#### maninthemirror

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Need some help here. Would prefer if you could provide solutions which primary school students can understand. Thank you

- Thread starter maninthemirror
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Need some help here. Would prefer if you could provide solutions which primary school students can understand. Thank you

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Please follow the rule of posting at this forum - enunciated at:

Need some help here. Would prefer if you could provide solutions which primary school students can understand. Thank you

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/

Are you sure that this problem needs to be solved "without algebra" (for primary school students)?

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My son hasn't learnt algebra yet. Is there any other way other than algebra???

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Need some help here. Would prefer if you could provide solutions which primary school students can understand. Thank you

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Yes... One of the ways would be "trial and error".My son hasn't learnt algebra yet. Is there any other way other than algebra???

Please share some of your thoughts/work here (In your last post - https://www.freemathhelp.com/forum/threads/pythagarous-question.116564/#post-464970 - you did not show any effort on your part)

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Oop! That's going the wrong way! We want to go down from 77% to 75%, not up. So instead of doubling, lets try halving. Suppose there are 500 marbles, 400 blue, 100 yellow. Adding 63 blue, 50 yellow gives 463 blue, 150 yellow for a total of 613. Now the blue marbles are 463/613, approximately 75.5%.

Alright! That's in the right direction and pretty close to 75%. Let's continue, trying 450 marbles, 360 blue, 90 yellow. Adding 63 blue and 50 yellow marbles gives 423 blue and 140 yellow for a total of 563 marbles. Now the blue marbles are 423/563, approximately 75.1%. We just need to go down a little more!

Let's try 440 marbles, 352 blue, 88 yellow. Adding 63 blue and 50 yellow gives 415 blue and 138 yellow for a total of 553 marbles. Now the blue marbles are 415/553, approximately 75.09%. That's very close but not exactly 75%.

Try 430 marbles, 344 blue, 86 yellow. Adding 63 blue and 50 yellow gives 407 blue and 136 yellow for a total of 543 marbles. Now the blue marbles are 407/543, approximately 74.9%. That's too low so now we know the answer is between 430 and 440 marbles, initially, so between 543 and 553 after the new marbles are added.

Now you can try 435 to see if that is too high or too low or "just right". If it is "just right" you are done. If it is too high, try every number between 430 and 435 (there are only four such numbers). If it is too low try every number between 435 and 440.

That's how this can be done without using algebra. If we could use algebra we might let M represent the total initial number of marbles. Then there are 0.8M blue marbles and 0.2M yellow marbles. Adding 63 blue and 50 yellow marbles give 0.8M+ 63 blue and 0.2M+ 50 yellow marbles for a total of M+ 113 marbles. The ratio of blue marbles to all marbles is (0.8M+ 63)/(M+ 113)= 0.75.

Multiply both sides by M+ 113 to get 0.8M+ 63= 0.75(M+ 113)= 0.75M+ 84.75. Subtract 0.75M and 63 from both sides to get 0.05M= 21.75. Finally, divide both sides to get M= 435. That was a lot faster and is a good reason to learn algebra!

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I am aware of some interesting visual techniques taught at the primary level that might apply here; I think this is a relatively challenging example of what can be done that way. It may help if you can tell me the context of the question: what methods your son has been taught, what grade level is involved, and whether this is from his specific curriculum or some other source (such as a contest). With the right facts, we may be able to find an answer that is not "merely" trial and error (though the latter is actually a powerful tool when done right).

Need some help here. Would prefer if you could provide solutions which primary school students can understand. Thank you

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Start with B blue marbles and Y yellow marbles.

We know that B / (B + Y) = 0.80 = 80%

You can demonstrate the arithmetic.

80 Blue + 20 Yellow ==> 80 / (80 + 20) = 80 / 100 = 0.80 = 80% -- See how that works?

NOw, go buy those extra marbles.

(80 + 63)/(80 + 63 + 20 + 50) = 143 / 213 = 0.67136 ==> 67% Whoops. That didn't work. Since we were just guessing at the 80 Blue and 20 Yellow, just to provide a numerical example, one really couldn't expect that to be a solution

There are lots of solutions if this is all we know:

8 / (8 + 2) = 8 / 10 = 0.80 = 80%

16 / (16 + 4) = 16 / 20 = 0.80 = 80%

The trick is to find the right one of these that ALSO satisfies the other criterion.

Back to this:

B / (B + Y) = 0.80 = 80%

Buy the extra marbles

(B+63) / ((B+63) + (Y+50)) = 0.75 = 75%

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Here's a version of the visual method I had in mind; it's actually not bad:

Need some help here. Would prefer if you could provide solutions which primary school students can understand. Thank you

Here's the thinking behind it: If 4/5 of the marbles initially are blue, then the ratio of blue:yellow is 4:1. So we draw a strip with 4 blue chunks, and one with one yellow chunk. Then we add 63 to the blue and 50 to the yellow.

Now we are told that 3/4 of the marbles are blue, so the ratio is now 3:1. That means the blue strip is 3 times the length of the yellow.

But now we have two ways to represent the same quantity. If we take away three blue chunks from each, we find that one chunk plus 63 is the same as 150. So the one chunk (the original number of yellow) is 150 - 63 = 87. The original total number of marbles was 5 times 87, or 435. And the final number of marbles is 113 more than that: 548.

This method using "strip" or "tape" diagrams, or "bar" models (which I've seen associated with Singapore math, and also in some Common Core-related materials) is very closely related to algebra. The work I did here also involves some important thinking about ratios and other number concepts, so it can be very educational.

I am still hoping to hear back about what methods these students have been learning; that would fit our request to show work and context, and would help me know whether working out this method was worth the effort. But I do enjoy figuring out non-algebraic ways to solve algebra-type word problems.

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1) Neither had the people who invented it.My son hasn't learnt algebra yet. Is there any other way other than algebra???

2) Not so. There's more algebra in that brain than you think. Feel free to broaden your definition.

3) No need to perpetuate the myth that algebra is hard so I need something else that I don't recognize as algebra but it really is.

4) I need five apples and I have only three. How many more apples do I need to buy? Algebra. No need to fear it.

5) What's a better deal for a loan, all other things being equal, 25% interest or 6% interest? Algebra.