[aside]More and more, the things I hear confirm the fact that in many areas of life “skill drives will” and not the other way around. That is, one likes something because they are good at it. This is opposed to the opposite, that is, one is good at something because they like it and work hard to become good at it (“will drives skill”). I agree that there is some feedback loop going on between these two, but I see that the biggest influence is the former.[/aside]
I’ve never had undue difficulties with word problems, FWIW. Proofs have always rubbed me the wrong way, at best. What counts as an explanation in math doesn’t really cut it in rhetoric and visa versa, typically.
In high school, I thought that if you don’t have an overall understanding of why c^2 = a^2 + b^2, then you don’t really “have it”. Intuitively, I understand the reasons for the 3/5 compromise (US Constitution) and there was a time that I could diagram a glucose molecule. But the Pythagorean theorem is another category of understanding altogether. That said, I can apply the Pythagorean theorem, once I just accept it as given. For me, knowing that it was advisable to accept something then move on helped.
Here’s another math tip. For dog’s sake, make sure you have more than 1 textbook covering the same material. A 2nd, 3rd or even 4th presentation can really help.
Actually, it’s possible to solve the problem about the yard that eleanorigby gives without algebra. This is true because the problem was designed to have answers that are integers. This is what happens on various sorts of multiple-choice tests like SAT’s and GRE’s, so it’s worthwhile to learn how to solve these problems without algebra, since it will save you time in many cases. Furthermore, you will learn what the problem is really about before you have to use the algebra, so this is a useful first step in any case. First, look at the diagram in Polerius’s post. As you can see, there is a smaller rectangle that’s within the larger 6 by 8 rectangle. The length of this smaller rectangle is x meters. The width of this rectangle is y meters.
Now, we know that the area of a rectangle is the length times the width or x times y in the case of the smaller rectangle. Notice that the smaller rectangle has to be placed symmetrically within the larger rectangle. That is, the top side of the new rectangle (in Polerius’s diagram) has to be w meters from the top side of the 6 by 8 rectangle, and the bottom side of the new rectangle also has to be w meters from the bottom side of the 6 by 8 rectangle. The right side of the rectangle also has to be w meters from the right side of the 6 by 8 rectangle and the left side has to be w meters from the left side.
O.K., so the length (x) of the smaller rectangle has to be 8 - w - w (or, in other words, 8 - 2w) meters, since it fits into the larger rectangle with w meters on either side. Similarly, the width (y) of the smaller rectangle has to be 6 - w - w (or, in other words, 8 - 2w) meters. To put it another way, you have to subtract the same number from 8 to find the new length as you subtract from 6 to find the new width.
Now, let’s try some integer answers. Suppose the length (x) of the smaller rectangle is 7 meters and the width (y) of the smaller rectangle is 5 meters. That would be true if w is 1/2 meters, since 7 + 1/2 + 1/2 = 8 and 5 + 1/2 + 1/2 = 6. The area of this rectangle would be 35 meters. This is too large, so let’s drop down by one meter in both length and width. Suppose the length is 6 meters and the width is 4 meters, which would make w be 1. The area is then 24 meters, so this is still too large.
Drop down to a length of 5 meters and a width of 3 meters, so that w is 1 1/2 (um, that’s one and a half, but trying to write it in the limited sort of symbols used here makes it hard to read). The area of the smaller rectangle is 15 square meters, which is what was wanted. So the answer to put down in the answer sheet is 1 1/2.
Remember, be sure you put down the answer they want, not some other variable that’s relevant. They asked for the value of w, so that’s what you put down. It’s easy to forget what variable they’re asking for. You might forget and think that they want the area instead and put down 15, or you might put down 5 or 3.
Now, if you haven’t understood that much, there’s no point going on to the algebra. Furthermore, in this case you don’t have to go on to the algebra. This is what happens in many multiple-choice questions. You can answer the question just by trying values for the variables. Don’t waste time by setting up the algebra equation if you don’t have to. SAT’s and GRE’s are timed tests, so don’t waste time doing things you don’t have to do. Please, you do not get extra credit for knowing how to do the algebra if you can find the answer without algebra, so don’t waste time on the algebra.
