Question 1

A cube has space diagonal 2√3 cm. Find the total surface area of the cube.

Accepted Answer

Introduction / Context: The diagonal that joins two opposite vertices of a cube (space diagonal) relates directly to its edge length. Once the edge is known, the total surface area follows from a standard formula. This problem checks recall and correct use of cube identities without needing advanced geometry. Given Data / Assumptions: Space diagonal d = 2√3 cm. For a cube of edge a: space diagonal d = a√3. Total surface area S = 6a^2. Concept / Approach: Invert the space-diagonal relation to find a = d/√3, then substitute into S = 6a^2 to get the surface area. Keep the expression exact; numbers are chosen so the radical cancels neatly. Step-by-Step Solution: a = d / √3 = (2√3) / √3 = 2 cmS = 6a^2 = 6 * 2^2 = 6 * 4 = 24 sq cm Verification / Alternative check: Rebuild the diagonal from a = 2: d = a√3 = 2√3 cm, matching the given. Therefore the derived surface area is consistent. Why Other Options Are Wrong: 15, 18, and 25 sq cm do not equal 6a^2 with a = 2. They arise if one confuses face area (a^2 = 4) or misapplies constants (e.g., 4a^2 = 16). Only 24 sq cm fits the cube identity. Common Pitfalls: Using the face diagonal formula a√2 instead of the space diagonal a√3; forgetting to square a in S = 6a^2; or rounding radicals prematurely. Final Answer: 24 sq cm

Question 2

A cube has total surface area 6 square units. Find the volume of the cube (in cubic units).

Accepted Answer

Introduction / Context: Surface area and volume of a cube are tied through the edge length a. Given the surface area, we first recover a from S = 6a^2 and then compute V = a^3. This reinforces correct manipulation of powers when moving between area (square units) and volume (cubic units). Given Data / Assumptions: Total surface area S = 6 square units. Cube identities: S = 6a^2 and V = a^3. Concept / Approach: Solve a from S = 6a^2 → a^2 = S/6; take square root to get a; then raise a to the third power to get volume. No approximations are needed here because values are exact. Step-by-Step Solution: a^2 = 6 / 6 = 1a = √1 = 1V = a^3 = 1^3 = 1 cubic unit Verification / Alternative check: From a = 1, S = 6 * 1^2 = 6 confirms the premise; V = 1 is therefore correct. Why Other Options Are Wrong: 2, 4, and 6 cubic units correspond to edges larger than 1 and would imply surface areas greater than 6. Common Pitfalls: Confusing S = 6a with S = 6a^2; skipping the square root step; or mixing up units. Final Answer: 1 cu unit

Question 3

A cube has volume 729 cubic centimetres. Find the length of its space diagonal (in centimetres).

Accepted Answer

Introduction / Context: The space diagonal of a cube connects opposite vertices and equals a√3, where a is the edge length. If the volume is known, we compute a from V = a^3 and then form the space diagonal directly. This problem tests chaining two standard cube identities correctly. Given Data / Assumptions: Volume V = 729 cm^3. Cube identities: V = a^3 and space diagonal d = a√3. Concept / Approach: Find a = ∛V. Then substitute a into d = a√3. Keep radicals exact; the number is crafted so the cube root is integral. Step-by-Step Solution: a = ∛729 = 9 cmd = a√3 = 9√3 cm Verification / Alternative check: Confirm volume from a = 9: 9^3 = 729 cm^3; thus the diagonal 9√3 cm is consistent. Why Other Options Are Wrong: 9√2 cm is the face diagonal for a = 9 (a√2), not the space diagonal; 18 cm is twice the edge, not applicable; 18√3 omits the division by 2 that would appear in some other contexts, not here. Common Pitfalls: Using face diagonal a√2 instead of space diagonal a√3; miscomputing the cube root of 729. Final Answer: 9√3 cm

Question 4

Find the total surface area of a cuboid with length 10 m, breadth 5 m, and height 3 m.

