CAT 2024 Slot 2 · DILR · Question 7

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.