# Stupid Math Word Problem **URL:** \ **Category:** Factual Questions\ **Created:** [March 18, 2009, 2:31pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885 "2009-03-18T14:31:21Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![bazarov5000](https://avatars.discourse-cdn.com/v4/letter/b/96bed5/32.png) [@bazarov5000](https://boards.straightdope.com/u/bazarov5000)\ **Post date:** [March 18, 2009, 2:31pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/1 "2009-03-18T14:31:21Z")
Like most Dopers (long time lurker, first time poster), I enjoy puzzles and riddles of all sorts which prompted my brother to buy me a “Use IT or Lose IT” (sic) desk calendar for Christmas. Each day brings a new challenge in various areas such as math, logic and language. There are days when the official answer seems a bit…odd, and the solution for March 10th seems flat out wrong. Here’s the problem: “The average of the number 10 and some unknown number, x, is divided by the sum of 10 and x. The result is 1/2. What is the value of x?” I came up with x=2. The average of 10 and 2 is 6, and 6 divided by 12 (the sum of 10 and 2) is 1/2. The official answer on the back of the calendar page is “Cannot be determined.” Am I missing something (did I perhaps Lose IT by not Using IT enough), or are they full of crap? ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [March 18, 2009, 2:35pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/2 "2009-03-18T14:35:59Z")
The average of X and 10 is (X+10)/2. Dividing this average by X+10 gives 1/2 for **any number X**. It works for 6, but it alway works for 0, 1 million and -4 (the average of 10 and 0 is 5, and 5 divided by (10+0) equals 1/2, etc.) ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [March 18, 2009, 3:10pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/3 "2009-03-18T15:10:22Z")
> [@Andy\_L](#): > > The average of X and 10 is (X+10)/2. Dividing this average by X+10 gives 1/2 for **any number X**. That should be any number except -10. So we do know something about X: it cannot be -10. ---
**Author:** ![magellan01](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@magellan01](https://boards.straightdope.com/u/magellan01)\ **Post date:** [March 18, 2009, 3:25pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/4 "2009-03-18T15:25:22Z")
> [@Andy\_L](#): > > The average of X and 10 is (X+10)/2. Dividing this average by X+10 gives 1/2 for **any number X**. It works for 6, but it alway works for 0, 1 million and -4 (the average of 10 and 0 is 5, and 5 divided by (10+0) equals 1/2, etc.) Wow. It’s been a really long time since I’ve done algebra, but I get that it works for any number: ((x + 10)/2)/(x + 10) = 1/2 multiplying both sides by x + 10 gives you: (x + 10)/2 = (x + 10)/2 Multiplying both sides by 2 results in: x + 10 = x + 10 So, x = anything and everything Unless, of course, I did something wrong. But I can’t see it if I did. Anyone? ---
**Author:** ![gigi](https://avatars.discourse-cdn.com/v4/letter/g/a587f6/32.png) [@gigi](https://boards.straightdope.com/u/gigi)\ **Post date:** [March 18, 2009, 3:26pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/5 "2009-03-18T15:26:25Z")
> [@Andy\_L](#): > > The average of X and 10 is (X+10)/2. Dividing this average by X+10 gives 1/2 for **any number X**. It works for 6, but it alway works for 0, 1 million and -4 (the average of 10 and 0 is 5, and 5 divided by (10+0) equals 1/2, etc.) Woot! I finally got one of these right! ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 18, 2009, 3:27pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/6 "2009-03-18T15:27:52Z")
> [@Giles](#): > > That should be any number except -10. So we do know something about X: it cannot be -10. This is significant and must be included in any answer. ---
**Author:** ![Winsling](https://avatars.discourse-cdn.com/v4/letter/w/8e8cbc/32.png) [@Winsling](https://boards.straightdope.com/u/Winsling)\ **Post date:** [March 18, 2009, 4:06pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/7 "2009-03-18T16:06:11Z")
> [@magellan01](#): > > Wow. It’s been a really long time since I’ve done algebra, but I get that it works for any number: > > ((x + 10)/2)/(x + 10) = 1/2 > [snip] > Unless, of course, I did something wrong. But I can’t see it if I did. Anyone? In your initial equation, you’re dividing by x + 10. If x = -10, you’re dividing by zero, and the math blows up. ---
**Author:** ![magellan01](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@magellan01](https://boards.straightdope.com/u/magellan01)\ **Post date:** [March 18, 2009, 4:14pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/8 "2009-03-18T16:14:36Z")
> [@Winsling](#): > > In your initial equation, you’re dividing by x + 10. If x = -10, you’re dividing by zero, and the math blows up. I see what you’re saying. But at the same time, I think my math is correct, and the result allows for x + -10. Which is correct? Both? That can’t be. Or did I do something wrong? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 18, 2009, 4:24pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/9 "2009-03-18T16:24:16Z")
