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I am teaching an elementary student. He has a homework as follows.

There are 16 students who use either bicycles or tricycles. The total number of wheels is 38. Find the number of students using bicycles.

I have 3 solutions as follows.

Using a single variable.

Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times 2 and the total number of tricycles times 3.

$$ 2\times x + 3 \times (16-x) = 38 $$

The solution is $x=10$.

Using 2 variables.

Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively. It implies that

\begin{align} x+y&=16\\ 2x+3y&=38 \end{align}

The solution is $x=10$ and $y=6$.

Using multiples

The multiples of 2 are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$

The multiples of 3 are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$

The possible wheel combinations with format (#bicycle wheels,#tricycle wheels):

(32,6) but there are 18 students

(26,12) but there are 17 students

(20,18) there are 16 students

(14, 24) there are 15 students

(8, 30) there are 14 students

(2,36) there are 13 students

Thus the correct combination is 10 bicycles and 6 tricycles.

My question

Is there any other simpler method?

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12  
Using differential equation is also welcome! :-) –  Oh my ghost Aug 28 '13 at 14:58
1  
If I were doing the teaching, I’d point him at Mark Lakata’s method. Then I’d use that as a springboard for explaining Robert Israel’s and Christian Fries’s answers and, depending on his background, the one-variable algebraic solution. –  Brian M. Scott Aug 28 '13 at 20:41
1  
I'd start the same as @BrianM.Scott, but afterwards I'd show the 2-variable method, and show how to substitute it into the other to create the 1-variable method. It's an extra step, but making it explicit like that can help. –  Izkata Aug 28 '13 at 23:14
2  
Unless it is a gifted or accelerated class, I can almost guarantee that an elementary student is being asked this question to see if they can guess-and-check. Start with 8 on each, find that is too many wheels, so try with more bikes and guess again. While it is valuable that a student might learn some of these other methods, you also don't want to confuse and frustrate them by not teaching the same method they are being taught in class or by teaching something too advanced. What is the rest of the unit on? What are other questions like? –  PeterL Aug 29 '13 at 20:42

18 Answers 18

up vote 240 down vote accepted

If everybody had bicycles, there would be $16 \times 2 = 32$ wheels. There are actually $38$, so that's $6$ additional wheels. Each one must be on a different tricycle. So there are $6$ tricycles, and the other $10$ students have bicycles.

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20  
This is the core of most "quick math tricks" I think - start at an easy to solve approximate answer, and refine it to the precise answer. Like multiplying by 12 - multiply by 10 and 2 and add them. –  xdumaine Aug 28 '13 at 17:44
4  
This isn't really starting at an easy to solve apporimate answer. If you take the first "single variable" equation and split the 3*(16-x) into (2*(16-x)+1*(16-x)) then combine the first part with the other part of the first equation then you get cancellation and 32 +(16-x) = 38 which is the approach described above. The core of most maths tricks is just using algebra to simplify otherwise more complicated equations. –  Chris Aug 28 '13 at 18:31
7  
The only thing I'd add to this answer is to check the answer by multiplying and adding the wheels of this mix of bikes and trikes to see that they match the total. –  KeithS Aug 28 '13 at 19:40
4  
I was about to post this same answer; now I won't. I was going to say that each time you trade a bicycle for a tricycle, you add one more wheel. –  Michael Hardy Aug 28 '13 at 23:42
1  
As an addition to Chris' commment, if we were to compare this solution to the more formal algebraic approaches mentioned in the question, then I'd say that it corresponds to multiplying the first equation in the second approach by 2 and subtracting it from the second. –  Carsten Schultz Aug 29 '13 at 10:33

I think the "simplest" (meaning least thinking, not least work) is to create a table of all possibilities.

bike trike wheels
 0    16     48
 1    15     47
 2    14     46
 3    13     45
 4    12     44
 5    11     43
 6    10     42
 7     9     41
 8     8     40
 9     7     39
10     6     38  <- answer
11     5     37
12     4     36
13     3     35
14     2     34
15     1     33
16     0     32

In fact, once you start filling out the table, you notice a pattern. The last column decreases by one on each line. That teaches the kid that there are hidden patterns, and if they discover the pattern after a few lines, then can visually extrapolate how many lines they'll need to calculate until they get close to the answer.

