Question 1

Surface area change when cube edge increases by 50%: If each edge of a cube is increased by 50%, by what percentage does the surface area increase?

Accepted Answer

Introduction / Context:Surface area scales with the square of the linear factor. Increasing each edge by 50% means the linear scale factor is 1.5. Square it to get the area factor, then convert to percentage increase. Given Data / Assumptions: Scale factor k = 1.5 Area factor = k^2 = 2.25 Concept / Approach:Percentage increase = (area factor − 1) * 100% = (2.25 − 1) * 100% = 125%. Step-by-Step Solution:k = 1 + 0.5 = 1.5k^2 = 2.25Increase = 2.25 − 1 = 1.25 → 125% Verification / Alternative check:Let original side be s; new side = 1.5s. Original area = 6s^2; new area = 6*(1.5s)^2 = 6*2.25s^2 = 13.5s^2, which is 125% more than 6s^2. Why Other Options Are Wrong:50% and 75% reflect linear or partial scaling; 100% would be doubling; 150% is an overestimate (area factor 2.5, not applicable here). Common Pitfalls:Applying 50% directly to area; forgetting to square the linear factor for area changes. Final Answer:125 %

Question 2

Two cubes: find product of edges from given differences: The difference of volumes is 152 m^3 and the difference of one-face areas is 20 m^2. If the sum of their edges (side lengths) is 10 m, find the product of their edges.

Accepted Answer

Introduction / Context:This problem links three algebraic relations for cube side lengths a and b (a > b): volume difference, face-area difference, and sum of edges. It tests recognition of factorization identities and system solving. Given Data / Assumptions: a^3 − b^3 = 152 a^2 − b^2 = 20 (difference of one-face areas) a + b = 10 Find a * b Concept / Approach:Use identities: a^3 − b^3 = (a − b)(a^2 + ab + b^2); and a^2 − b^2 = (a − b)(a + b). From a^2 − b^2 and a + b, get a − b, then solve a and b. Finally compute ab. Step-by-Step Solution:From a^2 − b^2 = 20 and a + b = 10 → (a − b) = 20 / 10 = 2Solve a and b: a = ( (a + b) + (a − b) ) / 2 = (10 + 2)/2 = 6; b = (10 − 2)/2 = 4Check volume difference: 6^3 − 4^3 = 216 − 64 = 152 (OK)Product ab = 6 * 4 = 24 Verification / Alternative check:Use a^2 + ab + b^2 with found a, b: 36 + 24 + 16 = 76; then (a − b)(a^2 + ab + b^2) = 2 * 76 = 152 equals the given difference, confirming a and b are correct. Why Other Options Are Wrong:21, 36, 48, 40 do not satisfy the three simultaneous conditions; only 24 is consistent with all constraints. Common Pitfalls:Misreading “sum of their edge” as sum of all 12 edges per cube; forgetting that “one-face area” is side^2; arithmetic slips solving a and b. Final Answer:24

Question 3

Cuboid from sum of edges and space diagonal — find surface area: The sum of the length, breadth, and depth (height) of a cuboid is 19 cm, and its space diagonal is 5√5 cm. Find the total surface area of the cuboid.

Accepted Answer

Introduction / Context:This problem asks for the surface area of a cuboid when the sum of its three edges (l + b + h) and its space diagonal are known. It relies on two identities that connect edges, diagonal, and pairwise products. Given Data / Assumptions: Let l, b, h be the edges in centimeters. l + b + h = 19 cm. Space diagonal d = 5√5 cm ⇒ d^2 = 125. Total surface area S = 2(lb + bh + hl). Concept / Approach:We use the identities: (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl), and d^2 = l^2 + b^2 + h^2. Eliminating l^2 + b^2 + h^2 allows us to solve directly for (lb + bh + hl) and hence S. Step-by-Step Solution:(l + b + h)^2 = 19^2 = 361d^2 = l^2 + b^2 + h^2 = 125361 = 125 + 2(lb + bh + hl) ⇒ lb + bh + hl = (361 − 125)/2 = 118Surface area S = 2(lb + bh + hl) = 2 * 118 = 236 cm2 Verification / Alternative check:No dimensions are individually needed; identities suffice and are standard for cuboids. Why Other Options Are Wrong:361 cm2 confuses (l + b + h)^2 with area; 125 cm2 uses d^2 incorrectly; 486 cm2 doubles a wrong intermediate. Common Pitfalls:Forgetting to multiply the pairwise sum by 2 at the end; mixing up d^2 with (l + b + h)^2. Final Answer:236 cm2