> [@magellan01](#): > > I see what you’re saying. But at the same time, I think my math is correct, and the result allows for x + -10. > > Which is correct? Both? That can’t be. Or did I do something wrong? You can’t divide by zero, so no, your math doesn’t allow for x = -10. ---
**Author:** ![magellan01](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@magellan01](https://boards.straightdope.com/u/magellan01)\ **Post date:** [March 18, 2009, 4:31pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/10 "2009-03-18T16:31:04Z")
> [@ultrafilter](#): > > You can’t divide by zero, so no, your math doesn’t allow for x = -10. I know you can’t divide by zero. But then what is wrong with what I laid out in my first post? ---
**Author:** ![Roland\_Orzabal](https://avatars.discourse-cdn.com/v4/letter/r/c2a13f/32.png) [@Roland\_Orzabal](https://boards.straightdope.com/u/Roland_Orzabal)\ **Post date:** [March 18, 2009, 4:36pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/11 "2009-03-18T16:36:59Z")
> [@magellan01](#): > > I know you can’t divide by zero. But then what is wrong with what I laid out in my first post? Think of it like a logic puzzle, with each new equation you derive being a clue. You started with ((x + 10)/2)/(x + 10) = 1/2. From that, we know that x cannot be -10. Each step of your work after that just generates a new clue. Each new equation is correct, but is done with the foreknowledge that x cannot be -10. That it could be -10 in your final clue, x + 10 = x + 10, is irrelevant, because we already know from previous clues that it cannot be. ---
**Author:** ![magellan01](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@magellan01](https://boards.straightdope.com/u/magellan01)\ **Post date:** [March 18, 2009, 4:48pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/12 "2009-03-18T16:48:06Z")
> [@Roland\_Orzabal](#): > > Think of it like a logic puzzle, with each new equation you derive being a clue. > > You started with ((x + 10)/2)/(x + 10) = 1/2. From that, we know that x cannot be -10. > > Each step of your work after that just generates a new clue. Each new equation is correct, but is done with the foreknowledge that x cannot be -10. That it could be -10 in your final clue, x + 10 = x + 10, is irrelevant, because we already know from previous clues that it cannot be. That makes sense, I guess. But let’s say this was a much more complicated equation, with multiple unknowns. Shouldn’t one be able to reduce the equation to its simplest form and not have to evaluate every step, except for the rules of algebra? It seems that this results in one equation being able to produce multiple, conflicting results. It doesn’t make sense to me. It’s as if math is failing us. I do understand why -10 is not a possibility at one point in the solving of the equation, but I thought that would mean that it would then also be an impossibility at all steps. Unless I did something wrong. Confused. But maybe that’s because I haven’t done this stuff in like 25 years. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [March 18, 2009, 4:56pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/13 "2009-03-18T16:56:33Z")
> [@magellan01](#): > > That makes sense, I guess. But let’s say this was a much more complicated equation, with multiple unknowns. Shouldn’t one be able to reduce the equation to its simplest form and not have to evaluate every step, except for the rules of algebra? Yes, you can reduce complex equations, but the rules of algebra follow you at every step. For example, if you have: a/b = c/b then you can deduce that a=c and that b is not 0. And if you have: a^2 = b^2 then you can deduce that either a=b or a=(-b). So, if you’re going through multiple stages, you’ve got to carry all the implications wiith you. ---
**Author:** ![magellan01](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@magellan01](https://boards.straightdope.com/u/magellan01)\ **Post date:** [March 18, 2009, 5:07pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/14 "2009-03-18T17:07:35Z")
> [@Giles](#): > > Yes, you can reduce complex equations, but the rules of algebra follow you at every step. For example, if you have: > a/b = c/b > then you can deduce that a=c and that b is not 0. > > And if you have: > a^2 = b^2 > then you can deduce that either a=b or a=(-b). > > So, if you’re going through multiple stages, you’ve got to carry all the implications wiith you. Sorry to be so dense, guys, but it seems to me that the process of simplifying an equation should always result in the same relationships. And maybe, in fact, -10 _can_ be considered an answer because when you plug it in the numerator becomes zero just as the denominator does, rendering 0/0. So the next question is: is it a numerator of 0 the one time it is permissible to have a denominator of 0. That would make sense and not have the equation generate two different answers during different stages of its simplification. If not, can anyone offer an equation in which the answer changes as one simplifies it? ---