Accepted Answer

Introduction / Context: The total surface area of a cuboid equals the sum of the areas of its six rectangular faces, compactly written as 2(lb + bh + hl). Substituting the three linear dimensions yields the answer in square units. This reinforces inserting correct pairs into the formula and avoiding arithmetic slips. Given Data / Assumptions: l = 10 m, b = 5 m, h = 3 m. Surface area S = 2(lb + bh + hl). Concept / Approach: Compute each face pair area (lb, bh, hl), add them, and multiply by 2. Check units are in metres so the result is in square metres. Step-by-Step Solution: lb = 10 * 5 = 50bh = 5 * 3 = 15hl = 3 * 10 = 30Sum = 50 + 15 + 30 = 95S = 2 * 95 = 190 sq m Verification / Alternative check: Compute all six faces explicitly: two of each pair give 2*50 + 2*15 + 2*30 = 100 + 30 + 60 = 190, same result. Why Other Options Are Wrong: 170 and 104–105 sq m arise from omitting a pair or misadding a term; only 190 satisfies the full formula. Common Pitfalls: Forgetting the factor 2; mixing dimensions; or computing perimeter-like sums instead of areas. Final Answer: 190 sq m

Question 5

A cube has total surface area 726 sq cm. Find the volume of the cube (in cubic centimetres).

Accepted Answer

Introduction / Context: Surface area S and volume V of a cube depend on the edge a. From S = 6a^2 we can determine a exactly, and then compute V = a^3. This question tests careful algebra and unit handling from area to volume. Given Data / Assumptions: S = 726 cm^2. S = 6a^2; V = a^3. Concept / Approach: Isolate a^2 = S/6, take its square root to get a, and then cube a. Numbers are chosen to yield an integer edge length. Step-by-Step Solution: a^2 = 726 / 6 = 121a = √121 = 11 cmV = a^3 = 11^3 = 1331 cm^3 Verification / Alternative check: Recompute surface area from a = 11: S = 6 * 121 = 726 cm^2, confirming consistency. Why Other Options Are Wrong: Other values are not perfect cubes corresponding to a whole-number edge derived from 726/6 = 121. Common Pitfalls: Forgetting the square root; computing 726/6 incorrectly; or mixing units. Final Answer: 1331 cm3

Question 6

A cuboid has volume 1989 cm^3 with length 17 cm and breadth 13 cm. Find the height of the cuboid (in centimetres).

Accepted Answer

Introduction / Context: Volume of a cuboid equals the product of its three mutually perpendicular edges. Given two edges and the volume, the remaining edge is found by simple division. This reinforces dimensional reasoning and correct unit arithmetic. Given Data / Assumptions: V = 1989 cm^3. l = 17 cm, b = 13 cm. V = l * b * h. Concept / Approach: Rearrange h = V / (l * b). Compute the product of the known sides first, then divide the volume to get an integer height. Step-by-Step Solution: l * b = 17 * 13 = 221h = 1989 / 221 = 9 cm Verification / Alternative check: Check: 17 * 13 * 9 = 221 * 9 = 1989 cm^3, which matches the given volume exactly. Why Other Options Are Wrong: 4, 7, and 14 cm would produce volumes 884, 1547, and 3094 cm^3 respectively with l = 17 and b = 13, none equal to 1989. Common Pitfalls: Arithmetic slip on 17 * 13; dividing by only one dimension; or mismatching units. Final Answer: 9 cm

Question 7

Three solid cubes with edges 1 cm, 6 cm, and 8 cm are melted and recast into a single cube. Find half of the total surface area of the new cube.

