This page provides CAT 2025 DILR Slot 1 with detailed solutions and analysis. Download the free PDF to practice real CAT questions and boost your preparation.
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Step 1:
From Condition (1), Asha, Bunty, and Chintu were in the
Elite at the beginning of Quarter 1. Therefore, Dolly,
Eklavya, and Falguni were in the Novice at the beginning
of Quarter 1. All three—Asha, Bunty, and Chintu—were
in the Novice at the beginning of Quarter 4. Hence, all
three—Dolly, Eklavya, and Falguni—were in the Elite at
the beginning of Quarter 4.
From Condition (3), based on the ratings given by Lalu,
Bunty was demoted to the Novice in Quarter 2. Therefore,
his rating in Quarter 1 was 1. Also, Asha and Dolly received
ratings of 1 and 2, respectively, in Quarter 3.
From Condition (2), the ratings of Dolly and Falguni were
2 and 3, respectively, in all three quarters. Hence, Falguni
was promoted to the Elite in Quarter 2.
Since Asha was demoted to the Novice in Quarter 3, her
cumulative score in Quarter 2 was less than or equal to
Chintu’s. Therefore, Asha’s and Chintu’s ratings in Quarter
1 were 2 and 3, in any order. Also, their ratings in Quarter
2 were 1 and 2, respectively.
Hence, the information can be shown in the table below.
Step 2:
The cumulative scores of Dolly and Eklavya in Quarter 2
were 4 each, but Eklavya’s rating in Quarter 2 was higher
than Dolly’s. Therefore, Eklavya was promoted to the Elite
in Quarter 3.
Since Chintu was demoted to the Novice in Quarter 4,
his cumulative score in Quarter 3 was less than or equal
to that of Eklavya. Therefore, the ratings of Chintu and
Eklavya in Quarter 3 were 1 and 2, respectively.
Hence, the final information is shown in the table below.
Eklavya’s score at the end of Quarter 2 was 4.
0
Step 1:
From Condition (1), Asha, Bunty, and Chintu were in the
Elite at the beginning of Quarter 1. Therefore, Dolly,
Eklavya, and Falguni were in the Novice at the beginning
of Quarter 1. All three—Asha, Bunty, and Chintu—were
in the Novice at the beginning of Quarter 4. Hence, all
three—Dolly, Eklavya, and Falguni—were in the Elite at
the beginning of Quarter 4.
From Condition (3), based on the ratings given by Lalu,
Bunty was demoted to the Novice in Quarter 2. Therefore,
his rating in Quarter 1 was 1. Also, Asha and Dolly received
ratings of 1 and 2, respectively, in Quarter 3.
From Condition (2), the ratings of Dolly and Falguni were
2 and 3, respectively, in all three quarters. Hence, Falguni
was promoted to the Elite in Quarter 2.
Since Asha was demoted to the Novice in Quarter 3, her
cumulative score in Quarter 2 was less than or equal to
Chintu’s. Therefore, Asha’s and Chintu’s ratings in Quarter
1 were 2 and 3, in any order. Also, their ratings in Quarter
2 were 1 and 2, respectively.
Hence, the information can be shown in the table below.
Step 2:
The cumulative scores of Dolly and Eklavya in Quarter 2
were 4 each, but Eklavya’s rating in Quarter 2 was higher
than Dolly’s. Therefore, Eklavya was promoted to the Elite
in Quarter 3.
Since Chintu was demoted to the Novice in Quarter 4,
his cumulative score in Quarter 3 was less than or equal
to that of Eklavya. Therefore, the ratings of Chintu and
Eklavya in Quarter 3 were 1 and 2, respectively.
Hence, the final information is shown in the table below.
No employee changed groups more than once up
to the beginning of Quarter 4.
5
Step 1:
From Condition (1), Asha, Bunty, and Chintu were in the
Elite at the beginning of Quarter 1. Therefore, Dolly,
Eklavya, and Falguni were in the Novice at the beginning
of Quarter 1. All three—Asha, Bunty, and Chintu—were
in the Novice at the beginning of Quarter 4. Hence, all
three—Dolly, Eklavya, and Falguni—were in the Elite at
the beginning of Quarter 4.
From Condition (3), based on the ratings given by Lalu,
Bunty was demoted to the Novice in Quarter 2. Therefore,
his rating in Quarter 1 was 1. Also, Asha and Dolly received
ratings of 1 and 2, respectively, in Quarter 3.