**Author:** ![Roland\_Orzabal](https://avatars.discourse-cdn.com/v4/letter/r/c2a13f/32.png) [@Roland\_Orzabal](https://boards.straightdope.com/u/Roland_Orzabal)\ **Post date:** [March 18, 2009, 5:14pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/15 "2009-03-18T17:14:20Z")
I think what may be throwing you is that the standard method writing out simplifications can be misleading. ((x + 10)/2)/(x + 10) = 1/2 is not really equivalent to (x + 10 = x + 10), for the very reason that in the first, x cannot be -10, whereas in the second, it can. A more accurate way to write the latter step might be (x + 10 = x + 10; x ≠ -10). Standard practice is to leave off the last bit because it’s already known, just as you might skip writing out the addition step when changing (x - 10 = 5) to (x = 15). ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [March 18, 2009, 5:14pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/16 "2009-03-18T17:14:57Z")
**magellan01** , In solving an equation you are allowed to use any legitimate algebraic operation to simplify until you get the final answer, BUT in doing so, you may introduce superfluous solutions at the end. The final step should be to try your “solutions” in the original equation to remove the extraneous ones. Of course, along the way, you can keep track of all the implications of your simplification process and remove the extraneous solutions as you go along. ---
**Author:** ![Tenebras](https://avatars.discourse-cdn.com/v4/letter/t/977dab/32.png) [@Tenebras](https://boards.straightdope.com/u/Tenebras)\ **Post date:** [March 18, 2009, 5:16pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/17 "2009-03-18T17:16:53Z")
> [@magellan01](#): > > I do understand why -10 is not a possibility at one point in the solving of the equation, but I thought that would mean that it would then also be an impossibility at all steps. Unless I did something wrong. You did something wrong. Algebra is nothing more than a way of doing arithmetic in general. When you multiply both sides of an equation by (x + 10), you’re doing it for all possible values of x at once. This works because multiplication is a very predictable thing. However, you still need to think about what you’re doing at every stage. Let’s break down what you’ve done. ```auto (x+10)/2 ---------- = 1/2 x+10 ``` This is fine. One number, divided by another number, is equal to 1/2. The first thing to notice is that if x = -10, this is not a well defined expression, because it is equivalent to 0/0, and 0/0 doesn’t have a value. Now you multiply both sides of the equation by (x+10). Very well, we get the following: ```auto (x+10)/2 ---------- * (x+10) = (x+10)/2 x+10 ``` which can be rewritten as ```auto x+10 (x+10)/2 * ------ = (x+10)/2 x+10 ``` At this point, you think to yourself “any number divided by itself is 1” so I can replace (x+10)/(x+10) by 1, which it is equal to. Of course, it isn’t true that any number divided by itself is 1, because 0/0 is not any number at all–it’s not defined. So, as long as x+10 is different from 0, you can move on to your next step. However, if x+10 is actually 0, you cannot. The key here is that algebra is a way to do arithmetic with unknown quantities. If you are careful, you aren’t “moving things from one side to the other” and you don’t ever “cancel the sevens.” That’s just shorthand for doing several steps in a row. Trying to “cancel” terms leads people to no end of trouble, because they have forgotten (or never learned) that what they are doing is replacing one fraction (x/x) by another fraction (1) which it is equal to. In the example above, the fraction (x+10)/(x+10) is the same as 1, so long as x+10=/=0. And that’s where your mistake lies. ---
**Author:** ![magellan01](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@magellan01](https://boards.straightdope.com/u/magellan01)\ **Post date:** [March 18, 2009, 5:31pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/18 "2009-03-18T17:31:33Z")
Thanks, **Tenebras** (and the rest of you), that makes sense. It does scare me a bit that all these engineer types are doing all this math for buildings and planes and bridges and they have to evaluate each and every step of complex equations with multiple unknowns. And in doing so have to know that something like “it can’t be -10” pops up along the way, but less obviously. But now I have a new question: algebraically, is 0/0 equal to 0 (0 divided by any number equals 0) or does that also “blow up” the math? ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [March 18, 2009, 5:35pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/19 "2009-03-18T17:35:42Z")
Nope. 0/0 is always undetermined. It is never zero and never allowable. ---
**Author:** ![tr0psn4j](https://avatars.discourse-cdn.com/v4/letter/t/958977/32.png) [@tr0psn4j](https://boards.straightdope.com/u/tr0psn4j)\ **Post date:** [March 18, 2009, 5:46pm UTC](https://boards.straightdope.com/t/stupid-math-word-problem/489885/20 "2009-03-18T17:46:41Z")
Although this problem was relatively easy, it reminded me of how much math I’ve forgotten since high school. ☹ [Next page](https://boards.straightdope.com/t/stupid-math-word-problem/489885.md?page=2)