Accepted Answer

Introduction / Context: When solids are melted and recast, volume is conserved (neglecting loss). The new cube’s edge is obtained from the total initial volume. With the new edge, compute the total surface area and then take half, as asked. This tests linking conservation to cube formulas. Given Data / Assumptions: Original edges: 1 cm, 6 cm, 8 cm. Volumes: 1^3 = 1, 6^3 = 216, 8^3 = 512 cm^3. Total volume V_total = 1 + 216 + 512 = 729 cm^3. New cube edge a satisfies a^3 = 729 ⇒ a = 9 cm. Concept / Approach: Compute TSA of the new cube: S = 6a^2. Then take half as requested. Keep arithmetic exact; numbers are chosen to be perfect cubes/squares. Step-by-Step Solution: a = ∛729 = 9 cmS = 6a^2 = 6 * 9^2 = 6 * 81 = 486 sq cmHalf of S = 486 / 2 = 243 sq cm Verification / Alternative check: Re-cubing 9 gives back 729, matching conserved volume; halving 486 is straightforward. Why Other Options Are Wrong: 486 sq cm is the full surface area, not half; 463 and 293 are distractors without geometric basis here. Common Pitfalls: Adding edges instead of volumes; forgetting to square the edge when computing surface area; or halving the edge rather than halving the surface area. Final Answer: 243 sq cm

Question 8

A rectangular box has internal dimensions 8 cm × 6 cm × 2 cm. Find the maximum length of a pencil that can fit inside (the space diagonal).

Accepted Answer

Introduction / Context: The longest straight object that fits inside a rectangular box lies along its space diagonal. This diagonal is found via the 3D Pythagoras relation using the three mutually perpendicular edges. The result often simplifies to a scaled square root. Given Data / Assumptions: a = 8 cm, b = 6 cm, c = 2 cm. Space diagonal d = √(a^2 + b^2 + c^2). Concept / Approach: Square each dimension, sum them, and take the square root. Factor the radicand if possible to a nice multiple. Step-by-Step Solution: d^2 = 8^2 + 6^2 + 2^2 = 64 + 36 + 4 = 104d = √104 = √(4 * 26) = 2√26 cm Verification / Alternative check: A face diagonal would be √(8^2 + 6^2) = 10 cm; including the third dimension 2 cm increases it appropriately to √104 ≈ 10.198, consistent. Why Other Options Are Wrong: 2√13 and 2√14 correspond to smaller sums; 10√2 ≈ 14.14 cm exceeds what can fit given the edges. Common Pitfalls: Using a face diagonal instead of space diagonal; arithmetic errors in squaring/adding; forgetting to simplify the radical. Final Answer: 2√26 cm

Question 9

A cube has volume 6240 cubic centimetres. How many rectangular boxes of internal dimensions 10 cm × 8 cm × 6 cm can fit in it by volume count (ignoring packing gaps)?

Accepted Answer

Introduction / Context: Sometimes questions approximate capacity by comparing volumes, ignoring geometric packing constraints. Here, we are effectively asked how many smaller boxes of given internal volume would “fill” a larger cubic capacity of 6240 cm^3. The integer quotient provides the count when a perfect pack is not required in the statement. Given Data / Assumptions: Large capacity (cube) V_big = 6240 cm^3. Each small box internal volume V_small = 10 * 8 * 6 = 480 cm^3. Interpretation: count by volume, i.e., floor(V_big / V_small). Concept / Approach: Divide total capacity by unit capacity: 6240 / 480. As the options are small integers, the result will be exact here. Step-by-Step Solution: V_small = 480 cm^3N = 6240 / 480 = 13 Verification / Alternative check: 13 * 480 = 6240 cm^3 exactly; thus the capacity-based count is an integer. Why Other Options Are Wrong: 12 gives only 5760 cm^3, leaving unused capacity; 15 and 17 exceed the given capacity. Common Pitfalls: Confusing a capacity (volume) question with a strict geometric packing/tiling question requiring divisibility along each dimension. The wording here supports capacity-based counting. Final Answer: 13

Question 10

A metal rectangular box has external dimensions 20 cm × 12 cm × 5 cm and uniform thickness 1 cm (closed box). Find the volume of metal used.

