What was the occupancy factor for segment D - E?

Assuming the tickets from A - C, A - D, and A - E to be $a,b$ and $c$, respectively. Also, assuming the tickets from C - D and C - E to be $x$ and $y$, respectively. We are also given that the number of tickets from A - C is equal to the number of tickets booked from A - E, so the value of $a=c$. We are also given that 40 tickets were booked from B to C, and 30 tickets were booked from B to E. Other information provided is that no tickets were booked from A to B, from B to D, or from D to E.
Putting all the given information in the table, we get,
From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,
A - B $=a+b+c=2a+b$
B - C $=a+b+c+30+40=2a+b+70$
C - D $=a+b+x+y+30$
D - E $=c+y+30=a+y+30$
We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,
Seats occupied in segment C - D $=\dfrac{95}{100}\times\ 200\ =\ 190$
We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.
We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.
Equating it with the above equation, we get,
$2a+b+70=200$
$2a+b=130$ --(1)
$a+b+x+y+30 = 190$
$a+b+x+y=160$ --(2)
We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.
Equating the expressions, we get,
$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$
$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$  ---(3)
So, the seats occupied during D - E $=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$
We know that this value has to be an integer.
We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.
We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.
The only value of a at which the equation (3) is an integer is when $a+30$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $a=50$, as at $a=40$ and 60, the value of $a+30$ is not a multiple of 4.
We can conclude that the value of $a=50$, and substituting in (1), we get the value of $b=30$.
Substituting in (3), we get,
$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$
$\ y\ +\ 80\ =\ 140$
$\ y\ =\ 60$
Substituting all the values in (2), we get,
$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$
$x\ =\ 20$
Putting all the values in the table, we get,
A - B $=a+b+c=2a+b=130$
B - C $=a+b+c+30+40=2a+b+70=200$
C - D $=a+b+x+y+30=190$
D - E $=c+y+30=a+y+30=140$
Occupancy factor of segment D - E can be calculated as,
Occupancy factor $=\dfrac{140}{200}\times\ 100\ =\ 70\%$
Hence, the correct answer is option B.