Another algebra problem - or is algebra needed?

Updated!

There are some marbles in Box A and Box B. If 50 marbles from Box A and 25 from Box B are removed each time, there will be 600 marbles left in Box A when all marbles are removed from Box B. If 25 marbles from Box A and 50 marbles from Box B are removed each time, there will be 1800 marbles left in Box A when all marbles are removed from Box B. How many marbles are there in each box?

Again, this is from Singapore and teachers have told students not to use algebra to solve this question. However, any form of heuristic tools are allowed to facilitate the students in solving the questions.

I'd like to point out that I feel it's a good problem, but students might benefit from some "preparation". You could set up a preparation problem like this:

Jar A has 100 marbles and jar B has 40 marbles. You will start removing marbles one by one from jar A, but by 2's from jar B. How many marbles are left in jar A when jar B is empty?
What if you remove 2 marbles at a time from jar A and one marble at a time from B, then how many marbles are left in A when B is empty?

Then we can vary the numbers, for example let jar A have 250 and B have 90. Or, let A have 250 and B have 400 and see what happens! Lastly, change the number of marbles removed to 25 and 50 as in the real problem.

Now, the information given is "reversed" in the original problem because there we know how many marbles will be left and don't know how many marbles were there in the beginning. This does make the problem more difficult, obviously.


Solution:

When I saw this problem, my head automatically "saw" another usable unknown as "how many times do we scoop out marbles from each box?" You see, in scenario 1 we scoop out 50 marbles at a time from A and 25 from B, but the scooping is done the same number of times. So I called that n.

That automatically headed me down the "algebra" route... I wanted to write stuff using n:

There are 50n + 600 marbles in A, and 25n in B.

Then in situation 2, we don't do the same amount of scoopings... so the n is not the same. This time, we take 50 marbles at a time from B until it's empty, so B got emptied in double time as compared to situation 1. So... we have n/2 or half as many scoopings taking place.

Thus there are (n/2)*25 + 1800 in A, and (n/2)*50 in B.

Now you can get an equation that solves it by setting the number of marbles in A equal to number of marbles in A from the two situations:

50n + 600 = (n/2)*25 + 1800

Some solving... n = 32.

Therefore there are 50*32 + 600 = 2200 marbles in A, and 25 * 32 = 800 marbles in B.

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Now, as far as solving it heuristically without using algebra... it was really hard for me at this point, since my head only wanted to consider the problem this one way. I started visualizing in my head two jars and two hands picking the marbles out... First one hand picks 50 each time out of the jar with more marbles and 25 out of the jar with less marbles until jar B runs out. Then the other way around: picking 25 out of A while taking 50 out of B, until B runs out.

Then I "saw" that the amount of marbles actually taken out from A in situation 1 was FOUR times the amount of marbles taken out from A in situation 2. (Just comparing how many marbles got taken out from A.)

This is because in 1, we took two times as many marbles out each time (50 vs 25), and also because in 1 it takes double that long (double the amount of scoopings) than in situation 2 (because we're timing all this by how quickly B runs out, and B runs out in half a time in situation 2).

We also know that the first time we took out 1200 more from jar A than in situation 2. So, that 1200 is 3/4 of the marbles taken out in sit 1. From which it's easy to get that 1200/3 * 4 = 1600 is the number of marbles taken out from A, in situation 1.

And that solves it then because now we know that there were 1600 + 600 = 2200 in A. And, since 1600 marbles were taken out from A by 50s, it means it was done 32 times. So B had 32 * 25 = 800 marbles.

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I definitely think algebra is the easier way to go ... less brain strain for sure!

You can find yet other solution methods in the comments.

An algebra problem

This question was set in one of the renowned primary school from Singapore. Given to me by "anonymous" to solve.

Andy has $200 more than Peter. Andy gives 60% of his money to Peter. Peter then gives 25% of his money to Andy. In the end, Peter has $200 more than Andy. How much did Andy have at first?

This is a great problem to solve with algebra. Why don't you try it first, before reading further? It sounds kind of interesting... first one guy has $200 more than the other, and in the end it's reversed.


Solution:

Let A be the initial amount Andy has, and P the initial amount Peter has.
Then we know that A = P + 200. We're going to use that later, but for now I'm going to write it all in terms of A and P.

Andy gives 60% of his money or 0.6A to Peter.

Peter has now P + 0.6A.
Andy has now 0.4A.

