Part A: each tricycle has three wheels, so with 48 wheels the number of tricycles was a =48/3=16 tricycles.
t=w/3 (the number of tricycles is the number of wheels divided by 3)

Part B:
The number of seats:
24=b+a (so b=24-a)
The number of seats is the sum of one seat per bicycle and one seat per a tricycle

also, 61=2a+3b (the number of wheels)

So we have:
24=b+a
b=24-a
We can substitute this for b:

61=2a+3(24-a)

and solve:
61=2a+3*24-3a
61=72-a
a=72-61
a=11

There were 11 bicycles!!
and there were 24-11 tricycles, so 13 tricycles.

Part C: each of the bikes has only one front-steering handlebar, so there were a total of 144 vehicles:

a+b+c=144

There were 378 pedals. And the number of pedals is:
2a+2b+4c=378 (the numbers 2,2,4 represent the number of pedals per vehicle)

divide by 2:
a+b+2c=189

Now, we have
a+b+2c=189
and

a+b+c=144
and we can subtract them from each other:
a+b+c-(a+b+2c)=144-189
-c=45
c=45, so there were 45 tandem bicycles!
(this also means that a+b=144-45, that is a+b=99)
now the wheels:
3a+2b+2c=320
Let's substitute c:
3a+2b+90=320

which is
3a+2b=240
We also know that a+b=99, so we can substract this from this equation:
3a+2b+-a-b=240-99
2a+b=141

and again:
2a+b-a-b=141-99
a=42 - there were 42 trycicles!!!

And the bicycles were the rest:
99-42=57 bycicles

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2 years ago