CAT 2024 Slot 2 · DILR · Question 8

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.