QUICK! Help me with this word problem!

[plain jane]

Abe, Barry, and Carlos have 256 marbles altogether. The ratio of Abe's marbles to Barry's marbles is 4 : 3. Baryy has 14 more marbles than Carlos. How many marbles does Abe have?

 

 

(I have the answer but not how to get it)

 

I don't know how to figure this out without know the ratio for all three. I've gone through all the example problems in the SM book and none of them have a problem where only one ratio is given. Aaaaaah. I know this probably isn't that hard and I'm going to kick myself but at the moment I am stumped.

 

If you could please show me how you solved this, I would be very grateful. Thanks!!

[jennynd]

4 part + 3 part + 3 parts - 14 is total ( Carlo is 3 parts -14)

 

(256+14)/10 is each part. Which is 27 . 27x4 is Abe

[Cosmos]

A = [------][------][------][------]

B = [------][------][------]

B = [----------------][14]

C = [----------------]

 

A + B + C + 14 = 10 u. (because C + 14 would be equal to 3 bar units)

 

So 256 + 14 = 10 u.

10 u. = 270

1 u. = 27

 

A = 4 x 27 = 108

B = 3 x 27 = 81

C = 81-14 = 67

 

ETA: jennynd beat me! I'll leave in case it helps.

[plain jane]

4 part + 3 part + 3 parts - 14 is total ( Carlo is 3 parts -14)

 

(256+14)/10 is each part. Which is 27 . 27x4 is Abe

 

 

 

Aaaaaaaaaa. Duh. Thank you. :blushing: Feeling dumb ATM. But thankful for the help!!

[kiwik]

A=1 1/3 B, C=B-14, A+B+C=256

 

The other methods are nicer though.

[mathwonk]

i think like kiwik: just translate the sentences into symbols: "ratio" is "divided by", "more than" is "Plus", "together" means "add them all up"

 

so A/B = 4/3. B = 14+C. A+B+C = 256. Then manipulate using rules for arithmetic.

 

i.e. A/B = 4/3 so multiplying by B gives A = 4B/3. But to make things doable we need to write replace all letters by just one of them. So since we have already A in terms of B, we write also C in terms of B.

 

I.e. B = 14 + C, so C = B-14. Now we can replace everything by B's in the big equation A+B+C = (4B/3) + B + (B - 14) = 256.

 

Whew! I'm tired, but let's push on.... now combining all that gives (4B/3) + (3B/3) + 3B/3 = 256 + 14 = 270.

 

Soooo,......let's see....hmmm, is that 10B/3 = 270? or 10B = 810? or B = 810/10 = 81?

 

Anyway we wanted A = 4(B/3) = 4(27) = 108. and just for completeness, C = B-14 = uhhh, ...81-14 = 67. (now I'm checking by looking back at others.)

 

I give this (my honest) solution just so you can compare your work to a professional mathematician's! Yours are MUCH better.

 

But this is a predictable method that will give the right answer, and that can be used without being at all clever (fortunately).

[garddwr]

I just skimmed the other answers after solving it my own way, and I think I did it slightly differently so thought I would post. I thought of it like this: Ration of A:B is 4:3, C+14=B; if I add 14 to the total I will then have a ration of 4:3:3 for a total of ten parts--which is really nice since 256+14=170, easily divisible into 10ths (17). Then I get A=17*4

[mathwonk]

beautiful!

[Cosmos]

I give this (my honest) solution just so you can compare your work to a professional mathematician's! Yours are MUCH better.

 

But this is a predictable method that will give the right answer, and that can be used without being at all clever (fortunately).

 

 

mathwonk, are you familiar with the Singapore Math program? It uses a method called "bar diagrams" that allow quite young students to solve problems like these before they have been exposed to the symbolic manipulation in algebra classes. I don't know what book plainjane took this problem from, but I would guess it's perhaps fifth grade level?

 

Of course, solving a system of linear equations the traditional way gives an algorithmic approach in a way that the bar diagrams cannot. But the visual representation is a tool that allows beginner students to solve surprisingly hard problems. I would guess that it also helps to develop modes of thinking that lay the foundation for algebra later.

