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