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RenB
Joined: 13 Jul 2022
Last visit: 06 Jul 2026
Posts: 388
Given Kudos: 302
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 388
10 Jul 2024, 06:37
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Bunuel
Attaching my working-
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WhatsApp Image 2024-07-10 at 7.05.23 PM.jpeg [ 44.15 KiB | Viewed 676 times ]
10 Jul 2024, 06:46
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IMO both children-
first find numbers of each-
SC is given 1
also given T < C,YA < MA
so only possibility -
T = 1
C = YA = 2
MA = 3
SC = 1
Given YA sits at leftmost, hence SC would be right most
YA - T - YA first 3 places
now since 2 of same type cant sit together, 4th, 6th and 8th has to be M
therefore seating-
YA - T - YA - MA - C - MA - C - MA - SC
first find numbers of each-
SC is given 1
also given T < C,YA < MA
so only possibility -
T = 1
C = YA = 2
MA = 3
SC = 1
Given YA sits at leftmost, hence SC would be right most
YA - T - YA first 3 places
now since 2 of same type cant sit together, 4th, 6th and 8th has to be M
therefore seating-
YA - T - YA - MA - C - MA - C - MA - SC
