X varies inversely as Y and Y varies inversely as Z. In a particular case, X = $$\frac{1}{16}$$ and Z = $$\frac{1}{32}$$. What will be the value of X when Z = 3? 

Correct option is A

Given:
X varies inversely as Y: X ∝ $$\frac 1Y$$​
Y varies inversely as Z: Y ∝ $$\frac 1Z$$​
Therefore, X ∝ Z (X is directly proportional to Z)
When X = $$\frac 1{16}$$​ and Z = $$\frac 1{32}$$​​
Formula Used:
If X ∝ Z, then $$\frac{X_1}{Z_1} = \frac{X_2}{Z_2}$$​​
Solution:
​$$\frac{1}{16} \div \frac{1}{32} = \frac{X}{3} \\\Rightarrow \frac{1}{16} \times \frac{32}{1} = \frac{X}{3} \\\Rightarrow 2 = \frac{X}{3} \\\Rightarrow X = 2 \times 3 = 6$$​​