My son got an extra-credit assignment in social studies that is a math word problem:
There’s a colony consisting of 90 people. There are four groups in the colony: farmers, misc. others, crossbowmen, and women. There are equal numbers of farmers and misc. others. The farmers and misc. others plus 14 equals the number of crossbowmen. The number of crossbowmen is 12.5 times the number of women. Women are the smallest group. How many are in each group?
I figured it out by finding the multiple of 12.5 that is an even number and less than 90.
But is there a way to solve it using an algebraic formula? I tried and got myself thoroughly confused.
If you start by assigning variables to each group of people, you can begin to construct equations that represent the situation. In this case, your variables could be something like
x = farmers
y = others
z = crossbowmen
w = women
You know that the total number of people is 90, and since no one belongs to more than one group at once, you can say that x+y+z+w = 90.
Then just construct equations for each of the next statements:
Equal numbers of farmers and others means x = y, for example.
I don’t particularly feel like solving it right now, but that’s the general procedure. You have 4 unknowns, and you have enough info to construct 4 equations. You can then start substituting equations into one another to start solving each of your unknowns separately.
Which leaves one wondering why the question included the information that women make up the smallest group. That isn’t necessary for the solution at all.
Word problems at that level often have information that’s not necessary in order for the solver to determine what is important and necessary to solve the problem, and what is frivolous, extra information. It’s also done at higher levels; in part because the necessary equation may be to obvious if only the exact necessary variables are included, and in part so that the solver must determine what is important and necessary to solve the problem, and what is frivolous, extra information.
FWIW, instead of using a, b, c, and d or w, x, y, and z, I like to use f, m, c, and w. It makes the problem a bit more fun, plus at the end you don’t have to remember which variable you assigned to which group.
Hey, women can’t even farm either, apparently.
This problem has an additional extra constraint that the numbers are all integers, since half a person can’t be a farmer, or even a woman. Occasionally tough extra credit problems will involve a type of question where that is important, since there isn’t a general method for coming up with a single solution. Usually there are a few possible numbers to test to make sure a possible solution works.
For example, consider if the problem did not give the information “equal numbers of farmers and misc.”. You wouldn’t be able to find the solution using typical algebra methods. You’d have to test a few numbers. In that case, the “women are the smallest group” would also be useful in finding the solution. You might start with 2 women and work up from there.
Dunno if this is commonly taught, but to solve a question with x variables, you need to have x formulas. Obviously, x=14 and 2x=28 only counts as one, so no cheating there. Is that common knowledge or just a random tidbit I picked up? So far, I’ve only heard one teacher say that (physics) but I can’t really say there’s been a lot of opportunity for it to come up in casual conversation.
It’s the foundation of the branch of mathematics called Linear Algebra which uses matrices to store the coefficients of a series of linear equations in order to efficiently and mechanically solve problems as Pazu has done above. If you can break a problem into “N equations / N unknowns” then you can get a computer running MATLAB (or C if you’re feeling plucky, or FORTRAN if you’re feeling really masochistic) to solve it for you.
Actually, to solve for n variables uniquely you will need n non-redundant equations. If the equations are all linear, then you will need n linearly independent equations. You can certainly solve for n variables with less than n equations. You just won’t have a unique solution. For example, two unknowns with one equation can have an infinite number of solutions.
‘Women are the smallest group’ also serves as a sanity check. If you crunch the numbers and come up with 40 women instead of 4, something went very wrong.
It’s a little strong to say that it’s the foundation of linear algebra (which, in fact, is not the study of matrices), but it would certainly come up in a beginning linear algebra class. I think I learned it in second year high school algebra.
There is another consideration. All the numbers must be integers. The OP twigged to this already apparently, as witness the statement that the number of women must be even. Anyway, this gets into diophantine equations which I am supremely unqualified to talk about, but shows up in a ton of puzzles, usually involving farmers buying livestock. I wouldn’t be surprised if you could vague up the other formulae and still get a puzzle with a unique solution. For example, replace “The farmers and misc. others plus 14 equals the number of crossbowmen” with “There are more crossbowmen than farmers and misc others combined.” I think that works. :dubious: