3. Overcrowded carriages. The histogram (Fig. 1) shows the distribution of passenger carriages by the number of passengers (in $4 \%$ of carriages, passengers are from 10 to 19, in $6 \%$ of carriages - from 20 to 29, etc.). If a carriage has 60 passengers or more, it is called overcrowded.

a) (for 6th grade, 1 point). Find the share of overcrowded carriages.

b) (for 6th grade, 2 points). Find the smallest possible share of passengers traveling in overcrowded carriages.

c) (for 6th grade, 2 points). Can the share of passengers traveling in overcrowded carriages be less than the share of overcrowded carriages?

![](https://cdn.mathpix.com/cropped/2024_05_06_142b35c881fc974fcd25g-02.jpg?height=1060&width=1670&top_left_y=204&top_left_x=270)
Answer: a) $40 \%$; b) approx. $49 \%$; c) cannot.

Solution. b) To make the share of passengers in overcrowded cars as small as possible, there should be as few passengers as possible in them, and as many as possible in non-overcrowded cars. Let the total number of cars be $N$. We find the maximum possible number of passengers in non-overcrowded cars:

$$
19 \cdot 0.04 N + 29 \cdot 0.06 N + 39 \cdot 0.12 N + 49 \cdot 0.18 N + 59 \cdot 0.2 N = 27.8 N ;
$$

and then the minimum possible number of passengers who traveled in overcrowded cars:

$$
60 \cdot 0.2 N + 70 \cdot 0.14 N + 80 \cdot 0.06 N = 26.6 N
$$

The share of passengers in overcrowded cars is

$$
\frac{26.6 N}{26.6 N + 27.8 N} = \frac{266}{544} \approx 0.49
$$

c) Suppose there are $N$ cars in total, and $n$ passengers are traveling in them, with $M$ of the cars being overcrowded, and $m$ passengers in them. Then on average, there are $\frac{n}{N}$ passengers per car, and $\frac{m}{M}$ passengers per overcrowded car, and the second number is not less (greater, if there is at least one non-overcrowded car) than the first:

$$
\frac{n}{N} \leq \frac{m}{M} \text{, hence } \frac{M}{N} \leq \frac{m}{n} \text{. }
$$