Question 4

Earthwork spread — rise of ground level: A rectangular tank 30 m long, 20 m wide, and 12 m deep is excavated in a field 500 m long and 30 m wide. If the excavated earth is spread evenly over the entire field, by how much (in meters) will the field l…

Accepted Answer

Introduction / Context:When excavated soil is uniformly spread over a larger area, the rise in ground level equals volume of excavation divided by the spread area. This is a direct volume-conservation application. Given Data / Assumptions: Tank: 30 m × 20 m × 12 m. Field: 500 m × 30 m. All excavated earth is spread across the whole field without loss. Concept / Approach:Let V be the excavated volume and A be the field area. Then rise h = V / A. Step-by-Step Solution:Excavation volume V = 30 * 20 * 12 = 7200 m3Field area A = 500 * 30 = 15000 m2Rise h = V / A = 7200 / 15000 = 0.48 m Verification / Alternative check:Unit analysis: m3 / m2 = m, consistent for a height rise. Why Other Options Are Wrong:0.5 m and 0.4 m are rounded or underestimated; 0.33 m assumes a different area or volume. Common Pitfalls:Mixing up field dimensions, omitting unit conversion, or restricting spreading only to some part of the field. Final Answer:0.48 m

Question 5

Right circular cone cut parallel to base — cone vs frustum volume ratio: A right circular cone of total height h is cut by a plane parallel to the base at a distance h/3 from the base. Find the ratio of the volumes of the smaller cone (toward the ve…

Accepted Answer

Introduction / Context:A plane parallel to the base of a cone slices off a top cone similar to the original. Similarity gives linear scale; volumes then scale by the cube of the linear factor. The remainder is a frustum. Given Data / Assumptions: Total cone height = h. Slice is at distance h/3 from the base, so the top small cone has height h − h/3 = 2h/3. Similarity ratio (small to original) = (2h/3)/h = 2/3. Concept / Approach:For similar solids, volume ratio = (linear ratio)^3. Hence V_small / V_full = (2/3)^3 = 8/27. The frustum volume is V_full − V_small = (27/27 − 8/27)V_full = 19/27 V_full. Therefore, V_small : V_frustum = 8 : 19. Step-by-Step Solution:Linear ratio k = 2/3Volume ratio small:full = k^3 = 8/27Frustum = 1 − 8/27 = 19/27Therefore, small:frustum = (8/27):(19/27) = 8:19 Verification / Alternative check:Pick any convenient dimensions satisfying similarity; the ratio remains 8:19. Why Other Options Are Wrong:4/7, 6/19, 18/19 mis-handle cubic scaling or subtract from 1 incorrectly. Common Pitfalls:Using area (square) scaling instead of volume (cube) scaling; misreading where h/3 is measured from. Final Answer:8/19

Question 6

Cylinder with unchanged radius — percent change in volume when height increases: If the height of a right circular cylinder is increased by 17.5% while the radius remains unchanged, by what percent does its volume increase?

