SOLUTION: I do not quite understand. My math equation says: Mike and his friends took their tricycles and wagons to the playground. Mike counted 36 wheels in all. How many tricycles and how

Question 573667: I do not quite understand. My math equation says: Mike and his friends took their tricycles and wagons to the playground. Mike counted 36 wheels in all. How many tricycles and how many wagons might there be at the playground? Write as many different possibilities as you can. Help!

Found 2 solutions by bucky, richard1234:
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
It may be obvious to you, but let's establish the rule that a tricycle has 3 wheels and a wagon has 4.
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Let's also establish that T represents the number of tricycles and W represents the number of wagons. If you multiply T by 3 that will be the number of wheels you have due to tricycles, and if you multiply W by 4 that will be the number of wheels due to wagons.
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The problem tells you that the total number of wheels is 36.
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Therefore we can write the equation:
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3T + 4W = 36 (the number of tricycle wheels plus the number of wagon wheels = 36)
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Let's recognize that the maximum number of wagons is 9. How do we know that? Because if you have 9 wagons, then you have 9 times 4 = 36 and all 36 wheels are on wagons. If you have 10 wagons, then you have 40 wagon wheels and that's not allowed. The maximum number of wheels is 36.
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Similarly, the maximum number of tricycles is 12. We know that because 12 tricycles will have a total of 36 wheels, the maximum allowed.
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Now let's solve our equation for one of the variables in terms of the other. Let's say we solve for T in terms of W by doing the following:
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Start with:
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3T + 4W = 36
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Subtract 4W from both sides to get:
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3T = 36 - 4W
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Solve for T by dividing both sides (all terms) by 3 to get:
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T = 12 - (4W)/3
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Now we can just assume values for the W, the number of wagons and see what this results in for T. Remember, we decided that there are at most 9 wagons, so we need to check for values of W equal to 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. All we need to do is substitute those values into our equation and see which of them result in a whole number for the number of tricycles (not a mixed number involving a fractional part of a tricycle. That wouldn't make sense.)
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So let's start:
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If W = 0 then (4W)/3 = 0 & T will be equal to 12 - 0 = 12. That works.
If W = 1 then (4W)/3 = 4/3 & T will be equal to 12 - 4/3 = 10&2/3. Won't work.
If W = 2 then (4W)/3 = 8/3 & T will be equal to 12 - 8/3 = 9&1/3. Won't work.
If W = 3 then (4W)/3 = 12/3 & T will be equal to 12 - 12/3 = 8. That works.
If W = 4 then (4W)/3 = 16/3 & T will be equal to 12 - 16/3 = 6&2/3. Won't work.
If W = 5 then (4W)/3 = 20/3 & T will be equal to 12 - 20/3 = 5&1/3. Won't work.
If W = 6 then (4W)/3 = 24/3 & T will be equal to 12 - 24/3 = 4. That works.
If W = 7 then (4W)/3 = 28/3 & T will be equal to 12 - 28/3 = 2&2/3. Won't work.
If W = 8 then (4W)/3 = 32/3 & T will be equal to 12 - 32/3 = 1&1/3. Won't work.
and finally,
If W = 9 then (4W)/3 = 36/3 & T will be equal to 12 - 36/3 = 0. That works.
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So we found that the possible combinations are:
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zero wagons and 12 tricycles
3 wagons and 8 tricycles
6 wagons and 4 tricycles
9 wagons and zero tricycles
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Each of these four possibilities will result in a total of 36 wheels.
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You could probably reason your way through this problem as fast as developing the equation as above. First recognize that the maximum number of tricycles is 12 because that would account for all 36 wheels. Then recognize that the number of tricycles must be an even amount because an odd number of tricycles will create an odd number of wheels and when you subtract an odd number of wheels from 36 you would end up with an odd number of wheels left for wagons. But wagons require an even number of wheels.
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So the possible numbers of tricycles would be 0, 2, 4, 6, 8, 10, and 12. Then all you have to do is for each of these 7 numbers, multiply by 3 wheels, subtract the answer from 36 and see if that result is exactly divisible by 4. If it is, then it is one of the answers.
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For example, let's try 2 tricycles. That means 3 times 2 or 6 wheels for tricycles. Subtract those 6 wheels from 36 and you have 30 wheels left for wagons. But that answer, 30 wheels, is not exactly divisible by 4, so you don't get a whole number of wagons. Therefore, 2 tricycles is not a good number.
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Now let's try 4 tricycles. That means 12 wheels for tricycles and that leaves 24 wheels for wagons. 24 wagon wheels is 6 wagons exactly, and so 4 tricycles and 6 wagons gives you 36 wheels.
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You can work out the remaining 5 numbers for the possible number of tricycles and you will see that you get the same four possible solutions as we did with the equation.
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I hope this helps you to understand the problem you were having difficulty with.
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Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Let t and w be the number of tricycles and wagons, respectively. Assuming a tricycle has 3 wheels and a wagon has 4 wheels, we have

This type of equation is called a "linear Diophantine equation." There are several techniques for solving them, but this one is pretty straightforward. Here, we can solve for t:

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Obviously, we want t and w to be nonnegative integers. Therefore 4w/3 must be an integer, which implies that w is a multiple of 3. We can list all of the possible solutions:

w = 0, t = 12
w = 3, t = 8
w = 6, t = 4
w = 9, t = 0

These are all of the possibilities. If there were tricycles *and* wagons, then the solutions w = 0, t = 12 and w = 9, t = 0 can be ignored.
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