length of the bar.

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Solution 2.



In either case, once you get that x = 9 (the amount of clips Holly had in the beginning), it's easy to solve that Laura had 28 clips left after giving 5 to Holly.

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Algebraically:

Initially Laura has L, Holly has L - 24. (Obviously you could also choose to use H and let Laura have H + 24.)

Then Laura gives 5 to Holly, so now Laura has L - 5 and Holly has L - 24 + 5 which is L - 19.

At this point Laura has double as many clips as Holly: L - 5 = 2(L - 19) and we can solve.

L - 5 = 2L - 38
L = 33.

Laura had 33 - 5 = 28 after she gave 5 to Holly.

There are probably other ways to solve this as well.

Introducing Make It Real Learning workbooks

I have recently had the pleasure to add Make It Real Learning workbooks to my site. These books contain real-life math activities with real-life data, companies, and situations. They are written by Frank Wilson.

Arithmetic I



Fractions, Percents, and Decimals I



Linear Functions I


Calculus I



Periodic and Piecewise Functions I

Some examples of the topics included in these activities are: cell phone plans, autism, population growth, cooking, borrowing money, credit cards, life spans, population growth, and music downloads. But there are many more, more than I can list here.

As students work through the problems, they can use the math skills and concepts they have learned in their math curriculum (such as the concept of average or graphing), and apply those to a situation from real life.

Each activity-lesson in the book contains several questions about the situation, starting with basics and going into more in-depth evaluations, and should be adequate for one-two complete class periods.

Why does that benefit you, the teacher? It will motivate the students far more than dull, boring word problems from the textbook. It will show students how mathematics is truly USEFUL. Also:

• The problems are written by an experienced math teacher (Frank Wilson)


• The problems are matched to the learning objectives of the National Council of Teachers of Mathematics (NCTM). This means that the concepts and skills required to complete the problems ARE found in typical middle and high school mathematics curriculum. You can simply replace some of the problems in your textbook with these real-life scenarios.
  
  
• These activities are excellent to be used in a small-group setting.
  
  
• Typically, the activities contain challenging parts and therefore allow students to practice real problem solving - not just apply knowledge from textbook examples to other almost identical problems.
  
  
• Gifted students can enjoy the challenge of solving all the questions on their own.

All books include complete solutions to all activities and problems. Please note that Make It Real Learning workbooks do not contain the instruction or explanations of the concepts.

List of available books:

Arithmetic I - for grades 3-6
Fractions, Percents, and Decimals I - for grades 4-8
Fractions, Percents, and Decimals II - for grades 6-11
Sets, Probability, and Statistics I - for grades 6-10

Linear Functions I - for algebra 1 and algebra 2
Linear Functions II - for algebra 1 and algebra 2
Quadratic Functions I - for algebra 1 and algebra 2

Exponential and Logarithmic Functions I
- for algebra 2/precalculus
Periodic and Piecewise Functions I
- for algebra 2/precalculus
Polynomial, Power, Logistic, and Rational Functions I
- for algebra 2/precalculus
Calculus I - for grade 12

Carnival picks

I have a collection of interesting links for you today; mostly some picks from Math Teachers At Play #2 blog carnival, which you SHOULD visit. Denise always has good stuff and really nice photos to spice up the reading.

Here's a neat way to practice multiplication tables by Math Mojo.

I also liked some simple dissection puzzles; you'll even get a free worksheet download.

Nick has posted two geometry "Wrapper" problems

This one is hilarious - lots of pictures of mathematical clocks!

Those were all from the carnival; please check out all the other posts as well; it is a good carnival!

Lastly, I visited this yesterday, and I think ALL of us should check out what a trillion dollars looks like.