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. 9:16
• B. 9:4
• C. 16:25
• D. 16:9
• E. 4:9

Hint:

When a quantity varies jointly as two other quantities, set up the proportionality equation $V = k \cdot d^2 \cdot t$ and use the given ratios.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The value $V$ of a coin is jointly proportional to the square of its diameter $d^2$ and its thickness $t$.
Step 2: Key Formula or Approach:
$V \propto d^2 \times t \implies \frac{V_1}{V_2} = (\frac{d_1}{d_2})^2 \times (\frac{t_1}{t_2})$.
Step 3: Detailed Explanation:
Given:
$\frac{d_1}{d_2} = \frac{4}{3}$ and $\frac{V_1}{V_2} = 4$.
Substitute these into the formula:
$4 = (\frac{4}{3})^2 \times (\frac{t_1}{t_2})$.
$4 = \frac{16}{9} \times \frac{t_1}{t_2}$.
$\frac{t_1}{t_2} = 4 \times \frac{9}{16}$.
$\frac{t_1}{t_2} = \frac{9}{4}$.
Step 4: Final Answer:
The ratio of their thickness is $9 : 4$.