It motives the concept of a linear progression ( -= 1 each time), and this begins to look like a downward slope of "something", which might be something that looks like a hill which maps to the abstract concept of a line graph with negative slope.

Also, the problem could be changed to something like 160 people, and 380 wheels and the table begins to get unwieldy. It is still the same concept, but at this point, the student will think "there must be a better way" which leads to using graphs and algebra to solve the same problem.

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10  
I really like this answer because it seems most appropriate for the younger audience. –  Squirtle Aug 28 '13 at 20:07
4  
My experience with members of such a "younger audience" is that the prospect of writing down that many numbers would make an approach like this pretty terrifying :) –  Pointy Aug 29 '13 at 17:24
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I'm certainly not a teacher, but my experience when working with kids is that grinding out answers using a formula is easy, coming up with the formula is hard. And for chrissakes, aren't the kids using spreadsheets already? –  Mark Lakata Aug 29 '13 at 20:46
2  
The style of your solution reminds me a known quote in computer science: "When in doubt, use brut force". Quote from Ken Thompson. I think it is the best first solution to come up with! :-) –  yves Baumes Aug 31 '13 at 13:12
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@MarkLakata Please teach them to write a quick Python script instead... –  Tobias Kienzler Sep 2 '13 at 11:52

Robert Israel's solution is (I believe) best suited towards an elementary school student if it is graphical:

  1. Sketch out 16 bicycles. Count the wheels. There are 32 wheels:

enter image description here

  1. Start adding wheels to make tricycles until you have 38 wheels (i.e., every time you add a wheel, count up from 32). Count the number of tricycles and bicycles. Now you know there are six trikes and ten bikes:

enter image description here

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:) and that's the graphical way to show what I answered and was "it's not different from Robert Israel answer"... graphical are always a good way to start –  woliveirajr Aug 29 '13 at 13:05
    
methods like this are good for understanding patterns and developing a method to solve the problem, but If you don't do it in a small scale and adjust to larger numbers then you wont really learn anything about the math. –  Grady Player Aug 29 '13 at 19:32
    
This is a brilliant way of showing the answer without using algebra, but I think it requires too much up-front brilliance to come up with this answer when starting with a blank sheet of paper. –  Mark Lakata Aug 29 '13 at 20:42
2  
@MarkLakata I think it could be used as one of those examples to show young students that a good tool to have in your problem-solving toolbox is to try to sketch out a solution. I would start out by saying, "hey, what if everyone had bicycles?" Then, "what if this kid had a tricycle? How many wheels are there now?" –  Chris Gregg Aug 30 '13 at 2:12
2  
If you are going to do it visually, give them a pile of 16 bicycles and 16 tricycles (little ones, or little cards showing them would also be fine), and then they can mix and match at will to get to the above answer. Some will start with all tricycles, some will start with all bicycles, and then they will make the leap. –  jmac Aug 30 '13 at 4:40

Remove 2 wheels from each vehicle, leaves you with 38-32=6 wheels, which belong to the 6 tricycles. Hence, there are 10 bicycles.

(short, and no "if") - Alternatively:

Remove 3 wheels from each vehicle, leaves you with 38-48=-10 wheels, which are missing on the 10 bikes.

