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

Radius of the base = r units
Curved surface area of a right cylinder = $$4\pi {\text{r}}h$$
Curved surface area of cylinder = $$2\pi {\text{RH}}$$
∴ According to the question,
$$2\pi {\text{rH = }}4\pi {\text{rh}}$$
⇒ Height of cylinder = 2h units

Length of rectangle paper = Circumference of the base of cylinder
If r is the radius of the cylinder :
$$\eqalign{ & \Rightarrow 44 = 2\pi r \cr & \Rightarrow r = \frac{{44 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 7\,cm \cr} $$

Let the breadth and height of the room be b and h metres respectively.
Then,
Area of the floor $$ = \left( {14b} \right)\,{m^2}$$
$$\eqalign{ & \therefore 14b = 2.2 \times 70 \cr & \Rightarrow b = \frac{{2.2 \times 70}}{{14}} \cr & \Rightarrow b = 11 \cr} $$
Volume of the room :
$$\eqalign{ & = \left( {14 \times 11 \times h} \right){m^3} \cr & = \left( {154h} \right){m^3} \cr} $$

$$\eqalign{ & \therefore 154h = 11 \times 70 \cr & \Rightarrow h = \frac{{11 \times 70}}{{154}} \cr & \Rightarrow h = 5 \cr} $$

Depth of the tank :
$$\eqalign{ & = \left( {\frac{{24}}{{4 \times 3}}} \right)m \cr & = 2\,m \cr} $$
Since the tank is open and thickness of material is to be ignored, we have
Sum of inner and outer surface :
$$\eqalign{ & = 2\left[ {\left\{ {2\left( {l + b} \right) \times h} \right\} + lb} \right] \cr & = 2\left[ {\left\{ {2\left( {4 + 3} \right) \times 2} \right\} + 4 \times 3} \right]{m^2} \cr & = 80\,{m^2} \cr} $$
∴ Cost of painting :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {80 \times 10} \right) \cr & = {\text{Rs}}{\text{. 800}} \cr} $$

Volume of water displaced :
$$\eqalign{ & = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr & = \frac{{18}}{5}{m^3} \cr} $$
Volume of water displaced by 1 man = 0.1 m³
∴ Number of men :
$$\eqalign{ & = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr & = \left( {\frac{{18}}{5} \times 10} \right) \cr & = 36 \cr} $$

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

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

Let the length of each side of the cube be a cm
Then, Volume of the part of cube outside water = Volume of the mass placed on it
⇒ 2a² = 0.2 × 1000
⇒ 2a² = 200
⇒ a² = 100
⇒ a = 10

Volume of the cylinder :
= Volume of the cube
= (11)³ cm³
= 1331 cm³
Let the radius of the base be r cm
Then,
$$\eqalign{ & \frac{{22}}{7} \times {r^2} \times 14 = 1331 \cr & \Rightarrow {r^2} = \frac{{1331}}{{44}} = \frac{{121}}{4} \cr & \Rightarrow r = \frac{{11}}{2} \cr & \Rightarrow r = 5.5 \cr} $$

Let their heights be 2h and 3h and radii be r and R respectively
Then,
$$\eqalign{ & \pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right) \cr & \Rightarrow \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2} \cr & \Rightarrow \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 \cr} $$