Martin had red toy cars and blue cars in a box. the ratio of the blue toy cars to the red toy cars is 5:2. When Martin 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 toy cars are now in the box?

Question

Martin had red toy cars and blue cars in a box. the ratio of the blue toy cars to the red toy cars is 5:2. When Martin 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 toy cars are now in the box?

Answer 1

see

http://www.jiskha.com/display.cgi?id=1473791919

same problem, different numbers.

Answer 2

To solve this problem, we can start by setting up equations based on the given information.

Let's say there are "x" red toy cars in the box and "y" blue toy cars in the box initially.
According to the information given:
The ratio of blue toy cars to red toy cars initially is 5:2, so we have y/x = 5/2.

When Martin put 15 more blue toy cars in the box, the new ratio of blue toy cars to red toy cars became 4:1. This means we now have (y+15)/x = 4/1.

Now we can solve these two equations to find the values of x and y.

From the first equation, we have y/x = 5/2. Rearranging it, we get y = (5/2)x.

Substituting this value of y in the second equation, we have (5/2)x + 15 = (4/1)x.

Multiplying the equation by 2 to eliminate the fraction, we get 5x + 30 = 8x.

Bringing the variables to one side, we have 8x - 5x = 30.

Simplifying, we find 3x = 30.

Dividing both sides by 3, we have x = 10.

Substituting this value of x in the first equation, we can find y:

y = (5/2)x
y = (5/2) * 10
y = 25

So, there are 10 red toy cars and 25 blue toy cars in the box now.