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16  
I dunno, removing wheels from bikes (especially the pesky derailleur back wheel) seems like a lot of work in order to simplify a math problem. And then you have to put all of them back on too. :) –  Mark Lakata Aug 28 '13 at 20:41
    
this 'feels' like the simplest/elegantest solution to me –  Ken Aug 28 '13 at 21:42
1  
Ummm... if you remove 2 wheels from a "bike", don't you just end up with a bike that has no wheels? –  Iszi Aug 28 '13 at 21:56
    
@Iszi: Ah, you are referring to the use of the word bike as a superset to bicycle and tricycle. Right. I edit the answer... –  Christian Fries Aug 29 '13 at 6:29
    
@Mark Lakata: Your are right. But I know people who favor this kind of a more practical solution ;-) –  Christian Fries Aug 29 '13 at 6:55

Based on Robert Israel's method:

If everybody had tricycles, there would be $16×3=48$ wheels. There are actually 38, so that's 10 fewer wheels. Each one must be on a different bicycle. "So there are 10 bicycles (and the other 6 students have tricycles)". The parenthesized part is optional, since that was not asked for.

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2  
I think you did copy-paste or typed too fast. The last sentence should be "So there are 10 bicycles (and the other 6 students have tricycles)". The parenthesised part is optional, since that was not asked for. –  Marc van Leeuwen Aug 28 '13 at 15:11
    
@MarcvanLeeuwen: Thank you! –  Oh my ghost Aug 28 '13 at 15:13
    
(+1) Since the question is for the number of bicycles, but not of the tricycles, this solution is yet a bit simpler than the original solution of Robert Israel. –  azimut Aug 29 '13 at 12:00
2  
"10 fewer wheels. Each one must be on a different bicycle." slightly mind bending, the wheels the bikes don't have are all on different bikes ;) –  jk. Aug 29 '13 at 14:02
    
Why the hell does a copy of another answer have upvotes? –  Miles Rout Aug 1 at 23:29

This is probably not the easiest for mental calculation, but it works.

The average number of wheels per student is $\frac{38}{16}=2+\frac38=\alpha$. So with $b,t$ the numbers of (bi/tri)cycles one has $b:t=(3-\alpha):(\alpha-2)=5:3$, and the number of bicycles is $\frac b{b+t}\times16=\frac 58\times16=10$.

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This is a little silly, perhaps, but here goes.

Since $38$ is an even number and since tricycles have an odd number of wheels (while bicycles have an even number), the number of tricycles must be even. Let's try bisection. If everyone rides a tricycle, you have $16\times3=48$ wheels, which is too many, while if no one does you have only $16\times2=32$, which is too small. So let's try $8$ and $8$, which gives $8\times3+8\times2=24+16=40$, which is too many again, so now try $4$ and $12$, giving $4\times3+12\times2=12+24=36$. That's too small. So we now know the number of tricycles must be an even number between $4$ and $8$. So it must be $6$, and the number of bicycles must be $10$.

Added 8/29/13: I can't resist adding a second solution which should also not be taken seriously as pedagogy for an elementary student but is, I think, worth pondering as a nice little exercise in logical analysis.

Make as many pairs of bicyclists and tricyclists as you can, leaving some number of unpaired cyclists. At this point we don't know whether the unpaired cyclists are on bikes or trikes, but, since the total number of cyclists ($16$) is an even number, we can say there is an even number of unpaired cyclists, who hence account for an even number of wheels. Now because the total number of wheels ($38$) is even, we can conclude there must be an even number of bike-trike pairs, since each pair accounts for an odd number ($2+3=5$) of wheels. This means the total number of paired cyclists is a multiple of $4$, which, since $16$ is also a multiple of $4$, means the number of unpaired cyclists is a multiple of $4$, so that the number of wheels they account for is a multiple of either $8$ (if they're on bikes) or $12$ (if they're on trikes).

Now let's get serious. Because there's an even number of paired cyclists, the number of wheels they account for is a multiple of $10$, ie. $10$, $20$, or $30$, leaving $28$, $18$, or $8$ wheels, respectively, for the unpaired cyclists. Neither $28$ nor $18$ is a multiple of either $8$ or $12$, leaving $8$ as the only possbility, which can only come from $4$ unpaired bicyclists and $6$ bike-trike pairs, for a total of $10$ bicyclists.

