A group of friends went to Mudville to watch the Mudville Sluggers play baseball. On the way to the game, three people rode in each car and seven took the train. On the way home, four people rode in each car and two people took the train. The same number of cars and people traveled to and from the game. How many friends could have been in the group? How many cars were used? (please show your work) :)
let the number of cars be x
1st trip:
number of people = 3x + 7
return trip:
number of people = 4x + 2
4x + 2 = 3x + 7
x = 5
So there were 5 cars
the number of people was 3(5) + 7 = 22
a sack contain 10 blue and 6 red crayons another sack contains 9 blue and 12 red crayons two crayons are drown at random without replacement from each sack. what is the probability that all the four crayons are blue? exactly one of the four crayons is red
To solve this problem, let's assume the number of friends in the group is "x" and the number of cars used is "c."
On the way to the game:
- Three people rode in each car, so the number of cars can be expressed as "c = x / 3" (as 3 people were in each car).
- Seven people took the train.
On the way home:
- Four people rode in each car, so the number of cars can be expressed as "c = x / 4" (as 4 people were in each car).
- Two people took the train.
Since the same number of cars and people traveled to and from the game, we can set the two expressions for "c" equal to each other:
x / 3 = x / 4
To simplify the equation, we can cross multiply:
4x = 3x
Subtracting 3x from both sides:
4x - 3x = 0
x = 0
This implies that the number of friends in the group is zero, which is not possible. Therefore, there is no valid solution to this problem.