A solid metal cylinder of $12\text{cm}$ height and $7\text{cm}$ radius is melted and recast into two cones in the proportion of $1:2$ (volume), keeping the height $12\text{cm}$. What would be the percentage change in the flat surface area before and after? (Use $\pi =\frac{22}{7}$)

The correct answer is 50%.

Let us solve the problem step-by-step.

Step 1: Calculate the volume and flat surface area of the initial solid metal cylinder.
Let the radius of the cylinder be $R=7\text{cm}$ and its height be $H=12\text{cm}$.
The volume of a cylinder is given by the formula:
$V_{\text{cylinder}}=\pi ·R^2·H$
The flat surface area of a cylinder consists of its two circular ends (top and bottom bases):
$A_{\text{flat, initial}}=2·\pi ·R^2$
Substituting the given values:
$A_{\text{flat, initial}}=2·\frac{22}{7}·7^2=2·22·7=308{\text{cm}}^2$

Step 2: Determine the volumes of the two recast cones.
The total volume of the metal remains constant after melting and recasting.
$V_{\text{total}}=\pi ·7^2·12=588\pi {\text{cm}}^3$
This total volume is divided into two cones in the volume ratio of 1:2.
Let the volumes of the two cones be $V_1$ and $V_2$ such that:
$V_1=\frac{1}{3}·V_{\text{total}}=\frac{1}{3}·588\pi =196\pi {\text{cm}}^3$
$V_2=\frac{2}{3}·V_{\text{total}}=\frac{2}{3}·588\pi =392\pi {\text{cm}}^3$

Step 3: Calculate the radius and base area of each cone.
Each cone has a height of $h=12\text{cm}$.
The volume of a cone is given by:
$V=\frac{1}{3}·\pi ·r^2·h$
For the first cone (with radius $r_1$):
$196\pi =\frac{1}{3}·\pi ·r_1^2·12$
$196=4·r_1^2$
$r_1^2=49\Rightarrow r_1$
The flat surface area of a cone is its circular base area, which is $\pi ·r^2$.
Thus, the base area of the first cone is:
$A_1=\pi ·r_1^2=49\pi$
For the second cone (with radius $r_2$):
$392\pi =\frac{1}{3}·\pi ·r_2^2·12$
$392=4·r_2^2$
$r_2^2=98$
Thus, the base area of the second cone is:
$A_2=\pi ·r_2^2=98\pi$

Step 4: Calculate the total flat surface area after recasting.
The flat surface area of the two cones together is the sum of their base areas: