‘x’ varies directly with the cube of ‘y’, and inversely with the square of ‘z’. If x = 1/36 when y = 2 and z = 3, then what is the value of 800x when y = 3 and z = 5?

Correct option is B

​$$\frac{1}{36} = k \times \frac{2^3}{3^2}$$​​

​$$\frac{1}{36} = k \times \frac{8}{9}$$​​

​$$\frac{9}{36} = 8k$$​​

​$$\frac{1}{4} = 8k$$​​

​$$k = \frac{1}{32}$$​​

​$$x = \frac{1}{32} \times \frac{3^3}{5^2}$$​​

​$$x = \frac{1}{32} \times \frac{27}{25}$$​​

​$$x = \frac{27}{800}$$​​