Accepted Answer

Introduction / Context:For a cylinder, volume V = πr^2h. If r is constant, volume varies directly with h. Therefore, any percentage change in height translates identically to volume. Given Data / Assumptions: r = constant h increases by 17.5% V ∝ h (with r fixed) Concept / Approach:Because r is unchanged, V_new / V_old = h_new / h_old = 1.175. The percentage increase in V equals 17.5%. Step-by-Step Solution:V_old = πr^2hV_new = πr^2(1.175h) = 1.175V_oldPercent increase = (1.175 − 1)*100% = 17.5% Verification / Alternative check:Pick r = 1, h = 100 (units); volumes scale exactly like heights. Why Other Options Are Wrong:18% and 19.8% are rounded or compounding misconceptions; 25% is unrelated. Common Pitfalls:Trying to apply compound effects without any radius change; mixing area vs. volume changes. Final Answer:17.5%

Question 7

Cuboid from sum of edges and diagonal (rephrased) — compute surface area: The sum of the length, breadth, and height of a cuboid is 19 cm, and its space diagonal is 5√5 cm. What is the total surface area of the cuboid?

Accepted Answer

Introduction / Context:Same setup as a classic identity problem: with l + b + h and the space diagonal known, compute the pairwise products and hence the total surface area. Given Data / Assumptions: l + b + h = 19 cm d = 5√5 cm ⇒ d^2 = 125 Surface area S = 2(lb + bh + hl) Concept / Approach:Use (l + b + h)^2 = d^2 + 2(lb + bh + hl). Rearranging gives the sum of pairwise products directly. Step-by-Step Solution:(l + b + h)^2 − d^2 = 361 − 125 = 2362(lb + bh + hl) = 236 ⇒ S = 236 cm2 Verification / Alternative check:Dimensions need not be individually determined; identity-based computation is sufficient. Why Other Options Are Wrong:361 and 125 are intermediate squares; “None” is unnecessary once identity is applied. Common Pitfalls:Forgetting to take 2 into account or mis-squaring 19. Final Answer:236 cm2

Question 8

Cuboid from sum of edges and diagonal — correct units for surface area: The sum of the length, breadth, and height of a cuboid is 19 cm, and its space diagonal is 5√5 cm. What is its total surface area (in cm2)?

Accepted Answer

Introduction / Context:This is the same identity-based calculation as earlier, but with units clarified: surface area must be in square centimeters, not square meters, as all linear data are in centimeters. Given Data / Assumptions: l + b + h = 19 cm d = 5√5 cm ⇒ d^2 = 125 S = 2(lb + bh + hl) in cm2 Concept / Approach:Compute pairwise products using (l + b + h)^2 − d^2, then double the sum to get surface area. Step-by-Step Solution:(l + b + h)^2 − d^2 = 19^2 − 125 = 361 − 125 = 236Thus S = 236 cm2 Verification / Alternative check:Unit consistency: cm inputs imply cm2 for area, not m2. Why Other Options Are Wrong:361 and 125 are misinterpreted squares; 486 is unrelated to the identity result. Common Pitfalls:Carrying units incorrectly (e.g., writing m2). Final Answer:236 cm2

Question 9

Cube with equal surface area and volume (numerically) — find edge length: If the numerical value of the volume (in unit^3) equals the numerical value of the surface area (in unit^2) of a cube, what is the edge length of the cube in units?

Accepted Answer

Introduction / Context:For a cube of edge a, the volume is a^3 and total surface area is 6a^2. The question states their numerical values are equal, which allows solving directly for a. Given Data / Assumptions: Volume V = a^3 Surface area S = 6a^2 Numerical equality: a^3 = 6a^2 with a > 0 Concept / Approach:Divide both sides by a^2 (a ≠ 0) to isolate a. Step-by-Step Solution:a^3 = 6a^2 ⇒ a = 6 Verification / Alternative check:Check: volume 216 and surface area 216 (units differ but numbers match) satisfies the condition. Why Other Options Are Wrong:No other integer satisfies a^3 = 6a^2 for positive a. Common Pitfalls:Treating units as needing to match; only numerical equality is required here. Final Answer:6

Question 10

Rectangular tank water depth — convert cm2 to m2 properly: A rectangular tank holds 2.6 m3 of water. If the area of its base is 6500 cm2, find the depth of water in meters.

