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

Sum of original dimension :
$$\eqalign{ & = 48 + 30 + 52 \cr & = 130 \cr} $$
Increase in sum :
$$\eqalign{ & = 156 - 130 \cr & = 26 \cr} $$
Since the dimensions have been increased proportionately,
So increase in shortest side :
$$ = \left( {26 \times \frac{{30}}{{130}}} \right)$$   "
$$ = 6$$ "

Let the length of each edge of small cube be a₁ and that of larger cube be a₂
Then,
$$\eqalign{ & 6a_1^2 = 96\, \cr & \Rightarrow a_1^2 = 16 \cr & \Rightarrow {a_1} = 4\,and \cr & 6a_2^2 = 384 \cr & \Rightarrow a_2^2 = 64 \cr & \Rightarrow {a_2} = 8 \cr} $$
∴ Number of cubes formed :
$$\eqalign{ & = \frac{{{\text{Volume of larger cube}}}}{{{\text{Volume of smaller cube}}}} \cr & = \left( {\frac{{8 \times 8 \times 8}}{{4 \times 4 \times 4}}} \right) \cr & = 8 \cr} $$

Total surface area :
$$\eqalign{ & = 2\pi r\left( {h + r} \right) \cr & = \left[ {2 \times \frac{{22}}{7} \times \frac{{35}}{{100}} \times \left( {1.25 + 0.35} \right)} \right]{{\text{m}}^2} \cr & = \left( {2 \times \frac{{22}}{7} \times \frac{{35}}{{100}} \times \frac{{16}}{{10}}} \right){{\text{m}}^2} \cr & = 3.52\,{{\text{m}}^2} \cr} $$
∴ Cost of the material :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {3.52 \times 80} \right) \cr & = {\text{Rs}}{\text{. 281}}{\text{.60}} \cr} $$

Volume of earth dug out :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 4 \times 4 \times 4} \right){{\text{m}}^3} \cr & = 704\,{{\text{m}}^3} \cr} $$
Area of embankment :
$$\eqalign{ & = \frac{{22}}{7} \times \left( {{7^2} - {4^2}} \right) \cr & = \left( {\frac{{22}}{7} \times 11 \times 3} \right){{\text{m}}^3} \cr & = \frac{{726}}{7}{{\text{m}}^3} \cr} $$
Height of embankment :
$$\eqalign{ & = \left( {\frac{{{\text{Volume}}}}{{{\text{Area}}}}} \right) \cr & = \left( {\frac{{704 \times 7}}{{726}}} \right){\text{m}} \cr & = \frac{{224}}{{33}}{\text{m}} \cr & = {\text{ 6}}\frac{{26}}{{33}}{\text{m}} \cr} $$

External radius, R = $$\frac{{25}}{2}$$ cm
Internal radius, r $$ = \left( {\frac{{25}}{2} - 1} \right){\text{cm}}$$   $$ = \frac{{23}}{2}{\text{ cm}}$$
Length, h = 20 cm
Whole surface area :
$$\eqalign{ & = 2\pi Rh + 2\pi rh + 2\pi \left( {{R^2} - {r^2}} \right) \cr & = 2\pi \left[ {\left( {R + r} \right)h + \left( {{R^2} - {r^2}} \right)} \right] \cr} $$
$$ = 2 \times \frac{{22}}{7} \times $$ $$\left[ {\left( {\frac{{25}}{2} + \frac{{23}}{2}} \right) \times 20 + \left( {\frac{{25}}{2} + \frac{{23}}{2}} \right)\left( {\frac{{25}}{2} - \frac{{23}}{2}} \right)} \right]$$        $${\text{ c}}{{\text{m}}^2}$$
$$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times 504} \right){\text{ c}}{{\text{m}}^2} \cr & = 3168{\text{ c}}{{\text{m}}^2} \cr} $$

Slant height of the cup, l = Radius of sheet = 14 cm
Circumference of the base :
= Circumference of the paper sheet
= $$\left( {\frac{{22}}{7} \times 14} \right)$$  cm
= 44 cm
Let the radius of the base of the cone be r cm
$$\eqalign{ & \therefore 2\pi r = 44 \cr & \Rightarrow r = \frac{{44 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 7 \cr & {\text{Height, h:}} \cr & = \sqrt {{l^2} - {r^2}} \cr & = \sqrt {{{\left( {14} \right)}^2} - {{\left( 7 \right)}^2}} \cr & = \sqrt {147} {\text{ cm}} \cr & = 7\sqrt 3 {\text{ cm}} \cr & = 12.12{\text{ cm}} \cr} $$

$$\eqalign{ & r = 7{\text{ cm, }}h{\text{ = 24 cm}} \cr & {\text{Now, }}\vartriangle {\text{AOB}} \sim \vartriangle {\text{COD}} \cr & {\text{So, }}\frac{{OA}}{{OC}} = \frac{{AB}}{{CD}} \cr & \Rightarrow \frac{h}{{\frac{h}{2}}} = \frac{r}{{CD}} \cr & \Rightarrow CD = \frac{r}{2} \cr} $$
∴ Volume of upper portion :
$$\eqalign{ & = \frac{1}{3}\pi {\left( {\frac{r}{2}} \right)^2}\left( {\frac{h}{2}} \right) \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12} \right){\text{ c}}{{\text{m}}^3} \cr & = 154{\text{ c}}{{\text{m}}^3} \cr} $$

Let their radii be R and r
Then,
$$\eqalign{ & \frac{{4\pi {R^2}}}{{4\pi {r^2}}} = \frac{4}{{25}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^2} = {\left( {\frac{2}{5}} \right)^2} \cr & \Rightarrow \frac{R}{r} = \frac{2}{5} \cr} $$
∴ Ratio of volumes :
$$\eqalign{ & = \frac{{\frac{4}{3}\pi {R^3}}}{{\frac{4}{3}\pi {r^3}}} \cr & = {\left( {\frac{R}{r}} \right)^3} \cr & = {\left( {\frac{2}{5}} \right)^3} \cr & = \frac{8}{{125}} \cr & = 8:125 \cr} $$

Let the radius of the cylinder be R
Then,
$$\eqalign{ & \pi \times {R^2} \times \frac{7}{3} = \frac{4}{3}\pi \times 7 \times 7 \times 7 \cr & \Rightarrow {R^2} = \left( {\frac{{4 \times 7 \times 7 \times 7}}{3} \times \frac{3}{7}} \right) \cr & \Rightarrow {R^2} = 196 \cr & \Rightarrow {R^2} = {\left( {14} \right)^2} \cr & \Rightarrow {R^2} = 14{\text{ cm}} \cr} $$
∴ Diameter = 2R = 28

Let the radius of the hemispherical bowl be r cm
Then,
$$\eqalign{ & 2\pi r = 176 \cr & \Rightarrow r = \frac{{176 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 28 \cr} $$
Volume of liquid in the bowl :
$$\eqalign{ & = \frac{1}{2} \times \left( {\frac{2}{3} \times \pi \times 28 \times 28 \times 28} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{2}{3} \times \pi \times 14 \times 28 \times 28} \right){\text{ c}}{{\text{m}}^3} \cr} $$
Volume of 1 glass :
$$ = \left( {\frac{2}{3} \times \pi \times 2 \times 2 \times 2} \right){\text{ c}}{{\text{m}}^3}$$
∴ Required number of person :
$$\eqalign{ & = \frac{{{\text{Volume of liquid in the bowl}}}}{{{\text{Volume of 1 glass}}}} \cr & = \left( {\frac{{14 \times 28 \times 28}}{{2 \times 2 \times 2}}} \right) \cr & = 1372 \cr} $$