I’ve known this riddle for years and I can figure out similar riddles in my head using trial and error, but how do I figure this out using an algebreic formula?
You bought two pencils in a stationery store, an expensve pencil and a cheap pencil. The expensive pencil costs a dollar more than the cheap pencil. All together, you spent $1.10. How much was the cheap pencil?
Let x be the price of the cheaper pencil, and y the price of the more expensive pencil.
Then the more expensive pencil is a dollar more expensive than the cheap one, so y = x+ 1. Also, the total price is $1.10, so x+y = 1.1.
Then substitute the first equation into the second to get:
x+y = 1.1
x+ (x + 1) = 1.1
2x = .1
x = .05
So the cheaper pencil costs $0.05 or 5 cents, and the more expensive pencil is $1.05.
x + y = 1.10 [All together, you spent $1.10.]
x = y + 1.00 [The expensive pencil costs a dollar more than the cheap pencil.]
substitute the second equation into the first:
y + 1.00 + y = 1.10
2y = .10
y = .05
x = 1.05 (from the first equation)
X + Y = 1.10
X - Y = 1.00
Since both equations are true, you can “add” them (Mathochist will be along with a proof I’m sure 
) and get
2X + 0Y = 2.10
And therefore X = $1.05 (clearly X is the expensive pencil). Plug that result into either of the first two equations to solve for Y. You can also subtract the equations from each other to solve for Y first.
I try not to respond to posts which smell too strongly of homework, creative backstories or not.
Whatever. Maybe you shouldn’t bother posting about any math questions then. What kind of math question can’t smell of homework?
Interesting. I saw the problem a little differently than everyone else. When I worked the problem in my head, since we are dealing with two like items (pencils), I treated them both as x:
x + (x+$1) = $1.10
2x = $1.10 - 1.00
x = .05
It never ocurred to me to treat one pencil as x and another as y as they are both the same items and can be directly related, thus eliminating the need for simultaneous equations as in **Jurph’s ** solution.
/hijack
what is the proper naming convention for variable assignment?
hijack/
There is no proper naming convention for variables any more. One of the early French algebraists had a system of always starting from the back of the alphabet for variables, and from the front of the alphabet for constants. In this crass and shoddy modern world, however, people have stopped holding strictly to those rules, and mathematicians instead simply pick variables that should be easy for the audience to remember.
Actually, you’re not treating them both as x, you can’t do that because they aren’t the same price. You’re just never formally declaring y but ‘inlining’ the expansion of the more expensive pencil as x + $1 In a sense, you’re doing the same work as solving the simultaneous equations, you just don’t notice it because the second equation (y = x + $1) is so simple you can substitute it into the other equation directly (and because the second equation is pretty simple too.)
Not sure there are any formal naming conventions for algebra word problems. Standard practice is to assign all important unknown quantities on the series x, y, z, w, v, u as they occur to you. If you realize you need more than six unknowns, it’s generally common to group them and use numbers to differentiate them… x1, x2, x3 and so on.
Did that make any sense??
But you just did the first step, substitution, in your head - it was the same exact technique.
You are really getting reamed by your office supply store. Ticonderoga #2 pencils go for $11.49 for 72 pencils!
Cite: http://www.staples.com/Catalog/Browse/Sku.asp?PageType=1&Sku=138180
Only if he buys the expensive pencils. That’s still more than three times the price of the cheap ones. 
Sorry, I forgot that to most people math is only good as a source of pointless exercises and “riddles” poseable as homework problems.
This is such a standard example of algebra that I have great difficulty believing that you can figure out “similar examples” in your head and have trouble with this one.
I did it similarly to Tanach. Basically, I thought, “well, the price of this pencil, plus (the price of this pencil plus a dollar), equals a buck ten.” I know that’s functionally equivalent to what the other solutions said, but it stayed a word problem in my head a little longer than for other folks :).
Daniel
Touche!
That’s what I get for failing to show my work;)
Happy Holidays:)
This is such a standard example of algebra that I have great difficulty believing that you can figure out “similar examples” in your head and have trouble with this one.
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Like I said, I use trial and error. I did it something like this:
Okay, the obvious answer is 10 cents. A dollar more than 10 cents is $1.10, $1.10 plus 10 cents is $1.20 - nope, that can’t be right, 10 cents is too much. Lets cut it in half and try 5 cents. 5 cents plus a dollar more ($1.05) is $1.10, yup, I got it.
This is the way I’ve always solved these sort of problems and I’m embarresed to say, it still seems easier to me than the above formulas. I know if the problem is more difficult, I would definitely need a formula. That’s why I asked. FWIW, I’ve been a member in good standing here for over two years and I wouldn’t use these boards to cheat on homework (and I haven’t been in school for years).
I don’t think I’ll be able to do other problems like this the way silverfish and John Mace solved it, but the way Jurph and Tenach explained it makes sense to me. Thanks everyone.
But it’s not really out of line for high-end drafting/drawing pencils at $10.77 a dozen.
Your roentgen tube has gone bad.
It is not a riddle. It is a word problem in mathematics!
As some have already noted, these are really all the same (or very nearly so) way of solving the problem. The standard technique to solve these problems is to simply assign variables (x, y, etc.) to the unknowns and then translate the wording into equations (I tried to highlight that by putting the exact wording next to the equations). Then you solve for one of the variables (in terms of the other) in one equation and substitute it into the other equation. At that point you only have one variable-- you solve for it, then substute the answer into the other equation and solve for the other variable.
Once you do a few that way it’s a piece of cake.
True. For myself, I find single-variable algebra to be far, far easier than multiple-variable algebra, and so I’ll always try to get to single-variable before putting it in equation form. But that’s just an idiosyncrasy of my head, I think.
Daniel