5. After rechecking the tickets of all passengers in three carriages, conductor Mirko concluded: If 55 more passengers had squeezed into the first carriage, then the number of passengers in the first carriage would be the same as in the second and third together. If 33 more passengers had squeezed into the second carriage, then the number of passengers in the second carriage would be the same as in the first and third together. The number of passengers in the first carriage is less than a quarter of the number of passengers in the third carriage. What is the maximum number of passengers that could have been in these three carriages?

No use of a pocket calculator or any manuals is allowed.

## Ministry of Science and Education of the Republic of Croatia  Agency for Education and Upbringing  Croatian Mathematical Society

## COUNTY COMPETITION IN MATHEMATICS March 1, 2023  7th grade - elementary school

Each task is worth 10 points. In addition to the final result, the procedure is also scored. To earn all points, it is necessary to find all solutions and determine that there are no others, write down the procedure, and justify your conclusions.
First solution.

Obviously, the number of passengers in the 1st carriage is less than the number of passengers in the second carriage.

Let's graphically represent the number of passengers in the 1st, 2nd, and 3rd carriages, noting that for this representation, it doesn't matter whether the number of passengers in the 3rd carriage is greater or smaller than the number in the larger of the remaining two.
I.

II. $\qquad$
III. - $\qquad$
$1 \text{ POINT}$

In accordance with the labels, it holds:
![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-23.jpg?height=258&width=1606&top_left_y=656&top_left_x=228)

This means that:

![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-23.jpg?height=189&width=1130&top_left_y=982&top_left_x=226)

from which we get:

![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-23.jpg?height=188&width=677&top_left_y=1314&top_left_x=244)

or

$$
(55+33): 2=44
$$

There were 44 passengers in the third carriage.

2 POINTS

Now we get:
![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-23.jpg?height=190&width=1322&top_left_y=1690&top_left_x=228)

$1 \text{ POINT}$

Since the number of passengers in the first carriage is less than $\frac{1}{4}$ of 44, which is 11, we conclude that there were at most 10 passengers in the first carriage.

This means that there were at most $10+11=21$ passengers in the 2nd carriage.

In these three carriages, there could have been at most $44+21+10=75$ passengers.