A train with some passengers reached Station A and 8 more people boarded the train. At Station B, 2/5 of the passengers alighted. Then at Station C, 2/3 of the passengers alighted and 3 people boarded the train. There were 17 people in the train when it left Station C. How many passengers were there in the train at the start?
Let x be the number of passengers at the start.
After 8 people boarded at Station A, the number of passengers became x + 8.
At Station B, (2/5)(x + 8) passengers alighted, leaving (3/5)(x + 8) passengers on the train.
At Station C, (2/3)(3/5)(x + 8) passengers alighted, leaving (1/3)(3/5)(x + 8) passengers on the train.
After 3 people boarded at Station C, the number of passengers became (1/3)(3/5)(x + 8) + 3 = 17.
Multiplying through the parentheses, we get (3/5)(x + 8) + 3 = 17
Multiplying through the parentheses again, we get (3/5)(x + 8) = 14
Multiplying through by 5, we get 3(x + 8) = 70
Distributing, we get 3x + 24 = 70
Subtracting 24 from both sides, we get 3x = 46
Dividing both sides by 3, we get x = 46/3
Since we can't have a fraction of a passenger, there were 15 passengers in the train at the start. Answer: \boxed{15}.
Let's assume the number of passengers on the train at the start is "x".
At Station A, 8 more people boarded the train, so the number of passengers increased to "x + 8".
At Station B, 2/5 of the passengers alighted, so the number of passengers reduced to (x + 8) - (2/5)(x + 8).
Simplifying, we have:
(x + 8) - (2/5)(x + 8) = (3/5)(x + 8).
At Station C, 2/3 of the passengers alighted and 3 people boarded the train, so the number of passengers reduced to (3/5)(x + 8) - (2/3)(3) = 17.
Simplifying, we have:
(3/5)(x + 8) - 2 = 17.
Now, let's solve for x:
(3/5)(x + 8) - 2 = 17
Multiplying both sides by 5 to clear the fraction:
3(x + 8) - 10 = 85
Distributing 3:
3x + 24 - 10 = 85
Combine like terms:
3x + 14 = 85
Subtracting 14 from both sides:
3x = 71
Divide both sides by 3:
x = 71/3
Therefore, there were 71/3 passengers on the train at the start.