CAT — Route Planning
9 questions, free to view. Click any question to see the answer and explanation.
The above is a schematic diagram of walkways (indicated by all the straight-lines) and lakes(3 of them, each in the shape of rectangles – shaded in the diagram) of a gated area. Different points on the walkway are indicated by letters (A through P) with distances being OP = 150 m, ON = MN = 300 m, ML = 400 m, EL = 200 m, DE = 400 m.
The following additional information about the facilities in the area is known.
1. The only entry/exit point is at C.
2. There are many residences within the gated area; all of them are located on the path AH and ML with four of them being at A, H, M, and L.
3. The post office is located at P and the bank is located at B.
One resident whose house is located at L, needs to visit the post office as well as the bank. What is the minimum distance (in m) he has to walk starting from his residence and returning to his residence after visiting both the post office and the bank?
[Note: In actual paper CAT 2024, two options (3) and (4) i.e., 3000 were same.]
3200
2700
3000
3400
3200
One person enters the gated area and decides to walk as much as possible before leaving the area without walking along any path more than once and always walking next to one of the lakes. Note that he may cross a point multiple times. How much distance (in m) will he walk within the gated area?
3200
3000
2800
3800
3800
One resident takes a walk within the gated area starting from A and returning to A without going through any point (other than A) more than once. What is the maximum distance (in m) she can walk in this way?
Visitors coming for morning walks are allowed to enter as long as they do not pass by any of the residences and do not cross any point (except C) more than once.What is the maximum distance (in m) that such a visitor can walk within the gated area?
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.
1. Segment C – D had an occupancy factor of 95%.
Only segment B – C had a higher occupancy factor.
2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.
3. Among the seats reserved on segment D – E,
exactly four-sevenths were from stations before C.
4. 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.
5. No tickets were booked from A to B, from B to D and from D to E.
6. The number of tickets booked for any segment was a multiple of 10.
What was the occupancy factor for segment D – E?
84%
35%
70%
77%
70%
Step 1:
From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.
Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.
From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.
Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.
Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.
An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.
Hence,
A to E + B to E = (4/7) × 140 = 80
Since B to E = 30,
A to E = 80 − 30 = 50
Therefore,
C to E = 140 − (50 + 30) = 60
Step 2:
From Conditions (1) and (6), the occupancy on segment C–D is
0.95 × 200 = 190
Also, the occupancy on segment B–C is 200.
For segment B–C,
AC + AD + AE + BC + BD + BE = 200
Substituting the known values,
50 + AD + 50 + 40 + 0 + 30 = 200
AD = 30
Similarly, for segment C–D,
AD + AE + BD + BE + CD + CE = 190
Substituting the known values,
30 + 50 + 0 + 30 + CD + 60 = 190
CD = 20
The completed booking table is shown below.
The occupancy on segment D–E is:
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 is
(140/200) × 100 = 70%.
How many tickets were booked from Station A to Station E?
Step 1:
From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.
Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.
From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.
Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.
Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.
An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.
Hence,
A to E + B to E = (4/7) × 140 = 80
Since B to E = 30,
A to E = 80 − 30 = 50
Therefore,
C to E = 140 − (50 + 30) = 60
Step 2:
From Conditions (1) and (6), the occupancy on segment C–D is
0.95 × 200 = 190
Also, the occupancy on segment B–C is 200.
For segment B–C,
AC + AD + AE + BC + BD + BE = 200
Substituting the known values,
50 + AD + 50 + 40 + 0 + 30 = 200
AD = 30
Similarly, for segment C–D,
AD + AE + BD + BE + CD + CE = 190
Substituting the known values,
30 + 50 + 0 + 30 + CD + 60 = 190
CD = 20
The completed booking table is shown below.
The number of tickets booked from Station A to Station E was 50.
How many tickets were booked from Station C?
Step 1:
From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.
Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.
From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.
Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.
Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.
An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.
Hence,
A to E + B to E = (4/7) × 140 = 80
Since B to E = 30,
A to E = 80 − 30 = 50
Therefore,
C to E = 140 − (50 + 30) = 60
Step 2:
From Conditions (1) and (6), the occupancy on segment C–D is
0.95 × 200 = 190
Also, the occupancy on segment B–C is 200.
For segment B–C,
AC + AD + AE + BC + BD + BE = 200
Substituting the known values,
50 + AD + 50 + 40 + 0 + 30 = 200
AD = 30
Similarly, for segment C–D,
AD + AE + BD + BE + CD + CE = 190
Substituting the known values,
30 + 50 + 0 + 30 + CD + 60 = 190
CD = 20
The completed booking table is shown below.
The number of tickets booked from Station C was = 20 + 60 = 80.
What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?
Step 1:
From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.
Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.
From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.
Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.
Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.
An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.
Hence,
A to E + B to E = (4/7) × 140 = 80
Since B to E = 30,
A to E = 80 − 30 = 50
Therefore,
C to E = 140 − (50 + 30) = 60
Step 2:
From Conditions (1) and (6), the occupancy on segment C–D is
0.95 × 200 = 190
Also, the occupancy on segment B–C is 200.
For segment B–C,
AC + AD + AE + BC + BD + BE = 200
Substituting the known values,
50 + AD + 50 + 40 + 0 + 30 = 200
AD = 30
Similarly, for segment C–D,
AD + AE + BD + BE + CD + CE = 190
Substituting the known values,
30 + 50 + 0 + 30 + CD + 60 = 190
CD = 20
The completed booking table is shown below.
Required difference = (50 + 40) – (30 + 20) = 40.
How many tickets were booked to travel in exactly one segment?
Step 1:
From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.
Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.
From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.
Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.
Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.
An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.
Hence,
A to E + B to E = (4/7) × 140 = 80
Since B to E = 30,
A to E = 80 − 30 = 50
Therefore,
C to E = 140 − (50 + 30) = 60
Step 2:
From Conditions (1) and (6), the occupancy on segment C–D is
0.95 × 200 = 190
Also, the occupancy on segment B–C is 200.
For segment B–C,
AC + AD + AE + BC + BD + BE = 200
Substituting the known values,
50 + AD + 50 + 40 + 0 + 30 = 200
AD = 30
Similarly, for segment C–D,
AD + AE + BD + BE + CD + CE = 190
Substituting the known values,
30 + 50 + 0 + 30 + CD + 60 = 190
CD = 20
The completed booking table is shown below.
The number of tickets booked to travel in exactly one segment = B to C + C to D = 40 + 20 = 60.
Want this as a timed attempt?
Log in free to attempt this topic with a real timer, analytics, streaks and bookmarks.