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Mensuration 3D Questions for SSC CGL

Free, AI-curated practice for the Mensuration 3D section of SSC CGL. We have 32+ verified questions in this bank. Below: 5 sample questions. Sign up free to unlock unlimited practice + AI explanations + per-topic analytics.

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Why this topic matters · 9 min read
Mensuration 3D is a high-weight topic in SSC CGL Quant, appearing 3-5 questions per exam across Tier 1 and Tier 2. Questions test surface area, volume, and conversion between shapes (melting/recasting). Cube, cuboid, cylinder, cone, sphere, and hemisphere are must-know. Expect one tricky question on combining two solids or converting one shape into another. Speed depends entirely on formula recall — no derivation needed.

Cube and Cuboid

A cube has all sides equal (side = a). A cuboid has length l, breadth b, height h. These are the simplest 3D shapes and often appear in questions about painting boxes, filling rooms, or cutting shapes. The diagonal of a cuboid is a commonly tested formula that many aspirants forget.

  • Cube: all edges equal, 6 equal square faces
  • Cuboid: 3 pairs of rectangular faces
  • Lateral Surface Area of cuboid = 2h(l+b), useful when top/bottom are open
  • Volume of cube = a^3; very common in questions about ratio of volumes after side changes
  • Diagonal of cube = a*sqrt(3); diagonal of cuboid = sqrt(l^2 + b^2 + h^2)
  • If side of cube is doubled, volume becomes 8 times — this ratio trick is directly asked
Key formulas
Cube TSA
TSA = 6a^2
When: Total surface area of cube
Cuboid TSA
TSA = 2(lb + bh + lh)
When: Total surface area of cuboid
Cuboid Volume
V = l x b x h
When: Volume of cuboid or rectangular box
Cuboid Diagonal
d = sqrt(l^2 + b^2 + h^2)
When: Longest diagonal inside the cuboid
Worked examples

A cube of side 6 cm is painted on all faces. It is then cut into 1 cm cubes. How many small cubes have exactly 2 faces painted? Answer: Small cubes with 2 painted faces lie on edges (not corners). Number = 12 edges x (6-2) = 12 x 4 = 48.

Ratio of volumes of two cubes with sides 3 cm and 6 cm = 3^3 : 6^3 = 27 : 216 = 1 : 8.

Cylinder

A cylinder has two circular faces and one curved surface. Think of a can of paint. SSC CGL loves questions where water flows through a pipe (cylinder) to fill a tank, or a roller (cylinder) paints a road — that uses Curved Surface Area. Always check if the question asks for open or closed cylinder.

  • Curved Surface Area (CSA) = 2*pi*r*h — used for pipes, rollers, road painting
  • Total Surface Area = 2*pi*r*(r+h) — used when both circular ends are included
  • Volume = pi*r^2*h
  • If height doubles and radius halves, volume becomes pi*(r/2)^2*(2h) = pi*r^2*h/2 — volume is halved
  • Hollow cylinder volume = pi*h*(R^2 - r^2) where R=outer radius, r=inner radius
Key formulas
Cylinder CSA
CSA = 2 x pi x r x h
When: Curved part only — pipes, rollers
Cylinder TSA
TSA = 2 x pi x r x (r + h)
When: Closed cylinder — tins, drums
Cylinder Volume
V = pi x r^2 x h
When: Filling, melting, recasting problems
Worked example

A cylindrical tank of radius 7 m and height 3 m is to be painted on the curved surface. Cost at Rs 5 per sq m = CSA x 5 = 2 x (22/7) x 7 x 3 x 5 = 2 x 22 x 3 x 5 = 660 Rs.

Cone

A cone has a circular base and a pointed apex. The slant height l connects the apex to the edge of the base. Remember: slant height is NOT the same as vertical height. Use Pythagoras: l^2 = r^2 + h^2. Most mistakes happen when a question gives vertical height but you use it directly as slant height in CSA formula.

  • Slant height l = sqrt(r^2 + h^2) — always calculate this first if not given
  • CSA = pi*r*l — think of it as unrolling the cone into a sector
  • TSA = pi*r*(r + l) — includes circular base
  • Volume = (1/3)*pi*r^2*h — one-third of cylinder with same base and height
  • Cone + hemisphere combo questions appear frequently in Tier 2
Key formulas
Slant Height
l = sqrt(r^2 + h^2)
When: Always needed for surface area; derive from r and h
Cone CSA
CSA = pi x r x l
When: Curved surface of cone — tent, funnel problems
Cone Volume
V = (1/3) x pi x r^2 x h
When: Volume of cone or melting cone into sphere etc.
Worked examples

A cone has radius 6 cm and height 8 cm. Slant height = sqrt(36+64) = sqrt(100) = 10 cm. CSA = pi x 6 x 10 = 60*pi sq cm.

A metallic cone of radius 6 cm and height 24 cm is melted into a sphere. Find sphere radius. Volume of cone = (1/3)*pi*36*24 = 288*pi. Volume of sphere = (4/3)*pi*r^3 = 288*pi. So r^3 = 216, r = 6 cm.

Sphere and Hemisphere

A sphere has no flat face. A hemisphere is exactly half a sphere but has one flat circular face at the bottom. This flat face is often forgotten when computing Total Surface Area of hemisphere. Think of a bowl: the inside curve plus the flat circular rim at the top.