Here’s how to do the algebra if you really want to know: The length of the smaller rectangle is 8 - 2w and the width of it is 6 - 2w. The area of the smaller rectangle is thus (8 - 2w)(6 - 2w). You can multiply this out to see that the area is thus 48 - 28w + 4(w2). (I will use w2 to mean w squared.) So we have the equation (since the area is 15):
48 - 28w + 4(w**2) = 15
Subtracting 15 from both sides and rearranging we get this equation:
4(w**2) - 28w + 33 = 0
In the standard quadratic equation, this means that a is 4, b is -28, and c is 33. Here’s the Wikipedia entry on the quadratic equation:
So in the quadratic equation, we have to first calculate (b**2) - 4ac. This equals 784 - 528 = 256. The square root of 256 is 16. So the two possible answers to this equation are (28 - 16) / (2 X 4) = 12/ 8 = 1 1/2 and (28 + 16) / (2 X 4) = 44/8 = 5 1/2. Now, the second answer, five and a half, is useless. This would mean that the length of the smaller rectangle is 8 minus two times 5 1/2, which is -3, and its width is 6 minus two times 5 1/2, which is -5. So the rectangle would be -3 times -5 meters. There isn’t any useful meaning to negative numbers when we’re talking about the width and length, so forget this answer. The answer is 1 1/2. This is a good example of why it’s not only useless to do the algebra, but the answers might even confuse you. Yes, it’s true that -3 times -5 also equals 15, but so what?
Well, she’s probably not got up to quadratics with complex roots just yet. 
I wrote:
> So the answer to put down in the answer sheet is 1 1/2.
Of course, tests today are on the computer, so you click on an answer and don’t write anything down. It’s hard for me to remember this.
I don’t believe you. Care to elaborate?
Yeah, when I saw the condition “The garden must contain exactly 15 square meters of area,” I thought, “The obvious answer would be for it to be 3 x 5 meters. Would that leave an even border all the way around? Yep, 1.5 meters on each side! So much for that problem!”
ETA: If the reason you’re working the problem is to learn/practice how to use algebra to set up and solve problems like this, that’s “cheating,” but if you’re just trying to get an answer as quickly as possible, that’s the smart way to go about it.
Disappointed in 770? I know they’ve changed the system since I took the SAT in 1975, but I remember 800 being the highest possible score. I got 640 math, 760 English, and that put me into the top percentile overall. I’d think 770 was pretty darned good.
I think everybody has a point where math stops being intuitive and they have to work at it. 2+2=4 is obvious for anyone. Differential equations are obvious for very few.
Too many people simply stop at the point where it stops being obvious to them, not realizing how little work it really is to move forward a bit if you only understand how these things apply to the real world.
I’ve been amazed at how many places the math I learned in school have applied in life, and how few teachers ever showed me those real uses. Just a few that come to mind:
[ul]
[li]I was selling a gas tank, which was a 3’ diameter cylinder 6’ long, laying on its side. There was a port in the top where I could put in a stick and see how much gas was in it (14", as it turned out). I needed to figure out how many gallons of gas were in it so I knew how much to charge.[/li][li]My brother was building a water tank. He needed it to have 1,000-gallon capacity. A metal shop said they’d fabricate him one: a basic cylinder with a top & bottom welded on. They’d charge by the square foot of steel. What ratio of height to diameter maximizes internal volume while minimizing the amount of metal?[/li][li]I’m building a clock with a hexagonal face. What angle do I set the miter saw to cut the triangles of wood that will comprise the face?[/li][li]I’m going to another town 1,000 miles away. Is it cheaper to take the car with good gas mileage and stay in a motel, or take the truck and camper (bad gas mileage) and stay in a cheap campground?[/li][li]And I can’t even list the number of times I’ve had to calculate the areas of odd shapes to figure out how much flooring (or whatever) to buy.[/li][/ul]
I know you’re joking but in case you were unaware, Barbie really did get in trouble for saying “math is hard.” I figured it was the inspiration for the title of this thread.
I’m bored, so here goes. Sorry if you find this annoying.
V = πr²h
V = π(18 in)²(14 in)
V = 4,536π in³
V = 4,536π in³(1 gal/231 in³)
V = (216π/11) gal
V ≈ 61.69 gal
Okay, this one was tricky since I haven’t done calculus in years. Let me know if it’s right.
V = πr²h
h = V/(πr²)
h = 1,000 gal/(πr²)
h = 1,000 gal/(πr²)
h = 231,000 in³/(πr²)
This gives us the height in terms of the radius in inches.
Now for what we really want, which is the surface area. Unroll the cylinder and add the top and the bottom.
