In a certain aquarium there began to become a visitors' line before opening and number of visitors are increasing in the line at a fixed rate per minute.
If three windows of admission ticket sale are opened simultaneously at the time of opening of the aquarium, the visitors' line will disappear in 15 minutes.
If six windows are opened, the line will disappear in 5 minutes.
If three windows of admission ticket sale are opened simultaneously at the time of opening of the aquarium, the visitors' line will disappear in 15 minutes.
If six windows are opened, the line will disappear in 5 minutes.
From how many minutes before opening did visitors begin to become in a line ?
The number of the person who the admission ticket were sold to at three windows in 15 minutes is <1> × 3 × 15 = <45> persons.
Considering the difference of <45> and <30>, it is found that in 15 - 5 =10 minutes <45> - < 30> = <15> persons joined the line.
In one minute, <15> / 10 = <1.5> persons joined the line.
Answer
15 minutes before opening
Solution
Suppose that an admission ticket is sold to <1> person in one minute at the one window.
The number of the person who the admission ticket were sold to at three windows in 15 minutes is <1> × 3 × 15 = <45> persons.
The number of the persons who the admission ticket were sold to at six windows in 5 minutes is <1> × 5 × 6 = <30> persons.
Considering the difference of <45> and <30>, it is found that in 15 - 5 =10 minutes <45> - < 30> = <15> persons joined the line.
In one minute, <15> / 10 = <1.5> persons joined the line.
The number of people that stood in the line at the time of opening is (<6> - <1.5>) × 5 = 22.5 persons.
Time to take that people forms a line is <22.5> / <1.5> = 15 minutes.
