Question 1

Open wooden box (painted inside) — find painting rate per cm2: A wooden box, open at the top, has wall thickness 0.5 cm and outside dimensions 21 cm × 11 cm × 6 cm. The inside surfaces are painted for a total expense of ₹ 70. What is the painting ra…

Accepted Answer

Introduction / Context: The painting cost equals the painted inside surface area multiplied by the rate. For an open-top box, the inside painted surfaces are the base and the four inner walls, computed using inner dimensions. Given Data / Assumptions: Outer: 21 × 11 × 6 cm; thickness = 0.5 cm. Inner dimensions: L_i = 21 − 2*0.5 = 20 cm; W_i = 11 − 2*0.5 = 10 cm; H_i = 6 − 0.5 = 5.5 cm (open top → subtract only bottom thickness). Cost = ₹ 70. Concept / Approach: Inside area A_i = base + two long walls + two short walls = (L_i*W_i) + 2(L_i*H_i) + 2(W_i*H_i). Rate = Cost / A_i. Step-by-Step Solution: Base = 20*10 = 200 cm2Two long walls = 2*(20*5.5) = 220 cm2Two short walls = 2*(10*5.5) = 110 cm2A_i = 200 + 220 + 110 = 530 cm2Rate = 70 / 530 ≈ ₹ 0.132 per cm2 Verification / Alternative check: Open top ensures no painting on the top face; inner height uses only bottom thickness. Why Other Options Are Wrong: ₹ 0.10 and ₹ 0.20 are rough-round values; ₹ 0.50 is far too high. Common Pitfalls: Using outer dimensions for area; subtracting thickness twice on height. Final Answer: ₹ 0.132 per cm2

Question 2

Packing small boxes in a large crate — convert meters to centimeters precisely: A wooden crate of inner dimensions 8 m × 7 m × 6 m is used to carry identical rectangular boxes of size 8 cm × 7 cm × 6 cm (no gaps). What is the maximum number of such …

Accepted Answer

Introduction / Context: This is a standard 3D packing by exact integer fit along each dimension. Convert the crateʼs dimensions to centimeters so they are commensurate with the small boxes, then multiply counts along each axis. Given Data / Assumptions: Crate: 8 m × 7 m × 6 m = 800 cm × 700 cm × 600 cm Box: 8 cm × 7 cm × 6 cm Assume perfect orientation and no wasted space. Concept / Approach: Count per dimension = floor(L/ℓ) × floor(W/w) × floor(H/h). Step-by-Step Solution: Along 800/8 = 100Along 700/7 = 100Along 600/6 = 100Total = 100 * 100 * 100 = 1,000,000 Verification / Alternative check: All divisions are exact integers; hence no leftover space in this idealized scenario. Why Other Options Are Wrong: They do not match the exact 100 × 100 × 100 fit. Common Pitfalls: Forgetting to convert meters to centimeters or mis-multiplying powers of 10. Final Answer: 1000000

Question 3

Cuboid dimensions from surface area and ratio: A cuboid has length : breadth : height in the ratio 6 : 5 : 4. If its total surface area is 33,300 cm2, find the actual dimensions (in cm) of length, breadth, and height.

Accepted Answer

Introduction / Context: We are given the ratio of the edges of a cuboid and its total surface area (TSA). Using the TSA formula and ratio scaling, we can determine the exact dimensions. This tests ratio scaling with geometric surface area. Given Data / Assumptions: Edge ratio: l : b : h = 6 : 5 : 4 Total surface area (TSA) = 33,300 cm2 Let l = 6k, b = 5k, h = 4k for some k > 0 Concept / Approach: Cuboid TSA = 2 * (l*b + b*h + h*l) Substitute proportional edges, solve for k, then compute actual dimensions. Step-by-Step Solution: Let l = 6k, b = 5k, h = 4k.TSA = 2 * (6k*5k + 5k*4k + 6k*4k) = 2 * (30k^2 + 20k^2 + 24k^2) = 2 * 74k^2 = 148k^2.Set 148k^2 = 33300 ⇒ k^2 = 33300 / 148 = 225 ⇒ k = 15.Thus l = 6*15 = 90 cm, b = 5*15 = 75 cm, h = 4*15 = 60 cm. Verification / Alternative check: Compute TSA with 90, 75, 60: l*b=6750, b*h=4500, h*l=5400; sum=16650; TSA=2*16650=33300 cm2, matches given. Why Other Options Are Wrong: 90, 85, 60: Breaks the ratio 6 : 5 : 4. 85, 75, 60: Not in a constant multiple of 6 : 5 : 4. 90, 75, 70: Violates the 4-part for height. Common Pitfalls: Forgetting to multiply all three pairwise products inside TSA. Using perimeter-like formula instead of surface area. Taking k = √(33300/148) but rounding incorrectly; here it is exactly 15. Final Answer: 90, 75, 60

Question 4

Cube from space diagonal: The space diagonal of a cube is 8√3 cm. Find the total surface area of the cube (in cm2).

