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.

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.

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.

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.

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.

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 1 curriculum advice

I have just finished writing a LONG article on homeschool algebra 1 recommendations and advice. It took me quite many hours to write and research.

Questions about what to do for algebra 1 have become one of the "frequently asked questions", so I decided to write down something that I can refer people to, from now on.

I realize you may have different opinions and even suggestions for algebra curriculum in homeschool, so if you let me know, I'm willing to look into other possibilities not mentioned in the article.

Algebra 1 curriculum advice

I have just finished writing a LONG article on homeschool algebra 1 recommendations and advice. It took me quite many hours to write and research.

Questions about what to do for algebra 1 have become one of the "frequently asked questions", so I decided to write down something that I can refer people to, from now on.

I realize you may have different opinions and even suggestions for algebra curriculum in homeschool, so if you let me know, I'm willing to look into other possibilities not mentioned in the article.

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.

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.

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".

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".

High school math videos - free

Today I wanted to highlight two companies that both offer lots of free high school level math materials in the form of videos. These are real, commercial companies, yet they have chosen to offer the videos online for free. Both have something else they sell, hoping to make some money obviously from that part of the business.

The video content can really be of help for all students, teachers, or parents who need additional help with high school math (algebra, geometry, calculus). And since I'm highlighting these two, if you have a topic you have trouble with (such as polynomials or factoring or inequalities) you can even check out videos for that topic in both places.

1) BrightStorm Math.

They currently have over 2,000 videos available for free - a free registration is required though. Algebra through calculus. What they are selling is test preparation courses.

2) MathTV.com


MathTV.com has over 6,000 free math videos, including some in Spanish. Prealgebra through calculus. What they are selling is their own textbooks, physical and online. However, their online textbooks are available for free for evaluation and class testing through June 2010.

High school math videos - free

Today I wanted to highlight two companies that both offer lots of free high school level math materials in the form of videos. These are real, commercial companies, yet they have chosen to offer the videos online for free. Both have something else they sell, hoping to make some money obviously from that part of the business.

The video content can really be of help for all students, teachers, or parents who need additional help with high school math (algebra, geometry, calculus). And since I'm highlighting these two, if you have a topic you have trouble with (such as polynomials or factoring or inequalities) you can even check out videos for that topic in both places.

1) BrightStorm Math.

They currently have over 2,000 videos available for free - a free registration is required though. Algebra through calculus. What they are selling is test preparation courses.

2) MathTV.com


MathTV.com has over 6,000 free math videos, including some in Spanish. Prealgebra through calculus. What they are selling is their own textbooks, physical and online. However, their online textbooks are available for free for evaluation and class testing through June 2010.

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.

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.

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.

Review of Mangahigh games

I got an advance notice of a new games website called Mangahigh.com. It's more than just simple math games, though. These games are designed extremely well, both from mathematical and "enjoyment" perspective. Mangahigh is led by a team of mathematicians, educationalists, and games designers so the games feature commercial-quality gameplay.

I wanted to highlight two of the free games here:

Flower Power
This is the one that I liked most... it's addictive! Grow flowers and harvest them to make money. Practice ordering decimals, fractions, and percentages. The game starts with ordering decimals (daisies), and proceeds into fractions (tulips or roses) and percents. Each time you get a full stem, you need to decide whether to pick the flowers to sell (earn money) or to let them be pollinated and thus get more flowers to grow. Grades 3-8.


Save Our Dumb Planet
Defend Earth from deadly meteorites using missiles. A team of dumb scientists are on hand to suggest possible trajectories. Practice drawing lines, quadratic curves, and some harder curves using their equations. The game has many levels, and you can stay at the easier levels if you so wish. Don't listen to the dumb scientists' talk - they mislead you! Grades 8-11 (algebra).

If I understood right, the games that are there now is just the beginning, and there are more to come.

---

Lastly, Mangahigh also has a powerful game called Prodigi. This is a math learning engine (or a math practice environment) that has thousands of maths problems with worked solutions and hints that adapt to each student's ability and learning speed. You can try Prodigi for free; however, only on a certain "easy" and set level. To enjoy the full possibilities of Prodigi, you need to subscribe.

Review of Mangahigh games

I got an advance notice of a new games website called Mangahigh.com. It's more than just simple math games, though. These games are designed extremely well, both from mathematical and "enjoyment" perspective. Mangahigh is led by a team of mathematicians, educationalists, and games designers so the games feature commercial-quality gameplay.

I wanted to highlight two of the free games here:

Flower Power
This is the one that I liked most... it's addictive! Grow flowers and harvest them to make money. Practice ordering decimals, fractions, and percentages. The game starts with ordering decimals (daisies), and proceeds into fractions (tulips or roses) and percents. Each time you get a full stem, you need to decide whether to pick the flowers to sell (earn money) or to let them be pollinated and thus get more flowers to grow. Grades 3-8.


Save Our Dumb Planet
Defend Earth from deadly meteorites using missiles. A team of dumb scientists are on hand to suggest possible trajectories. Practice drawing lines, quadratic curves, and some harder curves using their equations. The game has many levels, and you can stay at the easier levels if you so wish. Don't listen to the dumb scientists' talk - they mislead you! Grades 8-11 (algebra).

If I understood right, the games that are there now is just the beginning, and there are more to come.

---

Lastly, Mangahigh also has a powerful game called Prodigi. This is a math learning engine (or a math practice environment) that has thousands of maths problems with worked solutions and hints that adapt to each student's ability and learning speed. You can try Prodigi for free; however, only on a certain "easy" and set level. To enjoy the full possibilities of Prodigi, you need to subscribe.