Question

Retro Rides is a club for owners of vintage cars and motorcycles. Every year the club gets together for a ride. This year, 38 vehicles participated in the ride. The total number of tires of all the vehicles was 104. Assuming each car has 4 tires and each motorcycle has 2 tires, how many each of cars and motorcycles participated in the ride?

A. 18 cars; 20 motorcycles
B. 14 cars; 24 motorcycles
C. 11 cars; 27 motorcycles
D. 16 cars; 22 motorcycles

Answer 1

Answer: OPTION B

Step-by-step explanation:

Let's call:

c: the number of cars.

m: the number of motorcycles.

Based on the given information, you can set up the following system of equations:

$\left \{ {{c+m=38} \atop {4c+2m=104}} \right.$

You can solve it by the Elimination method:

- Multiply the first equation by -4.

- Add both equations to cancel out the variable "c".

- Solve for "m":

$\left \{ {(-4)(c+m)=38(-4)} \atop {4c+2m=104}} \right.\\\\\left \{ {-4c-4m=-152} \atop {4c+2m=104}} \right.\\-------\\-2m=-48\\m=24$

(24 motorcycles)

 - Substitute m=24 into any of the original equations ans solve for "c". Then:

$c+24=38\\c=38-24\\c=14$

(14 cars)

Answer 2

Hello!

The answer is:

Option B.

$Cars=14\\Motorcycles=24$

Why?

We know that the total of tires of all the vehicles was 104, and there were 38 vehicles with 4 tires and motorcyles with 2 tires, so, we can calculate how many cars and motorcyles participated in the ride using the following equation:

Let x be the cars and y be the motorcyles, so:

$x+y=38$

and,

$4x+2y=104$

So, isolating x in terms of y from the first equation, we have:

$x=38-y$

Then, substituting "x" into the second equation we have:

$4(38-y)+2y=104\\152-4y+2y=104\\152-2y=104\\2y=152-104\\2y=48\\y=\frac{48}{2}=24$

Now, substituting "y" into the first equation, we have:

$x+24=38$

$x=38-24=14$

So, there were a total of 14 cars and 24 motorcycles in the ride.

Have a nice day!