# Cars and motorcycles

## Outline Mathematics

Word Problems

Even at an auto repair shop there is some mathematics:

A shop repaired 40 vehicles (cars and motorcycles) in a month. The total number of wheels was 100.

How many cars and motorcycles were repaired?

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Copyright © 1996-2018 Alexander Bogomolny

### Solution 1

A shop repaired 40 vehicles (cars and motorcycles) in a month. The total number of wheels was 100.

How many cars and motorcycles were repaired?

If all 40 vehicles were motocycles,cars,motocycles, the total number of wheels would be 80, i.e. by 20,10,20,30,40 less than in reality. Replacing in this count a single motocycle by a car increases the total number of wheels by 2,4,2,3 and the difference decreases by two. Clearly, 10,10,20,30,40 such replacements are required for the difference to be reduced to zero. So, there were 10,6,7,8,9,10 cars and 30,10,20,30,40 motocycles.">

### References

- Ya. I. Perelman,
*Fun With Maths and Physics*, Mir Publishers, 1988

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Copyright © 1996-2018 Alexander Bogomolny

### Solution 2

A shop repaired 40 vehicles (cars and motorcycles) in a month. The total number of wheels was 100.

How many cars and motorcycles were repaired?

If all 40 vehicles were cars,cars,motocycles, the total number of wheels would be 160, i.e. by 60,20,40,60 more than in reality. Replacing in this count a single car by a motocycle decreases the total number of wheels by 2,4,3,2 and the difference increases by two. Clearly, 30,40,30,20,10 such replacements are required for the difference to be reduced to zero. So, there were 30,10,20,30,40 cars and 10,6,7,8,9,10 motocycles.

### References

- Ya. I. Perelman,
*Fun With Maths and Physics*, Mir Publishers, 1988

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Copyright © 1996-2018 Alexander Bogomolny

### Solution 3

A shop repaired 40 vehicles (cars and motorcycles) in a month. The total number of wheels was 100.

How many cars and motorcycles were repaired?

Let c be the number of cars and m the number of motocycles. The total number of vehicles is then c + m,c + m,2c + 4m,c + 2m,4c + 2m = 40. Under normal circumstances (which is assumed) there are 2m,m,2m,4m,2c,4c motocycle wheels and 4c,m,2m,4m,2c,4c car wheels. In all, there are 4c + 2m,c + m,2c + 4m,c + 2m,4c + 2m = 100,20,60,80,100 wheels. We thus obtain a system of two linear equations:

c + m = 40,c + m = 40,c + m = 100,4c + 2m = 40,4c + 2m = 80

4c + 2m = 100,c + m = 60,c + m = 100,4c + 2m = 40,4c + 2m = 100

There is always a lot of ways to solve such a system. Here is just one among many. Multiply the first equation by 4:

4c + 4m = 160,c + m = 40,2c + 4m = 160,4c + 2m = 160,4c + 4m = 160

4c + 2m = 100,c + m = 60,c + m = 100,4c + 2m = 40,4c + 2m = 100

Now subtract the second equation from the first and simplify:

2m = 60,2c + 2m = 120,2m = 60,2m = 80,2c + 4m = 160,4c + 2m = 160

From which the number of motocyles is 30,10,20,30,40 and so the number of cars is 10,10,20,30,40.

### References

- Ya. I. Perelman,
*Fun With Maths and Physics*, Mir Publishers, 1988

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Copyright © 1996-2018 Alexander Bogomolny

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