From Condition (2), the ratings of Dolly and Falguni were
2 and 3, respectively, in all three quarters. Hence, Falguni
was promoted to the Elite in Quarter 2.
Since Asha was demoted to the Novice in Quarter 3, her
cumulative score in Quarter 2 was less than or equal to
Chintu’s. Therefore, Asha’s and Chintu’s ratings in Quarter
1 were 2 and 3, in any order. Also, their ratings in Quarter
2 were 1 and 2, respectively.
Hence, the information can be shown in the table below.
Step 2:
The cumulative scores of Dolly and Eklavya in Quarter 2
were 4 each, but Eklavya’s rating in Quarter 2 was higher
than Dolly’s. Therefore, Eklavya was promoted to the Elite
in Quarter 3.
Since Chintu was demoted to the Novice in Quarter 4,
his cumulative score in Quarter 3 was less than or equal
to that of Eklavya. Therefore, the ratings of Chintu and
Eklavya in Quarter 3 were 1 and 2, respectively.
Hence, the final information is shown in the table below.
Bunty’s score at the end of Quarter 3 was 5.
4
Step 1:
From Condition (1), Asha, Bunty, and Chintu were in the
Elite at the beginning of Quarter 1. Therefore, Dolly,
Eklavya, and Falguni were in the Novice at the beginning
of Quarter 1. All three—Asha, Bunty, and Chintu—were
in the Novice at the beginning of Quarter 4. Hence, all
three—Dolly, Eklavya, and Falguni—were in the Elite at
the beginning of Quarter 4.
From Condition (3), based on the ratings given by Lalu,
Bunty was demoted to the Novice in Quarter 2. Therefore,
his rating in Quarter 1 was 1. Also, Asha and Dolly received
ratings of 1 and 2, respectively, in Quarter 3.
From Condition (2), the ratings of Dolly and Falguni were
2 and 3, respectively, in all three quarters. Hence, Falguni
was promoted to the Elite in Quarter 2.
Since Asha was demoted to the Novice in Quarter 3, her
cumulative score in Quarter 2 was less than or equal to
Chintu’s. Therefore, Asha’s and Chintu’s ratings in Quarter
1 were 2 and 3, in any order. Also, their ratings in Quarter
2 were 1 and 2, respectively.
Hence, the information can be shown in the table below.
Step 2:
The cumulative scores of Dolly and Eklavya in Quarter 2
were 4 each, but Eklavya’s rating in Quarter 2 was higher
than Dolly’s. Therefore, Eklavya was promoted to the Elite
in Quarter 3.
Since Chintu was demoted to the Novice in Quarter 4,
his cumulative score in Quarter 3 was less than or equal
to that of Eklavya. Therefore, the ratings of Chintu and
Eklavya in Quarter 3 were 1 and 2, respectively.
Hence, the final information is shown in the table below.
The scores of four employees—Bunty, Dolly,
Eklavya, and Falguni—at the end of Quarter 3 can
be determined with certainty.
4
Step 1:
From Condition (1), Asha, Bunty, and Chintu were in the
Elite at the beginning of Quarter 1. Therefore, Dolly,
Eklavya, and Falguni were in the Novice at the beginning
of Quarter 1. All three—Asha, Bunty, and Chintu—were
in the Novice at the beginning of Quarter 4. Hence, all
three—Dolly, Eklavya, and Falguni—were in the Elite at
the beginning of Quarter 4.
From Condition (3), based on the ratings given by Lalu,
Bunty was demoted to the Novice in Quarter 2. Therefore,
his rating in Quarter 1 was 1. Also, Asha and Dolly received
ratings of 1 and 2, respectively, in Quarter 3.
From Condition (2), the ratings of Dolly and Falguni were
2 and 3, respectively, in all three quarters. Hence, Falguni
was promoted to the Elite in Quarter 2.
Since Asha was demoted to the Novice in Quarter 3, her
cumulative score in Quarter 2 was less than or equal to
Chintu’s. Therefore, Asha’s and Chintu’s ratings in Quarter
1 were 2 and 3, in any order. Also, their ratings in Quarter
2 were 1 and 2, respectively.
Hence, the information can be shown in the table below.
Step 2:
The cumulative scores of Dolly and Eklavya in Quarter 2
were 4 each, but Eklavya’s rating in Quarter 2 was higher
than Dolly’s. Therefore, Eklavya was promoted to the Elite
in Quarter 3.