Accepted Answer

Introduction / Context:Depth equals volume divided by base area, but unit consistency is essential: convert the base area from cm2 to m2 before dividing. Given Data / Assumptions: Volume V = 2.6 m3 Base area A = 6500 cm2 = 6500/10000 = 0.65 m2 Concept / Approach:Depth h = V / A with consistent SI units. Step-by-Step Solution:h = 2.6 / 0.65 = 4 m Verification / Alternative check:Units: m3 / m2 = m, which is correct for depth. Why Other Options Are Wrong:3.5, 5, 8 arise from incorrect area conversion or arithmetic. Common Pitfalls:Using 6500 as m2 directly; forgetting that 1 m2 = 10000 cm2. Final Answer:4

Question 11

Room dimensions l, b, h — perimeter of ceiling vs area of walls (percentage): Let a room have length l, breadth b, and height h. Express the perimeter of the ceiling as a percentage of the total area of the four walls.

Accepted Answer

Introduction / Context:This is a ratio/percentage problem connecting linear measure (perimeter of the rectangle l × b) to area measure (four walls area), requiring careful unit handling. Given Data / Assumptions: Ceiling (top face) is l × b. Perimeter of ceiling P_ceiling = 2(l + b). Area of four walls A_walls = perimeter of base × height = 2(l + b) * h. Concept / Approach:Percentage = (P_ceiling / A_walls) * 100. Step-by-Step Solution:P_ceiling / A_walls = [2(l + b)] / [2(l + b)h] = 1/hPercentage = (1/h) * 100 = 100/h % Verification / Alternative check:Dimensional check: perimeter over area yields 1/length; multiplying by 100 gives percent per unit height, consistent with 100/h %. Why Other Options Are Wrong:100h% inverts the relationship; h% and h/100% misuse units. Common Pitfalls:Forgetting that four wall area uses the base perimeter times h, not 2lh + 2bh directly (though algebraically equivalent). Final Answer:100/h%

Question 12

Brickwork volume — number of bricks for a rectangular sump wall: A rectangular sump of inner dimensions 6 m × 5 m × 4 m is to be built so that the outer dimensions are 6.2 m × 5.2 m × 4.2 m. Approximately how many bricks of size 20 cm × 10 cm × 5 cm…

Accepted Answer

Introduction / Context:The number of bricks equals (volume of brickwork) / (volume of one brick). The brickwork volume is the difference between outer and inner rectangular prism volumes. Given Data / Assumptions: Outer: 6.2 m × 5.2 m × 4.2 m Inner: 6 m × 5 m × 4 m Brick: 20 cm × 10 cm × 5 cm = 0.2 m × 0.1 m × 0.05 m Ignore mortar, openings, and wastage for this idealized count. Concept / Approach:Compute ΔV = V_outer − V_inner, then divide by V_brick. Step-by-Step Solution:V_outer = 6.2*5.2*4.2 = 135.408 m3V_inner = 6*5*4 = 120 m3ΔV = 15.408 m3V_brick = 0.2*0.1*0.05 = 0.001 m3Count = 15.408 / 0.001 = 15408 Verification / Alternative check:Wall thickness implied: 0.1 m on each face; the computed ΔV matches that intuition. Why Other Options Are Wrong:Values differ by small offsets that do not arise from exact volume arithmetic. Common Pitfalls:Forgetting to convert brick dimensions from centimeters to meters; subtracting areas instead of volumes. Final Answer:15408

Question 13

Equal surface areas — sphere vs cube, find volume ratio: A sphere and a cube have the same surface area. What is the ratio of the volume of the sphere to the volume of the cube?

Accepted Answer

Introduction / Context:This is a comparison of two solids under an equal-surface-area constraint. Express both volumes in terms of a single parameter using the surface-area equality, then take the ratio. Given Data / Assumptions: Sphere: surface area S_s = 4πr^2, volume V_s = (4/3)πr^3 Cube: surface area S_c = 6a^2, volume V_c = a^3 Set 4πr^2 = 6a^2 ⇒ r^2/a^2 = 3/(2π) Concept / Approach:Write V_s/V_c = [(4/3)π r^3]/a^3 = (4/3)π (r^2/a^2)^{3/2} and substitute r^2/a^2 from the surface-area equality. Step-by-Step Solution:V_s/V_c = (4/3)π * (3/(2π))^{3/2}= (4/3)π * [3√3/(2√2 * π√π)]= √6/√π Verification / Alternative check:Numerically, √6/√π ≈ 2.449/1.772 ≈ 1.382, a reasonable constant ratio. Why Other Options Are Wrong:1:2 and 3:1 are arbitrary; 6:π confuses square roots and direct factors. Common Pitfalls:Forgetting the 3/2 power when converting from area to volume ratios. Final Answer:√6 : √π

