Question 1

A rectangular box measures 10 cm × 8 cm × 5 cm. What is the maximum length of a pencil that can fit inside it (i.e., the space diagonal)?

Accepted Answer

Introduction / Context: The longest segment inside a rectangular box is its space diagonal. For dimensions a, b, c, the space diagonal d satisfies d^2 = a^2 + b^2 + c^2. Substitute the given side lengths and simplify the radical if possible. Given Data / Assumptions: a = 10 cm, b = 8 cm, c = 5 cm. Space diagonal formula: d = √(a^2 + b^2 + c^2). Concept / Approach: Compute the squared sum and factor to simplify. Step-by-Step Solution: d^2 = 10^2 + 8^2 + 5^2 = 100 + 64 + 25 = 189d = √189 = √(9 * 21) = 3√21 cm Verification / Alternative check: N/A beyond arithmetic; 3√21 ≈ 13.747 cm makes sense for a box with maximum side 10 cm. Why Other Options Are Wrong: √150 and √98 correspond to smaller squared sums; 3√52 is too large (≈ 21.6 cm) and impossible for given box dimensions. Common Pitfalls: Using face-diagonal formula √(a^2 + b^2) instead of space diagonal; forgetting to include all three squared terms. Final Answer: 3√21 cm

Question 2

How many rectangular pieces of size 5 cm × 5 cm × 10 cm can be cut from a larger cake of size 5 cm × 30 cm × 30 cm (no waste by thickness mismatch)?

Accepted Answer

Introduction / Context: Cutting identical rectangular blocks from a larger rectangular block without waste is a packing-by-integers problem. The count equals the product of how many blocks fit along each dimension, using an orientation that matches the dimensions as integers. Given Data / Assumptions: Big cake: 5 × 30 × 30 (cm). Small piece: 5 × 5 × 10 (cm). Reorientation of the small piece dimensions is allowed. Concept / Approach: Choose orientation so each small dimension divides the corresponding big dimension. A natural fit is: along thickness 5 → 5, along one length 5 → 30, along the other 10 → 30. Step-by-Step Solution: Along 5: ⌊5/5⌋ = 1Along 30 with 5: ⌊30/5⌋ = 6Along 30 with 10: ⌊30/10⌋ = 3Total pieces = 1 * 6 * 3 = 18 Verification / Alternative check: Other orientations (e.g., 10 along thickness) would waste the 5-thickness dimension; the chosen orientation achieves exact fits on all sides. Why Other Options Are Wrong: 10 and 15 undervalue the packing; 30 would require fitting 6 along both 30-dimensions and multiple along thickness, which the 10-cm side cannot support. Common Pitfalls: Not considering rotation of the small block; forgetting to take floors of the ratios for each dimension. Final Answer: 18

Question 3

Cylinder height from volume and base area: Find the height h of a right circular cylinder whose volume is 551 m^3 and whose base area is 36.5 m^2.

Accepted Answer

Introduction / Context: When the base area of a cylinder is known, its height follows directly from the volume relation V = (base area) * (height). This problem tests unit awareness and one-step algebra. Given Data / Assumptions: Volume V = 551 m^3 Base area A_base = 36.5 m^2 Right circular cylinder; no rounding until the final step Concept / Approach: Use the linear relation V = A_base * h, hence h = V / A_base. With consistent SI units, the quotient gives meters directly. Step-by-Step Solution: h = V / A_base = 551 / 36.5 m36.5 * 15 = 547.5; remainder = 3.5 → h ≈ 15 + (3.5 / 36.5) ≈ 15.0959 mRounded sensibly, h ≈ 15.10 m (to 2 decimals) Verification / Alternative check: Back-calculate V using h = 15.10 m: V ≈ 36.5 * 15.10 = 551.15 m^3, matching to rounding tolerance, confirming that the exact quotient is ~15.096 m. Why Other Options Are Wrong: 7 m, 10.5 m, and 14 m are far from 15.096 m when multiplied by 36.5. Since none listed equals ~15.10 m, the correct choice is “None of these.” The distractor 15.1 m (if shown) illustrates the correct magnitude but is not among the original options. Common Pitfalls: Confusing diameter/radius formulas; mixing cm^2 with m^2; rounding too early leading to an option that seems to fit when it does not. Final Answer: None of these

Question 4

Longest pole across a rectangular room (space diagonal): A room measures 12 m (length) × 9 m (width) × 8 m (height). Find the length of the longest pole that can fit inside.

