Price of a diamond is directly proportional to square of its weight. A man broke the diamond accidentally in three pieces in the ratio of 3 : 5 : 7 and thus loses Rs 42600. What was the original price (in Rs) of the diamond?

Let weight of original diamond = $$15x$$ and price of diamond = Rs. $$C$$

According to ques, => $$C \propto (15x)^2$$

=> $$C=225kx^2$$ --------------(i)

Let weight of each broken piece be $$3x,5x$$ and $$7x$$ g

Thus, cost of first piece (from equation (i)) = $$k(3x)^2=9kx^2$$

Similarly, price of each of the remaining pieces = $$25kx^2$$ and $$49kx^2$$

=> Total cost = $$9kx^2+25kx^2+49kx^2= 83kx^2$$

Amount lost = $$225kx^2-83kx^2=42600$$

=> $$kx^2=\frac{42600}{142}=300$$

$$\therefore$$ Original price of diamond = $$225\times300=Rs.$$ $$67,500$$

=> Ans - (C)