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 A to Station E?

Entered answer:

✅ Correct Answer: 50

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).

From the solved table, $A\to E = 50$ .

This was fixed because $A\to C = A\to E$ , $A\to C > 30$ , and the value $50$ is the only candidate making $C\to E = \tfrac{3}{4}(A\to E + 30) = 60$ a valid multiple of $10$ .

Tickets from A to E $= 50$ .

Question 9

What was the occupancy factor for segment D – E?

Question 11

How many tickets were booked from Station C?

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 A to Station E?

Solution

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