CAT 2024 Slot 2 · DILR · Set 2: Walkways and Lakes in a Gated Area
The full set and all 4 questions (Q6–Q9) from CAT 2024 Slot 2, with answers and solutions.
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.
Question 6 · Quant Based Puzzles and Optimisation
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?
- A
3400
- B
2700
- C
3000
- D
3200
Answer & solution
Answer: Option D
Solving the set
Step 1 – Lengths of all paths. The diagram is a grid of 4 columns of points and 4 rows of points. Using the given distances (OP = 150, ON = MN = 300 for the widths; ML = 400, EL = 200, DE = 400 for the heights):
|
Line |
Segments and lengths (m) |
|
Columns (widths) |
A–B, H–G, I–J, P–O = 150; B–C, G–F, J–K, O–N = 300; C–D, F–E, K–L, N–M = 300 |
|
Rows (heights) |
A–H, B–G, C–F, D–E = 400; H–I, G–J, F–K, E–L = 200; I–P, J–O, K–N, L–M = 400 |
|
Diagonal I–G |
√(150² + 200²) = 250 |
|
Diagonal O–K |
√(300² + 400²) = 500 |
Step 2 – Special points. Entry/exit is only at C. Residences lie on paths A–H and M–L (and at A, H, M, L). The post office is at P and the bank at B.
Step 3 – Lake-side paths. The three shaded lakes are C-D-E-F, G-F-K-J and K-L-M-N. The paths along their sides are listed below.
|
Lake |
Paths next to it |
Length (m) |
|
Lake 1 (C, D, E, F) |
C–D 300, D–E 400, E–F 300, F–C 400 |
1400 |
|
Lake 2 (G, F, K, J) |
G–F 300, F–K 200, K–J 300, J–G 200 |
1000 |
|
Lake 3 (K, L, M, N) |
K–L 300, L–M 400, M–N 300, N–K 400 |
1400 |
|
Total |
12 paths |
3800 |
Step 4 – Useful shortest routes (used in the first question).
|
Leg |
Shortest route |
Distance (m) |
|
L → P (post office) |
L–K–O–P = 300 + 500 + 150 |
950 |
|
P → B (bank) |
P–I–G–B = 400 + 250 + 400 |
1050 |
|
B → L (home) |
B–C–F–K–L = 300 + 400 + 200 + 300 (or B–G–F–E–L) |
1200 |
|
Total |
3200 |
This question
The resident must go from L to P, from P to B, and from B back to L (or the reverse order, which gives the same total). The minimum is the sum of the three shortest legs (Step 4 table).
• L → P: the diagonal K–O (500) is shorter than going K–N–O (700) or K–J–O (700), so L–K–O–P = 950.
• P → B: the diagonal I–G (250) beats I–H–G (350) or I–J–G (350), so P–I–G–B = 1050.
• B → L: B–C–F–K–L = 1200 (B–G–F–E–L is also 1200).
Total = 950 + 1050 + 1200 = 3200 m.
Hence, the answer is option 4.
Question 7 · Quant Based Puzzles and Optimisation
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?
- A
3800
- B
2800
- C
3200
- D
3000
Answer & solution
Answer: Option A
Solving the set
Step 1 – Lengths of all paths. The diagram is a grid of 4 columns of points and 4 rows of points. Using the given distances (OP = 150, ON = MN = 300 for the widths; ML = 400, EL = 200, DE = 400 for the heights):
|
Line |
Segments and lengths (m) |
|
Columns (widths) |
A–B, H–G, I–J, P–O = 150; B–C, G–F, J–K, O–N = 300; C–D, F–E, K–L, N–M = 300 |
|
Rows (heights) |
A–H, B–G, C–F, D–E = 400; H–I, G–J, F–K, E–L = 200; I–P, J–O, K–N, L–M = 400 |
|
Diagonal I–G |
√(150² + 200²) = 250 |
|
Diagonal O–K |
√(300² + 400²) = 500 |
Step 2 – Special points. Entry/exit is only at C. Residences lie on paths A–H and M–L (and at A, H, M, L). The post office is at P and the bank at B.