Accepted Answer

Introduction / Context: The problem links a cube’s space diagonal to its edge length and then asks for surface area. This is a direct application of the cube diagonal relation and the surface area formula. Given Data / Assumptions: Space diagonal d = 8√3 cm For a cube of edge a: d = a√3 Concept / Approach: Compute a from d = a√3. Surface area SA = 6 * a^2. Step-by-Step Solution: Given d = 8√3 and d = a√3 ⇒ a = 8.SA = 6 * a^2 = 6 * 8^2 = 6 * 64 = 384 cm2. Verification / Alternative check: Recompute: a = d / √3 = (8√3)/√3 = 8; SA = 6*64 = 384. Why Other Options Are Wrong: 512 cm2: Uses 8 faces or wrong formula. 192 cm2: Halves the correct value. 786 cm2: Arbitrary, not from standard formula. Common Pitfalls: Confusing space diagonal with face diagonal (face diagonal = a√2). Forgetting to square a when computing SA. Final Answer: 384 cm2

Question 5

Room height from two diagonals: The longest rod that can lie flat on the floor of a rectangular room is 10 m (the floor diagonal). The longest rod that can fit inside the room is 10√2 m (space diagonal). Find the height of the room (in m).

Accepted Answer

Introduction / Context: This is a 3D Pythagoras application. The floor diagonal gives √(l^2 + b^2); the space diagonal gives √(l^2 + b^2 + h^2). Using both, we can isolate the height h of the room. Given Data / Assumptions: Floor diagonal = 10 m ⇒ √(l^2 + b^2) = 10 Space diagonal = 10√2 m ⇒ √(l^2 + b^2 + h^2) = 10√2 Concept / Approach: Use the identity: (space diagonal)^2 − (floor diagonal)^2 = h^2. Step-by-Step Solution: (10√2)^2 − (10)^2 = h^2200 − 100 = h^2 ⇒ h^2 = 100 ⇒ h = 10 m. Verification / Alternative check: Plug back: space diagonal = √(10^2 + h^2) with floor diagonal 10 gives √(100 + 100) = √200 = 10√2, consistent. Why Other Options Are Wrong: 8 m, 7.5 m, 6 m: Each gives space diagonal less than 10√2, contradicting the given value. Common Pitfalls: Confusing the floor diagonal with a wall diagonal. Not squaring correctly when applying Pythagoras in 3D. Final Answer: 10 m

Question 6

Cube volume from space diagonal: The space diagonal of a cube is (14 × √3) cm. Find the volume of the cube (in cm3).

Accepted Answer

Introduction / Context: Again we relate a cube’s space diagonal to its edge and then compute the volume. This is a direct geometry formula application. Given Data / Assumptions: Space diagonal d = 14√3 cm For a cube: d = a√3, volume V = a^3 Concept / Approach: Solve for edge a, then compute V = a^3. Step-by-Step Solution: d = a√3 ⇒ a = d / √3 = (14√3)/√3 = 14.V = a^3 = 14^3 = 2744 cm3. Verification / Alternative check: Check order: 14^2=196, 14^3=196*14=2744, consistent. Why Other Options Are Wrong: 2744√3 cm3: Incorrectly carries the √3 into volume. 588 cm3, 3528 cm3: Not equal to a^3 for any integer a here. Common Pitfalls: Mistaking face diagonal (a√2) for space diagonal (a√3). Forgetting to cube the edge length. Final Answer: 2744 cm3

Question 7

Longest pencil in a box (space diagonal): A rectangular box has dimensions 8 cm × 6 cm × 2 cm. What is the maximum length of a pencil that can fit inside it?

