Mark had red toy cars and blue toy cars in a box. The ratio of the blue toy cars to the red toycars was 5:2. when Mark put 15 more blue toy cars in the box, the ratio of the blue toy cars to the red toy cars became 4:1. How many red toy cars and blue t0y cars are now in the box?
I've done two of these. See what you can do with this one.
To solve this problem, we'll need to use algebra. Let's use the variable "x" to represent the initial number of red toy cars in the box.
According to the given information, the ratio of blue toy cars to red toy cars was initially 5:2. This means that for every 5 blue toy cars, there were 2 red toy cars. So the initial number of blue toy cars can be represented as "5/2 * x".
When Mark added 15 more blue toy cars, the ratio of blue toy cars to red toy cars became 4:1. This means that for every 4 blue toy cars, there is now 1 red toy car. So the new number of blue toy cars can be represented as "(5/2 * x) + 15".
We can now set up an equation based on the given information:
(5/2 * x + 15) / x = 4/1
To solve this equation, we'll need to cross-multiply:
5 * x + 30 = 8 * x
30 = 8 * x - 5 * x
30 = 3 * x
x = 10
So the initial number of red toy cars, represented by "x", is 10. Plugging this value into our expression for the initial number of blue toy cars, we get:
5/2 * 10 = 25
Therefore, initially, there were 10 red toy cars and 25 blue toy cars in the box.