Question

The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1:2:3:4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.

• A. Rs. 1.4 lakh
• B. Rs. 2 lakh
• C. Rs. 1 lakh
• D. Rs. 2.1 lakh

Hint:

When dealing with variations of quantities, remember to use proportionality to express the relationships and solve for unknowns.

Answer 1

The Correct Option is B

Let the original weight of the diamond be \( x \). The cost of the diamond varies as the square of its weight, so the cost is proportional to \( x^2 \). Let the cost of the original diamond be \( kx^2 \), where \( k \) is the constant of proportionality. The diamond breaks into four pieces with weights in the ratio 1:2:3:4. Let the weight of each piece be \( x_1 = 1x \), \( x_2 = 2x \), \( x_3 = 3x \), and \( x_4 = 4x \). The cost of each piece is proportional to the square of its weight: \[ \text{Cost of piece 1} = k(1x)^2 = kx^2 \] \[ \text{Cost of piece 2} = k(2x)^2 = 4kx^2 \] \[ \text{Cost of piece 3} = k(3x)^2 = 9kx^2 \] \[ \text{Cost of piece 4} = k(4x)^2 = 16kx^2 \] The total cost of the four pieces is: \[ kx^2 + 4kx^2 + 9kx^2 + 16kx^2 = 30kx^2 \] The merchant got Rs. 70,000 less for the pieces than the original price, so: \[ 30kx^2 = kx^2 + 70000 \] \[ 29kx^2 = 70000 \] \[ kx^2 = \frac{70000}{29} \approx 2413.79 \] Thus, the original price of the diamond is approximately \( 2 \) lakh Rs.