Step 3 – Lake-side paths. The three shaded lakes are C-D-E-F, G-F-K-J and K-L-M-N. The paths along their sides are listed below.
|
Lake |
Paths next to it |
Length (m) |
|
Lake 1 (C, D, E, F) |
C–D 300, D–E 400, E–F 300, F–C 400 |
1400 |
|
Lake 2 (G, F, K, J) |
G–F 300, F–K 200, K–J 300, J–G 200 |
1000 |
|
Lake 3 (K, L, M, N) |
K–L 300, L–M 400, M–N 300, N–K 400 |
1400 |
|
Total |
12 paths |
3800 |
Step 4 – Useful shortest routes (used in the first question).
|
Leg |
Shortest route |
Distance (m) |
|
L → P (post office) |
L–K–O–P = 300 + 500 + 150 |
950 |
|
P → B (bank) |
P–I–G–B = 400 + 250 + 400 |
1050 |
|
B → L (home) |
B–C–F–K–L = 300 + 400 + 200 + 300 (or B–G–F–E–L) |
1200 |
|
Total |
3200 |
This question
He may only walk on lake-side paths, each at most once, starting and ending at C. The 12 lake-side paths total 3800 m (Step 3).
To use all of them once and come back to C, every point must have an even number of these paths (so he can leave every point he enters).
|
Point |
Lake-side paths meeting there |
Count |
|
C, D, E, G, J, L, M, N |
2 each |
Even |
|
F |
F–C, F–E, F–G, F–K |
4 (even) |
|
K |
K–F, K–J, K–L, K–N |
4 (even) |
All counts are even and the paths are connected, so such a walk exists, for example C–D–E–F–G–J–K–L–M–N–K–F–C. It covers all 12 paths: 300 + 400 + 300 + 300 + 200 + 300 + 300 + 400 + 300 + 400 + 200 + 400 = 3800 m. He passes F and K twice, which is allowed.
Hence, the answer is option 1.
Question 8 · Quant Based Puzzles and Optimisation
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?
Answer & solution
Answer: 5100
Solving the set
Step 1 – Lengths of all paths. The diagram is a grid of 4 columns of points and 4 rows of points. Using the given distances (OP = 150, ON = MN = 300 for the widths; ML = 400, EL = 200, DE = 400 for the heights):
|
Line |
Segments and lengths (m) |
|
Columns (widths) |
A–B, H–G, I–J, P–O = 150; B–C, G–F, J–K, O–N = 300; C–D, F–E, K–L, N–M = 300 |
|
Rows (heights) |
A–H, B–G, C–F, D–E = 400; H–I, G–J, F–K, E–L = 200; I–P, J–O, K–N, L–M = 400 |
|
Diagonal I–G |
√(150² + 200²) = 250 |
|
Diagonal O–K |
√(300² + 400²) = 500 |
Step 2 – Special points. Entry/exit is only at C. Residences lie on paths A–H and M–L (and at A, H, M, L). The post office is at P and the bank at B.
Step 3 – Lake-side paths. The three shaded lakes are C-D-E-F, G-F-K-J and K-L-M-N. The paths along their sides are listed below.
|
Lake |
Paths next to it |
Length (m) |
|
Lake 1 (C, D, E, F) |
C–D 300, D–E 400, E–F 300, F–C 400 |
1400 |
|
Lake 2 (G, F, K, J) |
G–F 300, F–K 200, K–J 300, J–G 200 |
1000 |
|
Lake 3 (K, L, M, N) |
K–L 300, L–M 400, M–N 300, N–K 400 |
1400 |
|
Total |
12 paths |
3800 |
Step 4 – Useful shortest routes (used in the first question).
|
Leg |
Shortest route |
Distance (m) |
|
L → P (post office) |
L–K–O–P = 300 + 500 + 150 |
950 |
|
P → B (bank) |
P–I–G–B = 400 + 250 + 400 |
1050 |
|
B → L (home) |
B–C–F–K–L = 300 + 400 + 200 + 300 (or B–G–F–E–L) |
1200 |
|
Total |
3200 |
This question
She needs the longest loop from A that does not repeat any point. The best loop passes through all 16 points:
|
Step |
Path |
Length (m) |
Running total |
|
1 |
A–B |
150 |
150 |
|
2 |
B–G |
400 |
550 |
|
3 |
G–F |
300 |
850 |
|
4 |
F–C |
400 |
1250 |
|
5 |
C–D |
300 |
1550 |
|
6 |
D–E |
400 |
1950 |
|
7 |
E–L |
200 |
2150 |
|
8 |
L–M |
400 |
2550 |
|
9 |
M–N |
300 |
2850 |
|
10 |
N–K |
400 |
3250 |
|
11 |
K–J |
300 |
3550 |
|
12 |
J–O |
400 |
3950 |
|
13 |
O–P |
150 |
4100 |
|
14 |
P–I |
400 |
4500 |
|
15 |
I–H |
200 |
4700 |
|
16 |
H–A |
400 |
5100 |
Why the diagonals cannot help: in a loop through all 16 points every point uses exactly two paths, and each corner (A, D, M, P) has only two paths, so A–B, A–H, D–C, D–E, M–L, M–N, P–O and P–I are all used.