Accepted Answer

Introduction / Context: The longest stick that fits in a rectangular room aligns with the space diagonal of the cuboid. This classic 3D Pythagoras application appears often in aptitude tests. Given Data / Assumptions: L = 12 m, W = 9 m, H = 8 m Space diagonal d = √(L^2 + W^2 + H^2) Concept / Approach: Apply the 3D extension of Pythagoras: d = √(a^2 + b^2 + c^2) for a cuboid with edges a, b, c. Step-by-Step Solution: d = √(12^2 + 9^2 + 8^2) = √(144 + 81 + 64)d = √289 = 17 m Verification / Alternative check: Compute floor diagonal first: √(12^2 + 9^2) = √225 = 15; then √(15^2 + 8^2) = √(225 + 64) = √289 = 17. Same result. Why Other Options Are Wrong: 12 m and 14 m are below the space diagonal; 21 m exceeds any dimension; 15 m is only the floor diagonal, not the space diagonal. Common Pitfalls: Using only the floor diagonal; mixing up dimensions; arithmetic errors when squaring and adding. Final Answer: 17 m

Question 5

Longest rod inside a cube (space diagonal): What is the longest rod that can fit in a cubical vessel with side 10 cm?

Accepted Answer

Introduction / Context: For a cube, the longest segment entirely inside is the space diagonal, not the face diagonal or the edge. This checks formula recall and recognition of 3D geometry basics. Given Data / Assumptions: Cubic side s = 10 cm Space diagonal formula: d = s√3 Concept / Approach: Know three canonical lengths: edge (s), face diagonal (s√2), and space diagonal (s√3). The rod must align with the space diagonal. Step-by-Step Solution: d = s√3 = 10√3 cm Verification / Alternative check: Construct via 3D Pythagoras: face diagonal = s√2; then √((s√2)^2 + s^2) = √(2s^2 + s^2) = s√3, confirming the formula. Why Other Options Are Wrong: 10 cm (edge) and 10√2 (face diagonal) are shorter than the space diagonal; “None” is false since 10√3 is listed; 5√6 equals 10√(1.5), still smaller than 10√3. Common Pitfalls: Confusing face diagonal with space diagonal; forgetting the √3 factor for cubes. Final Answer: 10√3

Question 6

Longest rod in a 20 m × 16 m × 12 m room: Find the maximum length that can fit inside the room (the space diagonal).

Accepted Answer

Introduction / Context: The problem asks for the space diagonal of a rectangular room. This is another direct usage of the 3D Pythagoras theorem and light numeric handling. Given Data / Assumptions: L = 20 m, W = 16 m, H = 12 m Space diagonal d = √(L^2 + W^2 + H^2) Concept / Approach: Compute the sum of squares of the three edges and take the square root. Present the answer to one decimal place to match options. Step-by-Step Solution: d = √(20^2 + 16^2 + 12^2) = √(400 + 256 + 144)d = √800 = √(16 * 50) = 4√50 = 20√2 ≈ 28.284 mRounded to the listed precision → 28.2 m Verification / Alternative check: Since 20√2 ≈ 28.284, the nearest given option is 28.2 m (options do not include 28.3 m). Why Other Options Are Wrong: 20 m and 16.4 m are underestimates; 48 m is much too large; 28.3 m would be closer numerically but is not among the provided choices. Common Pitfalls: Using only two dimensions; arithmetic slips when adding squares; incorrect rounding. Final Answer: 28.2 m

Question 7

Sheet thickness from volume and spread area: Half a cubic meter of gold is hammered into a sheet that covers 1 hectare. Find the sheet’s thickness in cm.

