**Volume and Surface Area Aptitude Question and Answer – Formulas**

**Square**

Area = S^{2}

Perimeter = 4s

s = length of the sides, d = length of diagonal.

** Rectangle**

Area = base * height = b * h

Perimeter = 2 (b + h)

**Triangle**

Area = (1/2) * base * height

Perimeter = x + y + z (Summation of three sides of a triangle)

**Rhombus**

Area = (1/2) * Product of the diagonals between the sides * sine of the angle between the sides.

Perimeter = 4 * side (any side)

Diagonal = 2 * area / diagonal

**Parallelogram**

Area = Product of any two sides * sine of the included angle

Perimeter = 2 (a + b) (a and b are the two adjacent sides)

**Trapezium**

Area = (1 / 2) * Sum of the parallel sides * height

**Cuboid**

Let length = l, breadth = b, and height = h

- Volume of cuboid = (l * b * h) cubic units
- Whole surface area of cuboid = 2(lb + bh + hl) sq.units
- Diagonal of cuboid = units.

**Cube**

Let each edge of a cube = “a” units.

- Volume of the cube = a
^{3}cubic units. - Whole Surface area of cube = (6a
^{2}) sq. units. - Diagonal of the cube = units.

**Cylinder**

Let the radius of the base of a cylinder be r units and height of the cylinder be h units.

- Volume of the cylinder = (πr
^{2}h) cubic units. - Curved Surface area of the cylinder = (2πrh) sq.units
- Total Surface area of the cylinder = (2πrh + 2πr
^{2}) sq.units

**Sphere**

Let r be the radius of the sphere.

- Volume of the sphere = cubic units.
- Surface area of the sphere = sq. units.
- Volume of hemisphere = cubic units.
- Curved Surface area of the hemisphere = (2πr
^{2}) sq. units. - Whole Surface area of the hemisphere = (3πr
^{2}) sq. units.

**Right Circular Cone:**

Let r be the radius of the base, h is the height, and I is the slant height of the cone.

- Slant height I =
- Volume of the cone = cubic units.
- Curved surface area of the cone = (πrI) sq. units = sq. units.
- Total Surface area of the cone = (πrI + πr
^{2}) = πr(I + r) sq. units.

**Frustum of a right circular cone:**

Let the radius of the base of the frustum = R, the radius of top = r, height = h and slant height = l units.

- slant height, l = unit.
- Curved surface area = π (r + R) l sq. units.
- Total Surface area = π {(r + R) l + r
^{2}+ R^{2}} sq. units. - Volume = (πh / 3) (r
^{2}+ R^{2}+ rR) cubic units.

**Volume and Surface Area Aptitude Quick Method**

**For a Closed Wodden box:**

- Capacity = (external length – 2 * thickness) * (external breadth – 2 * thickness) * (external height – 2 * thickness).
- Volume of Material = External Volume – Capacity.
- Weight of Wood = Volume of Wood * Density of wood.

**Problems Involving Ratios:**

**Two Spheres**- (Ratio of radii)
^{2}= ratio of surface areas. - Ratio of Volumes = (ratio of radii)
^{3}. - (Ratio of surface areas)
^{3}= (ratio of volume)^{2}.

- (Ratio of radii)
**Two Cylinders:**- When the radii are equal:
- Ratio of volumes = ratio of heights.
- Ratio of Curved Surface areas = ratio of heights.
- Ratio of Volumes = (ratio of curved surface areas).

- When height are equal
- Ratio of volumes = (ratio of radii)
^{2}. - Ratio of curved surface areas = ratio of radii.
- Ratio of volumes =(ratio of curved surface areas)
^{2}.

- Ratio of volumes = (ratio of radii)
- When volumes are equal
- Ratio of radii =
- Ratio of curved surface areas =

- When curved surface areas are equal
- Ratio of volumes = ratio of radii.
- Ratio of volumes = inverse ratio of heights.
- Ratio of radii = inverse ratio of heights.

- When the radii are equal:

**Volume and Surface Area Aptitude Question and Answer With Detailed Explanation** – Quantitative Aptitude

## Volume and Surface Area Aptitude

Question 1 |

12 Kg | |

60 Kg | |

72 Kg | |

96 Kg |

**Answer: Option B**

__Explanation:__

Volume of Water displaced = ( 3 * 2 * 0.01)m

^{3}

=0.06m

^{3}

Therefore, Mass of man= Volume of water displaced * Density of water

=(0.06 * 1000)kg

**Mass of man = 60kg**

Question 2 |

^{3}) is:

4830 | |

5120 | |

6420 | |

8960 |

**Answer: Option B**

__Explanation:__

Volume of the box = Length * Breadth * Height

From the given,

Length = 48 * (2 * 8) (i.e., Two square formed side)

Length = 32m

Breadth = 36 - (2*8)

Breadth = 20m

Height = 8m, From the given

Therefore, Volume = 32 * 20 * 8

**Volume of the box = 5120m**

^{3}Question 3 |

^{2}and its volume is 924m

^{3}. Find the ratio of its diameter to its height.

3 : 7 | |

7 : 3 | |

6 : 7 | |

7 : 6 |

**Answer: Option B**

__Explanation:__

Question 4 |

^{2}. The Length of its diagonal is

**Answer: Option C**

__Explanation:__

Surface area of cube = 6a

^{2}

600 = 6a

^{2}

a

^{2}= 100

Question 5 |

**Answer: Option C**

__Explanation:__

Question 6 |

49m ^{2} | |

50m ^{2} | |

53.5m ^{2} | |

55m ^{2} |

**Answer: Option A**

__Explanation:__

Area of the Wet Surface = [2 * (lb + bh + lh) - lb]

= 2* (bh + lh) + lb

=[2* (4 * 1.25 + 6 * 1.25) + (6 * 4)]m

^{2}

= 49m

^{2}

Question 7 |

^{3}, then the rise in the water level in the tank will be:

20cm | |

25cm | |

35cm | |

50cm |

**Answer: Option B**

__Explanation:__

Total Volume of water displaced = (4 * 50)m

^{3}= 200m

^{3}

Therefore, Rise in water level = (200 / (40*20))m = 0.25m

**The rise in the water level in the tank = 25cm**

Question 8 |

150 | |

140 | |

120 | |

100 |

**Answer: Option D**

__Explanation:__

Length of the hall = 40m

Breadth of the hall = 25m

Height of the Hall = 20m

Volume of the hall = L * B * H

= 40 * 25 * 20 = 20,000m

^{3}

Space Occupied by each person = 200m

^{3}

No. of Person that can accommodate in the hall = (Volume of the hall / Space Occupied by one person)

=(20000/200)

**No. of Person that can accommodate in the hall = 100Persons**

Question 9 |

1500m ^{3} | |

1400m ^{3} | |

1200m ^{3} | |

1000m ^{3} |

**Answer: Option A**

__Explanation:__

1 Hectare = 10,000m

^{2}

1.5 Hectare = 1.5 * 10,000 = 15,000m

^{2}

Depth = 10cm of rainfall = (10/100)

Therefore, Volume of water = Area * Depth

=15,000 * (10/100)

**Volume of water = 1500m**

^{3}Question 10 |

4.5cm | |

4.57cm | |

4.67cm | |

4.7cm |

**Answer: Option C**

__Explanation:__