Subject: General Aptitude · Chapter: Ratio and Proportion · Exam: · Marks: · Difficulty:
The price of a diamond varies as the cube of its volume. A cubical variety of this diamond was worth ₹ 10,00,000. If this diamond accidentally broke into 8 equal cubical diamonds, then the total loss in value amounts to
(a)₹ 9,00,000
(b)₹ 9,47,532
(c)₹ 9,50,000
(d)₹ 9,84,375
Answer
Answer (as printed): D
Explanation
Let the volume of each small piece be $x$ cu . units. Then, original volume of the diamond $=(8 x) \mathrm{cu}$. units. Original price of the diamond $=k \times(8 x)^{3}=512 k x^{3}$, where $k$ is a constant. $$512 k x^{3}=1000000 \Rightarrow k x^{3}=\frac{1000000}{512} .$$ Price of each smaller piece $=k x^{3}$. Total price of the 8 pieces $=8 k x^{3}$. $$\begin{aligned} \therefore \text { Loss in value } & =\left(512 k x^{3}-8 k x^{3}\right)=504 k x^{3} \\ & =504 \times \frac{1000000}{512}=984375 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.