 

Since most of us didn't grow up using this method, though, it's sometimes tricky to figure out how to solve it without using algebra. The kids are actually at an advantage here, since they don't know any different. Hence posts like plainjane's!

 

Anyway, I just didn't want you to think none of us know how to do algebra!

[mathwonk]

Actually I thought your solutions were much better than mine. I just wanted to give mine in case it provided a useful answer to the original question. Now that you mention it I am reminded that Euclid uses bar diagrams in a sense in his geometry book to do algebra.

 

E.g. he remarks that any two lengths can be written as a given length plus or minus a fixed amount. and he shows this by a diagram as I recall. Lets see if I can remember this. ah yes. you lay off one length followed by the other. then you mark off the shorter one B on the longer one A. Then what is left over (A- B) you cut in half. Then you add that half (A-B)/2 to the shorter one B.

 

Now that gives you C = B + (A-B)/2 for your given length. Then the fixed amount you add or subtract is D = (A-B)/2. I.e. if you add it to that C you get B + (A-B)/2 + (A-B)/2 = B +(A- B) = A. And if you subtract it you get B+(A-B)/2 - (A-B)/2 = B.

 

Thus A = C+D and B = C-D. That also makes it easy to multiply them since then AB = (C+D)(C-D) = C^2 - D^2. which Euclid makes use of in some discussions involving Pythagoras theorem. Ah yes, it lets him take square roots. Since a right triangle with hypotenuse C and side D, will have another side E such that C^2 = D^2 + E^2. Thus E^2 = C^2 - D^2 = AB.

 

So this lets him use Pythagoras to take the square root of any product AB. For example A or B could be 1, so then he can take a square root of any number purely geometrically. This is Prop.II.5. From the equivalent result that he can construct a square of the same area as any given rectangle, he deduces that he can find a square equal in size to any rectilinear figure at all, Prop. II.14.

 

It may not be quite the same, but I think this suggests the bar diagram method of doing algebra without symbols is older than algebra, and at least 2,000 years old. This is one reason I like to recommend kids learn from Euclid. it really contains all other later math methods and ideas, not only algebra in a geometric form, but also limits and ideas underlying real numbers and integral calculus.

 

Book II of Euclid is where the geometric algebra is, if someone is interested. But in teaching it to 10 year old kids, I found it efficient to also show how symbolic methods could make things easier and quicker. The geometric methods on the other hand make things visually memorable, and I love them.

 

I still remember the first time I saw a geometric demonstration of the equation (a+ B)^2 = a^2 + 2ab + b^2, decomposing a square of side a+b into two squares of sides a and b and two ab rectangles. I was 21 years old and in a Harvard lecture by the great educational psychologist Jerome Bruner in about 1965. I wondered why no one had ever shown me that before! Then in my 60's I saw it in Euclid, Book II. Prop. II.4

 

Here is a link for an illustrated Euclid: (in English and also in Greek for hard core classical linguists out there! Anybody?)

 

http://farside.ph.ut...id/Elements.pdf

 

 

 

Forgive me for simple minded explanations. I don't mean any disrespect. That's just how I tend to explain things. I know nowadays many home schoolers teach their children things I only learned in college or later, even in math. I learned a lot also from my kids at camp, as well as from your posts here.

 

maybe someone can teach me why b or B followed by parenthesis is a smiley face!

[mathwonk]

This may be well known also but I only learned it last summer while preparing to teach epsilon camp. Namely Euclid shows how to solve the quadratic equation X^2 + X = 1, and more generally X^2 + cX = c^2, in Prop. II.11, and then shows that solving this equation suffices to construct an angle of 72 degrees, in Prop. IV.10, i.e. to construct a regular pentagon.

 

Indeed since he has shown how to take arbitrary square roots geometrically, one can solve, essentially by the quadratic formula, any quadratic equation geometrically. I hope this may interest someone who may not have seen it before. In general geometric ways of solving algebraic equations interest me. I just didn't realize they were still taught, as I had not learned them in school.