Accepted Answer

Introduction / Context: Hollow box problems ask for the volume of material by subtracting inner (hollow) volume from the outer solid volume. With uniform thickness t around all faces, inner dimensions reduce by 2t along each direction for a closed box. This problem builds skill in modeling shells and subtracting volumes. Given Data / Assumptions: External: 20 × 12 × 5 cm. Thickness t = 1 cm; closed rectangular box (has a lid). Inner dimensions: (20 − 2) × (12 − 2) × (5 − 2) = 18 × 10 × 3 cm. Concept / Approach: Volume of metal = V_outer − V_inner. Compute each cuboid volume and subtract. The numbers are exact integers. Step-by-Step Solution: V_outer = 20 * 12 * 5 = 1200 cm^3V_inner = 18 * 10 * 3 = 540 cm^3Volume of metal = 1200 − 540 = 660 cm^3 Verification / Alternative check: If the box were open at one end, inner reduction would differ for one face; here “box” with no contrary note implies closed, matching the subtraction used. Why Other Options Are Wrong: 550, 656, and 475 cm^3 do not equal the precise difference; they might come from reducing only some dimensions or arithmetic slips. Common Pitfalls: Forgetting to reduce by 2t in each dimension; treating it as an open box unintentionally; unit confusion. Final Answer: 660 cm3

Question 11

How many cubes of edge 3 cm can be cut from a larger cube of edge 18 cm (no wastage)?

Accepted Answer

Introduction / Context: When cutting equal cubes from a larger cube without waste, the count along each edge equals the integer ratio of edge lengths. The total number is the cube of that ratio because packing occurs in three independent directions. This tests spatial counting with simple integer ratios. Given Data / Assumptions: Large cube edge A = 18 cm. Small cube edge a = 3 cm. Perfect divisibility without kerf loss or waste. Concept / Approach: Number per edge = A / a = 18 / 3 = 6. Total cubes = 6^3 because stacking occurs in 3D (length, breadth, height). Step-by-Step Solution: Count per edge = 18 / 3 = 6Total = 6 * 6 * 6 = 216 Verification / Alternative check: Volume ratio: (18^3) / (3^3) = 5832 / 27 = 216, consistent with discrete packing since edges divide exactly. Why Other Options Are Wrong: 63 and 432 are common miscomputations (e.g., squaring instead of cubing or doubling an intermediate count). “None” is unnecessary because 216 is exact. Common Pitfalls: Multiplying only two dimensions (6^2 = 36); forgetting 3D packing; or misreading edge units. Final Answer: 216

Question 12

A cube has edge 2 cm and a cuboid has dimensions 1 cm × 2 cm × 3 cm. Paint available covers 54 cm^2. Which can be fully painted?

Accepted Answer

Introduction / Context: This is a budgeted-area problem: determine whether available paint is sufficient to cover the total surface areas. Compute each object’s required area and compare with the available coverage. The twist is checking if both together still fit the paint budget. Given Data / Assumptions: Cube edge a = 2 cm ⇒ S_cube = 6a^2 = 6 * 4 = 24 cm^2. Cuboid 1 × 2 × 3 ⇒ S_cuboid = 2(ab + bc + ca) = 2(2 + 6 + 3) = 22 cm^2. Available area = 54 cm^2. Concept / Approach: Sum the two required areas and compare with 54. If the sum ≤ 54, both can be painted; otherwise determine which individual one(s) fit. Step-by-Step Solution: S_cube = 24 cm^2S_cuboid = 22 cm^2Total needed = 24 + 22 = 46 cm^2 ≤ 54 cm^2Hence both can be painted fully. Verification / Alternative check: Individually each requires less than 54; together they still use only 46, leaving an 8 cm^2 margin. Why Other Options Are Wrong: “Only cube” or “Only cuboid” underestimate the available paint; “Neither” is false because 24 and 22 are both below 54. Common Pitfalls: Forgetting to double for the cuboid’s opposite faces; arithmetic slips in 2(2 + 6 + 3). Final Answer: Both cube and cuboid can be painted

Question 13

The whole surface area of a rectangular block is 8788 sq cm and its dimensions are in the ratio 4 : 3 : 2 (length : breadth : height). Find the length (in cm).

