Sixty-eight people are sitting in 20 cars and each car contains at mos

let, the maximum possible number of cars that could contain exactly 1 be x
so,
(20-x)4+x=68
calculated the equation
we get x=4

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If 8 cars contain 1 person each, then we have 68 - 8 = 60 people for the remaining 20 - 8 = 12 cars. This makes an average of 60/12 = 5 people per car, which is also impossible.

C. 4

If 4 cars contain 1 person each, then we have remaining 68 - 4 = 64 people for remaining 20 - 4 = 16 cars. This makes an average of 64/16 = 4 people per car, which is possible.

So the maximum number of cars that could contain exactly 1 person each is 4.

Alternate Solution:

To maximize the number of cars with only one person, let’s suppose all the remaining cars contain 4 people. Let n denote the number of cars with only one person. Then, the remaining 20 - n cars contain 4(20 - n) people. We are given that the 4(20 - n) people together with the n people each of whom ride alone must add up to 68; therefore we have:

4(20 - n) + n = 68

80 - 4n + n = 68

12 = 3n

n = 4

So the maximum number of cars that could contain exactly 1 person each is 4.

Answer: C

So each car contains at most 4 people

this means that, y = possible number of cars that could contain exactly 1

y*1 + (20-y) * 4 = 68

We can directly plug in from C

4 + 16 * 4 = 68

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