[garddwr]

This may be well known also but I only learned it last summer while preparing to teach epsilon camp. Namely Euclid shows how to solve the quadratic equation X^2 + X = 1, and more generally X^2 + cX = c^2, in Prop. II.11, and then shows that solving this equation suffices to construct an angle of 72 degrees, in Prop. IV.10, i.e. to construct a regular pentagon.

 

Indeed since he has shown how to take arbitrary square roots geometrically, one can solve, essentially by the quadratic formula, amy quadratic equation geometrically. I hope this may interest someone who may not have seen it before. In general geometric ways of solving algebraic equations interest me. I just didn't realize they were still taught, as I had not learned them in school.

 

 

From what I've read about epsilon camp I wish I could attend myself. We need a fun, exciting math camp for homeschool moms.

[mathwonk]

Speaking of elementary algebra. The Quadratic formula is something that always puzzled me as a high school student. I found it somewhat messy to derive as they did in my book, by "completing the square". But the same derivation, expressed as follows by the great 18th century Italian algebraist LaGrange I believe, seemed clear to me years later. Is this explanation a standard one today? Does anyone have an algebra book that gives it this way? More important, does it seem this explanation may have any advantages over the usual one? I like it because it tells me where the "4" comes from in the formula, and also the (1/2), and also explains the b^2-4c. of course I have also learned in old age that leaving out the "a" makes it look easier and conceptually clearer.

Problem: how to solve a quadratic equation X^2 - bX + c = 0.

solution: We know that if the roots are r,s, then the LHS of the equation factors as X^2 -bX+c = (X-r)(X-s) = X^2 - (r+s)X + rs.

Therefore b = r+s and c = rs. Now we want to recover r and s from b and c. Since we already know that r+s = b, it would suffice to find r-s in terms of b,c, since then b ± (r-s) = (r+s) ± (r-s) = 2r, 2s.

To do this we use the familiar relation between the squares of r+s and r-s.

I.e. (r+s)^2 = r^2 + 2rs + s^2, and (r-s)^2 = r^2 -2rs +s^2.

Hence (r-s)^2 = (r+s)^2 - 4rs = b^2 - 4c. Thus r-s = sqrt(b^2-4c).

Hence 2r, 2s = b ± sqrt(b^2-4c), and thus r,s = (1/2)(b ± sqrt(b^2-4c)).

 

If it had been taught to me this way, I probably wouldn't have had to write it in the cover of my book.

[mathwonk]

As to epsilon camp, my notes from the actual geometry classes are reproduced free on my web page, and anyone is welcome to copy them and use them as a guide for reading Euclid, e.g. in the beautiful green lion edition.

 

http://www.math.uga....camp2011/10.pdf

 

I wrote them in a conversational tone as much as possible, for the audience of brilliant 10 year olds.

 

The main obstacle of the camp is the expense, about $1,000/week per camper, which goes almost entirely for housing and food, plus travel. This money however is well spent, since actually meeting the other kids and parents in person is hugely valuable. Personal instruction is helpful, plus the participants get a valuable reassurance that there are others out there "like me".

 

But the purely mathematical part can be to some extent reproduced mainly by dedication and persistence, just organizing a group to read the book and notes together, and sticking to it.

 

There is some stuff in my notes, especially at the end, that one ordinarily only learns in upper level undergrad or grad courses, like why it is impossible to trisect some angles geometrically, and also how to find the volume of a ball in 4 dimensional space, using only Archimedes ideas and methods.

 

If you want an enrichment book for geometry written by one of the greatest mathematicians of all time, try "Geometry and the imagination" by David Hilbert, and Stefan Cohn-Vossen.

 

This is unfortunately apparently not available in a cheap paperback, but here are some used copies:

 

http://www.abebooks....the imagination

 

You should examine it in a library first to see if you like it.

[Dana]

maybe someone can teach me why b or B followed by parenthesis is a smiley face!

 

 

It's an emoticon for a smiley face with sunglasses. There's a check box to the right to enable emoticons? I wonder if turning it off will avoid that.... B)

 

I've found the bar models in Singapore to be really cool.