Question 14

Earth spread over remainder of field — compute level rise (corrected spread area): A rectangular pit (tank) measuring 5 m × 4.5 m × 2.1 m is dug at the center of a rectangular field measuring 13.5 m × 25 m. The dug earth is spread evenly over the re…

Accepted Answer

Introduction / Context:Here the earth from the pit is not spread over the entire field but only over the remaining part after excluding the pit footprint. This slightly increases the rise compared to spreading over the whole field. Given Data / Assumptions: Pit volume V = 5 * 4.5 * 2.1 m3 Field area A_field = 13.5 * 25 m2 Pit base area A_pit = 5 * 4.5 m2 Spread area A_spread = A_field − A_pit Concept / Approach:Level rise h = V / A_spread. Step-by-Step Solution:V = 47.25 m3A_field = 337.5 m2; A_pit = 22.5 m2A_spread = 337.5 − 22.5 = 315 m2h = 47.25 / 315 = 0.15 m Verification / Alternative check:If instead spread over the whole field, rise would be 47.25/337.5 = 0.14 m (smaller), confirming exclusion of the pit increases h slightly. Why Other Options Are Wrong:0.12 m and 0.18–0.20 m correspond to different spread areas or rounding beyond the given data. Common Pitfalls:Forgetting to exclude the pit base area from the spread area. Final Answer:0.15 m

Question 15

Cone scaled in radius and height — new volume ratio: The base radius and height of a cone are both doubled. What is the ratio of the new volume to the original volume?

Accepted Answer

Introduction / Context:Volume of a cone is V = (1/3)πr^2h. Scaling r and h scales volume by r^2h, i.e., by the square of radius scale times the height scale. Given Data / Assumptions:r → 2r and h → 2h. Concept / Approach:New/old volume factor = (2^2) * (2) = 8. Step-by-Step Solution:V_new / V_old = (r_new^2 h_new)/(r^2 h) = (4r^2 * 2h)/(r^2 h) = 8 Verification / Alternative check:Pick numbers (e.g., r=1, h=1) to see old V=(1/3)π, new V=(8/3)π → ratio 8:1. Why Other Options Are Wrong:Other ratios ignore the quadratic dependence on radius and the linear dependence on height. Common Pitfalls:Adding scale factors instead of multiplying; forgetting radius is squared in volume. Final Answer:8 : 1

Question 16

Percent decrease in sphere surface area when radius shrinks: If the radius of a sphere is decreased by 24%, by what percent does its surface area decrease?

Accepted Answer

Introduction / Context:Sphere surface area S = 4πr^2, so S scales with r^2. A percentage change in r results in a squared factor for S. Given Data / Assumptions:r_new = 0.76 r (since 24% decrease). Concept / Approach:Compute S_new/S_old = (0.76)^2 and convert to percent decrease. Step-by-Step Solution:S_new/S_old = 0.76^2 = 0.5776Percent decrease = (1 − 0.5776)*100% = 42.24% Verification / Alternative check:Small changes approximate as 2Δr, but exact method is squaring the factor. Why Other Options Are Wrong:49% and 44% are rough guesses; 46.2% corresponds to compounding a different percentage. Common Pitfalls:Applying linear decrease to area; forgetting the square dependence. Final Answer:42.24%

Question 17

Cube side reduced — percent decrease in surface area: If each side of a cube is decreased by 19%, find the percentage decrease in its total surface area.