Peter then gives 25% of his money to Andy. But this isn't 0.25P because Peter doesn't have P dollars anymore because Andy already gave him some. It's 0.25 (P + 0.6A) that he gives back to Andy.

Peter has now 0.75(P + 0.6A)
Andy has 0.4A + 0.25 (P + 0.6A)

In the end, Peter has $200 more than Andy. That gives us a way to write an equation:

0.75(P + 0.6A) = 200 + 0.4A + 0.25 (P + 0.6A)

It's now simple to solve this equation of two variables by first substituting A = P + 200. The rest of it is just basic manipulations.


0.75(P + 0.6(P + 200)) = 200 + 0.4(P + 200) + 0.25(P + 0.6(P + 200))

0.75(P + 0.6P + 120) = 200 + 0.4P + 80 + 0.25(P + 0.6P + 120)

0.75(1.6P + 120) = 280 + 0.4P + 0.25(1.6P + 120)

0.75(1.6P + 120) = 280 + 0.4P + 0.25(1.6P + 120)

1.2P + 90 = 280 + 0.4P + 0.4P + 30

0.4P = 220

P = 550

and so A = 750

Check:
First Peter had $550, Andy $750.
Then Andy gives $450 to Peter.

Peter has now $1000, Andy $300.
Then Peter gives $250 to Andy.

Peter has now $750, Andy $550.

Mean & mode freebie download

This free lesson about mean and mode will get you a foretaste for my upcoming 5-A Complete Curriculum from the LightBlue Series.

Download it here:
Mean, Mode, and Bar Graphs - lesson for 5th grade.

In the lesson I highlight the idea of mode versus mean (average) and when you can calculate the mean. Students also graph the data in bar graphs.

I didn't include the median because elementary lessons on mean, median, and mode tend to concentrate on the calculation aspect only, and I didn't want that. In this lesson they at least get to graph the data and think if mean (average) is "calculable". So I decided to postpone the median till 6th grade...

But here are some other lessons on these topics. Even with these you can see how much the actual calculations dominate the lessons.

Using and Handling Data
Simple explanations for finding mean, median, or mode.
www.mathsisfun.com/probability


Mode of a Set of Data
A very simple and clear lesson with examples and interactive quiz questions.
www.mathgoodies.com/lessons/vol8/mode.html


Finding the Mean, Median, and Mode
A great lesson with interactive quiz questions in the end. It also explains briefly the different uses for mean, median, and mode. After all, why do we have three different numbers for the central tendency of the data set?
www.algebralab.org/lessons/lesson.aspx?file=Algebra_StatMeanMedianMode.xml


Mean, Median, and Mode
Lesson on how to calculate mean, median, and mode for set of data given in different ways. Also has interactive exercises.
www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i5/bk8_5i2.htm


GCSE Bitesize Mean, mode and median lessons
Explanations with simple examples.
www.bbc.co.uk/schools/gcsebitesize/maths/data/measuresofaveragerev1.shtml


Measures Activity
Enter you own data and the program will calculate mean, median, mode, range and some other statistical measures.
www.shodor.org/interactivate/activities/Measures/


Landmark Shark Game
You're dealt five number cards, and using that as your data set you need to choose which of the range, median, or mode is the largest number.
media.emgames.com/emgames/demosite/playdemo.html?activity=M5A006&activitytype=dcr&level=3


Train Race Game
Calculate the median and range of travel times for four different trains, then choose a good train to take based on your results.
www.bbc.co.uk/education/mathsfile/shockwave/games/train.html

A quite long prime

Mathematicians at UCLA have verified now the largest known prime number, 13 million digits long!

It's a Mersenne prime, in the form 2^p − 1. This new prime is 2^43,112,609 − 1. (And no, you can't put it in your calculator and get an answer... )

Read the news story

This kind of prime hunting requires a network of computers running together to do the massive amounts of calculations.

Credit card math

Considering the crisis taking place within U.S. financial institutions, it's a good reminder of how we MUST teach our kids how to work out the math with credit cards, mortgages, and any type of loans.

Murray Bourne from squareCircleZ has taken this matter to heart and has written a good lesson about math that goes into taking a loan. He's also analyzed several misleading credit card ads and figured out the TRUE interest rate (which you find out after reading the fine print).

First go here:

Credit Cards - a simplified discussion on how banks make money on credit cards. This is a must read for all people (young or old) who even consider taking a credit.

Then check these posts and warn your youngsters:

Misleading Credit Card Advertising. This ad advertises "a attractive interest rate of 5% p.a." (p.a. means per annum or yearly interest rate), but has a "small" administration fee of 6%.