Accepted Answer

Introduction / Context: The longest object that fits inside a rectangular box equals its space diagonal. This tests 3D Pythagoras across three perpendicular edges. Given Data / Assumptions: Dimensions: 8 cm, 6 cm, 2 cm Max length L = √(8^2 + 6^2 + 2^2) Concept / Approach: Apply space diagonal formula for a cuboid. Step-by-Step Solution: L = √(8^2 + 6^2 + 2^2) = √(64 + 36 + 4) = √104.Since 104 = 4 * 26, L = √(4*26) = 2√26 cm. Verification / Alternative check: Approximate: √26 ≈ 5.099 ⇒ 2√26 ≈ 10.198 cm. Why Other Options Are Wrong: 2√13 cm: Uses half the sum inside the root incorrectly. 2√14 cm: Incorrect decomposition. 10√2 cm: Corresponds to √(8^2 + 6^2) only, ignoring the 2 cm height. Common Pitfalls: Using face diagonal instead of space diagonal. Arithmetic slips inside the square root. Final Answer: 2 √26 cm

Question 8

Discharge per minute in a rectangular river channel: A river 2 m deep and 45 m wide flows at 3 km/h. How much water (in m3) passes a section each minute?

Accepted Answer

Introduction / Context: Flow rate in a prismatic channel equals cross-sectional area times velocity. Convert the velocity to meters per minute and multiply by area to get discharge per minute. Given Data / Assumptions: Depth = 2 m, Width = 45 m ⇒ Area A = 2 * 45 = 90 m2 Speed v = 3 km/h = 3000 m/h = 50 m/min Concept / Approach: Discharge per minute Q = A * v (with consistent units). Step-by-Step Solution: A = 90 m2, v = 50 m/min.Q = 90 * 50 = 4500 m3/min. Verification / Alternative check: Per hour: 4500 * 60 = 270,000 m3/h; dividing by 60 returns 4500 m3/min. Why Other Options Are Wrong: 3000 m3: Ignores correct speed conversion. 27000 m3: Off by a factor of 6 (hour vs minute confusion). 2100 m3: Underestimates due to wrong area or speed. Common Pitfalls: Not converting km/h to m/min correctly. Using depth+width instead of depth×width for area. Final Answer: 4500 m3

Question 9

Cost of bricks for a wall from volume: Bricks cost Rs 750 per 1000. Each brick measures 25 cm × 12.5 cm × 7.5 cm. Find the cost of bricks to build a wall 200 m long, 1.8 m high, and 37.5 cm thick (ignore mortar gaps).

Accepted Answer

Introduction / Context: The number of bricks required equals (volume of wall) / (volume of one brick). Multiplying the brick count by the rate per 1000 gives the total cost. Unit consistency (meters vs centimeters) is essential. Given Data / Assumptions: Wall: 200 m × 1.8 m × 0.375 m Brick: 25 cm × 12.5 cm × 7.5 cm = 0.25 m × 0.125 m × 0.075 m Rate: Rs 750 per 1000 bricks Ignore mortar allowances (ideal packing). Concept / Approach: Compute both volumes in the same units. Number of bricks = V_wall / V_brick; Cost = (bricks / 1000) * rate. Step-by-Step Solution: V_wall = 200 * 1.8 * 0.375 = 135 m3.V_brick = 0.25 * 0.125 * 0.075 = 0.00234375 m3.Bricks needed = 135 / 0.00234375 = 57,600.Cost = 57.6 * 750 = Rs 43,200. Verification / Alternative check: In cm3: wall 135,000,000; brick 2343.75; 135,000,000 / 2343.75 = 57,600 (same count). Why Other Options Are Wrong: Rs. 42, 600 / 41, 860 / 40, 750: Result from rounding, wrong brick volume, or rate errors. Common Pitfalls: Mistaking cm for m in one dimension. Forgetting to divide by 1000 before multiplying by rate. Final Answer: Rs. 43, 200

Question 10

Surface area change after melting to a cube: A solid block (27 cm × 8 cm × 1 cm) is melted and recast into a cube. Find the difference between the surface areas of the original block and the new cube (in cm2).