Accepted Answer

Introduction / Context: Volume conservation gives thickness when a volume is spread uniformly over a known area. This tests conversions between meters and centimeters and hectares to square meters. Given Data / Assumptions: Volume V = 0.5 m^3 Area A = 1 hectare = 10,000 m^2 Thickness t = V / A (in meters), then convert to centimeters Concept / Approach: Use t = V / A. Convert meters to centimeters by multiplying by 100 (1 m = 100 cm). Step-by-Step Solution: t (in m) = 0.5 / 10000 = 0.00005 mt (in cm) = 0.00005 * 100 = 0.005 cm Verification / Alternative check: Back-check: Area * thickness = 10,000 m^2 * 0.00005 m = 0.5 m^3, confirming consistency. Why Other Options Are Wrong: 0.5 cm and 0.05 cm are 100× and 10× too thick; 0.0005 cm and 0.00005 cm are 10× and 100× too thin. Common Pitfalls: Confusing cm with m; using 1 hectare = 1000 m^2 (incorrect); forgetting to multiply by 100 when converting thickness to cm. Final Answer: 0.005 cm

Question 8

Rainfall volume over land area: In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground (in m^3).

Accepted Answer

Introduction / Context: Rainfall depth multiplied by area gives the water volume. The key is converting centimeters to meters and hectares to square meters correctly. Given Data / Assumptions: Depth h = 5 cm = 0.05 m Area A = 2 hectares = 20,000 m^2 Volume V = A * h Concept / Approach: Use direct multiplication with consistent SI units to obtain m^3. Step-by-Step Solution: V = 20,000 * 0.05 = 1,000 m^3 Verification / Alternative check: Per hectare volume = 10,000 * 0.05 = 500 m^3; for 2 hectares → 1000 m^3. Same result. Why Other Options Are Wrong: 100 and 10 m^3 miss a factor of 10–100; 10,000 m^3 is 10× too large; 500 m^3 is only for 1 hectare. Common Pitfalls: Leaving depth in cm; misremembering 1 hectare = 10,000 m^2. Final Answer: 1000 m^3

Question 9

Tank depth from base area and volume (unit conversion care): A rectangular tank has base area 6500 cm^2 and holds 2.6 m^3 of water. Find the depth (in meters).

Accepted Answer

Introduction / Context: This item checks careful unit conversion between cm^2 and m^2 before applying depth = volume / area. Mismatched units are a common trap. Given Data / Assumptions: A_base = 6500 cm^2 = 0.65 m^2 (since 1 m^2 = 10,000 cm^2) V = 2.6 m^3 Depth d = V / A_base Concept / Approach: Convert the base area to m^2, then divide the volume by this area to get meters for depth. Step-by-Step Solution: A_base = 6500 / 10000 = 0.65 m^2d = 2.6 / 0.65 = 4 m Verification / Alternative check: Back-check: 0.65 m^2 * 4 m = 2.6 m^3—exactly the stated volume. Why Other Options Are Wrong: 3.5 m and 5 m are off due to unit slips; 8 m doubles the correct value; 2 m halves it. Common Pitfalls: Converting cm^2 as if it were a linear unit; dividing 2.6 by 6500 (forgetting to convert area units first). Final Answer: 4 m

Question 10

Counting unit cubes in a larger cube: How many 10 cm × 10 cm × 10 cm cubes fit in a cube of edge 1 m?

