## Bicycle Puzzle

This week’s puzzle from Chris Smith is a nice contextualised simultaneous equations puzzle I intend to use next week.

It boils down to:

2X + B + T = 135

2X + 2B + 3T = 269

X + B + T = 118

Via elimination using 1 and 3 we can see that X (number of tandems) is 17.

That gives:

34 + B + T = 135

34 + 2B + 3T = 269

17 + B + T = 118

1 and 3 rearrange to the same leaving:

B + T = 101

2B + 3T = 235

Elimination gives T (number of Tricycles) to be 33 which leave B (number of Bicycles) to be 68.

A nice little problem.

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Just a thought:What does # of seats – # of front handlebars represent?

Bikes have the same number of each, as do trikes; but the tandems have an extra seat (2) for each set of front handlebars (1). When you carry out this subtraction, then, you are left with exactly the # of tandems. (Take a moment to pause with this observation, if it isn’t manifestly clear.)

So we can immediately conclude:

# of tandems = 135 – 118 = 17.

These 17 tandems account for 34 seats, 34 wheels, and 17 front handlebars. Let’s remove them from the problem entirely.

Tandemless problem:Givenonlybikes and trikes, there are 101 seats, 235 wheels, and 101 front handlebars.In this case, we ask:

What does # of wheels – twice the # of seats represent?

For bikes, these numbers are the same; for each trike, there is an extra 1 left over. (Again: Pause if this is unclear!)

In particular, we can see that:

# of trikes = 235 – 2(101) = 235 – 202 = 33.

And since the total number of seats for bikes & trikes is 101, and each has exactly one seat, we compute:

# of bikes + # of trikes = 101, i.e.,

# of bikes + 33 = 101, i.e.,

# of bikes = 68.

All of this can be translated into algebraic notation, but it’s a bit interesting (I think) to interpret the problem in this more “arithmetical” manner.

Final answer for # of tandems, bikes, and trikes, respectively: 17, 68, and 33.

(This answer indicates the one at which you arrived suffers from a slight error/typo; I’m sure you can spot it.)

I really must start proof reading! And I love your algebra free version!

The “algebra free version” occurs to me because I saw (about 7 years ago) a problem in which the number of wheels and pedals are given for a collection of bikes and trikes. As always, you can write out various equations and solve using algebra. The “arithmetical” approach is to observe:

# of wheels – # of pedals = # of trikes.

Once you know the # of trikes, figuring out the # of bikes is not too tough.

Similar problems can be constructed for, e.g., unicycles and bicycles given the # of seats and # of wheels.

Even the classic problems around animals can be solved in this way. For example, consider:

The park has people and dogs. The total number of heads is 25, and the total number of legs is 80. How many people (resp. dogs) are there altogether?

Probably you have seen this sort of problem before. But let us modify it by considering

leftlegs. Why?Because now the problem is:

25 heads

40 left legs

From here, note that each person has 1 head and 1 left leg, whereas each dog has 1 head and 2 left legs.

So:

# of left legs – # of heads = # of dogs.

That is, # of dogs = 40 – 25 = 15.

Since everyone has just one head, we also have:

# of dogs + # of people = 25, i.e.,

15 + # of people = 25, i.e.,

# of people = 10.

Algebraically, this is equivalent to taking the equations:

P + D = 25 and 2P + 4D = 40,

and dividing the latter equation by 2 (this is “considering only left legs”) then subtracting it from the former equation to find the number of dogs. Finally, one reuses the first equation to find the number of people.

Similar approaches solve most problems posed within this category.

If the number of front handlebars is 118, then the total number of cycles must be 118. I agree that there are 17 tandems and 33 trikes but not 33 bikes. Surely there should be 68 bikes?

Yes, a dilly typo. Have ammend!