How many houses are vacant in Block AA?
Answer and solution
Correct answer: 3
Solution
Answer: 3
1. Use the schematic to count only same-block occupied neighbours. In Block AA, only Q 2 can have a quoted price exactly 24: base 10 + road 5 + three neighbours ×3. All other houses’ possible prices fail to reach exactly 24.
2. Therefore Q 2 is vacant, has no parking, and P 2, Q 1, R 2 are occupied. Exactly one Row-1 house in AA is occupied, so P 1 and R 1 are vacant. AA has exactly three vacant houses.
3. In BB, T 2 cannot be the ₹15-lakh house: it already has road contribution 5, and at least one occupied neighbour. Otherwise S 2, U 2 would both be vacant, forcing S 1, U 1 both occupied and violating the one-occupied-house rule for Row 1.
4. Thus T 1 is priced at 15. It has no road contribution; 12+3×1=15 forces parking and exactly one occupied neighbour. Since T 2 is vacant, one of S 1, U 1 is occupied.
5. If S 1 were occupied, U 1 would be vacant and U 2 occupied. U 1, without parking or road adjacency, would cost 10+3=13, below BB’s minimum 15. Reject this case. Hence U 1 is occupied and S 1 vacant; S 2 is occupied to satisfy Column S.
6. U 2 is free to be vacant or occupied. BB vacant houses include S 1, T 1, T 2 and possibly U 2. The Row-2 vacant count is therefore 2 or 3. All these configurations satisfy the price constraints.
7. AA vacant houses are P 1, Q 2, R 1. TITA answer 3.
Verification: 7. AA vacant houses are P 1, Q 2, R 1. TITA answer 3.
1. Use the schematic to count only same-block occupied neighbours. In Block AA, only Q 2 can have a quoted price exactly 24: base 10 + road 5 + three neighbours ×3. All other houses’ possible prices fail to reach exactly 24.
2. Therefore Q 2 is vacant, has no parking, and P 2, Q 1, R 2 are occupied. Exactly one Row-1 house in AA is occupied, so P 1 and R 1 are vacant. AA has exactly three vacant houses.
3. In BB, T 2 cannot be the ₹15-lakh house: it already has road contribution 5, and at least one occupied neighbour. Otherwise S 2, U 2 would both be vacant, forcing S 1, U 1 both occupied and violating the one-occupied-house rule for Row 1.
4. Thus T 1 is priced at 15. It has no road contribution; 12+3×1=15 forces parking and exactly one occupied neighbour. Since T 2 is vacant, one of S 1, U 1 is occupied.
5. If S 1 were occupied, U 1 would be vacant and U 2 occupied. U 1, without parking or road adjacency, would cost 10+3=13, below BB’s minimum 15. Reject this case. Hence U 1 is occupied and S 1 vacant; S 2 is occupied to satisfy Column S.
6. U 2 is free to be vacant or occupied. BB vacant houses include S 1, T 1, T 2 and possibly U 2. The Row-2 vacant count is therefore 2 or 3. All these configurations satisfy the price constraints.
7. AA vacant houses are P 1, Q 2, R 1. TITA answer 3.
G Strategy shortcut
Method: Start with exact prices, not occupancy guesses. ₹24 forces Q 2 with three occupied neighbours; ₹15 in T forces T 1 with parking. Then eliminate S 1 occupied because U 1 would cost only ₹13.Verification: 7. AA vacant houses are P 1, Q 2, R 1. TITA answer 3.
G Strategy: Method: Start with exact prices, not occupancy guesses. ₹24 forces Q 2 with three occupied neighbours; ₹15 in T forces T 1 with parking. Then eliminate S 1 occupied because U 1 would cost only ₹13. Verification: 7. AA vacant houses are P 1, Q 2, R 1. TITA answer 3.