Whew!

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Graph theory in action here. –  Oh my ghost Aug 29 '13 at 2:20

If all students were using just a wheel, there would be 16 wheels.

If all students were using two wheels, they would be on bicycles, and there would be 32 wheels.

We have 38 wheels, so some students are using 3 wheels. 38 -32 = 6, so 6 students are using 3 wheels, and the remaining (16 - 6) will be using 2 wheels.

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7  
how does it differ from Robert Israel's answer? –  Oh my ghost Aug 28 '13 at 15:49
3  
step by step explanation, trying first with one wheel per student, since I don't how "elementary school" is the child... –  woliveirajr Aug 28 '13 at 16:01
    
I think most children no matter how young would know intuitively that everybody on bicycles would be less wheels than some on bicycles and some on tricycles. Also the question doesn't raise the possibility of a student only having one wheel so introducing it in the answer seems superfluous... –  Chris Aug 28 '13 at 18:36
    
Except that you don't need to multiply any variable by 2 or 3, you just keep adding up... Specially because the question asked for a "simpler method". Just adding up does this. –  woliveirajr Aug 28 '13 at 18:57

Marc's idea for dummies: $38/16=2\frac 38$. So we are $3/8$ of the way from all bicycles to all tricycles (the general idea of linearity should be known if one wants to follow this approach). $\frac 38\cdot 16=6$ tricycles.

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Don't discount guess-and-check.

Guess-and-check can be a valid way to solve problems like this because:

  • If he guesses right the first time, he will have solved the problem much faster than he would otherwise.
  • It will develop arithmetic skills.
  • Guess-and-check, i.e., brute force, generally suffers from being terribly inefficient. I'd argue that this will force him to do one of two things:

    1. Develop an intuition about the problem. In this case, he might notice the pattern mentioned by Mark Lakata about a linear progression. Beyond that he might think of doing a bisection, which all of a sudden becomes efficient and ceases to be brute-force. Or he might just solve the problem the algebraic way without even realizing it.

    2. He realizes the value in solving the problem mathematically.

    Either way, a lesson is learned.

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I have a better way to explain this to a elementary student. XD

Like the problem, I'll say there are 16 monsters, several of them are amphisbaenas (a snake have two heads) and the remains are cerberus (a dog have three heads). They all raise their heads and I'm the master.

After my first whistle, all monsters lower one head, that's of course 16 heads. And after my seconde whistle, all monsters lower one more head, that's 32 heads in total now.

So only the 6 cerberus still raise one head. And that's the answer.

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Sounds like: cut two heads from every monster (32) and count the remaining heads. But why are monsters with different heads better than vehicles? Might depend on the kids you are teaching ;-). (Note however that the question was "Find the number of students using bicycles.") –  Christian Fries Aug 29 '13 at 7:04
2  
After telling my monster story, I'll say "Now let's come to the bicycle problem..." XD –  DonPope Aug 29 '13 at 7:16
    
Actually I would do it straight in the original setup and make a shortcut: my magic whistle would grow a third wheel on each bicycle so I'll get $48-38=10$ bicycles at once and avoid any destructive thinking in the process as a bonus (though I admit that I also prefer to think in terms of chopping heads rather than in terms of growing wheels). However I agree that this is a nice explanation. –  fedja Aug 31 '13 at 15:17

Simplest

Every kid has at least two wheels, $16 \times 2 = 32$ in total. The remaining $38 - 32 = 6$ wheels belong to tricycles, so there are $16 - 6 = 10$ kids on bicycles.

Trial-and-error

Choose a random integer $x$ between $0$ and $16$ repeatedly until $2x + 3(16 - x)$ equals $38$. If the equality holds, $x$ is the solution. This can be done by hand, using dice or a random number table, but the easiest way is to write a computer program (see the code here).