And then check another misleading credit card ad which advertises a 0% p.a.*. Notice it has an asterisk... and the info in the asterisk can lead up to 24% annual interest!

Two new books in the Blue Series

64 pages
(of which 37 are lesson pages)
Price: $2.50 download

Buy at Kagi (PDF download)

Sample pages (PDF)

Contents and Introduction
Shapes
Right Angles
Line Symmetry
Perimeter
Solids

1. Math Mammoth Early Geometry covers geometry topics for the early elementary grades (approximately grades 1-3).

The first lessons in this book have to do with shapes - that is where geometry starts. Children learn the names of the common shapes, and also put several shapes together to form new ones, or divide an existing shape into new ones. They practice using a ruler to draw various shapes and are introduced to tilings.

Next children learn the concepts of parallel lines and lines that are at a right angle (perpendicular lines). The book also has beginner lessons about symmetry, area, perimeter, and solids.

After studying these early geometry lessons, you can continue the study of geometry with Math Mammoth Geometry 1 book. In it, children will learn to classify figures (during grades 4-6) according to their sides and angles, and learn much more about area, perimeter, and volume.

52 pages
(of which 40 are lesson pages)
Price: $3.00 PDF download



Sample pages (PDF)
Contents
Counting One, Two,and Five-Cent Coins
Counting Coins Review
Euros, part 2
Practising Shopping
Mental Math and Money Problems

2. Math Mammoth European Money is a worktext that covers money-related topics usually encountered during grades 1-3. The book contains both textbook explanations and exercises, and is designed to be very easy to teach from, requiring very little teacher preparation (you do need to find practice coins before the
lessons).

The book starts with first-grade topics such as counting coins with cent-amounts and easy problems about
change.

From there, the lessons advance toward second-grade, and finally to third grade topics, such as practicing with euro amounts, and figuring out total bills and change. Therefore, you can also let your child work the pages of this book in different time periods, and not go through it all at once, depending on your child's
current level.

A simple ratio problem

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

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

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

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

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

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

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

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

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

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

Useful links

These are just some links sent my way over the last few months that I thought you might enjoy.

• Teach Banzai is an online system for financial education. It sounds really interesting! Students can set up various "jars" (budgets), enter transactions (real or fictional), and continue using the system for life if they so wish. It's all free and there are lesson plans offered.


• SpaceTime TV
  Watch educational videos online from full length documentaries to short video clips featuring shows from PBS, National Geographic, and the History Channel. They've done a good job of collecting the best free science and history videos. I can just imagine how difficult and time-consuming it'd be to find those if you went searching on YouTube an other video sites yourself.
  


• Dimensions
  Interesting movie - which is available totally free online - about how mathematicians can visualize the fourth dimension. Some of that stuff will give you "mathematical vertigo", but some of the chapters are easy enough for middle schoolers on up, teaching about stereographic projection, the difference between 2nd and 3rd dimensions, complex numbers, and more.
  


• Math Problem Generator
  On this site you need to browse through a collection of problems and add them one by one to your worksheet so it is slow - BUT the problem collection includes prealgebra, algebra, high school geometry, proportions, statistics, probability, algebra 2, and SAT (inc. word problems in all of the categories), many of which you won't find on many of the other worksheet sites.

Carnival of Homeschooling

Carnival of Homeschooling number 141 is now online at Why Homeschool. Go visit and read it! My submission was the review I recently did of Danica McKellar's books.

Here's one pick from the carnival I enjoyed: Hands-on ABC Order Activities which has some neat ideas for helping kids learn their ABC order.

A word problem

Another word problem was sent my way.

Todd, Chris, and Rod have 30 birds. Rod has five times as many as Todd. Todd has one fourth the number Chris has. How many birds do they each have?

The numbers in this problem are so small that I won't make any equations -- it might be QUICKER to use guess and check!

Just notice Todd has the least amount, and once you know (or guess) how many birds Todd has, then you immediately know how many Chris and Rod have. Rod will have 5 times the amount Todd has, and Chris will have four times the amount Todd has.

We will start by giving Todd 1 bird, and checking if the condition is met that the total becomes 30. Then we will give Todd 2 birds and check again.

Rod    5    10    15
Chris  4     8    12
Todd   1     2     3
---------------------
Total  10   20    30

So Todd has 3 birds, Chris has 12, and Rod has 15.

