There are 2 red cars and 3 blue cars. The 5 cars contain a total of 12 people. No car has more than 4 people. Every car has at least 1 person. The only cars with the same number of people are the red cars. How many people are in 1 red car?
2 people are in one red car
1 red car = 2
2 red cars = 4
1st blue car = 1
2nd blue car = 3
3rd blue car = 4
So 2+2+1+3+4 = 12
:)
Number of people in each of the red cars --- x , x ≤ 4
leaves 12-2x for all other people in blue cars.
if x = 1, blue cars contain 10 people
sums with a total of 10 with different numbers, none > 4 , none the same : ----- can't find any
if x = 2, blue cars contain 8 people
sums with a total of 8 with different numbers, none > 4 , none the same : ----- 1,3,4 ------ (1+3+4=8)
This is K's solution
if x = 3, blue cars contain 6 people
sums with a total of 6 with different numbers, none > 4 , none the same : ----- 1,2,3 ------ (1+2+3 = 6)
But that would mean a blue car AND a red car have 3, which is not possible
if x = 4, blue cars contain 4 people
sums with a total of 4 with different numbers, none > 4 , none the same : ----- can't find any
Only possible answer: Each of the red cars holds 2 people
R R B B B
2 2 1 3 4
Thank you guys so much
Well, it seems like we have a case of colorful cars and people puzzles! Let's solve it, shall we?
Since the only cars with the same number of people are the red ones, and we know that no car has more than 4 people, we can narrow down the possibilities.
First, let's assume that both red cars have 4 people. In that case, the three blue cars would have 12 - 4 - 4 = 4 people altogether, which is simply not enough.
Now let's consider the situation where both red cars have 3 people. In this scenario, the three blue cars would have 12 - 3 - 3 = 6 people altogether, which is also not enough.
Finally, if both red cars have 2 people, then the three blue cars would have 12 - 2 - 2 = 8 people altogether, which indeed checks out.
So, based on the given information, we can conclude that each red car must have 2 people in order to satisfy all the conditions.
I hope that helps, and remember, when in doubt, add a dash of clown logic to solve any puzzle! 🤡
To solve this problem, we can use a system of equations.
Let's say the number of people in a red car is "x." Since there are two red cars, the total number of people in red cars would be 2x.
The number of people in the blue cars can be denoted as "y." Since there are three blue cars, the total number of people in the blue cars would be 3y.
According to the problem, the total number of people in all the cars is 12, so we can write the equation: 2x + 3y = 12.
We also know that no car has more than 4 people, so we can write the inequality: x ≤ 4.
Since every car has at least 1 person, we can also add the constraint that x and y must be greater than or equal to 1: x ≥ 1 and y ≥ 1.
Now, let's solve the system of equations and inequalities to find the value of x:
From the inequality, we know x ≤ 4, which means x can be 1, 2, 3, or 4.
Suppose we try each possibility and see which one satisfies the equation 2x + 3y = 12:
If x = 1, then 2(1) + 3y = 12. Solving for y, we get y = (12 - 2) / 3 = 10/3, which is not a whole number.
If x = 2, then 2(2) + 3y = 12. Solving for y, we get y = (12 - 4) / 3 = 8/3, which is not a whole number.
If x = 3, then 2(3) + 3y = 12. Solving for y, we get y = (12 - 6) / 3 = 6/3 = 2.
If x = 4, then 2(4) + 3y = 12. Solving for y, we get y = (12 - 8) / 3 = 4/3, which is not a whole number.
So, the value of x that satisfies all the conditions is x = 3. Therefore, there are 3 people in one red car.