Unfold Prep
CAT 2025 · SLOT 1
Passage
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.
- Segment C – D had an occupancy factor of 95%. Only segment B – C had a higher occupancy factor.
- Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.
- Among the seats reserved on segment D – E, exactly four-sevenths were from stations before C.
- 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.
- No tickets were booked from A to B, from B to D and from D to E.
- The number of tickets booked for any segment was a multiple of 10.
Q3.How many tickets were booked from Station C?
In the exam, the best way to solve this is to draw a boarding strictly against destination matrix, alongside the route map.
Let's first fill in the direct clues: • Rule 2 says exactly 40 tickets were B→C and 30 tickets were B→E. • Rule 5 says zero tickets were A→B, B→D, and D→E.
Next, let's look at Rule 4: • A→C equaled A→E. Let's call them both 's' (so q = s). • 's' must be strictly greater than B→E (which we know is 30).
Now we tackle Rule 3. On segment D-E, 4/7 of the total passengers boarded before C. Let's highlight the physical D-E stretch and those passing through it.
Cross-multiplying our fraction ( s+30 / x = 4/3 ) gives us a core working algebraic equation for 's' and 'x'.
Using Rule 6 (every ticket total is a multiple of 10): • We test multiples > 30 in our equation 3s + 90 = 4x. • For example, s=40 gives $x=52.5$ ❌. • At s=50, we get $x=60$ ✅.
If we tried s=90, total seats on D-E would exceed our max train capacity of 200. Thus, s=50 and x=60 are our definitive answers.
Looking at Rule 1 for segment C-D: • Segment C-D has 95% occupancy (190 total capacity). • Who rolls over the C-D tracks? Anyone boarding on or before C, going to D or E!
Summing those highlighted passengers passing over C-D (50 + r + 30 + 0 + 60 + w = 190).
Using the second part of Rule 1: • ONLY segment B-C had higher occupancy than segment C-D. • Since C-D hit 190 (95%), B-C must logically be 200 (completely packed at 100%!).
Summing the highlighted passengers passing through segment B-C (50 + r + 50 + 30 + 0 + 40 = 200) effortlessly gives us r=30.
And since our C-D segment equation was r + w = 50, and r is 30, w smoothly resolves to 20! The journey logic is totally solved.
Our entire ticket table is mapped out. Combining abstract equations with physical train segments ensures you never miss a passenger!