1) Persons X, Y, Z and Q live in red, green, yellow or blue-coloured houses placed in a sequence on a street. Z lives in a yellow house. The green house is adjacent to the blue house. X does not live adjacent to Z. The yellow house is in between the green and red houses. The colour of the house, X lives in is
[CAT 2000]
(1) blue
(2) green
(3) red
(4) Not possible to determine
Solution
“Persons X, Y, Z and Q live in red, green, yellow or blue-coloured houses placed in a sequence on a street.” From this statement, we can infer that this is a
Linear Arrangement case.
Clue 1: “Z” lives in the yellow house, this is a direct clue.
Clue 2: the green house is adjacent to the blue house. This clue has two possibilities as the order isn't specified. The two possibilities are:
(i) (Green, Blue)
(ii) (Blue, Green)
Clue 3: “X” does not live adjacent to “Z”. From this clue, we can say, “X” is not to the immediate left or right of “Z” in other words “X” is not the immediate neighbour of “Z”.
Clue 4: The yellow house is in between the green and red house. Since the order of the green and red house is not given, there are two possibilities with this clue:
(i) (Green, Yellow, Red)
(ii) (Red, Yellow, Green)
Case Solution
From clues 2 and 4, we can conclude the green house cannot be the first or the last house in the sequence.
If green was the first house, blue has to be the second house (from clue 2) but yellow should be in-between the green and red house. (from clue 4)
Similarly, green cannot be the last house either as it does not satisfy both clue 2 and clue 4.
Hence the green house is either the 2nd or the 3rd house in the sequence. Based on this we get two different cases:
Case 1:
Case 2:
From clue 3, “X” is not the immediate neighbour of “Z”. Hence, from cases 1 and 2 we can conclude that “X” cannot be living in Red and Green coloured houses.
Case 1:
Case 2:
Hence from both the cases, we can conclude that X lives in the Blue house.
Answer: (1) blue