Accepted Answer

Introduction / Context: Melting conserves volume. We first find the cube’s side from equal volume, then compute both surface areas and take the difference. This tests volume conservation and area formulas. Given Data / Assumptions: Rectangular solid: 27 × 8 × 1 cm Volume is conserved. Concept / Approach: V_block = l*b*h. Let cube side = a with a^3 = V_block. Compute SA_block and SA_cube and subtract. Step-by-Step Solution: V_block = 27*8*1 = 216 cm3 ⇒ a = 6 cm (since 6^3 = 216).SA_block = 2*(27*8 + 8*1 + 27*1) = 2*(216 + 8 + 27) = 502 cm2.SA_cube = 6*a^2 = 6*36 = 216 cm2.Difference = 502 − 216 = 286 cm2. Verification / Alternative check: Recompute each product to avoid arithmetic slips; sums match. Why Other Options Are Wrong: 296/300/284 cm2: Small arithmetic deviations, not matching exact computations. Common Pitfalls: Using perimeter-like ideas instead of surface area. Forgetting both small faces 8*1 and 27*1 contribute to SA. Final Answer: 286 cm2

Question 11

Wire length change with radius scaled: A wire is stretched so that its radius becomes one-third of the original while volume remains constant. By what factor does its length increase?

Accepted Answer

Introduction / Context: For a wire modeled as a cylinder, volume V = π r^2 L. If radius changes but volume is fixed, length must adjust inversely with r^2. This is a pure proportionality problem. Given Data / Assumptions: Initial radius r, length L; final radius r/3; volume constant. Concept / Approach: V_initial = π r^2 L, V_final = π (r/3)^2 L_new. Equate V_initial = V_final and solve for L_new / L. Step-by-Step Solution: π r^2 L = π (r^2 / 9) * L_new.Cancel π r^2: L = (1/9) L_new ⇒ L_new = 9 L. Verification / Alternative check: Area scales with r^2; decreasing radius by factor 3 decreases area by 9; length must grow by 9 to keep volume constant. Why Other Options Are Wrong: 6 times / 2 times / 1 time: Do not satisfy V constant with r → r/3. Common Pitfalls: Thinking length scales inversely with r, not r^2. Forgetting cylindrical volume formula. Final Answer: 9 times

Question 12

Find breadth from wall volume relation: A wall is 5 times as high as it is broad and 8 times as long as it is high. If its volume is 12.8 m3, find the breadth (in cm).

Accepted Answer

Introduction / Context: The wall’s three dimensions are given in chained ratios in terms of breadth b. Express height and length in b, form the volume equation, and solve for b. Finally convert meters to centimeters. Given Data / Assumptions: Breadth = b, Height = 5b, Length = 8*(Height) = 40b Volume = 12.8 m3 Concept / Approach: Volume V = b * 5b * 40b = 200 b^3. Solve for b^3 = V / 200, then take cube root. Step-by-Step Solution: 200 b^3 = 12.8 ⇒ b^3 = 12.8 / 200 = 0.064.b = cube_root(0.064) = 0.4 m.Convert to cm: 0.4 m = 40 cm. Verification / Alternative check: Check: 0.4 * 2.0 * 16.0 = 12.8 m3 (since 5b=2.0, 40b=16.0); product is 12.8. Why Other Options Are Wrong: 22.5/25/30 cm: Do not satisfy 200 b^3 = 12.8 when converted back. Common Pitfalls: Using 8 times the breadth instead of 8 times the height for length. Incorrect meter–centimeter conversion. Final Answer: 40 cm

Question 13

Sphere from volume to curved surface area: The volume of a sphere is 38,808 cm3. Find its curved surface area (in cm2) using π = 22/7.

Accepted Answer

Introduction / Context: Given volume, we find the radius, then compute curved surface area (CSA). Using π = 22/7 simplifies to integer-friendly values in many classic problems. Given Data / Assumptions: V = 38,808 cm3, π = 22/7 Sphere: V = (4/3) * π * r^3, CSA = 4 * π * r^2 Concept / Approach: Solve r from V, then compute CSA. Step-by-Step Solution: (4/3) * (22/7) * r^3 = 38808 ⇒ r^3 = 38808 * 3 / (4 * 22/7) = 9261.r = cube_root(9261) = 21 cm.CSA = 4 * (22/7) * r^2 = 4 * (22/7) * 441 = (88/7) * 441 = 88 * 63 = 5544 cm2. Verification / Alternative check: Compute 21^3 = 9261; back-substitution gives the original volume. Why Other Options Are Wrong: 1386/4158/8316: Do not equal 4πr^2 for r = 21 and π = 22/7. Common Pitfalls: Using 3.14 for π here yields small rounding differences; the question expects 22/7. Mixing up CSA with total surface area (same for sphere) is fine, but arithmetic must match. Final Answer: 5544 sq. cm

Question 14

How many cones equal one cylinder (same r and h): A right circular cylinder is completely filled with water. How many right circular cones with the same radius and height are needed to store the same water?

