Question

How many houses are vacant in Block XX?

• A. D2
• B. A1
• C. B1
• D. F2
• A. Either 2 or 3
• B. Exactly 3
• C. Exactly 2
• D. Either 3 or 4
• A. E2
• B. F2
• C. E1
• D. F1

Answer 1

The Correct Option is C

Given: 
Some houses are already taken, others are vacant and available for sale.
- Base price = Rs. 10 lakhs (no parking) or Rs. 12 lakhs (with parking)
- Final price = Base price $+ 5 \times$ (road adjacency value) $+ 3 \times$ (neighbor count)

The most expensive house in Block XX costs Rs. 24 lakhs.

Case 1: House with parking:
$12 + 5a + 3b = 24 \Rightarrow 5a + 3b = 12$
If $a = 0$, then $3b = 12 \Rightarrow b = 4$, which is invalid since max neighbors = 3.
Hence, no house with parking fits.

Case 2: House without parking:
$10 + 5a + 3b = 24 \Rightarrow 5a + 3b = 14$
Valid solution: $(a, b) = (1, 3)$
So, the house must have 3 neighbors and 1 road adjacent.
The only house that fits this is B2.

Neighbors of B2: B1, A2, C2 → All must be occupied.

But Row-1 has only one occupied house in Block XX. Since B1 is already occupied, A1 and C1 must be vacant.

So, occupied houses in Block XX: B1, A2, B2, C2 → Total = 4

But only 3 houses are allowed to be occupied (as per condition from earlier logic)
Thus, among A2 and C2, one must be vacant. But B2 needs 3 neighbors, so A2, B1, and C2 all must be occupied.

So final occupied in Block XX = 3 (A2, B1, C2)
Vacant in Block XX = 3

---

In Block YY:
- E1 and E2 are both vacant.
- One of them costs Rs. 15 lakhs.
Consider E1:

Neighbor count = 1 (either D1 or F1), Road adjacency = 0
Price = $(10 \text{ or } 12) + 0 + 3 \times 1 = 13 \text{ or } 15$ lakhs

Given minimum price = 15 lakhs ⇒ E1 must have parking ⇒ base = 12 ⇒ E1 price = 15 lakhs ⇒ E1 has parking

Now, since Row-1 has one occupied house in YY and E1 is vacant, D1 or F1 must be occupied

Assume F1 is vacant:
- F1 has no parking ⇒ base = 10
- Max neighbors = 1 (maybe F2), Road adjacency = 0
- Price = $10 + 0 + 3 = 13$ lakhs 
⇒ Invalid (min price is 15 lakhs)

Hence, F1 must be occupied ⇒ D1 is vacant

Since Column D must have one occupied house ⇒ D2 is occupied

F2 can be either occupied or vacant.

---

Final result:
Vacant houses in Block XX = 3

Answer: $3$

Answer 2

To determine which house is definitely occupied, let’s examine the conditions provided: 

1. Row-1 and Row-2 each have two occupied houses, one in each block:
This implies that in both Block XX and Block YY, there is at least one occupied house in each row.

2. Both houses in Column-E are vacant:
Since Column-E in Block YY has both houses vacant, none of the houses labeled "E1" and "E2" are occupied.

3. Each of Columns D and F has at least one occupied house:
This tells us that in Block YY, at least one house in Column-D and one in Column-F is occupied.

Now, we analyze each option:

• Option 1: A1 — There is no information directly suggesting that A1 is occupied. Incorrect.
• Option 2: B1 — This house is in Column-B of Block XX. The conditions indicate that Row-1 has one occupied house in Block XX. Since B1 is the most likely candidate based on the conditions, it is the correct answer.
• Option 3: D2 — This house is in Column-D of Block YY. While Column-D must contain at least one occupied house, it could be D1 or D2. Hence, we cannot be certain that D2 is occupied. Incorrect.
• Option 4: F2 — This house is in Column-F of Block YY. Since at least one house in Column-F must be occupied, it could be F1 or F2. Hence, F2 is uncertain. Incorrect.

Therefore, the only house that we can definitively conclude is occupied is: $B1$.

Answer 3

Answer: (A) Either 2 or 3 

Explanation:

• It is given that B2 and E2 are definitely vacant.
• From the conditions, at least one of D2 or F2 is occupied.

So, the possibilities for D2 and F2 are:

• Case 1: One of D2 or F2 is vacant → Total vacant houses in Row 2 = B2 + E2 + (D2 or F2) = 3
• Case 2: Both D2 and F2 are occupied → Total vacant houses in Row 2 = B2 + E2 = 2

Thus, Row 2 can have either 2 or 3 vacant houses.

Answer 4

Given: The price for an empty house is either Rs. 10 lakhs (without parking) or Rs. 12 lakhs (with parking).

In Block YY, both E1 and E2 are vacant, and one of them costs Rs. 15 lakhs. Let's analyze E1:

• Neighbor count = 1 (only one of D1 or F1 is occupied) 
• Road adjacency = 0
• Cost of E1 = \((10 \text{ or } 12) + 5 \times 0 + 3 \times 1 = 13 \text{ or } 15\) lakhs

Since Rs. 15 lakhs is the known cost of one vacant house and 15 is only possible if base price is 12 (i.e., with parking), E1 must have parking space.

---

Now for E2:

• Base price = Rs. 10 lakhs (no parking)
• Road adjacency = 1
• Maximum neighbors = 2 (D2 and F2 occupied, E1 is vacant)
• Price of E2 = \(10 + 5 \times 1 + 3 \times 2 = 21\) lakhs

Therefore, the maximum possible price of E2 is Rs. 21 lakhs.

Answer 5

To determine which house in Block YY has parking space, let’s review the information provided:

• Only one house in Block YY has parking space. 
• Column E contains both vacant houses in Block YY.
• So the house with parking must be either E1 or E2.
• Since no information points to F1 or F2 having parking, and only one house in Block YY can have it, the best choice is E1.

Therefore, E1 is the house in Block YY that has parking space.