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Quantitative Aptitude

Surface Area & Volume

Types of Numbers2 / 51Divisibility & Remainders
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Concepts (2)

Mastering cone and sphere surface areas and volumes, including melting/recasting, is vital for SSC CGL. Focus on formulas, shortcuts, and common mistakes for speed and accuracy.

Core Formulas

Understanding the fundamental formulas for cones and spheres is non-negotiable for SSC CGL. Memorize these for quick recall:

Cone:

  • Volume: V = (1/3)πr²h
  • Slant Height (l): l = √(r² + h²)
  • Curved Surface Area (CSA): CSA = πrl
  • Total Surface Area (TSA): TSA = πr(l + r)

Sphere:

  • Volume: V = (4/3)πr³
  • Surface Area (SA): SA = 4πr²

Hemisphere:

  • Volume: V = (2/3)πr³
  • Curved Surface Area (CSA): CSA = 2πr²
  • Total Surface Area (TSA): TSA = 3πr² (includes the circular base)

Worked Example 1

Q: A cone has a radius of 7 cm and a height of 24 cm. Calculate its volume and curved surface area. Solution:

  1. Find Slant Height (l): Using l = √(r² + h²), l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.
  2. Calculate Volume: V = (1/3)πr²h = (1/3) * (22/7) * 7² * 24 = (1/3) * (22/7) * 49 * 24 = 22 * 7 * 8 = 1232 cm³.
  3. Calculate CSA: CSA = πrl = (22/7) * 7 * 25 = 22 * 25 = 550 cm².

Worked Example 2

Q: A metallic sphere of radius 3 cm is melted and recast into a cone of height 9 cm. Find the radius of the cone. Solution:

  1. Principle: When a solid is melted and recast, its volume remains constant.
  2. Volume of Sphere: V_sphere = (4/3)πr³ = (4/3)π(3)³ = (4/3)π * 27 = 36π cm³.
  3. Volume of Cone: Let the radius of the cone be R. V_cone = (1/3)πR²h = (1/3)πR²(9) = 3πR² cm³.
  4. Equate Volumes: V_sphere = V_cone 36π = 3πR² 36 = 3R² R² = 12 R = √12 = 2√3 cm. The radius of the cone is 2√3 cm.

Shortcuts & Tricks

  • Pythagorean Triplets: Memorize common triplets like (3,4,5), (5,12,13), (7,24,25), (8,15,17) for (r, h, l) to quickly find the slant height l of a cone without calculation.
  • Ratio Method: If dimensions (radius, height) are scaled by a factor k:
    • Surface Area scales by k².
    • Volume scales by k³. Example: If sphere radius doubles, SA becomes 2² = 4 times, Volume becomes 2³ = 8 times.
  • π Value: Use 22/7 when dimensions are multiples of 7. Otherwise, keep π in calculations or use 3.14 if options demand it. Often, π cancels out in melting/recasting problems.
  • Melting/Recasting: Always equate volumes. This is the core principle.

Common Mistakes

  1. Confusing CSA and TSA: Especially for hemispheres, students often use 2πr² (CSA) instead of 3πr² (TSA) when the question asks for total surface area.
  2. Incorrect Slant Height: Forgetting to calculate l or using h instead of l in the cone's CSA formula (πrl).
  3. Calculation Errors: Mistakes in squaring, cubing, or multiplying, particularly when dealing with fractions like (1/3) or (4/3).
  4. Unit Inconsistency: Not converting all dimensions to the same unit before calculation, leading to incorrect answers.

Derivation (brief)

While full derivations aren't needed for SSC CGL, a brief understanding helps recall.

  • Cone Volume: The volume of a cone is (1/3) the volume of a cylinder with the same base radius and height. This can be intuitively understood by imagining three identical cones fitting perfectly inside a cylinder.
  • Sphere Volume/Surface Area: These are typically derived using integral calculus. For competitive exams, direct memorization is key, but knowing they stem from advanced mathematical principles can reinforce their validity.

