CAT Exam Model Paper 2 with solutions for free online practice

The given metal cylinder is as follows Here, radius(r) = $\frac{14}{2}=7\text{ cm}$ height (h) =10 cm ∴ Volume of cylinder = $\pi r^{2} h = \frac{22}{7} \times 7^{2} \times 10$ =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 $r_{1}$ 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{V_{1}}{V_{2}} = \frac{3}{4}$$\Rightarrow \frac{\frac{1}{3}\pi r_{1}^{2} h}{\frac{1}{3}\pi r_{2}^{2} h} = \frac{3}{4}$$\Rightarrow \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{3}{4} \Rightarrow r_{1}^{2} = \frac{3}{4} r_{2}^{2}$ ...(i)where, h = Height$r_1$ = Radius of 1st cone, $r_2$ = Radius of 2nd cone$l_1$ = Slant height of 1st cone$l_2$ = Slant height of 2nd cone$\text{NOW},\ V_{1} + V_{2} = 1540\ \text{cm}^{3}$⇒$\frac{1}{3}\pi r_{1}^{2}h + \frac{1}{3}\pi r_{2}^{2}h = 1540$⇒$\frac{1}{3}\pi h (r_{1}^{2} + r_{2}^{2}) = 1540$⇒$\frac{1}{3} \times \frac{22}{7} \times 10 (r_{1}^{2} + r_{2}^{2}) = 1540$⇒$r_{1}^{2} + r_{2}^{2} = \frac{154 \times 21}{22}$⇒$r_{1}^{2} + r_{2}^{2} = 147$...(ii)⇒$\frac{3}{4} r_{2}^{2} + r_{2}^{2} = 147$....[from Eq. (i)]⇒$7 r_{2}^{2} = 147 \times 4$⇒$r_{2}^{2} = \frac{147 \times 4}{7}$⇒$r_{2}^{2} = 84$ $\therefore r_{2} = 2\sqrt{21}\ \text{cm}$From Eq. (i), $r_{1}^{2} = \frac{3}{4} \times 84 = 63$∴$r_{1} = \sqrt{63}\ \text{cm}$Now, flat surface area of the cylinder $= 2\pi r^{2} = 2\pi \times (7)^{2} = 98\pi\ \text{cm}^{2}$and flat surface area of the cones $= \pi r_{1}^{2} + \pi r_{2}^{2} = 63\pi + 84\pi$$= 147\pi\ \text{cm}^{2}$∴ Required percentage change $= \frac{147\pi - 98\pi}{98\pi} \times 100 = \frac{4900}{98} = 50\%$