CAT 1996 Question Paper Question 17

Question 17

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 = $$100k^2C = 100000$$


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