Accepted Answer

Introduction / Context: This tests the known volume relationship between a cylinder and a cone with identical radius and height. It is a standard ratio result used often in mensuration. Given Data / Assumptions: Same radius r and height h for both solids. V_cylinder = π r^2 h; V_cone = (1/3) π r^2 h. Concept / Approach: Compare the formulas to find how many cones equal one cylinder. Step-by-Step Solution: V_cylinder / V_cone = (π r^2 h) / ((1/3) π r^2 h) = 3.Therefore, 3 cones of the same r and h equal the cylinder volume. Verification / Alternative check: Pick any r,h (e.g., r=1, h=1): V_cyl=π; V_cone=π/3; three cones give π. Why Other Options Are Wrong: 2/4/5: Do not match the derived exact ratio of 3. Common Pitfalls: Using 1/2 instead of 1/3 for the cone volume. Forgetting the factor 1/3 attached to the cone formula. Final Answer: 3

Question 15

Height ratio for equal volumes (cylinder vs cone): A right cylinder and a right circular cone have the same radius and the same volume. Find the ratio of the height of the cylinder to the height of the cone.

Accepted Answer

Introduction / Context: With the same radius, equal volumes force a relationship between heights using V_cyl = π r^2 h_c and V_cone = (1/3) π r^2 h_k. Equate and solve for h_c : h_k. Given Data / Assumptions: Same radius r; equal volumes. V_cyl = π r^2 h_c, V_cone = (1/3) π r^2 h_k. Concept / Approach: Set π r^2 h_c = (1/3) π r^2 h_k ⇒ h_c = h_k / 3. Step-by-Step Solution: h_c = (1/3) h_k ⇒ h_c : h_k = 1 : 3. Verification / Alternative check: Pick r=1, h_k=3: V_cone=(1/3)π*1*3=π; with h_c=1, V_cyl=π. Why Other Options Are Wrong: 3:1 / 3:5 / 2:5: Inconsistent with the cone’s 1/3 volume factor. Common Pitfalls: Inverting the ratio (writing 3:1 rather than 1:3). Final Answer: 1 : 3

Question 16

Rise in water level due to a submerged sphere: A cylinder of radius 4 cm contains water. A solid sphere of radius 3 cm is fully immersed. By how many cm does the water level rise?

Accepted Answer

Introduction / Context: Volume displaced equals the volume of the sphere. The rise in height equals displaced volume divided by the cylinder’s cross-sectional area. This checks volume formulas and unit consistency. Given Data / Assumptions: Cylinder radius R = 4 cm ⇒ area A = π R^2 = 16π cm2 Sphere radius r = 3 cm ⇒ volume V_sph = (4/3) π r^3 = (4/3) π * 27 = 36π cm3 Concept / Approach: Height rise h = displaced volume / area = V_sph / A. Step-by-Step Solution: h = (36π) / (16π) = 36 / 16 = 9 / 4 = 2.25 cm. Verification / Alternative check: Cancel π; arithmetic 36/16 reduces to 9/4 = 2.25, consistent. Why Other Options Are Wrong: 4.5 cm: Doubles the correct rise. 4/9 cm or 2/9 cm: Misplaced numerator/denominator during division. Common Pitfalls: Using cylinder volume formula instead of area in the denominator. Forgetting to cube the sphere radius when computing volume. Final Answer: 2.25 cm

Question 17

Cylinder height from sphere volume equality: A circular cylinder has the same radius as a sphere and their volumes are equal. The height of the cylinder is equal to what multiple of its radius?

Accepted Answer

Introduction / Context: Equating cylinder and sphere volumes with common radius yields a direct relationship between cylinder height and radius. This is a standard mensuration identity. Given Data / Assumptions: Common radius r. V_cylinder = π r^2 h; V_sphere = (4/3) π r^3; volumes equal. Concept / Approach: Set π r^2 h = (4/3) π r^3 and solve for h in terms of r. Step-by-Step Solution: π r^2 h = (4/3) π r^3 ⇒ cancel π r^2 (r>0) to get h = (4/3) r. Verification / Alternative check: Pick r=3 ⇒ sphere V = (4/3)π*27 = 36π; cylinder with h=4 gives V=π*9*4=36π, equal. Why Other Options Are Wrong: 2/3 r or r: Too short to produce equal volume. Diameter (2r): Too tall; volume would exceed the sphere volume. Common Pitfalls: Dropping the 1/3 in sphere’s volume. Confusing diameter and radius when interpreting options. Final Answer: 4/3 times its radius

Question 18

Volume of iron in a hollow roller: A hollow cylindrical garden roller is 63 cm long with girth (outer circumference) 440 cm. Iron thickness is 4 cm. Find the volume of iron used (in cubic cm). Take π = 22/7.

