Question

What was the occupancy factor for segment D–E? 

• A. 35%
• B. 70%
• C. 77%
• D. 84%

Answer 1

The Correct Option is B

Approach: Lean entirely on the 4 : 3 split. Once you know the "before C" half of the D-E load, the whole D-E load is just that half scaled up by 7/4 - no need to chase A-D or C-D.

The before-C load on D-E: Tickets crossing D-E that originate before C are exactly the A-to-E and B-to-E tickets. We are told B-to-E $ = 30 $. The A-to-E count equals the A-to-C count, which is larger than $ 30 $ and a multiple of $ 10 $; matching the $ 95\% $ (i.e. $ 190 $-seat) load on C-D forces A-to-E $ = 50 $. So the before-C part of D-E is $ 50 + 30 = 80 $ seats.

Scale up: Clue 3 says this $ 80 $ is four-sevenths of the whole D-E load. Therefore total D-E seats $ = 80 \times \dfrac{7}{4} = 140 $. (The leftover three-sevenths, $ 60 $ seats, are the C-to-E tickets.)

Occupancy: $ \dfrac{140}{200} \times 100 = 70\% $. Notice the A-D versus C-D ambiguity never enters, because neither of those ticket types touches the D-E segment. The occupancy factor for D-E is 70%.

Answer 2

Step 1: A-C = A-E = \(x\), and \(x\) must be a multiple of 10 greater than 30, so try 40, 50, 90, 130, …
Step 2: For each candidate, C-E = \(\tfrac{3(x+30)}{4}\) must also be a whole multiple of 10, this already eliminates \(x=40\) (gives 52.5) and confirms \(x=50\) gives C-E = 60.
Step 3: Check segment C-D = 190: with \(x=50\), C-E=60, B-E=30, we need A-D + C-D = 190 - 50 - 30 - 60 = 50, which is achievable (non-negative); every larger candidate for \(x\) makes this sum negative, ruling them out.
\[ \boxed{50} \]

Answer 3

Step 1: Tickets from C = C-D + C-E (a ticket can't run backward).
Step 2: C-E = 60 (fixed by the four-sevenths D-E split with A-E = 50, B-E = 30).
Step 3: A-D + C-D = 50 from the segment C-D total (190); using the baseline A-D = 0 gives C-D = 50.
\[ \boxed{110} \]

Answer 4

Step 1: Tickets to C = A-C + B-C = 50 + 40 = 90.
Step 2: Tickets to D = A-D + B-D = A-D + 0; baseline A-D = 0.
Step 3: Difference = 90 - 0 = 90.
\[ \boxed{90} \]

Answer 5

Step 1: One-segment routes are A-B, B-C, C-D, D-E only.
Step 2: A-B = 0 and D-E = 0 (given); B-C = 40 (given).
Step 3: C-D = 50 under the baseline split (A-D = 0), so total = 0 + 40 + 50 + 0 = 90.
\[ \boxed{90} \]