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Explanation:

Consider the solution to first question of this set.

E1 costs 15 laks.

Let us calculate the maximum possible price of E2.
E2's base price is 10 lakh as it cannot have a parking space. (Only 1 house in YY i.e., E1 has a parking space.)
Road adjancecy for E2 = 1 and maximum neighbor count of E2 will be 2 (Both D2 and F2 are occupied and E1 is vacant)

∴ E2's price = 10 + 5 × 1 + 3 × 2 = 21 lakhs

Hence, 21.

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