  • Sphere TSA = 4*pi*r^2 — there is no separate CSA vs TSA for sphere
  • Hemisphere CSA = 2*pi*r^2 (curved part only)
  • Hemisphere TSA = 3*pi*r^2 (curved + flat circular base)
  • Volume of sphere = (4/3)*pi*r^3
  • Volume of hemisphere = (2/3)*pi*r^3
  • If radius is doubled, surface area becomes 4 times and volume becomes 8 times
Key formulas
Sphere SA
SA = 4 x pi x r^2
When: Surface area of full sphere — ball, planet problems
Sphere Volume
V = (4/3) x pi x r^3
When: Volume of sphere — melting/recasting to cone or cylinder
Hemisphere TSA
TSA = 3 x pi x r^2
When: Total area of bowl including circular base
Worked example

How many small spheres of radius 1 cm can be made from a large sphere of radius 4 cm? Volume of large = (4/3)*pi*64. Volume of small = (4/3)*pi*1. Number = 64/1 = 64 spheres.

Melting and Recasting (Conversion Problems)

These are the most common SSC CGL 3D problems. One solid is melted and recast into another. The key principle: Volume remains constant (assuming no wastage). Set Volume1 = Volume2 and solve for the unknown dimension. Wastage, if mentioned, means final volume = original volume x (1 - wastage%).

  • Core rule: Volume of original = Volume of new shape (no wastage)
  • With wastage: Volume of new = Volume of original x (1 - wastage/100)
  • Number of small shapes = Volume of large / Volume of one small shape
  • Always convert units to the same before computing
  • Common combo: cylinder melted into cones, sphere melted into small spheres, cone into hemisphere
Key formulas
Recasting Rule
V1 = V2 (no wastage); V2 = V1 x (1 - w/100) with wastage w%
When: Any melting or recasting problem
Number of pieces
n = Volume of original / Volume of one small piece
When: How many small shapes from one large shape
Worked example

A cylinder of radius 4 cm and height 9 cm is melted into cones of radius 2 cm and height 3 cm. Number of cones = Volume of cylinder / Volume of one cone = (pi*16*9) / ((1/3)*pi*4*3) = 144*pi / 4*pi = 36 cones.

⚠ Common mistakes to avoid
  • Using vertical height h instead of slant height l in the cone CSA formula — always calculate l = sqrt(r^2+h^2) first
  • Forgetting the flat circular base of hemisphere in TSA — hemisphere TSA is 3*pi*r^2, not 2*pi*r^2
  • Confusing Lateral/Curved Surface Area with Total Surface Area when the question asks for cost of painting only the sides
  • Not converting units — mixing cm and m in the same problem leads to wrong volume by a factor of 10^6
  • In cube painting and cutting problems, miscounting edge cubes — corners have 3 painted faces, edges have 2, faces have 1, and inner cubes have 0
🧠 Memory aids
  • Cone volume = one-third of cylinder: CONE is a CUT cylinder — cut away two-thirds, keep one-third
  • Hemisphere TSA = 3*pi*r^2: think 3 as Three-part bowl: 2 curved + 1 flat = 3
  • Slant height mnemonic: SLOPE needs Pythagoras — l^2 = r^2 + h^2, just like a right triangle on the side of the cone
  • For recasting: VOLUME IS IMMORTAL — it never changes when you melt and recast (unless wastage is given)
  • Sphere surface area = 4 x circle area: one sphere wraps around exactly 4 circles of the same radius
🎯 SSC CGL exam tips
  • SSC CGL Tier 1 typically has 2-3 direct formula-application questions on cylinder and cone. Memorize formulas cold — no time to derive.
  • Tier 2 Paper 1 (Math) often has a combined solid question like a cone placed on a hemisphere or cylinder — practice finding total surface area by adding CSA of each part (do not double-count shared base).
  • Recasting problems (melting one shape into another) appear almost every year. These are 1-2 minute questions if you remember Volume1 = Volume2 — do not overthink.
  • Cube painting and cutting is a pure logic question — no formula needed, just visualize layers. Corner=3 faces, Edge=2 faces, Face=1 face, Inside=0 faces. Draw it once and remember for life.
  • Use pi = 22/7 when radius is a multiple of 7, and pi = 3.14 otherwise — SSC questions are designed for clean answers with the right pi value.

Sample questions

Q1 · medium · AI-verified
The slant height of a frustum of a cone is 4 cm and the radii of its circular bases are 3 cm and 6 cm. What is the curved surface area of the frustum (in cm²)? (Use π = 22/7)
  1. 132 cm²
  2. 113 cm²
  3. 198 cm²
  4. 792/7 cm²
Q2 · hard · AI-verified
A cube of side 4 cm is painted on all faces and then cut into cubes of side 1 cm. How many smaller cubes have exactly two faces painted?
  1. 24
  2. 32
  3. 8
  4. 16
Q3 · medium · AI-verified
The radius of the base of a right circular cone is 14 cm and its height is 24 cm. What is the volume of the cone (in cm³)? (Use π = 22/7)
  1. 3696 cm³
  2. 4928 cm³
  3. 7392 cm³
  4. 5544 cm³
Q4 · medium · AI-verified
If the radius of a cylinder is doubled and its height is halved, then by what factor does its volume change?
  1. Volume becomes 2 times
  2. Volume remains unchanged
  3. Volume becomes 4 times
  4. Volume becomes 8 times
Q5 · medium · AI-verified
A cylinder and a cone have the same base radius of 7 cm. If the height of the cylinder is 14 cm and that of the cone is 14 cm, what is the ratio of their volumes?
  1. 2 : 1
  2. 1 : 3
  3. 1 : 1
  4. 3 : 1
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