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

Let the depth of the cistern be h metres Then,
$$\eqalign{ & 4.5 \times 3 \times h = 50 \cr & \Rightarrow h = \frac{{50}}{{13.5}} \cr & \Rightarrow h = \frac{{100}}{{27}} \cr} $$
Area of sheet required :
$$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = lb + 2h\left( {l + b} \right) \cr & = \left[ {4.5 \times 3 + 2 \times \frac{{100}}{{27}} \times \left( {4.5 + 3} \right)} \right]{{\text{m}}^2} \cr & = \left( {13.5 + \frac{{200}}{{27}} \times 7.5} \right){\text{ }}{{\text{m}}^2} \cr & = \left( {\frac{{27}}{2} + \frac{{500}}{9}} \right){{\text{m}}^2} \cr & = \frac{{1243}}{{18}}{\text{ }}{{\text{m}}^2} \cr} $$
∴ Weight of lead :
$$\eqalign{ & = \left( {27 \times \frac{{1243}}{{18}}} \right)kg \cr & = \left( {\frac{{3729}}{2}} \right)kg \cr & = 1864.5\,kg \cr} $$

Let the original length, breadth and height of the cuboid be x, 2x and 3x units respectively
Then, original volume = (x × 2x × 3x) cu.units = 6x³ cu.units
New length = 200% of x = 2x
New breadth = 300% of 2x = 6x
New height = 300% of 3x = 9x
∴ New volume :
= (2x × 6x × 9x) cu.units
= 108x³ cu.units
Increase in volume :
= (108x³ - 6x³) cu.units
= (102x³) cu.units
∴ Required ratio :
$$\eqalign{ & = \frac{{102{{\text{x}}^3}}}{{6{{\text{x}}^3}}} \cr & = 17{\text{ }}\left( {{\text{Times}}} \right) \cr} $$

Since the tank is open at the top, we have :
Area of sheet required = Surface area of the tank
$$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = \left[ {30 \times 20 + 2\left( {20 \times 12 + 30 \times 12} \right)} \right]{{\text{m}}^2} \cr & = \left( {600 + 1200} \right){{\text{m}}^2} \cr & = 1800{\text{ }}{{\text{m}}^2} \cr} $$
Length of sheet required :
$$\eqalign{ & = \left( {\frac{{{\text{Area}}}}{{{\text{Width}}}}} \right) \cr & = \frac{{1800}}{3}m \cr & = 600\,m \cr} $$
∴ Cost of the sheet
= Rs. (600 × 10)
= Rs. 6000

Since a cube has 4 diagonals, we have :
Length of a diagonal
$$\eqalign{ & = \left( {\frac{{12}}{4}} \right)cm \cr & = 3\,cm \cr} $$
Let the length of each edge of the cube be a cm
Then,
$$\eqalign{ & \sqrt 3 a = 3 \cr & or,a = \sqrt 3 \cr} $$
∴ Total length of the edges of the cube = $$12\sqrt 3\, $$ cm

Let original edge = a
Then, original volume = a³
New edge :
$$\eqalign{ & = \frac{{150}}{{100}}a \cr & = \frac{{3a}}{2} \cr} $$
New volume :
$$\eqalign{ & = {\left( {\frac{{3a}}{2}} \right)^3} \cr & = \frac{{27{a^3}}}{8} \cr} $$
Increase in volume :
$$\eqalign{ & = \left( {\frac{{27{a^3}}}{8} - {a^3}} \right) \cr & = \frac{{19{a^3}}}{8} \cr} $$
∴ Increase % :
$$\eqalign{ & = \left( {\frac{{19{a^3}}}{8} \times \frac{1}{{{a^3}}} \times 100} \right)\% \cr & = 237.5\% \cr} $$

$$\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & \Rightarrow S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & \Rightarrow S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \Rightarrow \frac{{dS}}{{dr}} = \frac{{ - 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr} $$
∴ S is minimum when :
$$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \Rightarrow \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & \Rightarrow V = 2\pi {r^3} \cr & \Rightarrow \pi {r^2}h = 2\pi {r^3} \cr & \Rightarrow h = 2r \cr} $$

Volume of water flown into the tank in 5 min :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 100 \times 100 \times 7} \right){\text{cu}}{\text{.feet}} \cr & = 220000\,{\text{cu}}{\text{.feet}} \cr} $$
∴ Rate of flow of water :
$$\eqalign{ & = \left( {\frac{{220000}}{{5 \times 60}}} \right){\text{cu}}{\text{.feet/sec}} \cr & = 733.3 \approx 700\,{\text{cu}}{\text{.feet/sec}} \cr} $$

h = 15 cm, r = 8 cm
So,
$$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\,\,\, = \sqrt {{8^2} + {{\left( {15} \right)}^2}} \cr & \,\,\,\,\,\, = 17\,cm \cr} $$
∴ Curved surface area :
$$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 8 \times 17} \right){\text{ c}}{{\text{m}}^2} \cr & = 136\pi {\text{ c}}{{\text{m}}^2} \cr} $$

Let their radius and height be 5x and 12x respectively
Slant height of the cone,
$$l = \sqrt {{{\left( {5x} \right)}^2} + {{\left( {12x} \right)}^2}} = 13x$$
$$\eqalign{ & \frac{{{\text{Total surface area of cylinder}}}}{{{\text{Total surface area of cone}}}} \cr & = \frac{{2\pi r\left( {h + r} \right)}}{{\pi r\left( {l + r} \right)}} \cr & = \frac{{2\left( {h + r} \right)}}{{\left( {l + r} \right)}} \cr & = \frac{{2 \times \left( {12x + 5x} \right)}}{{\left( {13x + 5x} \right)}} \cr & = \frac{{34x}}{{18x}} \cr & = \frac{{17}}{9}\,Or\,17:9 \cr} $$

$$\eqalign{ & 4\pi {r^2} = 5544 \cr & \Rightarrow {r^2} = \left( {5544 \times \frac{1}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^2} = 441 \cr & \Rightarrow r = 21 \cr} $$
∴ Volume :
$$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21} \right){\text{ c}}{{\text{m}}^3} \cr & = 38808{\text{ c}}{{\text{m}}^3} \cr} $$