I'm guessing this might be a problem from algebra course, and that the author/teacher intends algebra to be used. But I would say that use the most effective and quick method which isn't always algebra...

With algebra, you'd assign Todd to have x birds, then Chris has 4x birds and Rod has 5x birds. Add those up and let the sum be 30:

x + 4x + 5x = 30

10x = 30

x = 3.

So it's pretty quick this way too. I just chose the guess and check because the total was only 30 and so I knew it wouldn't take long.

Abacus and basic division concept

Let's be reminded first of all that there are two basic "interpretations" of division:

1) Sharing division. In it, you think that 16 / 2 means "If there are 16 beads divided between 2 people, how many does each one get?

2) Quotative division. In it, you think how many groups of the same size you can form. Or, "how many times does the divisor fit into the dividend?" For example, 16 / 2 is interpreted as: "If you have 16 beads, how many groups of 2 beads can you make?"

Sharing division is easy to understand as a concept, but it's hard to do without the knowledge of multiplication tables. Just imagine, if a child who doesn't know their times tables is trying to find the answer to the question, "If there are 56 strawberries and 8 people, how many will each one get?" It'll take some guesswork, trying and checking.

But quotative division is easy for children to do with the help of manipulatives (or by drawing pictures). Just make a group of 56, and start forming groups of 8 out of it. Once you're done, check how many groups you got. There's no guessing. It's just repeated subtraction - each group the child forms is "subtracted" or set apart.

For this reason I have decided to start off with quotative division in my Division 1 book. Children can first just partition off groups of certain size from the picture problems and count how many groups they got.

And this activitity lends itself perfectly to the 100-bead abacus, as well.

For example, we have 20 / 4.

Take 20 beads:

| 0 0 0 0 0 0 0 0 0 0    |
| 0 0 0 0 0 0 0 0 0 0    |

Group them in groups of 4.

| 0000  0000   00    |
| 0000  0000   00    |

Note the one group is formed from 2 beads on one bar and 2 on the other. Answer: 5 groups.

Another example: 27 / 9

First take 27 beads.

| 0 0 0 0 0 0 0 0 0 0                |
| 0 0 0 0 0 0 0 0 0 0                |
| 0 0 0 0 0 0 0               0 0 0  |

Form one group of 9 from the first bar (green), another from the second bar (blue). The third group is the purple ones.

| 0     0 0 0 0 0 0 0 0 0            |
| 0     0 0 0 0 0 0 0 0 0            |
| 0 0 0 0 0 0 0               0 0 0  |

Groups of fives are particularly interesting since they look so "orderly" on the abacus.

Ask your student, like I did today, what's 90 / 5?

Well, you organize your 90 beads like this, and answer is easily counted as 18 groups.

| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |
| 0 0 0 0 0        0 0 0 0 0  |

You can, of course, use abacus for sharing division, too. But make easy problems where the sharing is easy to do, such as 20 / 2 or 24 / 2 or 15 / 3.

Once the concepts of quotative division and sharing division are clear, you need to spend a lot of time with multiplication/division connection, because that is what children will use to solve most division problems - they'll need to get away from using manipulatives after the initial stages.

Timed drills for math facts?

Are timed math tests (drills) necessary and for what purpose? If necessary, how often should facts be tested? Are timed math tests recommended for students who have "math issues." , such as low confidence and slower progress.

I simply feel that timed drills are a tool among many, when it comes to learning math facts. Some kids will "thrive" on them, or in other words quickly learn when they are used. Perhaps they like racing against the clock or like the challenge. There exist computer games that are timed that can work very well for drilling facts. Check for example:

Math Magician games has a simple 2-minute countdown, and if you answer 20 questions in that time, you get an award.

Some of the games at Sheppard Software don't time you but give you more points the faster you go. That site is actually filled with several types of games just for math facts practice.

Yet for other kids timed drills may be counterproductive and end up in tears and frustration. The proof is in the pudding though. Just try them and see how it goes.

I feel timed drills are a good tool but not absolutely necessary. Kids can learn their facts so many ways. For example, some learn to use clever shortcuts such as "Since 7 + 7 = 14, then 7 + 8 is one more or 15" and actually perform that in their heads so quickly that you can't tell the difference if they memorized it or did some calculation. Some learn them by heart from much usage in problems, without being timed.

You're welcome to use timed drills with low-confidence students because you never know, perhaps a student like that has decent memory and can succeed with it. However, if you already know that a student cannot easily remember facts without context, then it might not work.