Accepted Answer

Introduction / Context: Partitioning a big cube into equal smaller cubes reduces to counting how many fit along each edge, then cubing that count (since the arrangement is 3D). Given Data / Assumptions: Big cube edge = 1 m = 100 cm Small cube edge = 10 cm Number per edge = 100 / 10 = 10 Concept / Approach: Total small cubes = (number per edge)^3 = 10^3 = 1000. Step-by-Step Solution: Along each edge: 10 cubesTotal: 10 * 10 * 10 = 1000 Verification / Alternative check: Volume ratio method: (100^3) / (10^3) = 1,000,000 / 1000 = 1000, consistent. Why Other Options Are Wrong: 10 and 100 count 1D/2D tilings, not 3D; 10000 is 10,000 which is 10^4, too large; 500 is arbitrary. Common Pitfalls: Forgetting the third dimension; mixing cm with m without converting; dividing volumes but forgetting cubes must tile exactly (they do here). Final Answer: 1000

Question 11

Sphere where volume equals surface area (numerically): If a sphere’s volume equals its surface area numerically, find its radius (units consistent).

Accepted Answer

Introduction / Context: Setting two formulas equal tests algebraic manipulation. For spheres: Volume V = (4/3)πr^3 and Surface area S = 4πr^2. Equating V and S “numerically” means the numbers match in the same unit system. Given Data / Assumptions: V = (4/3)πr^3 S = 4πr^2 V = S (numerically) Concept / Approach: Divide both sides by 4πr^2 (positive for r > 0) to isolate r. Step-by-Step Solution: (4/3)πr^3 = 4πr^2Divide by 4πr^2 → r/3 = 1r = 3 (units) Verification / Alternative check: Compute both sides at r = 3: V = (4/3)π * 27 = 36π; S = 4π * 9 = 36π. Equal indeed. Why Other Options Are Wrong: 1, 2, and 4 do not satisfy r/3 = 1; 6 makes V triple S. Only 3 works. Common Pitfalls: Dropping π incorrectly; treating equality as dimensional rather than numerical; accepting r = 0 (invalid radius). Final Answer: 3 unit

Question 12

Curved surface area from given sphere volume: A sphere has volume V = (88/21) × 14^3 cm^3. Find its curved surface area (cm^2).

Accepted Answer

Introduction / Context: Recognize that (4/3)π matches (88/21) when π is taken as 22/7, a standard competition convention. This makes solving for r immediate, then CSA = 4πr^2 follows. Given Data / Assumptions: V = (88/21) * 14^3 cm^3 Sphere formulas: V = (4/3)πr^3; CSA = 4πr^2 Concept / Approach: Note (4/3)π = (4/3)*(22/7) = 88/21. Therefore, (88/21)r^3 = (88/21)*14^3 ⇒ r = 14 cm. Then compute CSA. Step-by-Step Solution: r = 14 cmCSA = 4πr^2 = 4π * 196 = 784πWith π = 22/7 → 784 * 22/7 = 112 * 22 = 2464 cm^2 Verification / Alternative check: Back-substitute r into V: (4/3)π * 14^3 = (88/21)*14^3, consistent with the given V. Why Other Options Are Wrong: Values 2424, 2446, 2484 deviate from exact 2464 with π = 22/7; 2400 is a rounded guess, not exact. Common Pitfalls: Trying to compute π more precisely breaks the designed integer match; mixing diameter and radius when squaring. Final Answer: 2464 cm^2

Question 13

Cylinder vs cone volume (same base radius and height): A cylinder and a cone share the same radius and height. Find the ratio of their volumes (cylinder : cone).

Accepted Answer

Introduction / Context: This is a direct ratio comparison using standard volume formulas for solids sharing the same base and height. Given Data / Assumptions: Cylinder volume V_cyl = πr^2h Cone volume V_cone = (1/3)πr^2h Concept / Approach: Compute V_cyl : V_cone = (πr^2h) : ((1/3)πr^2h) = 1 : 1/3 = 3 : 1. Step-by-Step Solution: V_cyl : V_cone = (πr^2h) : ((1/3)πr^2h) = 3 : 1 Verification / Alternative check: Pick r = h = 1 for a numerical check: cylinder 1π; cone (1/3)π → ratio 3 : 1. Why Other Options Are Wrong: 1:3 reverses the order; 1:2 and 2:1 do not match the 1/3 factor; 2:3 is arbitrary. Common Pitfalls: Forgetting the 1/3 in cone volume; mixing up which solid is first in the ratio. Final Answer: 3 : 1

Question 14

Diagonal of a cube from side length: A cube has side 9 cm. Find the length of its space diagonal (in cm).