Since Chintu was demoted to the Novice in Quarter 4,
his cumulative score in Quarter 3 was less than or equal
to that of Eklavya. Therefore, the ratings of Chintu and
Eklavya in Quarter 3 were 1 and 2, respectively.
Hence, the final information is shown in the table below.
Only Statement II is necessarily true.
7
Step 1:
One tap means “Yes”, two taps mean “No”, and three
taps mean “Maybe”.
Each question received at least one “Yes”, one “No”, and
one “Maybe’’. So each person received more than 6 taps.
From conditions (1) and (4), Alia tapped 2 Yes and 2 No.
From condition (6), Badal and Clive tapped = 40 – (6 + 11
+ 9) = 14 times and Clive tapped more times in total than
Badal.
Also, from condition (4), Badal tapped 6 times and Clive
tapped 8 times.
From condition (3), Let the number of taps received by
Badal or Dilshan or Ehsaan be x.
Then, the number of taps received by Clive = 40 – 9 – 3x
= 31 – 3x.
But 31 – 3x > 6 or x < 8.33 i.e., x = 7 or 8
For x = 7, the number of taps received by Clive will be 31
– 21 = 10, which is not possible. Since Badal tapped Yes
or No for Clive.
For x = 8, then the number of tabs received by Clive will
be 31 – 24 = 7, which is possible.
The information can be shown in the table below.
Step 2:
From condition (5), No one’s answer to Alia’s question
matched the answer that Alia gave to that person’s
question. This was also true for Ehsaan.
The final information can be shown in the table below.
Clive received 7 taps for his question.
1
Step 1:
One tap means “Yes”, two taps mean “No”, and three
taps mean “Maybe”.
Each question received at least one “Yes”, one “No”, and
one “Maybe’’. So each person received more than 6 taps.
From conditions (1) and (4), Alia tapped 2 Yes and 2 No.
From condition (6), Badal and Clive tapped = 40 – (6 + 11
+ 9) = 14 times and Clive tapped more times in total than
Badal.
Also, from condition (4), Badal tapped 6 times and Clive
tapped 8 times.
From condition (3), Let the number of taps received by
Badal or Dilshan or Ehsaan be x.
Then, the number of taps received by Clive = 40 – 9 – 3x
= 31 – 3x.
But 31 – 3x > 6 or x < 8.33 i.e., x = 7 or 8
For x = 7, the number of taps received by Clive will be 31
– 21 = 10, which is not possible. Since Badal tapped Yes
or No for Clive.
For x = 8, then the number of tabs received by Clive will
be 31 – 24 = 7, which is possible.
The information can be shown in the table below.
Step 2:
From condition (5), No one’s answer to Alia’s question
matched the answer that Alia gave to that person’s
question. This was also true for Ehsaan.
The final information can be shown in the table below.
Alia and Badal tapped an equal number of times in
total.
1
Step 1:
One tap means “Yes”, two taps mean “No”, and three
taps mean “Maybe”.
Each question received at least one “Yes”, one “No”, and
one “Maybe’’. So each person received more than 6 taps.
From conditions (1) and (4), Alia tapped 2 Yes and 2 No.
From condition (6), Badal and Clive tapped = 40 – (6 + 11
+ 9) = 14 times and Clive tapped more times in total than
Badal.
Also, from condition (4), Badal tapped 6 times and Clive
tapped 8 times.
From condition (3), Let the number of taps received by
Badal or Dilshan or Ehsaan be x.
Then, the number of taps received by Clive = 40 – 9 – 3x
= 31 – 3x.
But 31 – 3x > 6 or x < 8.33 i.e., x = 7 or 8
For x = 7, the number of taps received by Clive will be 31
– 21 = 10, which is not possible. Since Badal tapped Yes
or No for Clive.
For x = 8, then the number of tabs received by Clive will
be 31 – 24 = 7, which is possible.
The information can be shown in the table below.
Step 2:
From condition (5), No one’s answer to Alia’s question
matched the answer that Alia gave to that person’s
question. This was also true for Ehsaan.
The final information can be shown in the table below.
Clive’s response to Ehsaan’s question was ‘No’.
6
Step 1:
One tap means “Yes”, two taps mean “No”, and three
taps mean “Maybe”.
Each question received at least one “Yes”, one “No”, and
one “Maybe’’. So each person received more than 6 taps.
