### 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.

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.