Accepted Answer

Introduction / Context: When only the surface area and the ratio of edges are given, introduce a common factor x to express the three dimensions and substitute into the surface-area formula. Solve for x, then recover each dimension as a multiple of x. This is a common technique for ratio-based solids questions. Given Data / Assumptions: Ratios: l : b : h = 4 : 3 : 2 ⇒ l = 4x, b = 3x, h = 2x. Total surface area S = 8788 cm^2. Formula: S = 2(lb + bh + hl). Concept / Approach: Compute lb, bh, hl in terms of x, sum, multiply by 2, set equal to 8788, and solve x. Then find l = 4x. Step-by-Step Solution: lb = (4x)(3x) = 12x^2bh = (3x)(2x) = 6x^2hl = (2x)(4x) = 8x^2S = 2(12 + 6 + 8)x^2 = 2 * 26x^2 = 52x^252x^2 = 8788 ⇒ x^2 = 169 ⇒ x = 13Length l = 4x = 4 * 13 = 52 cm Verification / Alternative check: Compute b = 39 cm and h = 26 cm from x = 13; recomputing S gives 8788 cm^2, confirming consistency. Why Other Options Are Wrong: 26 and 13 are smaller multiples (h and x); 104 cm doubles l mistakenly. Common Pitfalls: Dropping the factor 2 in S; mishandling squared ratios; misadding 12 + 6 + 8. Final Answer: 52 cm

Question 14

A cuboid has areas of three mutually adjacent faces equal to 120 cm^2, 72 cm^2, and 60 cm^2. Find the volume of the cuboid.

Accepted Answer

Introduction / Context: If ab, bc, and ca (areas of three adjacent faces) are known for a cuboid with edges a, b, c, then the volume abc can be found without individually solving for a, b, c using the identity abc = √[(ab)(bc)(ca)]. This is a standard shortcut in 3D mensuration. Given Data / Assumptions: ab = 120 cm^2, bc = 72 cm^2, ca = 60 cm^2. Volume V = abc. Concept / Approach: Multiply the three face areas and take the square root: (ab)(bc)(ca) = (abc)^2. Then V = √(product). This avoids solving three equations for the three unknowns. Step-by-Step Solution: (ab)(bc)(ca) = 120 * 72 * 60120 * 60 = 7200; 7200 * 72 = 518400V = √518400 = 720 cm^3 Verification / Alternative check: Optional factorization: 518400 = (720)^2 confirms the square root cleanly. Why Other Options Are Wrong: 820, 750 are arbitrary nearby integers; only 720 is the exact square root of the computed product. Common Pitfalls: Forgetting that the product equals (abc)^2; arithmetic errors when multiplying three numbers; unit confusion. Final Answer: 720 cm3

Question 15

A cuboid water tank has capacity 50,000 litres (i.e., 50 m^3). If its length is 2.5 m and depth is 10 m, find the breadth (in metres).

Accepted Answer

Introduction / Context: Unit conversion between litres and cubic metres is essential: 1000 L = 1 m^3. With the tank’s capacity in litres, convert to m^3 and then use V = l * b * d for a cuboid to find the missing breadth. This emphasizes dimensional consistency in fluid-capacity problems. Given Data / Assumptions: Capacity = 50,000 L = 50 m^3. l = 2.5 m, depth (height) d = 10 m. V = l * b * d. Concept / Approach: Rearrange b = V / (l * d), substitute the numbers, and compute exactly. Step-by-Step Solution: b = 50 / (2.5 * 10) = 50 / 25 = 2 m Verification / Alternative check: Check: 2.5 * 2 * 10 = 50 m^3 ⇒ 50,000 L, which matches the given capacity. Why Other Options Are Wrong: 4, 6, or 9 m would yield 100, 150, or 225 m^3 capacities with the same l and d, contradicting the 50 m^3 requirement. Common Pitfalls: Forgetting 1000 L = 1 m^3; placing 10 in the numerator instead of the denominator; arithmetic slips with decimals. Final Answer: 2 m

