Question

In a township, the price of each house was 25,00,000 (in Indian Rupees) last month. The number of houses sold in a month (Q in thousands) is sensitive to the price of the house (P in Indian Rupees) and establishes a relationship as \[ Q = 6685 - 0.00158P. \] If the price of each house increases by 20% in the current month, then the decrease in sale of the houses (in percentage, rounded off to two decimal places) compared to last month will be \(\underline{\hspace{1cm}}\).

Hint:

To calculate percentage decrease, find the difference between old and new values, then divide by the old value and multiply by 100.

Answer 1

Correct Answer: 27 - 30

Let the initial price of the house be \( P_0 = 25,00,000 \) INR. From the equation given, the initial number of houses sold, \( Q_0 \), is: \[ Q_0 = 6685 - 0.00158 \times 25,00,000 = 6685 - 3950 = 2735 \text{ (in thousands)}. \] Now, the price increases by 20%, so the new price \( P_1 \) is: \[ P_1 = 1.2 \times P_0 = 1.2 \times 25,00,000 = 30,00,000 \text{ INR}. \] The new number of houses sold \( Q_1 \) is: \[ Q_1 = 6685 - 0.00158 \times 30,00,000 = 6685 - 4740 = 1945 \text{ (in thousands)}. \] The percentage decrease in the sale of houses is: \[ \text{Percentage decrease} = \frac{Q_0 - Q_1}{Q_0} \times 100 = \frac{2735 - 1945}{2735} \times 100 \approx 29.00%. \]