Question

Roy had 46 more toy cars than Wayne. If Wayne and Roy were to give away 3/5 and 6/7 of their toy cars respectively, Roy would have twice as many toy cars left as Wayne. How many toy cars did Roy have?

Answer 1

Let's assume Wayne has "x" toy cars.

Given that "Roy had 46 more toy cars than Wayne," we can say that Roy has "x+46" toy cars.

Now, if Wayne gives away 3/5 of his toy cars, he would be left with (2/5)x toy cars.

And if Roy gives away 6/7 of his toy cars, he would be left with (1/7)(x+46) toy cars.

Given that "Roy would have twice as many toy cars left as Wayne," we can write the following equation:

(1/7)(x+46) = 2(2/5)x

Simplifying the equation:

(x+46)/35 = 4x/5

Cross-multiplying the equation:

5(x+46) = 140x/5

5x + 230 = 28x

Subtracting 5x from both sides:

230 = 23x

Therefore, x = 10

Thus, Wayne has 10 toy cars and Roy has x+46 = 56 toy cars.

Answer 2

Let's start by using algebra to solve the problem.

Let's call the number of toy cars that Wayne has "W" and the number that Roy has "R".

From the first sentence, we know that:

R = W + 46

Next, we know that if Wayne gives away 3/5 of his toys and Roy gives away 6/7 of his toys, then Roy will have twice as many toys left as Wayne.

We can set up an equation to represent this:

2(W - 3/5W) = R - 6/7R

Simplifying this equation:

2/5W = 1/7R

We can use the first equation (R = W + 46) to substitute for R in the second equation:

2/5W = 1/7(W + 46)

Simplifying this equation:

14/5W = W + 46

9/5W = 46

W = 23/5

We know that W must be a whole number (since we can't have a fraction of a toy car), so we can round up to the nearest whole number:

W = 5

Using the first equation (R = W + 46), we can solve for R:

R = 5 + 46

R = 51 + 5 = 56

Therefore, Roy has 56 toy cars.