From conditions (1) and (4), Alia tapped 2 Yes and 2 No.
From condition (6), Badal and Clive tapped = 40 – (6 + 11
+ 9) = 14 times and Clive tapped more times in total than
Badal.
Also, from condition (4), Badal tapped 6 times and Clive
tapped 8 times.
From condition (3), Let the number of taps received by
Badal or Dilshan or Ehsaan be x.
Then, the number of taps received by Clive = 40 – 9 – 3x
= 31 – 3x.
But 31 – 3x > 6 or x < 8.33 i.e., x = 7 or 8
For x = 7, the number of taps received by Clive will be 31
– 21 = 10, which is not possible. Since Badal tapped Yes
or No for Clive.
For x = 8, then the number of tabs received by Clive will
be 31 – 24 = 7, which is possible.
The information can be shown in the table below.
Step 2:
From condition (5), No one’s answer to Alia’s question
matched the answer that Alia gave to that person’s
question. This was also true for Ehsaan.
The final information can be shown in the table below.
Six ‘’Yes” responses were received across all the
questions.
4
Step 1:
In the initial position, Aslam and Chhavi are sitting next
to each other, while both Bashir and Davies have empty
chairs on either side of them. Since the arrangement is
circular, the initial order of Aslam and Chhavi does not
matter.
From condition (1), the four friends occupy adjacent chairs
only at the end of Turns 2 and 6. If Aslam moves
counterclockwise, then they will not occupy adjacent
chairs at the end of Turn 2. Therefore, Aslam must move
clockwise, and in Turn 1, Bashir moves counterclockwise.
From condition (2), Davies occupies Chair 2 after Turn 1,
and Chhavi occupies Chair 7 after Turn 2.
Step 2:
In Turn 3, Chhavi moves in counterclockwise and occupies
chair No. 6.
From condition (2), Davies occupies Chair 4 after Turn 5.
So in Turn 4, Davies moves clockwise and occupies Chair
4.
The number of the chair initially occupied by Bashir
is 4.
4
Step 1:
In the initial position, Aslam and Chhavi are sitting next
to each other, while both Bashir and Davies have empty
chairs on either side of them. Since the arrangement is
circular, the initial order of Aslam and Chhavi does not
matter.
From condition (1), the four friends occupy adjacent chairs
only at the end of Turns 2 and 6. If Aslam moves
counterclockwise, then they will not occupy adjacent
chairs at the end of Turn 2. Therefore, Aslam must move
clockwise, and in Turn 1, Bashir moves counterclockwise.
From condition (2), Davies occupies Chair 2 after Turn 1,
and Chhavi occupies Chair 7 after Turn 2.
Step 2:
In Turn 3, Chhavi moves in counterclockwise and occupies
chair No. 6.
From condition (2), Davies occupies Chair 4 after Turn 5.
So in Turn 4, Davies moves clockwise and occupies Chair
4.
No one sits on the chair numbered 4 at the end of
Turn 3.
1
Step 1:
In the initial position, Aslam and Chhavi are sitting next
to each other, while both Bashir and Davies have empty
chairs on either side of them. Since the arrangement is
circular, the initial order of Aslam and Chhavi does not
matter.
From condition (1), the four friends occupy adjacent chairs
only at the end of Turns 2 and 6. If Aslam moves
counterclockwise, then they will not occupy adjacent
chairs at the end of Turn 2. Therefore, Aslam must move
clockwise, and in Turn 1, Bashir moves counterclockwise.
From condition (2), Davies occupies Chair 2 after Turn 1,
and Chhavi occupies Chair 7 after Turn 2.
Step 2:
In Turn 3, Chhavi moves in counterclockwise and occupies
chair No. 6.
From condition (2), Davies occupies Chair 4 after Turn 5.
So in Turn 4, Davies moves clockwise and occupies Chair
4.
Chairs numbered 4, 5, 6, and 7 are occupied at
the end of Turn 6.
3
Step 1:
In the initial position, Aslam and Chhavi are sitting next
to each other, while both Bashir and Davies have empty
chairs on either side of them. Since the arrangement is
circular, the initial order of Aslam and Chhavi does not
matter.
From condition (1), the four friends occupy adjacent chairs
only at the end of Turns 2 and 6. If Aslam moves
counterclockwise, then they will not occupy adjacent
chairs at the end of Turn 2. Therefore, Aslam must move
clockwise, and in Turn 1, Bashir moves counterclockwise.