A = 2πr² + 2πrh
Replace h.
A = 2πr² + (2πr)(231,000 in³)/(πr²)
A = 2πr² + 462,000 in³/r
This is what we want to minimize, so we take the derivative.
A’ = 4πr – 462,000 in³/r²
Set it to 0 to find the minimum:
4πr – 462,000 in³/r² = 0
4πr³ – 462,000 in³ = 0
r³ = 462,000 in³/(4π)
r ≈ 33.25 in
And so, h ≈ 66.50 in.
This gives a ratio of height to diameter of 1:1 to minimize surface area of a cylinder for a given volume. Which makes me feel like an idiot.
If we assume a regular hexagon, the sum of the interior angles will be 180(n – 2), or 180(6 – 2) or 720. There are of course, six angles in a regular hexagon, and they are all the same, so each angle is 120. Since you will split each angle in half to create the isosceles triangles that make up the clock face, you want each triangle angle to be 60°. How to use this information to set up a miter saw, I don’t know.
Let:
P = price of gasoline in /gal
*C* = gas mileage of the car in mi/gal
*T* = gas mileage of the truck in mi/gal
*H* = cost of the hotel in /d
G = cost of the campground in $/d
D = number of days you will stay
The total cost of the car-hotel combo is
(2,000P)/C + HD,
and for the truck-camping combo is
(2,000P)/T + GD.
Calculate and choose the smaller. (Incidentally, I have new-found respect for mathematical typographers.)
I think you missed the “laying [sic] on its side” part, which makes it trickier.
Yeah. I just realized that. So, what you have to do is take a circular slice of the cylinder, determine how much of that circle is “filled” with gasoline, multiply that by the length of the cylinder, and then figure out how many gallons the volume is.
So to do that, first we figure out the area of the relevant sector of the circle and then subtract out the triangle to give the area of everything “beyond” the chord. (It would be good to have a diagram to help explain this.)
To do that we need to find x, where cos x = (18 in – 14 in)/(18 in). The value of x here is about 77.16°. Then we double it to get 154.32°. This means that 154.32°/360° is the proportion of the circle containing the gasoline that we want. So, we find the area of the circle, π(18 in)², which gives about 1,017.88 in², and then use the proportion to get (154.32°/360°)(1,017.88 in²) or 436.33 in².
Now we have to subtract out the triangle to get the gas only “beyond”. It’s an isosceles triangle, which can be separated into two right triangles each with a hypotenuse of 18 in and a leg of 4 in. To figure out the other leg, we have (4 in)² + b² = (18 in)², which means that b² = 308 in², and b ≈ 17.55 in. The area of the triangle is therefore (4 in)(17.55 in)/2, or 35.10 in. There are two such triangles, so we get a total of 70.20 in².
So the area of the circle covered with gasoline is the sector minus the triangle, or 436.33 in² – 70.20 in², which gives 366.13 in².
To find the total volume of gasoline, we multiply this by the “height” of the cylinder to give V = (366.16 in²)(72 in), which gives 26,361.51 in³. There are 231 in³ in a gallon so there are (26,361.51 in³)(1 gal/231 in³) or 114.12 gallons of gasoline in the tank.
What has always bugged me about math is that the teacher would always, always pull something out of her ass. How to balance this equation: 65x=y? or some such. Why, didn’t you know that the whole number 65 is really a fraction, 65/1? And guess what? If you multiply 65/1 X 1/65, you get… 1!* And if we use it as a fraction (something we had never, ever done before–never even so much hinted that it was possible to do so–whole numbers were most definitely NOT fractions and vice versa), then we can solve all our problems.
It drove me nuts. First they would tell me that a whole number isn’t a fraction and then (when it was convenient for them) they’d say it really WAS a fraction afterall, but then they wouldn’t use it AS A FRACTION ever again. It seemed unreliable to me (ironic, no?). It happened in trig as well. Well, let’s just IMAGINE this unit circle, ok everyone? Well, no–because I can IMAGINE a circle that has sides and corners–my imagination is unlimited. Anything is possible in imagination.**
*Yes, I get that 1/65 x 65/1= 1. I can do the cross multiplying and see that they do “cancel each other out”. But no one in their right mind would ever have thought of such a thing except on some impulse to bend the rules to make it work for them. It seemed to me that they made the rules up as they went along. Bugged me no end. Still does, really.
