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

$$\eqalign{ & \frac{{\pi \,{r^2}h}}{{2\pi \,rh}} = \frac{{924}}{{264}}\, \cr & \Rightarrow r = {\frac{{924}}{{264}} \times 2} = 7m \cr & {\text{And}},\,2\pi rh = 264 \cr & \Rightarrow h = {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} = 6m \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3 \cr} $$

Let the thickness of the bottom be x cm
Then,
$${\mkern 1mu} {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} = $$         $$8000 \times $$ $$1000$$
$$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = 8000 \times 1000$$
$$\eqalign{ & \Rightarrow \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} = 100 \cr & \Rightarrow x = 10\,{\text{cm}} = 1\,{\text{dm}} \cr} $$

$$\eqalign{ & h = 14\,cm,\,r = 7\,cm \cr & {\text{So}},\,l = \sqrt {{{\left( 7 \right)}^2} + {{\left( {14} \right)}^2}} = \sqrt {245} = 7\sqrt 5 \,cm \cr & \therefore {\text{Total}}\,{\text{surface}}\,{\text{area}} \cr & = \pi \,rl + \pi \,{r^2} \cr & = \left( {\frac{{22}}{7} \times 7 \times 7\sqrt 5 + \frac{{22}}{7} \times 7 \times 7} \right)c{m^2} \cr & = \left[ {154\left( {\sqrt 5 + 1} \right)} \right]c{m^2} \cr & = \left( {154 \times 3.236} \right)c{m^2} \cr & = 498.35\,c{m^2} \cr} $$

$$\eqalign{ & {\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}} \cr & = \left( {{3^3} + {4^3} + {5^3}} \right) = 216\,c{m^3} \cr & {\text{Let}}\,{\text{the}}\,{\text{edge}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}}\,{\text{be}}\,a \cr & So,\,{a^3} = 216\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 6\,cm \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{{6 \times \left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}}} \cr & = \frac{{50}}{{36}} \cr & = 25:18 \cr} $$

$$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{bricks}} \cr & = \frac{{{\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{wall}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{1}}\,{\text{brick}}}} \cr & = {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \cr & = 6400 \cr} $$

Surface area :
$$\eqalign{ & = {2\left( {7 \times 11 + 11 \times 13 + 7 \times 13} \right)} \cr & = {2 \times 311} \cr & = 622{\text{ c}}{{\text{m}}^2} \cr} $$

let l, b and h denote the length, breadth and depth of Meeta's lunch box
Then, length of Rita's lunch box :
$$ = 110\% {\text{ of }}l = \frac{{11l}}{{10}}$$
Breadth of Rita's lunch box :
$$ = 110\% {\text{ of }}b = \frac{{11b}}{{10}}$$
Depth of Rita's lunch box :
$$ = 80\% {\text{ of }}h = \frac{{4h}}{5}$$
∴ Ratio of the capacities of Rita's and Meeta's lunch boxes :
$$\eqalign{ & = \frac{{11l}}{{10}} \times \frac{{11b}}{{10}} \times \frac{{4h}}{5}:lbh \cr & = \frac{{121}}{{125}}:1 \cr & = 121:125 \cr} $$

Volume of the tank :
$$\eqalign{ & = \left( {8 \times 100 \times 6 \times 100 \times 2.5 \times 100} \right)c{m^3} \cr & = 120000000\,c{m^3} \cr & = \left( {\frac{{120000000}}{{1000}}} \right){\text{litres}} \cr & = 120000{\text{ litres}} \cr} $$

Surface area :
$$\eqalign{ & = \left[ {6 \times {{\left( {\frac{1}{2}} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2} \cr & = \frac{3}{2}{\text{ c}}{{\text{m}}^2} \cr} $$

Volume of the larger cube :
$$\eqalign{ & = \left( {{3^3} + {4^3} + {5^3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 216{\text{ c}}{{\text{m}}^3} \cr} $$
Let the edge of the larger cube be a cm
$$\eqalign{ & \therefore {a^3} = 216 \cr & \Rightarrow a = 6 \cr} $$
Required ratio :
$$\eqalign{ & = \frac{{6\left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}} \cr & = \frac{{6 \times 50}}{{6 \times 36}} \cr & = \frac{{25}}{{18}}\,Or\,25:18 \cr} $$