1. Answer option 1: 70%

Reasoning: Seats on D–E come only from tickets A→E, B→E and C→E (D→E = 0). Let A→E = x and C→E = y; B→E is given as 30. Clue 3 says exactly four-sevenths of D–E seats were from stations before C, so x + 30 = (4/7)(x + 30 + y). Rearranging gives y = (3x + 90)/4. Counts must be multiples of 10 and x > 30 (clue 4). Trying multiples of 10 yields x = 50 and y = 60. Thus seats on D–E = 50 + 30 + 60 = 140. Capacity is 200, so occupancy = 140/200 = 0.70 = 70%. D–E = A→E (50) + B→E (30) + C→E (60) = 140 seats → 70% occupancy. 

2. Answer: 50 tickets.

Reasoning: This is x from the equation above. From the 4/7 condition and the multiple-of-10 constraint we found x = 50, and clue 4 requires A→C = A→E = x, so A→E = 50.

3. Answer: 80 tickets.

Reasoning: Tickets originating at C are C→D and C→E. From the solved counts C→D = 20 and C→E = 60, so total from C = 20 + 60 = 80.

4. Answer: 40 tickets.

Reasoning: Tickets to station C are A→C and B→C. A→C = 50 (equal to A→E) and B→C = 40, so total to C = 50 + 40 = 90. Tickets to station D are A→D and C→D (B→D = 0). Using the consistent solution A→D = 30 and C→D = 20 gives total to D = 30 + 20 = 50. Difference = 90 − 50 = 40.

5. Answer: 60 tickets.

Reasoning: Exactly-one-segment trips are the adjacent-station bookings: A→B, B→C, C→D, D→E. Given A→B = 0 and D→E = 0, and the solved values B→C = 40 and C→D = 20, the total = 40 + 20 = 60.

Step by Step Table Making

Step 1: Clue 2 gives B→C = 40 and B→E = 30.

Step 2: Clue 3 states that among seats reserved on D–E, exactly 4/7 were from stations before C. Seats on D–E come only from A→E, B→E, C→E (D→E is zero). Let A→E = x and C→E = y. Then seats on D–E = x + 30 + y and seats from before C = x + 30. Condition: x + 30 = (4/7)(x + 30 + y).

Step 3: Clue 4 says A→C = A→E and that value is higher than B→E (30). So x is a multiple of 10 and x > 30. Also A→C = x.

Step 4: Clue 5 sets A→B = 0, B→D = 0, D→E = 0.

Step 5: Clue 6 restricts counts to multiples of 10.

Step 6: From the equation in Step 2 and the multiple-of-10 constraint, solving yields x = 50 and y = 60. (Detailed algebra: x+30 = (4/7)(x+30+y) → 7(x+30)=4(x+30+y) → 7x+210 = 4x+120+4y → 3x +90 = 4y → y = (3x+90)/4. Trying x values (40,50,60,…) and enforcing y integer multiple of 10 gives x=50 → y=(150+90)/4=240/4=60.)

Step 7: Clue 4 also requires A→C = A→E = 50 and A→E > B→E (50>30 satisfied). We still need A→D. Observing segment C–D must reach 190 seats (given 95% occupancy). Using current contributions from B→E (30 contributes to C–D), C→E (60), A→E (50) and unknown A→D and C→D, solving for A→D and C→D with multiples of 10 yields A→D = 30 and C→D = 20 (these satisfy totals and other constraints).

Step 8: Populate the tables with these values and compute segment sums.

Train bookings — Working table and calculations

Final route-wise ticket counts (all multiples of 10)

Total tickets booked (sum of routes) = 280.

Seat reservations contributed by each route to each segment

Segments are: A–B, B–C, C–D, D–E. A ticket from X to Y reserves one seat on every segment between X and Y.