Practice MCQ Questions and Answer on Volume and Surface Area

71.

Rita and Meeta both are having lunch boxes of a cuboid shape. Length and breadth of Rita's lunch box are 10% more than that of Meeta's lunch box, but the depth of Rita's lunch box is 20% less than that of Meeta's lunch box. The ratio of the capacity of Rita's lunch box to that of Meeta's lunch box is :

• (A) 11 : 15
• (B) 15 : 11
• (C) 121 : 125
• (D) 125 : 121

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 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 : $$\eqalign{ & = \frac{{11l}}{{10}} \times \frac{{11b}}{{10}} \times \frac{{4h}}{5}:lbh \cr & = \frac{{121}}{{125}}:1 \cr & = 121:125 \cr} $$

72.

A metallic cone of radius 12 cm and height 24 cm is ,melted and made into spheres of radius 2 cm each. How many spheres are there ?

• (A) 108
• (B) 120
• (C) 144
• (D) 180

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 $$\eqalign{ & {\text{Number of spheres :}} \cr & = \frac{{{\text{Volume of cone}}}}{{{\text{Volume of 1 sphere}}}} \cr & = \frac{{\frac{1}{3}\pi \times 12 \times 12 \times 24}}{{\frac{4}{3}\pi \times 2 \times 2 \times 2}} \cr & = 108 \cr} $$

73.

A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical ball. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is :

• (A) 2.5 cm
• (B) 2.66 cm
• (C) 3 cm
• (D) 3.5 cm

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 Let the radius of the third ball be R cm Then, $$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$   $$ + \frac{4}{3}\pi \times {R^3}$$   $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$ $$\eqalign{ & \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr & \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow R = \frac{5}{4} \cr} $$ ∴ Diameter of the third ball : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$

74.

A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?

• (A) 2 : 1
• (B) 3 : 2
• (C) 25 : 18
• (D) 27 : 20

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

75.

A rectangular water tank is 8 m high, 6 m long and 2.5 m wide. How many litres of water can it hold ?

• (A) 120 litres
• (B) 1200 litres
• (C) 12000 litres
• (D) 120000 litres

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

76.

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the ratio of its diameter to its height.

• (A) 3 : 7
• (B) 7 : 3
• (C) 6 : 7
• (D) 7 : 6

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

77.

The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is :

• (A) $$6\pi $$
• (B) $$12$$
• (C) $$36$$
• (D) $$36\pi $$

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 Let the length of each pipe be $$l$$ inches Then, volume of water in thinner pipe : $$\eqalign{ & = \left[ {\pi \times {{\left( {\frac{1}{2}} \right)}^2} \times 1} \right] \text{cu.inch} \cr & = \left( {\frac{{\pi l}}{4}} \right)\text{cu.inch} \cr} $$ Volume of water in thinker pipe : $$\eqalign{ & = \left( {\pi \times {3^2} \times l} \right)\text{cu.inch} \cr & = \left( {9\pi l} \right)\text{cu.inch} \cr} $$ ∴ Required number of pipes : $$\eqalign{ & = \frac{{9\pi l}}{{\left( {\frac{{\pi l}}{4}} \right)}} \cr & = 36 \cr} $$

78.

Each side of a cube is decreased by 25%. Find the ratio of the volume of the original cube and the resulting cube = ?

• (A) 64 : 1
• (B) 27 : 64
• (C) 64 : 27
• (D) 8 : 1

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 Let the side of cube = 10 cm ∴ Original volume = 10 × 10 × 10 = 1000 cm3 Now, side of new cube : = 10 - 25% of 10 = 7.5 ∴ New volume = 7.5 × 7.5 × 7.5 = 421.875 cm3 ∴ Required ratio : $$\eqalign{ & = \frac{{1000}}{{421.875}} \cr & = \frac{{1000000}}{{421875}} \cr & = \frac{{64}}{{27}}\,Or\,64:27 \cr} $$

79.

If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to from a single sphere, the diameter of the new sphere will be :

• (A) 12 cm
• (B) 24 cm
• (C) 30 cm
• (D) 36 cm

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 Volume of new sphere : $$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( 6 \right)}^3} + \frac{4}{3}\pi \times {{\left( 8 \right)}^3} + \frac{4}{3}\pi \times {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \left\{ {{{\left( 6 \right)}^3} + {{\left( 8 \right)}^3} + {{\left( {10} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3}\pi \times 1728} \right){\text{c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \times {{\left( {12} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr} $$ Let the radius of the new sphere be R Then, $$\eqalign{ & \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {\left( {12} \right)^3} \cr & \Rightarrow R = 12\,cm \cr} $$ ∴ Diameter : $$\eqalign{ & = 2R \cr & = 2 \times 12 \cr & = 24\,cm \cr} $$

80.

If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :

• (A) Remains unaltered
• (B) Decrease by 25%
• (C) Increase by 25%
• (D) Increase by 50%

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 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 : $$\eqalign{ & = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr & = 25\% \cr} $$