DIRECTIONS for questions: Read the information given below and answer the question that follows.

A train travels from Station A to Station E, passing through stations B, C, and D, in that order. The train has a seating capacity of 200. A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station.

A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B – C, C – D, and D – E.

The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200.

The following information is known.

1. Segment C – D had an occupancy factor of 95%. Only segment B – C had a higher occupancy factor.

2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.

3. Among the seats reserved on segment D – E, exactly four-sevenths were from stations before C.

4. The number of tickets booked from A to C was equal to that booked from A to E, and it was higher than that from B to E.

5. No tickets were booked from A to B, from B to D and from D to E.

6. The number of tickets booked for any segment was a multiple of 10.

How many tickets were booked from Station C?

The correct answer is 80.

Let's define the segments of the route as A-B, B-C, C-D, and D-E. The seating capacity is 200. We can represent the number of tickets booked from station i to station j as T_ij.

From the given rules, we know the following direct values and relations:

• Rule 5 states: T_AB = 0, T_BD = 0, and T_DE = 0.
• Rule 2 states: T_BC = 40 and T_BE = 30.
• Rule 4 states: T_AC = T_AE. It also states T_AC > T_BE, so T_AC > 30. Let's denote T_AC = T_AE = x.
• Let T_AD = y.
• Let T_CD = p and T_CE = q.
• Rule 6 states all ticket numbers are multiples of 10.

Next, let's analyze the occupancy factors for the segments. The capacity is 200. From Rule 1, segment C-D had an occupancy of 95%, which is $0.95\times 200=190$ seats. It also states only segment B-C had a higher occupancy factor. Since ticket counts are multiples of 10, the occupancy of B-C must be exactly 200 seats (100% capacity).

The number of seats reserved on segment B-C is the sum of all tickets that pass through it: T_AC + T_AD + T_AE + T_BC + T_BD + T_BE. Setting this equal to 200:

$x+y+x+40+0+30=200$

$2x+y=130$

Since x is a multiple of 10 and x > 30, the possible values for x are 40, 50, or 60. This gives corresponding y values of 50, 30, or 10.

Now, let's look at segment D-E. The total seats reserved on D-E are the tickets that pass through it: T_AE + T_BE + T_CE + T_DE. Substituting our variables, this is $x+30+q+0$.

Rule 3 states that exactly four-sevenths of the seats on D-E were from stations before C. The tickets on D-E from stations before C are T_AE and T_BE, which sum to $x+30$. We can set up the equation:

$x+30=\frac{4}{7}(x+30+q)$

Solving for q:

$\frac{7}{4}(x+30)=x+30+q$

$q=\frac{3}{4}(x+30)$

We know q must be an integer and a multiple of 10. Let's test our possible values for x:

• If x = 40, $q=\frac{3}{4}\left(70\right)=52.5$ (Not an integer)
• If x = 50, $q=\frac{3}{4}\left(80\right)=60$ (Valid, multiple of 10)
• If x = 60, $q=\frac{3}{4}\left(90\right)=67.5$ (Not an integer)

Therefore, we must have x = 50, which means y = 30, and q = 60. This establishes that T_CE = 60.

Finally, we analyze segment C-D. The seats reserved on C-D are T_AD + T_AE + T_BD + T_BE + T_CD + T_CE. We established this equals 190. Substituting our known values:

$30+50+0+30+p+60=190$

$170+p=190$

$p=20$

This means T_CD = 20. The question asks for the total number of tickets booked from Station C, which is T_CD + T_CE = $20+60=80$.