Solve the following problem. The owner of a toy store is taking inventory of bicycles and tricycles in an unusual way. He said he counted 50 wheels and 40 pedals.How many bicycles and how many tricycles did he have?
wheels = 3 t + 2 b = 50
pedals = 2 t + 2 b = 40
so subtract second from first
1 t + 0 b = 10
t = 10
then b = 10
check
3 t = 30
2 b = 20 so 50 wheels
2 t = 20
2 b = 20 so 40 pedals
To solve this problem, let's assign variables to the number of bicycles and tricycles. Let's say the number of bicycles is represented by 'b' and the number of tricycles is represented by 't'.
Since each bicycle has 2 wheels and 2 pedals, and each tricycle has 3 wheels and 2 pedals, we can create the following equations based on the information given:
Number of wheels: 2b + 3t = 50
Number of pedals: 2b + 2t = 40
To find the values of 'b' and 't', we can solve this system of equations.
First, let's multiply the second equation by 3 to eliminate 't':
6b + 3t = 120
Now we have the following equations:
2b + 3t = 50
6b + 3t = 120
By subtracting the first equation from the second equation, we can solve for 'b':
(6b + 3t) - (2b + 3t) = 120 - 50
4b = 70
b = 70/4
b = 17.5
Since we cannot have a fraction of a bicycle, we'll round 'b' to the nearest whole number. So, the owner has 18 bicycles.
Now, substitute the value of 'b' back into one of the original equations to solve for 't':
2(18) + 2t = 40
36 + 2t = 40
2t = 40 - 36
2t = 4
t = 4/2
t = 2
So, the owner has 2 tricycles.
Therefore, the owner has 18 bicycles and 2 tricycles.