Volume And Surface Area - Aptitude MCQ Questions and Solutions with Explanations

Volume of pyramid :
$$\eqalign{ & = \frac{1}{3} \times {\text{Area of base}} \times {\text{Height}} \cr & = \left( {\frac{1}{3} \times 25 \times 9} \right){\text{c}}{{\text{m}}^3} \cr & = 75\,{\text{c}}{{\text{m}}^3} \cr} $$

Volume of cylinder = $$\pi {r^2}h$$
∴ Curved surface area of cylinder = $$2\pi rh$$
$$\eqalign{ & \therefore \frac{{\pi {r^2}h}}{{2\pi rh}} = \frac{{616}}{{352}} \cr & \Rightarrow r = \frac{{2 \times 616}}{{352}} \cr & \Rightarrow r = 3.5\,m \cr} $$
∴ Volume of cylinder :
$$\eqalign{ & \Rightarrow \pi {r^2}h = 616 \cr & \Rightarrow \frac{{22}}{7} \times 3.5 \times 3.5 \times h = 616 \cr & \Rightarrow 11 \times 3.5 \times h = 616 \cr & \Rightarrow h = \frac{{616}}{{11 \times 3.5}} \cr & \Rightarrow h = 16 \cr} $$
∴ Total surface area of the cylinder :
$$\eqalign{ & = 2\pi rh + 2\pi {r^2} \cr & = 2\pi r\left( {h + r} \right) \cr & = 2 \times \frac{{22}}{7} \times 3.5\left( {16 + 3.5} \right) \cr & = 2 \times \frac{{22}}{7} \times 3.5\left( {19.5} \right) \cr & = 22 \times 19.5 \cr & = 429\,{\text{sq}}{\text{.m}} \cr} $$

Let the radius of a right circular cine be R cm and height be H cm
Volume of right circular cone $$ = \frac{1}{3}\pi {R^2}H{\text{ cu}}{\text{.cm}}$$
When height of right circular cone is increased by 200% and radius of the base is reduce by 50%
New volume :
$$\eqalign{ & {\text{ = }}\frac{1}{3}\pi {\left( {\frac{R}{2}} \right)^2}.3H \cr & = \frac{1}{3}\pi \frac{{{R^2}4}}{4}.3H \cr & = \frac{{\pi {R^2}H}}{4} \cr} $$
Difference :
$$\eqalign{ & = \pi {R^2}H\left( {\frac{1}{3} - \frac{1}{4}} \right) \cr & = \frac{1}{{12}}\pi {R^2}H \cr} $$
Decrease percentage :
$$\eqalign{ & = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr & = 25\% \cr} $$

$$\eqalign{ & \pi {r^2} = 346.5 \cr & {r^2} = \left( {346.5 \times \frac{7}{{22}}} \right) \cr & {r^2} = \frac{{441}}{4} \cr & {r^2} = \frac{21}{2} \cr & \therefore l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\,\,\,\, = \sqrt {\frac{{441}}{4} + {{\left( {14} \right)}^2}} \cr & \,\,\,\,\,\,\, = \sqrt {\frac{{1225}}{4}} \cr & \,\,\,\,\,\,\, = \frac{{35}}{2} \cr} $$
So, area of canvas needed :
$$\eqalign{ & \pi rl = \left( {\frac{{22}}{7} \times \frac{{21}}{2} \times \frac{{35}}{2}} \right){{\text{m}}^2} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{33 \times 35}}{2}} \right){{\text{m}}^2} \cr} $$
∴ Length of canvas :
$$\eqalign{ & = \left( {\frac{{33 \times 35}}{{2 \times 1.1}}} \right){\text{ m}} \cr & = 525{\text{ m}} \cr} $$

Radius of hemisphere bowl = 6 cm
∴ Volume of hemisphere :
$$\eqalign{ & = \frac{2}{3}\pi {r^3} \cr & = \frac{2}{3} \times \frac{{22}}{7} \times 6 \times 6 \times 6 \cr & = \frac{{9504}}{{21}} \cr & = 452.57{\text{ c}}{{\text{m}}^3} \cr} $$

Let the radius of the cylinder be r
Then, radius of the sphere = 2r
$$\eqalign{ & \frac{{{\text{Volume of cylinder}}}}{{{\text{Volume of sphere}}}} = \frac{3}{2} \cr & \Rightarrow \frac{{\pi {r^2}h}}{{\frac{4}{3}\pi {{\left( {2r} \right)}^3}}} = \frac{3}{2} \cr & \Rightarrow \frac{h}{r} = 16 \cr & \Rightarrow h = 16r \cr} $$
∴ Required ratio :
$$\eqalign{ & \frac{{{\text{Total surface area of cylinder}}}}{{{\text{Surface area of sphere}}}} \cr & = \frac{{2\pi r.\left( {16r} \right) + 2\pi {r^2}}}{{4\pi {{\left( {2r} \right)}^2}}} \cr & = \frac{{34\pi {r^2}}}{{16\pi {r^2}}} \cr & = \frac{{17}}{8}\,Or\,17:8 \cr} $$

If R is the radius of sphere, volume of the sphere = $$\frac{4}{3}\pi {R^3}$$
When radius of sphere is increased by 10%
New volume :
$$\eqalign{ & = \frac{4}{3}\pi {\left( {1.1R} \right)^3} \cr & = \frac{4}{3}\pi {R^3}\left( {1.331} \right) \cr} $$
Difference :
$$\eqalign{ & = \frac{4}{3}\pi {R^3}\left( {1.331} \right) - \frac{4}{3}\pi {R^3} \cr & = \frac{4}{3}\pi {R^3}\left( {1.331 - 1} \right) \cr & = \frac{4}{3}\pi {R^3}\left( {0.331} \right) \cr} $$
Increase % :
$$\eqalign{ & = \frac{{\frac{4}{3}\pi {R^3}\left( {0.331} \right)}}{{\frac{4}{3}\pi {R^3}}} \times 100 \cr & = 33.1\% \cr} $$

Volume flown in 5 hours :
$$\eqalign{ & = \left( {225 \times 162 \times \frac{{20}}{{100}}} \right){{\text{m}}^3} \cr & = 7290\,{{\text{m}}^3} \cr} $$
Volume flown in 1 hour :
$$\eqalign{ & = \left( {\frac{{7290}}{5}} \right){{\text{m}}^3} \cr & = 1458\,{{\text{m}}^3} \cr} $$
∴ Required speed :
$$\eqalign{ & = \left( {\frac{{1458}}{{0.60 \times 0.45}}} \right){\text{m/hr}} \cr & = 5400\,{\text{m/hr}} \cr} $$

Volume of the cube = 8³ cu.m = 512 cu.m

Volume of the cuboid :
$$\eqalign{ & = \left( {9 \times 8 \times 6} \right){\text{ c}}{{\text{m}}^3} \cr & = 432{\text{ c}}{{\text{m}}^3} \cr} $$
Volume of the cube :
$$\eqalign{ & = \left( {\frac{1}{2} \times 432} \right){\text{ c}}{{\text{m}}^3} \cr & = 216{\text{ c}}{{\text{m}}^3} \cr} $$
$$\eqalign{ & {a^3} = 216 \cr & a = \root 3 \of {216} \cr & a = 6 \cr & 6{a^2} = 6 \times {6^2} \cr & 6{a^2} = 216 \text{ cm}^2 \cr} $$