Question 16

A cuboid has volume 720 cm^3, total surface area 484 cm^2, and base area 72 cm^2. Find its dimensions (length, breadth, height) in cm.

Accepted Answer

Introduction / Context: Given base area lb, volume V, and total surface area S, a cuboid’s three dimensions can be recovered systematically. From V = lb * h we get height; then S = 2(lb + lh + bh) yields l + b. With l + b and lb known, l and b are roots of a quadratic. This integrates multiple cuboid identities coherently. Given Data / Assumptions: lb = 72 cm^2. V = 720 cm^3 ⇒ h = V / lb = 10 cm. S = 2(lb + lh + bh) = 484 cm^2. Concept / Approach: Use h = 10 in S to solve for l + b. Then solve t^2 − (l + b)t + lb = 0 for t, whose roots are l and b. Pick the positive pair producing the given base area. Step-by-Step Solution: 484 = 2(72 + 10l + 10b)Divide by 2: 242 = 72 + 10(l + b) ⇒ 10(l + b) = 170 ⇒ l + b = 17Solve t^2 − 17t + 72 = 0 ⇒ (t − 9)(t − 8) = 0Thus (l, b) = (9, 8) in some order; with h = 10 Verification / Alternative check: Check: lb = 72, V = 72 * 10 = 720, S = 2(72 + 90 + 80) = 2 * 242 = 484, all consistent. Why Other Options Are Wrong: Other triples give incorrect base area or surface computations; only 9, 8, 10 fits all three conditions. Common Pitfalls: Forgetting to halve S; forming the wrong quadratic; mixing the roles of base and height. Final Answer: 9, 8 and 10 cm

Question 17

For a cube, suppose the numerical value of its volume equals the numerical sum of all its edges. Find the total surface area (square units).

Accepted Answer

Introduction / Context: This functional condition equates a^3 (volume) to 12a (sum of edges) for a cube with edge a. Solving the resulting equation determines a, then the total surface area 6a^2 is immediate. It practices translating a verbal condition into algebra and following through with correct power handling. Given Data / Assumptions: Volume V = a^3. Sum of edges = 12a. Equation: a^3 = 12a with a > 0 (physical edge). Concept / Approach: Divide both sides by a (a ≠ 0) to get a^2 = 12. Then compute S = 6a^2 directly from this value without first extracting a, which avoids radicals in the surface area. Step-by-Step Solution: a^3 = 12a ⇒ a^2 = 12S = 6a^2 = 6 * 12 = 72 (square units) Verification / Alternative check: If desired, a = √12 = 2√3; then V = (2√3)^3 = 8 * 3√3 = 24√3 and 12a = 24√3 match numerically, consistent with the condition. Why Other Options Are Wrong: 12 and 36 are too small (correspond to using a or a^2 directly); 144 doubles the correct surface area. Common Pitfalls: Cancelling by a when a = 0 (edge cannot be zero); computing surface area as 6a instead of 6a^2. Final Answer: 72

Question 18

Each edge of a cube is reduced by 19%. Find the percentage decrease in its total surface area.

