Question

The value of each of a set of coins varies as the square of its diameter if its thickness remains constant, and it varies as the thickness if the diameter remains constant. If the diameter of two coins are in the ratio $4 : 3$, what should be the ratio of their thickness if the value of the first is four times that of the second?

• A. 16 : 9
• B. 9 : 4
• C. 9 : 16
• D. 4 : 9

Hint:

Always separate variation effects for each dimension and then combine for the total proportionality.

Answer 1

The Correct Option is B

Value $\propto d^2 \times t$. Given $\frac{d_1}{d_2} = \frac{4}{3}$, $\frac{V_1}{V_2} = 4$: $\frac{d_1^2 t_1}{d_2^2 t_2} = 4 \Rightarrow \frac{(16/9) t_1}{t_2} = 4 \Rightarrow \frac{t_1}{t_2} = \frac{4 \times 9}{16} = \frac{9}{4}$.