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Trickier Word Problems / Advanced Applications

Introduction

This pattern covers composite mensuration and layered word problems - shaded regions, solids with parts removed or added, embedded solids, and multi-step conversion problems. These questions combine several formulas and require careful interpretation of the wording, correct ordering of operations, and unit consistency.

Mastering this pattern trains you to translate real-world descriptions into mathematical steps and avoid common traps (hidden subtractions, leftover material, or mistaken use of radii/diameters).

Pattern: Trickier Word Problems / Advanced Applications

Pattern

Key concept: Break the problem into parts - model each part with an exact formula, add/subtract volumes or areas as required, and delay numeric π substitution until cancellation is possible.

Common strategies:
• Draw a clear labelled diagram before starting.
• Identify which parts are added and which are removed (shaded region = outer - inner).
• Keep π symbolic until final step if multiple π terms will cancel.
• Watch for implied units and convert before combining results.

Step-by-Step Example

Question

A solid right circular cone of height 24 cm and base radius 7 cm is hollowed out by removing a smaller similar cone from its tip so that the remaining solid is a frustum. The smaller cone removed is similar to the original and its height is one-third of the original cone's height. Find the volume of the frustum. (Use π symbolically.)

Solution

  1. Step 1: Understand the parts and draw a diagram.

    The frustum volume = Volume(original cone) - Volume(smaller similar cone removed).

  2. Step 2: Compute scaling for the smaller cone.

    Smaller cone height = (1/3) × 24 = 8 cm. Similarity ratio (linear) = smaller height / original height = 8/24 = 1/3. So smaller cone radius = (1/3) × 7 = 7/3 cm.

  3. Step 3: Write volumes using formulas (keep π symbolic).

    V_original = (1/3)πR³_h? No - use (1/3)πR²H.
    V_original = (1/3)π × 7² × 24 = (1/3)π × 49 × 24 = 392π.
    V_small = (1/3)π × (7/3)² × 8 = (1/3)π × (49/9) × 8 = (392/27)π.

  4. Step 4: Subtract to get frustum volume.

    V_frustum = V_original - V_small = 392π - (392/27)π = 392π × (1 - 1/27) = 392π × (26/27) = (10192/27)π.

  5. Final Answer:

    V_frustum = (10192/27)π cm³ (exact).
    Quick numeric check (optional): (10192/27) ≈ 377.48 so ≈ 377.48π ≈ 1186.6 if π ≈ 3.1416.

  6. Quick Check:

    Confirm similarity scaling: small radius = 7/3, small volume = (1/3)π(7/3)²×8 = 392π/27; subtraction yields correct fraction. Units consistent (cm³) ✅

Quick Variations

1. A cylindrical tank with a hemispherical bottom - combine cylinder and hemisphere volumes.

2. A cube with a spherical cavity - subtract sphere volume from cube volume.

3. Shaded area between inscribed and circumscribed polygons/circles - outer area minus inner area.

4. Material lost during melting (e.g., 5% loss) - multiply initial volume by (1 - loss%) before recasting.

5. Composite solids with shared surfaces - ensure not to double-count overlapping volumes when adding parts.

Trick to Always Use

  • Step 1 → Draw and label - mark radii, heights, and which parts are removed/added.
  • Step 2 → Express every part with an exact formula (keep π, fractions symbolic).
  • Step 3 → Use similarity ratios to find missing lengths when solids are similar (linear ratio → squared for areas, cubed for volumes).
  • Step 4 → Add or subtract volumes/areas as the problem states; only evaluate numerically at the end.
  • Step 5 → For "how many" recast problems, divide and take the integer part (floor); report leftover material if required.

Summary

Summary

Trickier mensuration problems are solved by careful decomposition:

  • Translate the word problem into a labelled diagram and list the parts.
  • Write exact formulas for each part and use similarity ratios where necessary.
  • Add or subtract parts correctly; keep π symbolic until simplification is possible.
  • Convert units consistently and perform numeric evaluation only after simplifying algebraically.
  • Always perform a quick dimensional or numeric check to catch arithmetic or conceptual mistakes.

Practice

(1/5)
1. A cylindrical tank has a radius of 3.5 m and height 10 m. A hemispherical dome (same radius) is added on top of the cylinder. Find the total volume of the structure. (Use π = 22/7)
easy
A. 474.83 m³
B. 550.35 m³
C. 494.66 m³
D. 600.10 m³

Solution

  1. Step 1: Volume of cylinder.

    V_cyl = πr²h = (22/7) × (3.5)² × 10 = 22 × 17.5 = 385 m³.

  2. Step 2: Volume of hemisphere.

    V_hemi = (2/3)πr³ = (2/3) × (22/7) × (3.5)³ = (2/3) × 22 × 6.125 = 89.833333… m³.

