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.
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. Correct Answer Rs. 1 lakh
Given:
The cost of diamond varies directly as the square of its weight.
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.
Concept Used:
If A ∝ B, then we can write A = kB, for some constant k.
Calculation:
It is given that:
Price ∝ (Weight)2
⇒ Price = k(Weight)2
Let the weights of the four pieces now be x, 2x, 3x and 4x respectively.
New price = k(x)2 + k(2x)2 + k(3x)2 + k(4x)2
= 30kx2
Original price of unbroken diamond = k(x + 2x + 3x + 4x)2
= 100kx2
Reduction in price = 100kx2 - 30kx2 = 70kx2 = Rs. 70,000.
⇒ kx2 = 1000
And, original price = 100kx2 = Rs. 1,00,000.
