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.

Name: 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.
Uploaded: Oct. 19, 2021, 12:01 p.m.
Duration: 1 min 35 s
Description: 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.

Solution

Let the original weight of the diamond be equal to $$10k$$. So, after breaking into 4 pieces, the parts of the diamond weight $$k, 2k, 3k,4k$$

The price of the diamond varies directly in proportion to the weight. Let us assume, the $$P=C*W^2$$ where $$C$$ is a constant and $$W$$ is the weight of the diamond.

Therefore, the original price is $$C*10k*10k = 100k^2*C$$
The new weight is $$Ck^2 + C(2k)^2 + C(3k)^2 + C(4k)^2 = 30k^2C$$

The decrease in the price equals 70,000. So, $$100k^2C-30k^2C = 70000$$

Or, $$k^2C = 1000$$

Therefore the original price = $$100k^2C = 100000$$