**And no “but then it wouldn’t be a circle” cracks–this is imaginary!
Wendell–I figured there would be a way to solve it sans algebra. My way would be to lay out string or the garden hose until I got the plot I wanted and then I’d start digging. YMMV. 
Invisible–I should have gotten higher. I scored consistently higher in the practice exams, but like the posted analogy, the GRE has its own version of what words mean.
Polerius: I agree, skill does drive will. Its mirror image is that lack of skill drives lack of will, which is a vicious cycle, especially for a building block skill such as math. Couple with poor teaching and you have people like me who are math haters (and those who are downright math phobic). It’s too bad.
That makes perfect sense to me. I think I probably do some math in my head unconsciously, so no wonder I got mixed up when I was supposed to do it “their” way.
I can see where word problems would need modeling skills, but I don’t see where I lack them except in math. My issue (my biggest issue) was struggling to solve “what exactly is this problem asking me to solve for and how do I set that up”?
It sounds to me like you had a bad teacher. When a teacher says something new, your reaction should be “Oh, I understand where you got that from, although I didn’t think of it myself.” If your reaction is “Where in the name of all that’s good and true in the world did you get that from?”, you’re being badly taught.
I don’t understand how you could solve this problem using string or a hose. The important thing here is to draw the picture in Polerius’s post. You’ve got a smaller rectangle inside a larger rectangle. The larger rectangle is 6 meters by 8 meters, so its area is 48 square meters. You want to put a smaller rectangle neatly inside it. The answers to these questions are usually integers, so stick with integers to start with. You make its measurement one meter less in each direction. This 5 by 7 rectangle has an area of 35. That’s not small enough, so make it a 4 by 6 rectangle. This rectangle has an area of 24, so that’s still not small enough. Make it a 3 by 5 rectangle. This has an area of 15, so that’s the rectangle you want. Then look back at the statement of the problem. It asks for the width of the path around the small rectangle (that separates it from the large rectangle). It’s 1 1/2, since 3 + 1 1/2 + 1 1/2 = 6 and 5 + 1 1/2 + 1 1/2 = 8 (because on each side of the rectangle you have a 1 1/2 path).
Well, I’m not being completely honest here. I am telling you what I would do* if in RL I was faced with making a garden plot in my own courtyard.* Concretely, IOW, instead of just on paper. Not real fair, but that is how I would have solved the problem…
It’s too late for me. I’m 47 and have no interest in going back to really figure out binomials or whatever. Trust me, I know I was badly taught–I am the generation that was experimented on, from phonics to the New Math. Every year it was either schwa “e” or base 8 and then the next year it would all change again.
You know how in 1st grade kids are given worksheets with drawings on them of bricks or logs and asked to count how many are there? It’s supposed to help teach them place and counting, I guess. I can’t code it to draw it for you, but there’d be maybe 3 logs in the hundreds place, then 6 logs in the tens, then 8 logs in the ones place and you were supposed to write 368. But for me I counted not only the sides of the logs, but the ends or “faces” of them as well, so I would get 6 for the hundreds, 12 for tens etc. I didn’t know how to explain what I was doing and the teacher never figured it out–I got a lot of red ink in first grade. My math career didn’t start well and it ended with a whimper. But I “got” reading and English. Poetry and metaphor, symbolism and wordplay–it all made sense. I gave math up as a bad job. Too bad, because I really loved biology and micro. The math in chemistry drove me nuts as well–even though I could see that there was a sense to it and a pattern–it had its own poetry, in a way.
Like learning to read music (something else I never have done), I envy those with strong math ability. It is just recently that I am able to add/subtract double digits in my head and even multiple double digits in my head. To figure out a 20% tip, I divide the total by 10 and then double the result. No one taught me to do that–I realized I could do it that way and not have to ask someone else to figure it out for me or scribble on an old receipt in my purse. I get by.
eleanorigby writes:
> To figure out a 20% tip, I divide the total by 10 and then double the result.
That’s how people who are good at math do it too. If you’re practicing math, you might want to put in more work to understand the underlying idea, but if you’re doing it in a situation where you have to do it quickly, use anything that works. You’re not being graded on how much more work than necessary you put in.
I like the way you think! And I think my kids are getting a better math education than I did. I know that algebra is integrated into the lessons by about 3rd grade, as is some simple geometry. To me, this makes sense, instead of treating each aspect of math as a separate entity (there are limits to this, of course).