If the sum of the ages of Krishna and Radha is 30 years, and after X years, the age of X will be twice the age of Radha, then find the value of X.

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Question 5: Given: Radius of the base, $r = 14 \text{ m}$ Total height of the solid, $H = 50 \text{ m}$ Height of the conical part, $h_c = 30 \text{ m}$

Step 1: Calculate the height of the cylindrical part. The height of the cylindrical part ($h_{cy}$) is the total height minus the height of the conical part. $h_{cy} = H - h_c = 50 m - 30 m = 20 \text{ m}$

a) Find the volume of the cylindrical part. Step 2: Calculate the volume of the cylindrical part ($V_{cy}$). The formula for the volume of a cylinder is $V_{cy} = \pi r^2 h_{cy}$. $V_{cy} = \pi (14 m)^2 (20 m)$ $V_{cy} = \pi (196 m^2) (20 m)$ $V_{cy} = 3920 \pi m^3$ Using $\pi \approx \frac{22}{7}$: $V_{cy} = \frac{22}{7} \times 196 \times 20 m^3$ $V_{cy} = 22 \times 28 \times 20 m^3$ $V_{cy} = 12320 m^3$ The volume of the cylindrical part is $\boxed{12320 m^3}$.

b) Find the total volume of the solid object. Step 3: Calculate the volume of the conical part ($V_c$). The formula for the volume of a cone is $V_c = \frac{1}{3} \pi r^2 h_c$. $V_c = \frac{1}{3} \pi (14 m)^2 (30 m)$ $V_c = \frac{1}{3} \pi (196 m^2) (30 m)$ $V_c = \pi (196 m^2) (10 m)$ $V_c = 1960 \pi m^3$ Using $\pi \approx \frac{22}{7}$: $V_c = \frac{1}{3} \times \frac{22}{7} \times 196 \times 30 m^3$ $V_c = 22 \times 28 \times 10 m^3$ $V_c = 6160 m^3$ Step 4: Calculate the total volume of the solid object ($V_{total}$). The total volume is the sum of the volume of the cylindrical part and the volume of the conical part. $V_{total} = V_{cy} + V_c$ $V_{total} = 12320 m^3 + 6160 m^3$ $V_{total} = 18480 m^3$ The total volume of the solid object is $\boxed{18480 m^3}$.

c) How many times less is the volume of the conical part than the volume of the cylindrical part? Step 5: Find the