CAT 2024 Slot 2 · DILR · Question 6

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.