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: Value $V \propto d^2$ when thickness $t$ constant, and $V \propto t$ when $d$ constant.
Step 2: Combining, $V \propto d^2 \cdot t$. So $V = k \cdot d^2 \cdot t$ for some constant $k$.
Step 3: Let $d_1 : d_2 = 4 : 3$. So $d_1 = 4x$, $d_2 = 3x$.
Step 4: Given $V_1 = 4V_2$. So $k \cdot (4x)^2 \cdot t_1 = 4 \cdot k \cdot (3x)^2 \cdot t_2$.
Step 5: Cancel $k$ and $x^2$: $16 t_1 = 4 \cdot 9 t_2 = 36 t_2$.
Step 6: $t_1 = \frac{36}{16} t_2 = \frac{9}{4} t_2$.
Step 7: Ratio $t_1 : t_2 = 9 : 4$.
Step 8: Final Answer: The ratio of their thickness is 9:4.