CAT 2015 | Question: 75

Question

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

Answer 1

Given that, the ratio of weights of four pieces $ = \text{P}_{1} : \text{P}_{2} : \text{P}_{3} : \text{P}_{4} : = 1:2:3:4 $

so weights are : 

• $\text{P}_{1} = k$
• $\text{P}_{2} = 2k$
• $\text{P}_{3} = 3k$
• $\text{P}_{4} = 4k$

The cost of a diamond varies directly as the square of its weight.

So costs are :

• $\text{P}_{1} = k^{2}$
• $\text{P}_{2} = (2k)^{2} = 4k^{2}$
• $\text{P}_{3} = (3k)^{2} = 9k^{2}$
• $\text{P}_{4} = (4k)^{2} = 16k^{2}$

The weight of original diamond $ = k+2k+3k+4k = 10k$

The cost of original diamond $ =  (10k)^{2} = 100k^{2}$

Now, $100k^{2} – (k^{2} + 4k^{2} + 9k^{2} + 16k^{2}) = 70000 $

$\Rightarrow 100k^{2} – 30k^{2} = 70000$

$\Rightarrow 70k^{2} = 70000$

$\Rightarrow \boxed{k^{2} = 1000}$

$\therefore$ The price of original diamond $ = 100k^{2} = 100 \times 1000$

Correct Answer $:100000  =$ ₹ $100000$