Logical Reasoning

Math Puzzles

Hard

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.

How many tickets were booked from Station C?

Entered answer:

✅ Correct Answer: 80

Slide 1/7

The route is $A \to B \to C \to D \to E$ .

A ticket may be booked only from an earlier station to a later station, and it occupies a seat on every segment it spans.

We track the number of tickets for each origin–destination (OD) pair.

Main table: rows are the origin, columns are the destination. Cells where the destination is not ahead of the origin are marked $-$ .

From \ To	B	C	D
A
B	$-$
C	$-$	$-$
D	$-$	$-$	$-$

A second table tracks the seats reserved on each segment (capacity $200$ ) and its occupancy factor.

Segment	Reserved	Occupancy
A–B
B–C
C–D
D–E

Which tickets load each segment:

A–B: $A\to B,\ A\to C,\ A\to D,\ A\to E$

B–C: $A\to C,\ A\to D,\ A\to E,\ B\to C,\ B\to D,\ B\to E$

C–D: $A\to D,\ A\to E,\ B\to D,\ B\to E,\ C\to D,\ C\to E$

D–E: $A\to E,\ B\to E,\ C\to E,\ D\to E$

Goal: fill every OD cell.

From \ To	B	C	D	E
A	0
B	$-$	40	0	30
C	$-$	$-$
D	$-$	$-$	$-$	0

Segment	Reserved	Occupancy
A–B
B–C
C–D
D–E

From clue (5): no tickets $A\to B$ , $B\to D$ , $D\to E$ , so those three cells are $0$ .

From clue (2):

$B\to C = 40$

$B\to E = 30$

From \ To	B	C	D	E
A	0
B	$-$	40	0	30
C	$-$	$-$
D	$-$	$-$	$-$	0

Segment	Reserved	Occupancy
A–B
B–C	200	100%
C–D	190	95%
D–E

From clue (1), C–D has occupancy $95\%$ :

$\text{C–D reserved} = 0.95 \times 200 = 190$

B–C is the only segment higher than C–D, so it is the maximum.

Every OD count is a multiple of $10$ , so each segment load is a multiple of $10$ , and cannot exceed $200$ .

The only multiple of $10$ above $190$ is $200$ .

$\text{B–C reserved} = 200$

From \ To	B	C	D	E
A	0
B	$-$	40	0	30
C	$-$	$-$
D	$-$	$-$	$-$	0

Segment	Reserved	Occupancy
A–B	130	65%
B–C	200	100%
C–D	190	95%
D–E

The B–C load gives:

$AC + AD + AE + 40 + 0 + 30 = 200$

$AC + AD + AE = 130$

Since $A\to B = 0$ , the A–B load is:

$\text{A–B reserved} = 0 + (AC + AD + AE) = 130$

Occupancy $= 130/200 = 65\%$ .

From clue (4): $A\to C = A\to E$ and $A\to C > 30$ .

So $2\cdot AC + AD = 130$ with $AC \in \{40, 50, 60\}$ (as $AD \ge 0$ ).

From \ To	B	C	D	E
A	0	50	30	50
B	$-$	40	0	30
C	$-$	$-$
D	$-$	$-$	$-$	0

Segment	Reserved	Occupancy
A–B	130	65%
B–C	200	100%
C–D	190	95%
D–E

Candidates from $2\cdot AC + AD = 130$ :

$AC=40 \Rightarrow AD=50$

$AC=50 \Rightarrow AD=30$

$AC=60 \Rightarrow AD=10$

From clue (3), on D–E the seats from stations before C (that is A and B) are $\tfrac{4}{7}$ of the total, leaving $\tfrac{3}{7}$ from C.

Before-C seats $= AE + B\to E = AE + 30$ , and from-C seats $= CE$ .

So $CE : (AE+30) = 3 : 4$ , i.e. $CE = \tfrac{3}{4}(AC+30)$ .

Testing:

$AC=40 \Rightarrow CE = \tfrac{3}{4}(70) = 52.5$ (not a multiple of $10$ ) — reject

$AC=50 \Rightarrow CE = \tfrac{3}{4}(80) = 60$ — valid

$AC=60 \Rightarrow CE = \tfrac{3}{4}(90) = 67.5$ — reject

So $A\to C = 50$ , $A\to E = 50$ , $A\to D = 30$ .

From \ To	B	C	D	E
A	0	50	30	50
B	$-$	40	0	30
C	$-$	$-$		60
D	$-$	$-$	$-$	0

Segment	Reserved	Occupancy
A–B	130	65%
B–C	200	100%
C–D	190	95%
D–E	140	70%

From the previous slide, $C\to E = 60$ .

The D–E load is:

$\text{D–E reserved} = AE + B\to E + CE + 0 = 50 + 30 + 60 = 140$

Occupancy $= 140/200 = 70\%$ .

Verify clue (3): before-C seats $= 50 + 30 = 80$ , and $80/140 = \tfrac{4}{7}$ . This matches.

From \ To	B	C	D	E
A	0	50	30	50
B	$-$	40	0	30
C	$-$	$-$	20	60
D	$-$	$-$	$-$	0

Segment	Reserved	Occupancy
A–B	130	65%
B–C	200	100%
C–D	190	95%
D–E	140	70%

The C–D load gives:

$AD + AE + B\to D + B\to E + CD + CE = 190$

$30 + 50 + 0 + 30 + CD + 60 = 190$

$170 + CD = 190$

$C\to D = 20$

All OD values are now uniquely determined.

Occupancy order: B–C $100\%$ > C–D $95\%$ > D–E $70\%$ > A–B $65\%$ , so only B–C exceeds C–D, consistent with clue (1).

Tickets booked from Station C are those originating at C: $C\to D$ and $C\to E$ .

$C\to D = 20$

$C\to E = 60$

Total $= 20 + 60 = 80$ .

Question 9

What was the occupancy factor for segment D – E?

Question 10

How many tickets were booked from Station A to Station E?

Question 12

What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?

Question 13

How many tickets were booked to travel in exactly one segment?

How many tickets were booked from Station C?

Solution

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