Advanced Examples

Q: A hollow sphere has an outer radius of 6 cm and an inner radius of 3 cm. What is the volume of the material in the sphere? Solution: Volume of material = Volume of outer sphere - Volume of inner sphere V = (4/3)π(R_outer³ - R_inner³) = (4/3)π(6³ - 3³) = (4/3)π(216 - 27) = (4/3)π(189) = 4 * π * 63 = 252π cm³.

Q: A solid metallic sphere of radius 10 cm is melted and recast into 100 identical small solid spheres. What is the surface area of one small sphere? Solution:

  1. Volume of large sphere: V_large = (4/3)π(10)³ = (4/3)π(1000) cm³.
  2. Volume of one small sphere: V_small = V_large / 100 = (4/3)π(1000) / 100 = (4/3)π(10) cm³.
  3. Let r be the radius of a small sphere. (4/3)πr³ = (4/3)π(10). So, r³ = 10, r = ³√10 cm.
  4. Surface Area of one small sphere: SA = 4πr² = 4π(³√10)² = 4π(10^(2/3)) cm².

Variation Types

  • Combined Shapes: Problems involving a cylinder with hemispherical ends, a cone mounted on a cylinder, or a sphere inscribed in a cube. Break down the composite shape into simpler components.
  • Water Displacement: When an object is submerged in water, the volume of water displaced is equal to the volume of the object. This is a common application of volume calculations.
  • Percentage Change: If a dimension (e.g., radius) changes by a certain percentage, calculate the percentage change in surface area or volume. Use the ratio method shortcut for efficiency.
  • Frustum of a Cone: While less frequent, be aware of the frustum (a cone with its top cut off). Its formulas are more complex, but sometimes asked. V = (1/3)πh(R² + r² + Rr) and CSA = πl(R + r). l = √(h² + (R-r)²).

Time-Saving Methods

  • Option Elimination: Often, options will have π or be multiples of 22/7. If your calculated value doesn't match the form, you can eliminate options. Also, check for unit consistency in options.
  • Approximation: If the options are widely spaced, you can approximate π ≈ 3 or π ≈ 3.14 to quickly estimate the answer and eliminate choices.
  • Memorize Common Values: Know πr² for r=7 (154), r=14 (616), etc., as these appear frequently. Also, πr³ for small r values can save time.
  • Work Backwards: Sometimes, if the answer is given and you need to find a dimension, plugging options into the formula can be faster than solving algebraically.
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Master surface area and volume for cuboids, cubes, and cylinders. Focus on quick formula recall, precise calculations, and smart shortcuts to ace SSC CGL Mensuration questions.

Core Formulas

Prisms and Cylinders are fundamental 3D shapes in Mensuration. A prism is a polyhedron comprising an n-sided polygonal base, a second base which is a translated copy of the first, and n other faces (necessarily all parallelograms) joining corresponding sides of the two bases. A cylinder is a solid geometric figure with straight parallel sides and a circular or oval cross-section.

Cuboid (Rectangular Prism)

  • Volume (V): l × b × h
  • Lateral Surface Area (LSA) (Area of 4 walls): 2h(l + b)
  • Total Surface Area (TSA): 2(lb + bh + hl)
  • Diagonal (d): √(l² + b² + h²)

Cube (Square Prism)

  • Volume (V): a³
  • Lateral Surface Area (LSA) (Area of 4 walls): 4a²
  • Total Surface Area (TSA): 6a²
  • Diagonal (d): a√3

Cylinder (Right Circular Cylinder)

  • Volume (V): πr²h
  • Curved Surface Area (CSA) / Lateral Surface Area (LSA): 2πrh
  • Total Surface Area (TSA): 2πr(r + h) or 2πrh + 2πr²

Worked Example 1

Q: A cuboid has length 12 cm, breadth 8 cm, and height 5 cm. Find its volume and total surface area. A:

  • Volume (V) = l × b × h = 12 × 8 × 5 = 480 cm³
  • Total Surface Area (TSA) = 2(lb + bh + hl) = 2(12×8 + 8×5 + 5×12) = 2(96 + 40 + 60) = 2(196) = 392 cm²

Worked Example 2

Q: The radius of a cylinder is 7 cm and its height is 10 cm. Find its curved surface area and volume. (Take π = 22/7) A:

  • Curved Surface Area (CSA) = 2πrh = 2 × (22/7) × 7 × 10 = 2 × 22 × 10 = 440 cm²
  • Volume (V) = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cm³

Shortcuts & Tricks

  1. Ratio Method: If dimensions are in a certain ratio, calculate area/volume for a unit dimension and then scale. E.g., if radius is doubled, volume becomes 4 times (2²), CSA becomes 2 times. If all dimensions of a cuboid are doubled, volume becomes 8 times (2³), surface area becomes 4 times (2²).
  2. Divisibility by 11: For cylinder problems involving π = 22/7, the final answer for CSA, TSA, or Volume will often be a multiple of 11 (due to 22). Use this to eliminate options quickly, especially if calculations are complex.
  3. Pythagorean Triplets: For problems involving diagonals or finding height/radius from given slant height, recognize common Pythagorean triplets (e.g., 3-4-5, 5-12-13, 7-24-25) to save time.
  4. Unit Digit Check: Sometimes, you can eliminate options by just checking the unit digit of the calculated answer, especially for multiplication-heavy formulas.

Common Mistakes

  1. Confusing LSA/CSA with TSA: Students often use the formula for LSA/CSA when TSA is required, forgetting to add the area of the top and bottom bases.
  2. Incorrect Units: Always ensure units are consistent (e.g., all cm or all m) and the final answer has the correct unit (cm² for area, cm³ for volume).
  3. Calculation Errors with π: Forgetting to use π = 22/7 or 3.14 or making arithmetic mistakes during multiplication/division involving π.
  4. Forgetting Square/Cube: In formulas like πr²h or a³, sometimes r or a is not squared/cubed, leading to incorrect results.

Derivation (brief)

  • Volume of a Prism/Cylinder: The fundamental concept is that the volume of any prism (and a cylinder, which can be thought of as a prism with an infinite-sided polygonal base) is the Area of its Base × Height. This holds true for cuboids (lb × h), cubes (a² × a), and cylinders (πr² × h).
  • Lateral Surface Area (LSA) of a Prism/Cylinder: The LSA is the sum of the areas of its side faces. For a prism, if you unroll the side faces, they form a rectangle. The length of this rectangle is the perimeter of the base, and the width is the height of the prism. Hence, LSA = Perimeter of Base × Height. For a cylinder, the perimeter of the circular base is 2πr, so LSA = 2πr × h.

Advanced Examples

1. Hollow Cylinder:

  • If a hollow cylinder has outer radius R, inner radius r, and height h:
    • Volume of material: π(R² - r²)h
    • Total Surface Area: 2π(R + r)h + 2π(R² - r²) (CSA of inner + CSA of outer + 2 * area of ring bases)

2. Melting and Recasting:

  • When a solid is melted and recast into another shape, its volume remains constant. This is a crucial concept for many problems. E.g., if a metallic cylinder is melted and recast into n smaller spheres, Volume_cylinder = n × Volume_sphere.

Variation Types

  • Cost Problems: Calculating the cost of painting the surface (area × cost per unit area) or filling a container (volume × cost per unit volume).
  • Open/Closed Shapes: An open cylinder (without a top or bottom) will have a TSA of 2πrh + πr² (CSA + 1 base area).
  • Finding Dimensions: Given the volume or surface area, finding an unknown dimension (e.g., height, radius, side).

Time-Saving Methods

  1. Approximation: When options are far apart, approximate π as 3 or 3.14 instead of 22/7 for quicker mental calculations. Only use 22/7 for precise calculations or when r or h is a multiple of 7.
  2. Visualisation: Quickly sketch the figure. This helps in correctly identifying which surface areas are relevant (e.g., for an open box, the top surface is not included in TSA).
  3. Option Elimination: Always look at the options. Often, you can eliminate 2-3 options based on unit digits, divisibility rules (especially for π), or approximate magnitude, reducing the calculation burden significantly.
  4. Formula Manipulation: Practice rearranging formulas to directly solve for an unknown variable (e.g., h = V / (πr²) for a cylinder) rather than plugging in values and then solving.
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