  3. Step 3: Total volume.

    Total = 385 + 89.833333… = 474.833333… m³ ≈ 474.83 m³.

  4. Final Answer:

    474.83 m³ → Option A.

  5. Quick Check:

    Cylinder (≈385) + hemisphere (≈89.83) = 474.83 ✅
Hint: Total = cylinder volume + (2/3)πr³ for the hemisphere.
Common Mistakes: Using full-sphere volume instead of hemisphere or mixing radius/diameter.
2. A cube of side 10 cm has a cylindrical hole of radius 3 cm drilled straight through its center, perpendicular to one face. Find the volume of the remaining solid. (Use π = 3.14)
easy
A. 1000.0 cm³
B. 717.4 cm³
C. 972.0 cm³
D. 900.8 cm³

Solution

  1. Step 1: Volume of cube.

    V_cube = a³ = 10³ = 1000 cm³.

  2. Step 2: Volume of cylindrical hole.

    V_hole = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³.

  3. Step 3: Remaining volume.

    Remaining = 1000 - 282.6 = 717.4 cm³.

  4. Final Answer:

    717.4 cm³ → Option B.

  5. Quick Check:

    Subtract cylinder volume from cube volume: 1000 - 282.6 = 717.4 ✅
Hint: Remaining = cube volume - (πr² × cube side) when hole passes fully through.
Common Mistakes: Using cube side as radius or forgetting to multiply by hole height (10 cm).
3. A solid metal sphere of radius 7 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 4 cm. How many cones are formed? (Use π = 22/7)
easy
A. 10
B. 12
C. 28
D. 16

Solution

  1. Step 1: Volume of sphere.

    V_sphere = (4/3)πR³ = (4/3) × (22/7) × 7³ = (4/3) × 22 × 49 = 4312/3 ≈ 1437.333… cm³.

  2. Step 2: Volume of one cone.

    V_cone = (1/3)πr²h = (1/3) × (22/7) × (3.5)² × 4 = (1/3) × (22/7) × 49 = 154/3 ≈ 51.333… cm³.

  3. Step 3: Number of cones = total ÷ one cone.

    1437.333… ÷ 51.333… = 28.

  4. Final Answer:

    28 cones → Option C.

  5. Quick Check:

    28 × 51.333… = 1437.333… (matches sphere volume) ✅
Hint: Cancel π and common factors early; divide total volume by single-item volume and take integer.
Common Mistakes: Forgetting the 1/3 in cone volume or rounding too early.
4. A hemisphere of radius 7 cm is placed on top of a closed cylinder of the same radius and height 10 cm. Find the total external surface area of the combined solid (take π = 22/7).
medium
A. 792 cm²
B. 968 cm²
C. 1155 cm²
D. 902 cm²

Solution

  1. Step 1: Identify exposed surfaces.

    Exposed parts: curved surface of cylinder, curved surface of hemisphere, and bottom base of cylinder.

  2. Step 2: Compute each area.

    CSA_cyl = 2πrh = 2 × (22/7) × 7 × 10 = 140π = 440 cm².
    CSA_hemi = 2πr² = 2 × (22/7) × 7² = 98π = 308 cm².
    Base area = πr² = (22/7) × 49 = 49π = 154 cm².

  3. Step 3: Add areas.

    Total = 140π + 98π + 49π = 287π = 287 × (22/7) = 902 cm².

  4. Final Answer:

    902 cm² → Option D.

  5. Quick Check:

    Sum (440 + 308 + 154) = 902 ✅
Hint: Total exposed area = 2πrh (cylinder) + 2πr² (hemisphere) + πr² (bottom base) = (2rh + 3r²)π.
Common Mistakes: Including the circular join between cylinder and hemisphere as exposed or double-counting bases.
5. A solid cube of side 12 cm is cut into 8 equal smaller cubes. Find the total surface area of all the smaller cubes combined.
medium
A. 1728 cm²
B. 1152 cm²
C. 864 cm²
D. 1536 cm²

Solution

  1. Step 1: Side of each small cube.

    Divide side by 2 (since 8=2³): small side = 12 ÷ 2 = 6 cm.

  2. Step 2: Surface area of one small cube.

    SA_one = 6a² = 6 × 6² = 6 × 36 = 216 cm².

  3. Step 3: Total surface area of 8 cubes.

    Total = 8 × 216 = 1728 cm².

  4. Final Answer:

    1728 cm² → Option A.

  5. Quick Check:

    8×216 = 1728 ✅
Hint: When cutting a cube into 2×2×2 smaller cubes, each small side = half the original; total SA = 8 × 6 × (half-side)².
Common Mistakes: Using original cube’s surface area instead of summing new cubes’ surfaces.

Mock Test

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