Root finding

The number of kids on bicycles, $x$, is the root of the first order polynomial $f(x) = 2x + 3(16 - x) - 38 = 10 - x$. The latter form shows directly that the root is $10$.

It's worth deriving the general formula. Let $m$ and $n$ be the number kids and the number of wheels, respectively. With this notation, $f(x) = 2x + 3(m - x) - n = 3m - n - x$, that is, the root $x = 3m - n$.

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It looks like Bogo sort. –  Oh my ghost Aug 29 '13 at 16:42

Each student's cycle has at least two wheels. If we account for that, 38 - 2 * 16 = 6 students have a cycle with an additional wheel.

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Take variable $x$ for bicycles and $(16-x)$ for tricycles.

Now, consider the equation:

$$ 2x+3(16-x)=38$$

On solving this, we get $x=10.$ So, $10$ bicycles and 6 tricycles.

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OK. You just copied my question then. :-) –  Oh my ghost Aug 29 '13 at 4:57
1  
i just answered my own...from inbox, i saw the ques and directly commented. I didn't see ur ans otherwise I wouldn't hv answered it. sorry!! –  PrashantKalia Aug 29 '13 at 5:07

Consider the total wheel function $f$ of two variables $x$ the number of bicyclists and $y$ the number of "tricyclists". Since every time we add a bicyclist we increase the number of wheels by two, we have $$ f_x = 2. $$ Similarly for case of tricylists, $$ f_y = 3. $$ So we have the solutions $$ f(x,y) = 2x + A(y) \\ f(x,y) = 3y + B(x) $$ which give $$ f(x,y) = 2x + 3y + C. $$ However, $f(0, 0) = 0$ as we have no wheels when we have no riders, so $C = 0$.

Since the total number of cyclists is $16$, we are only interest in the portion of $f$ that lies on the line $y = 16 - x$, that is $f(x, 16 - x)$. Morever, we what to find when the total number of wheels on this line is $38$, so $$ \begin{aligned}f(x, 16 - x) & = 38 \\ 2x + 3(16-x) & = 38 \\ -x + 48 & = 38 \\ -x & = -10 \\ x & = 10. \end{aligned} $$ Thus the solution is the point $(x,y) = (10, 6)$.

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1  
Can you apply Lagrange multiplier or calculus of variation here to be more sophisticated? :-) –  Oh my ghost Aug 29 '13 at 13:15

Draw a linear graph with $x$ the number of trike riders and $y$ the number of wheels. Then just check the $x$ value at $y=38$.

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How to draw it without calculating some values ? –  Dominic Michaelis Aug 29 '13 at 9:37
    
@DominicMichaelis: You need to calculate both endpoints, of course. I left that out as a minor detail. ;-) –  Mike Hartl Aug 29 '13 at 9:40

16 (students) X 3 (tricycle wheels) - 38 (total number of wheels) = 10 (bicycles).

10 bicycles X 2 wheels = 20 wheels 6 tricycles x 3 wheels = 18 wheels

Number of students multiplied by the number of tricycle wheels minus the total number of wheels equals the number of bicycles.

:P

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We can answer that recursively ,for example: If we have an even number we will subtract the number by 2 and if the number is even we will subtract him by 3,basically that's how we think .

This way maybe isn't so good to solve all the question but it's good way to explain the situation intuitively ,and if you really want to see the recursive function I'll show you.

The recursive function:

So as I said before if we have bicycle we will subtract by 2 and if we have tricycle we will subtract by 3 , and from that we get our recursive function:

n represent the number of cycles and k the number of kids.

f(n,k)=f(n-2,k-1)+f(n-3,k-1)

Base cases:

f(1,3) =1 f(1,2) =1

For each other value we will get zero.

The answer will give us the number of ways,but this was a little demonstration of our mind solve problem of this kind.

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