Accepted Answer

Introduction / Context: Area scales with the square of linear dimensions. If each edge changes by a factor k, total surface area changes by k^2. Here the linear decrease is 19%, so k = 1 − 0.19 = 0.81. We then compute the area factor and translate to a percentage decrease. Given Data / Assumptions: Edge factor k = 0.81. Surface area factor = k^2. Concept / Approach: Compute k^2 and then determine the decrease: % decrease = (1 − k^2) * 100%. Exact arithmetic avoids rounding issues because 0.81^2 is a standard decimal square. Step-by-Step Solution: k^2 = (0.81)^2 = 0.6561Decrease = 1 − 0.6561 = 0.3439 = 34.39% Verification / Alternative check: Using percentage compounding: Area decreases by 19% twice (not additive), giving 34.39% overall, consistent with squaring. Why Other Options Are Wrong: 40% and 35% are rough guesses; 38.4% treats it like 2*19% − (19%)^2/100 but applied to the wrong quantity; only 34.39% correctly squares the linear factor. Common Pitfalls: Subtracting 2 * 19% directly (linear thinking) instead of squaring the scale factor; rounding prematurely. Final Answer: 34.39%

Question 19

Each edge of a cube is increased by 12%. By what percentage does its volume increase?

Accepted Answer

Introduction / Context: Volume scales with the cube of linear dimensions. If each edge increases by 12%, the linear factor is k = 1.12 and the volume factor is k^3. Converting this factor to a percentage increase gives the final answer, illustrating compound scaling in three dimensions. Given Data / Assumptions: Edge factor k = 1.12. Volume factor = k^3. Concept / Approach: Compute k^3 exactly (or to the precision given in options), then subtract 1 and express as a percentage. This reflects multiplicative growth in all three independent directions. Step-by-Step Solution: k^2 = 1.12^2 = 1.2544k^3 = 1.2544 * 1.12 = 1.404928Percentage increase = (1.404928 − 1) * 100% = 40.4928% Verification / Alternative check: Small-increase approximation 3 * 12% = 36% underestimates because it ignores compounding; the exact is 40.4928%, as computed. Why Other Options Are Wrong: 50.5240% and 60.3292% are too large; 30.4928% is too small and confuses squared with cubed growth. Common Pitfalls: Tripling 12% without compounding; rounding mid-calculation; mixing area and volume scaling. Final Answer: 40.4928%

Question 20

Right circular cylinder – compute volume: Find the volume V of a right circular cylinder whose height (length) is 80 cm and whose base has diameter 14 cm (so radius = 7 cm). Give the answer in cubic centimeters.

Accepted Answer

Introduction / Context: This problem asks for the volume of a right circular cylinder from its basic dimensions. It reinforces correct identification of radius from diameter and direct substitution into the standard formula for cylindrical volume. Given Data / Assumptions: Height h = 80 cm Diameter d = 14 cm → radius r = d/2 = 7 cm Use V = π * r^2 * h Use π ≈ 3.14 (or 22/7) for a numerical answer in cm^3 Concept / Approach: The volume of a right circular cylinder is proportional to both the cross-sectional area of the base (πr^2) and the height (h). After converting diameter to radius, compute r^2 and multiply by h and π. Step-by-Step Solution: r = 14 / 2 = 7 cmr^2 = 49 cm^2V = π * 49 * 80 = 3920π cm^3Using π ≈ 3.14 → V ≈ 3920 * 3.14 ≈ 12308.8 cm^3 (rounds to about 12309–12320)Using π = 22/7 → V = 3920 * (22/7) = 560 * 22 = 12320 cm^3 Verification / Alternative check: Reverse check: 12320 / 80 = 154 cm^2, which equals π * 7^2 ≈ 153.94 cm^2 using π = 3.14. Consistency verified within rounding. Why Other Options Are Wrong: 1400 and 1553 cm^3 are too small by an order of magnitude; 13320 cm^3 overshoots the exact 22/7 result; 12000 cm^3 is an imprecise round number not matching π-based computation. Common Pitfalls: Forgetting to halve the diameter to get radius; mixing units; dropping π inadvertently; arithmetic slips squaring 7 or multiplying by 80. Final Answer: 12320 cm^3

Volume and Surface Area Questions