Accepted Answer

Introduction / Context: For a hollow cylinder, volume of material equals π (R^2 − r^2) L. The girth gives the outer radius. Subtracting thickness yields the inner radius. Substitute and compute with π = 22/7. Given Data / Assumptions: Length L = 63 cm Outer circumference C = 440 cm ⇒ R = C / (2π) Thickness t = 4 cm ⇒ inner radius r = R − 4 π = 22/7 Concept / Approach: Compute R, r, then V = π (R^2 − r^2) L = π (R − r)(R + r)L. Step-by-Step Solution: R = 440 / (2 * 22/7) = 440 * 7 / 44 = 70 cm; r = 70 − 4 = 66 cm.R^2 − r^2 = (70 − 66)(70 + 66) = 4 * 136 = 544.V = (22/7) * 544 * 63 = 22 * 544 * 9 = 107,712 cm3. Verification / Alternative check: Unit check: cm * cm2 = cm3; arithmetic matches stepwise factors. Why Other Options Are Wrong: 54982 / 57636 cm3: Arithmetic slips. 56372 cubic m: Wrong unit (m vs cm). Common Pitfalls: Using diameter instead of radius after circumference. Forgetting to subtract thickness from the radius (not diameter). Final Answer: 107712 cubic cm

Question 19

Wire length from melted sphere: A solid copper sphere of diameter 18 cm is melted and drawn into a wire of radius 0.2 mm. Find the length of the wire (in meters).

Accepted Answer

Introduction / Context: Volume is conserved when a solid is melted and redrawn. Compute the sphere’s volume and equate it to the cylindrical wire’s volume to solve for the wire length. Unit consistency (mm to cm) is crucial. Given Data / Assumptions: Sphere diameter = 18 cm ⇒ radius r_s = 9 cm Wire radius r_w = 0.2 mm = 0.02 cm V_sphere = (4/3) π r_s^3; V_wire = π r_w^2 * L Concept / Approach: Set V_sphere = V_wire and solve for L. Step-by-Step Solution: V_sphere = (4/3)π * 9^3 = (4/3)π * 729 = 972π cm3.V_wire = π * (0.02)^2 * L = π * 0.0004 * L.Equate: 972π = 0.0004π L ⇒ L = 972 / 0.0004 = 2,430,000 cm = 24,300 m. Verification / Alternative check: Convert cm to m by dividing by 100: 2,430,000 / 100 = 24,300 m. Why Other Options Are Wrong: 243 m / 2430 m / 24.3 m: Off by powers of 10 due to unit conversion or squaring radius mistakes. Common Pitfalls: Using radius 0.2 cm instead of 0.2 mm = 0.02 cm. Forgetting to square the wire radius in π r^2 L. Final Answer: 24300 m

Question 20

Volume of cylinder from curved surface area and height: A cylinder has height 14 cm and curved surface area 264 cm2. Find its volume (in cm3). Use π = 22/7.

Accepted Answer

Introduction / Context: From curved surface area (CSA) we can find the radius, and then compute the cylinder’s volume. This checks correct use of CSA and volume formulas with π = 22/7. Given Data / Assumptions: Height h = 14 cm CSA = 264 cm2 CSA = 2π r h; Volume V = π r^2 h π = 22/7 Concept / Approach: Find r from r = CSA / (2πh); then compute V. Step-by-Step Solution: r = 264 / (2 * (22/7) * 14) = 264 / (88) = 3 cm.V = (22/7) * 3^2 * 14 = (22/7) * 9 * 14 = 22 * 18 = 396 cm3. Verification / Alternative check: Check CSA with r=3: 2*(22/7)*3*14 = 264 (consistent). Why Other Options Are Wrong: 308/1232/1448 cm3: Result from wrong r or using total surface area mistakenly. Common Pitfalls: Using total surface area formula instead of CSA. Not simplifying the fraction with π = 22/7 cleanly. Final Answer: 396 cm3

Volume and Surface Area Questions