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.



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.



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.

Comments

Anonymous said…
This can be solved by drawing model to give the students a better visualisation and understanding of the problem.

Notice the question involve two conditions using the "If" clause. When questions that involved two "If" clauses, you have to use two cases or senerios that involve the original amount to make a comparison.

My working is as such:

Solution:

Case 1:

Box A : Box B
50 : 25
2 : 1 (to the lowest term)

From here, we can come to a conclusion below:

From Case 1, Box A : 2 units + 60 left = Total number of marbles in Box A

Case 2:

Box A : Box B
25 : 50
1 : 2 (to the lowest term)

From here, we can come to another conclusion below:


From Case 2, Box A : 1 unit + 1800left = Total number of marbles in Box A

Thinking process : Notice that the total number of marbles for box A should be the same for both cases. Likewise for Box B.

3 units -----> 1800 - 600
= 1200

1 unit -----> 1200/3
= 400

From either Case 1 or 2,

From Case 2: Box A = 1800 + 400
= 2200

From case 2: Box B = 400 * 2
= 800

Therefore, There are 2200 and 800 marbles in Box A and Box B respectively at first.

Hope that my method is correct.

Any other methods other than the above method! If ther is, please share in this thread for the benefit of the students as well as the parents.
Anonymous said…
I am confused by anonymous's solution, although I got the same answer. I think what is confusing me is the "units." If we set the equations equal, that would give:

A in Case 1 = A in Case 2
2 units + 600 = 1 unit + 1800
1 unit = 1200
Marbles in A = 3000

In reality, the "units" in Case 1 are twice as big as the "units" in Case 2. Think about it this way: ALL of the marbles in Box B are taken away each time. In Case 1, all the marbles in B make 1 unit. In Case 2, the same number of marbles make 2 units. Therefore, the units in Case 1 are double the size of the units in Case 2.

I suppose this must be what anonymous meant by that "Thinking process" in the middle of the solution, because he (or she) obviously corrected for this problem.

Let's put both situations into "Case 2 units," which would make the equation look like this:

A in Case 1 = A in Case 2
4 units + 600 = 1 unit + 1800
3 units = 1200
1 unit = 400
Marbles in A = (4 x 400) + 600
= 2200
Anonymous said…
At any rate, I worked the problem differently. I decided to see if I could solve it in my head...

First thought: Set aside the extra 600 marbles from A. Remember them for later, or the final numbers won't come out right --- but for now, they are just in the way.

Case 1: How are the marbles left in A related to the marbles in B? Notice, we are taking exactly twice as much from A as from B each time, and they both end up empty. So A has twice as many marbles as B.

Case 2: How many marbles are left now? Remember, we set aside 600 marbles from A, so don't count them! There would be 1200 left.

Case 2, continued: This time, we only took away half as many from A as from B. Since A had twice as many as B, that is the same as having 4 halves of B, and we took away 1 half of B. Therefore, there are 3 units of marbles left in A, each of which is half the number of marbles that were in B.

That seems terribly convoluted when I write it, but it was easy to do in my head. A had 1200 left (remember, we set 600 aside, and we haven't put them back yet!), which I divide by 3. That tells me that half of the number of marbles in Box B is 400.

And from there it's pretty easy, as long as I remember to add the extra 600 back to A before I give my final answer.
Anonymous said…
Hi Letsplaymaths!,
Basically your method is the same as mine just that it is impossible for me to draw model in this thread. Perhaps, i can simplify it by using representation.

I shall use the Singapore model drawing approach.


To note: "O" represent 1 unit of a rectangle.

Case 1:

Box A : O O + 600 left
Box B : O + 0 left

Notice that Box A has twice of the marbles taken out.

Case 2:

Box A : O + 1800 left
Box B : O O + 0 left

Now, take a look at the two cases and made a distinctive comparison.

Box B in Case 1 and Box B in Case 2 should have the same amount of marbles in the beginning. Same goes to Box A.

Compare Box B in Case 2 to Box B in Case 1, can you tell the difference?

The number of marbles left in Box B in both cases is zero. Box B in Case 2 is 2 units while Box A in Case 1 is 1 unit.

The number of marbles that were taken out are also the same. If Box B is 2 units in Case 2,
Box A will be 2*2 = 4 units + 600
left. (part-whole unit)

Since the number of marbles in both cases for box A are the same,

we can derive and conclude as such:

Box A: 4 units + 600 (Case 1) = 1 unit + 1800 (Case 2)

3 units -----> 1200
1 unit -----> 400

Case 2:
Box A : 400 + 1800 = 2200
Box B : 400 * 2 = 800

Case 1:
Box A : 400 * 4 + 600 = 2200
Box B : 400 * 2 = 800
Maria Miller said…
Just a quick note to anonymous...
You wrote

Box A: 4 units + 600 (Case 1) = 1 unit + 1800 (Case 2)

That is definitely using algebra. You wrote an equation with an unknown called "unit". It's the same as writing

4x + 600 = x + 1800

By the way, I feel algebra is an excellent tool for solving this problem. But if the teachers tell students not to use algebra, then would they accept your solution?

On a similar note, your earlier representations used O for one unit like this:

To note: "O" represent 1 unit of a rectangle.

Case 1:

Box A : O O + 600 left
Box B : O + 0 left

Notice that Box A has twice of the marbles taken out.

Case 2:

Box A : O + 1800 left
Box B : O O + 0 left

I would be careful of using this, because you use the same symbol in case 2 and in case 1, yet the O in case 2 does not stand for the same number as in case 1. See, we know A is 2200 and B is 800. So in case 1, your 0 is 800, and in case 2 it ends up being 400.
Anonymous said…
The main difference in the way we worked was that I set aside the extra marbles, to make a problem I could hold in my head.

I find square brackets useful for making units when diagrams won't show. It's easy to fill the brackets with dashes to make different sized boxes:

Case 1
A = [---][---][600]
B = [---]

Case 2
A = [-][---1800---]
B = [-][-]

And then it's not hard to see how the units all fit together.
Anonymous said…
To Note: [ ] represent 1 unit of big rectangle while [ ] represent 1 unit of small rectangle.

Case 1

Box A [ ][ ][ 600 ]

Box B [ ]

Case 2

Box A [ ][ 1800 ]

Box B [ ][ ]


Rearrange the big rectangle to become a small rectangle:

Case 1

Box A [ ][ ][ ][ ][ 600 ]

Box B [ ][ ]


Case 2

Box A [ ][ 1800 ]

Box B [ ][ ]


From the above, it should be clearer now.
Anonymous said…
Oops! You can't use spaces to make your bars big or small, because most webpage programs automatically erase or ignore extra spaces. That's why I use dashes.
MathsGeek said…
Letsplaymath!,

Base on your thread and your concise explanation, i have come to a conclusion that you are a genius and you like Maths a lot. Perhaps, solving Maths problems are part of your forte. :)

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