Accepted Answer

Introduction / Context: The space diagonal in a cube is a standard formula application: d = s√3. This checks quick recall and substitution without overthinking. Given Data / Assumptions: s = 9 cm d = s√3 Concept / Approach: Apply the space diagonal formula for cubes. Step-by-Step Solution: d = 9√3 cm Verification / Alternative check: 3D Pythagoras from two orthogonal face diagonals also yields s√3, reinforcing correctness. Why Other Options Are Wrong: Other multiples of √3 correspond to different side lengths; only 9√3 matches s = 9 cm. Common Pitfalls: Confusing face diagonal (s√2) with space diagonal (s√3). Final Answer: 9√3

Question 15

Surface area of a sphere with radius 4 cm: Compute the total surface area (in square centimeters).

Accepted Answer

Introduction / Context: This is a straightforward substitution into the sphere surface area formula. The test is about using 4πr^2, not 2πr or 4/3πr^3. Given Data / Assumptions: r = 4 cm Surface area S = 4πr^2 Concept / Approach: Compute r^2 and multiply by 4π to keep the answer in exact π form as options are given with π. Step-by-Step Solution: S = 4π * 16 = 64π sq cm Verification / Alternative check: Numeric check: with π ≈ 3.1416, S ≈ 201.06 cm^2, consistent with 64π. Why Other Options Are Wrong: 69π, 35π, 48π are unrelated; 32π is half the correct value (confusing with 2πr^2, which is not a standard formula). Common Pitfalls: Using circumference or volume formulas instead of surface area; mis-squaring 4 as 4 rather than 16. Final Answer: 64π sq cm

Question 16

Volume of a hemisphere (radius 3 cm): Calculate the volume in cubic centimeters.

Accepted Answer

Introduction / Context: The hemisphere’s volume is half that of a sphere with the same radius. It checks recall of V_sphere and the 1/2 factor for a hemisphere. Given Data / Assumptions: r = 3 cm V_sphere = (4/3)πr^3 V_hemisphere = (1/2) * V_sphere Concept / Approach: Plug r = 3 into the sphere formula and then take half for the hemisphere. Step-by-Step Solution: V_sphere = (4/3)π * 27 = 36π cm^3V_hemisphere = 36π / 2 = 18π cm^3 Verification / Alternative check: Known template: hemisphere volume = (2/3)πr^3; substituting r = 3 gives (2/3)π * 27 = 18π cm^3, same result. Why Other Options Are Wrong: 4/3 π cm^3 omits r^3; 13π is arbitrary; 19/π is dimensionally wrong; 9π is half the hemisphere’s volume (quarter of a sphere). Common Pitfalls: Forgetting to halve the sphere’s volume; using surface area instead of volume. Final Answer: 18π cm^3

Question 17

Total surface area of a square pyramid: Slant height = 8 cm; base is a square of side 4 cm. Find the total surface area (cm^2).

Accepted Answer

Introduction / Context: For a right square pyramid, total surface area equals base area plus the sum of areas of four congruent triangular faces. Given slant height (not altitude), lateral area uses 1/2 * perimeter * slant height. Given Data / Assumptions: Base side a = 4 cm → base area A_base = a^2 = 16 cm^2 Slant height l = 8 cm Lateral area A_lat = (1/2) * (perimeter) * l = (1/2) * (4a) * l Concept / Approach: Compute perimeter then lateral area; add to base area for TSA. Step-by-Step Solution: Perimeter = 4 * 4 = 16 cmA_lat = (1/2) * 16 * 8 = 64 cm^2TSA = A_base + A_lat = 16 + 64 = 80 cm^2 Verification / Alternative check: Each triangular face area = (1/2) * base * slant height_along_face = (1/2) * 4 * 8 = 16; four faces → 64; adding base 16 → 80. Why Other Options Are Wrong: 64 ignores the base; 72, 84, 88 miscompute the lateral contribution or include extras not present. Common Pitfalls: Confusing slant height with triangular altitude to the apex; forgetting the base area in TSA. Final Answer: 80

Question 18

Cone height from volume equality with a sphere: A sphere of radius r has the same volume as a cone whose base radius is also r. Find the height of the cone.

