CAT Exam Model Paper 2 with solutions for free online practice

The given metal cylinder is as follows
Here, radius(r) = $\frac{14}{2}=7\mathrm{cm}$

height (h) =10 cm
∴ Volume of cylinder = $\mathrm{\pi }{\mathit{r}}^2\mathit{h}=\frac{22}{7}\times 7^2\times 10$ =1540cm${}^3$
This cylinder is melted and recast into two cones in the proportion of 3: 4 (volume), keeping the height 10 cm. Let radius of first cone be ${\mathit{r}}_1$ cm and radius of another cone be r${}_2$ cm and volume of first cone be V${}_1$ cm${}^3$ and volume of second cone be V${}_2$ cm${}^3$
Then, $\frac{{\mathit{V}}_1}{{\mathit{V}}_2}=\frac{3}{4}$
$\Rightarrow \frac{\frac{1}{3}\mathrm{\pi }{\mathit{r}}_1^2\mathit{h}}{\frac{1}{3}\mathrm{\pi }{\mathit{r}}_2^2\mathit{h}}=\frac{3}{4}$
$\Rightarrow \frac{{\mathit{r}}_1^2}{{\mathit{r}}_2^2}=\frac{3}{4}\Rightarrow {\mathit{r}}_1^2=\frac{3}{4}{\mathit{r}}_2^2$ ...(i)
where, h = Height
${\mathit{r}}_1$ = Radius of 1st cone, ${\mathit{r}}_2$ = Radius of 2nd cone
${\mathrm{l}}_1$ = Slant height of 1st cone
${\mathrm{l}}_2$ = Slant height of 2nd cone
$\mathrm{NOW},{\mathit{V}}_1+{\mathit{V}}_2=1540{\mathrm{cm}}^3$
⇒$\frac{1}{3}\mathrm{\pi }{\mathit{r}}_1^2\mathit{h}+\frac{1}{3}\mathrm{\pi }{\mathit{r}}_2^2\mathit{h}=1540$
⇒$\frac{1}{3}\mathrm{\pi }\mathit{h}\left({\mathit{r}}_1^2+{\mathit{r}}_2^2\right)=1540$
⇒$\frac{1}{3}\times \frac{22}{7}\times 10\left({\mathit{r}}_1^2+{\mathit{r}}_2^2\right)=1540$
⇒${\mathit{r}}_1^2+{\mathit{r}}_2^2=\frac{154\times 21}{22}$
⇒${\mathit{r}}_1^2+{\mathit{r}}_2^2=147$...(ii)
⇒$\frac{3}{4}{\mathit{r}}_2^2+{\mathit{r}}_2^2=147$....[from Eq. (i)]
⇒$7{\mathit{r}}_2^2=147\times 4$
⇒${\mathit{r}}_2^2=147\times 47$
⇒${\mathit{r}}_2^2=84$
$\therefore {\mathit{r}}_2=2\sqrt{21}\mathrm{cm}$
From Eq. (i), ${\mathit{r}}_1^2=\frac{3}{4}\times 84=63$
∴${\mathit{r}}_1=\sqrt{63}\mathrm{cm}$
Now, flat surface area of the cylinder $=2\mathrm{\pi }{\mathit{r}}^2=2\mathrm{\pi }\times {\left(7\right)}^2=98\mathrm{\pi }{\mathrm{cm}}^2$
and flat surface area of the cones $=\mathrm{\pi }{\mathit{r}}_1^2+\mathrm{\pi }{\mathit{r}}_2^2=63\mathrm{\pi }+84\mathrm{\pi }$
$=147\mathrm{\pi }{\mathrm{cm}}^2$
∴ Required percentage change $=147\mathrm{\pi }-98\mathrm{\pi }98\mathrm{\pi }\times 100=\frac{4900}{98}=50\%$