• If O–K is used: O is full (P, K). N must use N–M and N–K, so K is full (O, N). J can then only use J–G and J–I, so I is full (P, J) and H must use H–A and H–G, making G full (J, H). L must use L–E, so E is full (D, L). F is left with only one free neighbour (C) – impossible.
• If I–G is used: I is full (P, G). H must use H–A and H–G, so G is full (I, H). J can then only use J–K and J–O, so O is full (P, J), N must use N–M and N–K, and K is full (J, N). E is full (D, L) again, and F is left with only C – impossible.
So a loop through all 16 points uses only grid lines, and the loop above gives 5100 m. Skipping points to use a diagonal gives a shorter loop.
Hence, the answer is 5100.
Question 9 · Quant Based Puzzles and Optimisation
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?
Answer & solution
Answer: 3500
Solving the set
Step 1 – Lengths of all paths. The diagram is a grid of 4 columns of points and 4 rows of points. Using the given distances (OP = 150, ON = MN = 300 for the widths; ML = 400, EL = 200, DE = 400 for the heights):
|
Line |
Segments and lengths (m) |
|
Columns (widths) |
A–B, H–G, I–J, P–O = 150; B–C, G–F, J–K, O–N = 300; C–D, F–E, K–L, N–M = 300 |
|
Rows (heights) |
A–H, B–G, C–F, D–E = 400; H–I, G–J, F–K, E–L = 200; I–P, J–O, K–N, L–M = 400 |
|
Diagonal I–G |
√(150² + 200²) = 250 |
|
Diagonal O–K |
√(300² + 400²) = 500 |
Step 2 – Special points. Entry/exit is only at C. Residences lie on paths A–H and M–L (and at A, H, M, L). The post office is at P and the bank at B.
Step 3 – Lake-side paths. The three shaded lakes are C-D-E-F, G-F-K-J and K-L-M-N. The paths along their sides are listed below.
|
Lake |
Paths next to it |
Length (m) |
|
Lake 1 (C, D, E, F) |
C–D 300, D–E 400, E–F 300, F–C 400 |
1400 |
|
Lake 2 (G, F, K, J) |
G–F 300, F–K 200, K–J 300, J–G 200 |
1000 |
|
Lake 3 (K, L, M, N) |
K–L 300, L–M 400, M–N 300, N–K 400 |
1400 |
|
Total |
12 paths |
3800 |
Step 4 – Useful shortest routes (used in the first question).
|
Leg |
Shortest route |
Distance (m) |
|
L → P (post office) |
L–K–O–P = 300 + 500 + 150 |
950 |
|
P → B (bank) |
P–I–G–B = 400 + 250 + 400 |
1050 |
|
B → L (home) |
B–C–F–K–L = 300 + 400 + 200 + 300 (or B–G–F–E–L) |
1200 |
|
Total |
3200 |
This question
The visitor starts and ends at C and may not use A, H, M, L or the paths A–H and M–L. The usable points are B, C, D, E, F, G, I, J, K, N, O, P (12 points).
D can only be reached from C or E, and E (without L) only from D or F, so the loop must contain C–D–E–F. The longest loop through all 12 points is:
|
Step |
Path |
Length (m) |
Running total |
|
1 |
C–B |
300 |
300 |
|
2 |
B–G |
400 |
700 |
|
3 |
G–J |
200 |
900 |
|
4 |
J–I |
150 |
1050 |
|
5 |
I–P |
400 |
1450 |
|
6 |
P–O |
150 |
1600 |
|
7 |
O–N |
300 |
1900 |
|
8 |
N–K |
400 |
2300 |
|
9 |
K–F |
200 |
2500 |
|
10 |
F–E |
300 |
2800 |
|
11 |
E–D |
400 |
3200 |
|
12 |
D–C |
300 |
3500 |
Hence, the answer is 3500.
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