From condition (2), Davies occupies Chair 2 after Turn 1,
and Chhavi occupies Chair 7 after Turn 2.
Step 2:
In Turn 3, Chhavi moves in counterclockwise and occupies
chair No. 6.
From condition (2), Davies occupies Chair 4 after Turn 5.
So in Turn 4, Davies moves clockwise and occupies Chair
4.
At the end of Turn 7, only Davies is sitting on a
chair adjacent to the one occupied by Bashir.
1
From the radar graph, the table below shows the import
tariffs (in %) imposed by each country on other countries.
From the bar graph, the table below shows the import
tariffs (in Billion USD) imposed by each country on other
countries.
India charged an import tariff of 3.5 billion USD on
imports from Japan, which is 50% of the total
imports.
Hence, Japan’s exports to India are worth
7.0 billion USD.
4
From the radar graph, the table below shows the import
tariffs (in %) imposed by each country on other countries.
From the bar graph, the table below shows the import
tariffs (in Billion USD) imposed by each country on other
countries.
Option (1): Exports by Japan to UK = 6 × 1/0.4
= 15 Billion USD
Option (2): Exports by France to Japan = 3 × 1/
0.3 = 10 Billion USD
Option (3): Imports by France from India = 6.5 × 1/
0.4 = 16.25 Billion USD
Option (4): Imports by US from France = 6 × 1/0.2
= 30 Billion USD
Hence, option (4) is the correct answer.
1
From the radar graph, the table below shows the import
tariffs (in %) imposed by each country on other countries.
From the bar graph, the table below shows the import
tariffs (in Billion USD) imposed by each country on other
countries.
Import by India from UK = 5 × 1/0.2 = 25 Billion
USD
Export from India to UK = 3 × 1/0.3 = 10 Billion
USD
Hence, trade deficit of India with UK = 25 – 10
= 15 Billion USD
4
From the radar graph, the table below shows the import
tariffs (in %) imposed by each country on other countries.
From the bar graph, the table below shows the import
tariffs (in Billion USD) imposed by each country on other
countries.
Import by France from US = 5.5 × 1/0.3 = 18.33
Billion USD
Export by France to US = 6 × 1/0.2 = 30 Billion
USD
So trade surplus of France with US = 30 – 18.33 =
11.66 Billion USD
Import by UK from US = 2.5 × 1/0.2 = 12.5 Billion
USD
Export from UK to US = 3 × 1/0.3 = 10 Billion USD
So trade deficit of UK with US = 12.5 – 10 = 2.5
Billion USD
Hence, only France has trade surplus with US.
3
Step 1:
From condition (2), Exactly 40 tickets were booked from
B to C and 30 tickets were booked from B to E.
From condition (5), No tickets were booked from A to B,
from B to D and from D to E.
From condition (4), The number of tickets booked from A
to C = The number of tickets booked from A to E > 30
From condition (3), Among the seats reserved on segment
D – E, exactly four-sevenths were from stations before C
i.e., from A and B.
The factors of both 7 and 10 are 70 and 140.
70 is not possible, since A to E + B to E > 60
So A to E + B to E = 140 × 4/7 = 80 and C to E = 140 –
80 = 60
Step 2:
From conditions (1) and (6), Segment C – D had an
occupancy = 0.95 × 200 = 190 and segment B – C had
an occupancy = 200.
AC + AD + AE + BC + BD + BE = 200
⇒ 50 + AD + 50 + 40 + 0 + 30 = 200
⇒ AD = 30
AD + AE + BD + BE + CD + CE = 190
⇒ 30 + 50 + 0 + 30 + CD + 60 = 190
⇒ CD = 20
Hence, the final information can be shown in the table
below.
The occupancy for segment D – E was = A to E +
B to E + C to E + D to E
= 50 + 30 + 60 + 0 = 140
Hence, the occupancy factor for segment D – E
was = 140/200 × 100 = 70%.
50
Step 1:
From condition (2), Exactly 40 tickets were booked from
B to C and 30 tickets were booked from B to E.
From condition (5), No tickets were booked from A to B,
from B to D and from D to E.
From condition (4), The number of tickets booked from A
to C = The number of tickets booked from A to E > 30
From condition (3), Among the seats reserved on segment
D – E, exactly four-sevenths were from stations before C
i.e., from A and B.