Accepted Answer

Introduction / Context: Equating volumes tests recognition of standard formulas and simple algebra. Equal base radii remove one variable immediately. Given Data / Assumptions: V_sphere = (4/3)πr^3 V_cone = (1/3)πr^2h V_sphere = V_cone Concept / Approach: Cancel common factors and solve for h in terms of r. Step-by-Step Solution: (4/3)πr^3 = (1/3)πr^2hMultiply both sides by 3/(πr^2): 4r = hTherefore, h = 4r Verification / Alternative check: Substitute h = 4r back: V_cone = (1/3)πr^2 * 4r = (4/3)πr^3 = V_sphere, confirming equality. Why Other Options Are Wrong: 2r, r/3, and (2/3)r understate the needed height; 3r still falls short; only 4r equates the volumes. Common Pitfalls: Forgetting the 1/3 factor in cone volume; mixing radius of sphere with diameter in cone base. Final Answer: 4r

Question 19

Rise in ground level from excavated soil: A tank 3 m × 2 m × 1.5 m is dug in a field 22 m × 14 m. If the soil is spread evenly on the field, by how much does the level rise (in cm)?

Accepted Answer

Introduction / Context: Conservation of volume implies the excavated soil volume equals the added “slab” volume over the field. This problem checks multiplication/division with consistent units and final unit conversion to centimeters. Given Data / Assumptions: Tank volume V = 3 * 2 * 1.5 = 9 m^3 Field area A = 22 * 14 = 308 m^2 Rise h (in meters) satisfies A * h = V Concept / Approach: Compute h = V / A; then convert meters to centimeters by multiplying by 100. Step-by-Step Solution: h (m) = 9 / 308 ≈ 0.02922 mh (cm) = 0.02922 * 100 ≈ 2.922 cm ≈ 2.98 cm (to 2 d.p.) Verification / Alternative check: Reverse: 308 * 0.02922 ≈ 8.99976 ≈ 9 m^3, confirming roundoff consistency. Why Other Options Are Wrong: 0.29–0.299 cm are ten times too small; 4.15 cm is too large; 1.5 cm halves the correct value. Common Pitfalls: Converting to cm^3 wrongly; using perimeter instead of area; rounding prematurely before unit conversion. Final Answer: 2.98 cm

Question 20

Surface area ratio from cube volume ratio: Two cubes have volumes in the ratio 1 : 27. Find the ratio of their surface areas.

Accepted Answer

Introduction / Context: For similar solids, volumes scale as the cube of the linear scale factor, and surface areas scale as its square. Extract the edge ratio first from the volume ratio, then square for surface areas. Given Data / Assumptions: V1 : V2 = 1 : 27 Edge ratio s1 : s2 = ∛(1) : ∛(27) = 1 : 3 Concept / Approach: Surface area ratio = (s1 : s2)^2 = 1^2 : 3^2 = 1 : 9. Step-by-Step Solution: Edge ratio = 1 : 3Surface area ratio = 1^2 : 3^2 = 1 : 9 Verification / Alternative check: Pick sample edges 1 and 3; areas are 6*1^2 and 6*3^2 → 6 : 54 → 1 : 9 after canceling 6. Why Other Options Are Wrong: 1:3 confuses linear with area; 1:8 belongs to cubes vs squares; 1:18 mixes factors; 1:27 is the volume ratio, not area. Common Pitfalls: Forgetting to take cube root for edge ratio before squaring for area. Final Answer: 1 : 9

Volume and Surface Area Questions