To solve the problem, we first need to establish how many people can ride the G-Force roller coaster in one complete trip.
Each train of the roller coaster has:
- 15 forward-facing seats
- 9 backward-facing seats
Calculating the total number of seats per ride:
\[ 15 + 9 = 24 \text{ total seats} \]
Next, we analyze how many people are in line for each direction:
- People in line to ride forward: 130
- People in line to ride backward: 100
Let \( f \) be the number of rides for the forward-facing seats, and \( b \) the number of rides for the backward-facing seats.
Each ride can accommodate:
- \( 15 \) forward-facing passengers per ride,
- \( 9 \) backward-facing passengers per ride.
Thus, the number of people served after \( f \) rides in the forward direction is:
\[ 15f \]
And the number served after \( b \) rides in the backward direction is:
\[ 9b \]
Initially, we have:
- Forward-facing passengers left: \( 130 - 15f \)
- Backward-facing passengers left: \( 100 - 9b \)
We want to find \( f \) and \( b \) such that the number of people remaining in both lines is the same. Therefore, we set up the equation:
\[ 130 - 15f = 100 - 9b \]
Rearranging the equation:
\[ 130 - 100 = 15f - 9b \] \[ 30 = 15f - 9b \]
We can simplify this equation by dividing everything by 3:
\[ 10 = 5f - 3b \]
Next, we can express \( b \) in terms of \( f \):
\[ 5f - 10 = 3b \] \[ b = \frac{5f - 10}{3} \]
Since \( b \) must be a non-negative integer, \( 5f - 10 \) must be a non-negative multiple of 3. From this, we can deduce that \( 5f - 10 \geq 0 \) leads to:
\[ 5f \geq 10 \implies f \geq 2 \]
Also, checking for integers that satisfy \( b \geq 0 \):
\[ 5f - 10 \equiv 0 \mod 3 \]
Calculating \( 5f - 10 \mod 3\):
\[ 5 \equiv 2 \mod 3 \implies 5f \equiv 2f \mod 3 \] Thus, we need: \[ 2f - 10 \equiv 0 \mod 3 \]
Calculating \( -10 \mod 3 \) gives:
\[ -10 \equiv 2 \mod 3 \]
So we have:
\[ 2f \equiv 2 \mod 3 \]
This simplifies to:
\[ f \equiv 1 \mod 3 \]
Hence, \( f \) can take values of \( 1, 4, 7, \ldots \), but since \( f \geq 2 \), we check values starting from \( f = 4 \):
- If \( f = 4 \):
\[ b = \frac{5(4) - 10}{3} = \frac{20 - 10}{3} = \frac{10}{3} \text{ (not an integer)} \]
- If \( f = 5 \):
\[ b = \frac{5(5) - 10}{3} = \frac{25 - 10}{3} = \frac{15}{3} = 5 \]
- If \( f = 6 \):
\[ b = \frac{5(6) - 10}{3} = \frac{30 - 10}{3} = \frac{20}{3} \text{ (not an integer)} \]
- If \( f = 7 \):
\[ b = \frac{5(7) - 10}{3} = \frac{35 - 10}{3} = \frac{25}{3} \text{ (not an integer)} \]
- If \( f = 8 \):
\[ b = \frac{5(8) - 10}{3} = \frac{40 - 10}{3} = \frac{30}{3} = 10 \]
Now we have \( f = 5 \) and \( b = 10 \). Thus, after 5 rides forward and 10 rides backward, the lines will have equal numbers of remaining people.
Finally, calculate remaining people:
For forward:
\[ 130 - 15 \times 5 = 130 - 75 = 55 \]
For backward:
\[ 100 - 9 \times 10 = 100 - 90 = 10 \]
After each completed ride, both lines will have the same number of passengers waiting, which occur at \( \text{number of rides} = 5 + 10 = 15 \).
Thus, the answer is:
\[ \boxed{15} \] rides.
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