The factors of both 7 and 10 are 70 and 140.
70 is not possible, since A to E + B to E > 60
So A to E + B to E = 140 × 4/7 = 80 and C to E = 140 –
80 = 60
Step 2:
From conditions (1) and (6), Segment C – D had an
occupancy = 0.95 × 200 = 190 and segment B – C had
an occupancy = 200.
AC + AD + AE + BC + BD + BE = 200
⇒ 50 + AD + 50 + 40 + 0 + 30 = 200
⇒ AD = 30
AD + AE + BD + BE + CD + CE = 190
⇒ 30 + 50 + 0 + 30 + CD + 60 = 190
⇒ CD = 20
Hence, the final information can be shown in the table
below.
The number of tickets booked from Station A to
Station E was 50.
80
Step 1:
From condition (2), Exactly 40 tickets were booked from
B to C and 30 tickets were booked from B to E.
From condition (5), No tickets were booked from A to B,
from B to D and from D to E.
From condition (4), The number of tickets booked from A
to C = The number of tickets booked from A to E > 30
From condition (3), Among the seats reserved on segment
D – E, exactly four-sevenths were from stations before C
i.e., from A and B.
The factors of both 7 and 10 are 70 and 140.
70 is not possible, since A to E + B to E > 60
So A to E + B to E = 140 × 4/7 = 80 and C to E = 140 –
80 = 60
Step 2:
From conditions (1) and (6), Segment C – D had an
occupancy = 0.95 × 200 = 190 and segment B – C had
an occupancy = 200.
AC + AD + AE + BC + BD + BE = 200
⇒ 50 + AD + 50 + 40 + 0 + 30 = 200
⇒ AD = 30
AD + AE + BD + BE + CD + CE = 190
⇒ 30 + 50 + 0 + 30 + CD + 60 = 190
⇒ CD = 20
Hence, the final information can be shown in the table
below.
The number of tickets booked from Station C was
= 20 + 60 = 80.
40
Step 1:
From condition (2), Exactly 40 tickets were booked from
B to C and 30 tickets were booked from B to E.
From condition (5), No tickets were booked from A to B,
from B to D and from D to E.
From condition (4), The number of tickets booked from A
to C = The number of tickets booked from A to E > 30
From condition (3), Among the seats reserved on segment
D – E, exactly four-sevenths were from stations before C
i.e., from A and B.
The factors of both 7 and 10 are 70 and 140.
70 is not possible, since A to E + B to E > 60
So A to E + B to E = 140 × 4/7 = 80 and C to E = 140 –
80 = 60
Step 2:
From conditions (1) and (6), Segment C – D had an
occupancy = 0.95 × 200 = 190 and segment B – C had
an occupancy = 200.
AC + AD + AE + BC + BD + BE = 200
⇒ 50 + AD + 50 + 40 + 0 + 30 = 200
⇒ AD = 30
AD + AE + BD + BE + CD + CE = 190
⇒ 30 + 50 + 0 + 30 + CD + 60 = 190
⇒ CD = 20
Hence, the final information can be shown in the table
below.
Required difference = (50 + 40) – (30 + 20) = 40.
60
Step 1:
From condition (2), Exactly 40 tickets were booked from
B to C and 30 tickets were booked from B to E.
From condition (5), No tickets were booked from A to B,
from B to D and from D to E.
From condition (4), The number of tickets booked from A
to C = The number of tickets booked from A to E > 30
From condition (3), Among the seats reserved on segment
D – E, exactly four-sevenths were from stations before C
i.e., from A and B.
The factors of both 7 and 10 are 70 and 140.
70 is not possible, since A to E + B to E > 60
So A to E + B to E = 140 × 4/7 = 80 and C to E = 140 –
80 = 60
Step 2:
From conditions (1) and (6), Segment C – D had an
occupancy = 0.95 × 200 = 190 and segment B – C had
an occupancy = 200.
AC + AD + AE + BC + BD + BE = 200
⇒ 50 + AD + 50 + 40 + 0 + 30 = 200
⇒ AD = 30
AD + AE + BD + BE + CD + CE = 190
⇒ 30 + 50 + 0 + 30 + CD + 60 = 190
⇒ CD = 20
Hence, the final information can be shown in the table
below.
The number of tickets booked to travel in exactly
one segment = B to C + C to D = 40 + 20 = 60.