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 to travel in exactly one segment?

The correct answer is 60.

Let's break down the problem step-by-step by defining the variables for the number of tickets booked between any two stations. Let $T_{AB}$ represent the number of tickets from station A to B, and so on.

The segments of the route are A to B, B to C, C to D, and D to E. The total seating capacity of the train is 200.

From the given information, we can establish the following facts:

1. The occupancy factor for segment C to D is 95%. This means 95% of the 200 seats are reserved.
$0.95\times 200=190$
seats reserved on segment C to D. Only segment B to C had a higher occupancy factor. Since all ticket numbers are multiples of 10, the seats reserved on B to C must be 200 (100% occupancy).

2. We are given the exact number of some tickets:
$T_{BC}=40$
$T_{BE}=30$

3. Let's denote the number of tickets from A to C as $x$. We are told $T_{AC}=T_{AE}=x$, and that this number is higher than $T_{BE}$ (which is 30). Therefore, $x>30$. Since it must be a multiple of 10, $x\ge 40$.

4. We are also told no tickets were booked for A to B, B to D, and D to E:
$T_{AB}=0$
$T_{BD}=0$
$T_{DE}=0$

Now, let's analyze the occupancy of each segment.

Segment B to C:
The tickets that use segment B to C are those starting at or before B and ending at or after C. These are: $T_{AC},T_{AD},T_{AE},T_{BC},T_{BD},T_{BE}$.
The sum of these must be 200:
$x+T_{AD}+x+40+0+30=200$
$2x+T_{AD}+70=200$
$2x+T_{AD}=130$

Segment D to E:
The total seats reserved on D to E are $T_{AE}+T_{BE}+T_{CE}+T_{DE}$ =
$x+30+T_{CE}+0$.
We are told that exactly four-sevenths ($\frac{4}{7}$) of these were from stations before C (which are A and B). So, the tickets from A and B to E are $T_{AE}+T_{BE}=x+30$.
Setting up the equation:
$x+30=\frac{4}{7}(x+30+T_{CE})$
$7(x+30)=4(x+30+T_{CE})$
$7x+210=4x+120+4T_{CE}$
$3x+90=4T_{CE}$

Since all ticket counts are multiples of 10, and $x\ge 40$, let's test possible values for $x$:
If $x=40$, then
$3\left(40\right)+90=210$
, so $T_{CE}=52.5$ (Not a multiple of 10).
If $x=50$, then
$3\left(50\right)+90=240$
, so $T_{CE}=60$ (Valid).
If $x=60$, then
$3\left(60\right)+90=270$
(Not divisible by 4).
If $x=70$, then
$3\left(70\right)+90=300$
, so $T_{CE}=75$ (Not a multiple of 10).
If $x=90$, then
$2x+T_{AD}=130$
would mean
$180+T_{AD}=130$
, which gives a negative number of tickets (Invalid).
Thus, the only valid solution is $x=50$.

Substituting $x=50$ into the B to C segment equation:
$2\left(50\right)+T_{AD}=130$
$100+T_{AD}=130$
$T_{AD}=30$

Segment C to D:
The tickets using segment C to D are $T_{AD},T_{AE},T_{BD},T_{BE},T_{CD},T_{CE}$.
We know their sum is 190:
$30+50+0+30+T_{CD}+60=190$
$170+T_{CD}=190$
$T_{CD}=20$

Finding the final answer:
The question asks for the number of tickets booked to travel in exactly one segment. These are tickets from A to B, B to C, C to D, and D to E.
$T_{AB}=0$
$T_{BC}=40$
$T_{CD}=20$
$T_{DE}=0$

Summing these up gives:
$0+40+20+0=60$

Therefore, 60 tickets were booked to travel in exactly one segment.