Accepted Answer

Introduction / Context:Total surface area of a cube is proportional to a^2 (since TSA = 6a^2). Reducing the side by p% reduces area by the square of the side factor. Given Data / Assumptions:a_new = 0.81 a (19% decrease). Concept / Approach:Area factor = (0.81)^2; percent decrease = (1 − factor) * 100%. Step-by-Step Solution:(0.81)^2 = 0.6561Decrease = (1 − 0.6561)*100% = 34.39% Verification / Alternative check:Matches the rule: for small p, area decreases by approximately 2p (here ~38%), but exact is 34.39% because 19% is not very small. Why Other Options Are Wrong:40.4% and 38.4% overestimate by linear approximation; 35.6% is still off. Common Pitfalls:Using 2 * 19% exactly instead of squaring the factor. Final Answer:34.39%

Question 18

Cone volume change — height tripled, radius halved: If the height of a right circular cone is increased by 200% (i.e., tripled) and the base radius is reduced by 50%, what is the net change in the cone’s volume?

Accepted Answer

Introduction / Context:Volume V of a cone is (1/3)πr^2h. Scale factors multiply: radius affects volume quadratically; height affects it linearly. Given Data / Assumptions:h_new = 3h (200% increase), r_new = 0.5r (50% decrease). Concept / Approach:Compute factor f = (r_new^2 h_new)/(r^2 h) = (0.5^2)*3 = 0.25*3 = 0.75. Step-by-Step Solution:V_new = 0.75 V_oldChange = −25% (a decrease of one quarter) Verification / Alternative check:Pick r=2, h=2 ⇒ V_old = (1/3)π*4*2 = (8/3)π; new r=1, h=6 ⇒ V_new = (1/3)π*1*6 = 2π; 2π/(8/3)π = 0.75. Why Other Options Are Wrong:They do not align with the combined quadratic and linear scaling. Common Pitfalls:Adding the percentages instead of multiplying the factors. Final Answer:decreases by 25%

Question 19

Displacement in a cylinder by a sphere — find sphere radius: A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. When a solid sphere is dropped in and completely submerged, the water level rises by 6.75 cm. Find the radius of the s…

Accepted Answer

Introduction / Context:Submerging a solid displaces an equal volume of water. In a vertical cylinder, the rise in level converts readily to a displaced volume: V_displaced = πR_cyl^2 * Δh. Given Data / Assumptions: Cylinder radius R_cyl = 12 cm Rise Δh = 6.75 cm Sphere volume V_sphere = (4/3)πr^3 Concept / Approach:Set displaced volume equal to the sphere volume and solve for sphere radius r. Step-by-Step Solution:V_displaced = π*(12^2)*6.75 = π*144*6.75 = 972π cm3(4/3)πr^3 = 972π ⇒ r^3 = 972*(3/4) = 729r = 9 cm Verification / Alternative check:9^3 = 729 confirms r = 9 cm exactly. Why Other Options Are Wrong:6 cm and 8 cm under-displace; “None” is unnecessary with exact equality. Common Pitfalls:Forgetting π cancels; mis-multiplying 144 by 6.75. Final Answer:9 cm

Question 20

Sphere immersed in cylindrical vessel — compute water rise: A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is completely immersed. By how much (in cm) does the water level rise?

Accepted Answer

Introduction / Context:Again use volume displacement: the water level rise times the cross-sectional area of the cylinder equals the volume of the immersed sphere. Given Data / Assumptions: Cylinder radius R = 4 cm Sphere radius r = 3 cm Concept / Approach:πR^2 * h_rise = (4/3)πr^3 ⇒ h_rise = (4/3) r^3 / R^2. Step-by-Step Solution:h_rise = [(4/3)*27] / 16 = 36/16 = 9/4 cm Verification / Alternative check:Numerically 9/4 = 2.25 cm; units are consistent. Why Other Options Are Wrong:4/9 cm and 2/3 cm invert or mis-square radii; 4/5 cm is unrelated. Common Pitfalls:Dropping the π factors prematurely without keeping track (they cancel, but after setup). Final Answer:9/4 cm

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