Question 1

Quadrilateral Angles — Ratio to Actual Measure: Angles of a quadrilateral are in the ratio 3 : 4 : 5 : 8. Find the smallest angle (in degrees).

Accepted Answer

Introduction / Context: The interior angles of any quadrilateral sum to 360°. When a ratio is given, convert the ratio to parts, find the value per part, and multiply by each term to get actual angles. This checks comfort with ratio scaling and polygon-angle sums. Given Data / Assumptions: Angle ratio: 3 : 4 : 5 : 8 Sum of interior angles of quadrilateral = 360° Concept / Approach: Total parts = 3 + 4 + 5 + 8 = 20. One part equals 360° / 20 = 18°. The smallest angle corresponds to the smallest ratio term, which is 3 parts. Multiply to obtain the measure. Step-by-Step Solution: Compute one part: 360° / 20 = 18°.Smallest angle = 3 * 18° = 54°. Verification / Alternative check: Other angles: 4*18° = 72°, 5*18° = 90°, 8*18° = 144°; sum = 54 + 72 + 90 + 144 = 360°. Why Other Options Are Wrong: 40°, 36°, 18° do not match 3 parts at 18° each. 72° is the second-smallest (4 parts), not the smallest. Common Pitfalls: Using triangle sum (180°) instead of quadrilateral sum (360°). Final Answer: 54°.

Question 2

Tiling a Square Floor — Count of 50 cm Tiles: A square room of side 10 m is to be fully tiled with square tiles of side 50 cm. What is the smallest number of tiles needed?

Accepted Answer

Introduction / Context: Counting tiles requires dividing the floor area by the area of one tile, keeping units consistent. Here, dimensions mix metres and centimetres, so converting to the same unit avoids mistakes. Given Data / Assumptions: Room: square, side = 10 m Tile: square, side = 50 cm = 0.5 m No wastage assumed; exact tiling Concept / Approach: Room area = 10 * 10 = 100 m^2. Tile area = 0.5 * 0.5 = 0.25 m^2. Number of tiles = 100 / 0.25 = 400. This is a simple area division once units are aligned in metres. Step-by-Step Solution: Convert tile side: 50 cm = 0.5 m.Room area = 10^2 = 100 m^2.Tile area = 0.5^2 = 0.25 m^2.Tiles needed = 100 / 0.25 = 400. Verification / Alternative check: Alternatively, along each side we place 10 / 0.5 = 20 tiles; total tiles = 20 * 20 = 400. Why Other Options Are Wrong: 200 and 300 undercount; 500 and 600 overcount. Common Pitfalls: Leaving tile side in centimetres and mixing units. Final Answer: 400.

Question 3

Isosceles Triangle — Perimeter and Equal Sides Given: The perimeter of an isosceles triangle is 26 cm and the two equal sides together measure 20 cm. Find the third side and each equal side (respectively).

Accepted Answer

Introduction / Context: An isosceles triangle has two equal sides. When the perimeter and the sum of the equal sides are known, the third side follows by subtraction. This problem tests reading precision and simple arithmetic within geometric context. Given Data / Assumptions: Perimeter P = 26 cm Equal sides together = 20 cm ⇒ each equal side = 10 cm Third side = P − (sum of equal sides) Concept / Approach: Let the equal sides be a and a; given a + a = 20 ⇒ a = 10. The third side b = P − 2a. Report the ordered pair: third side first, then each equal side, as requested in the stem. Step-by-Step Solution: a = 20 / 2 = 10 cm.b = 26 − 20 = 6 cm.Therefore: 6 cm and 10 cm. Verification / Alternative check: Perimeter check: 10 + 10 + 6 = 26 cm (matches). Why Other Options Are Wrong: Other pairs do not sum to the required perimeter with two equal sides of 10 cm. Common Pitfalls: Reporting the equal side first despite the requested order. Final Answer: 6 cm and 10 cm.

Question 4

Rectangle — Ratio 5 : 3 and Length 8 m More: The ratio of length to breadth is 5 : 3, and length is 8 m more than breadth. Find the area (in square metres).

Accepted Answer

Introduction / Context: Rectangles with a given side ratio and an additive difference can be solved with a single variable. Convert the ratio to algebra, use the difference to find the scale, then compute area as length times breadth in square metres. Given Data / Assumptions: L : B = 5 : 3 L = B + 8 Units in metres Concept / Approach: Let L = 5k and B = 3k. Then L − B = 2k = 8 ⇒ k = 4. Substitute back to get actual dimensions and multiply to obtain area. This linear system avoids guesswork and ensures consistency. Step-by-Step Solution: 2k = 8 ⇒ k = 4.L = 5k = 20 m; B = 3k = 12 m.Area = 20 * 12 = 240 m^2. Verification / Alternative check: Difference check: 20 − 12 = 8 (satisfied). Ratio check: 20 : 12 = 5 : 3. Why Other Options Are Wrong: 300 and 250 assume other k values. 185 and 196 do not match the derived dimensions. Common Pitfalls: Mistaking which term in the ratio corresponds to length vs breadth. Final Answer: 240 sq m.

Question 5

Rectangle — Length : Breadth = 2 : 1 and Breadth 5 cm Less: The ratio L : B is 2 : 1. If breadth is 5 cm less than length, find the perimeter (in cm).

Accepted Answer

Introduction / Context: This is a straightforward system using a ratio and a difference. After finding the actual sides, compute the perimeter using P = 2(L + B). Units remain in centimetres throughout. Given Data / Assumptions: L : B = 2 : 1 ⇒ L = 2k, B = k B = L − 5 Perimeter P = 2(L + B) Concept / Approach: From B = L − 5 and L = 2k, B = k, we get k = 5. Substitute to find sides and then compute the perimeter. Step-by-Step Solution: k = 5 ⇒ L = 10 cm, B = 5 cm.Perimeter P = 2(10 + 5) = 30 cm. Verification / Alternative check: Difference check: 10 − 5 = 5 cm; ratio check: 10 : 5 = 2 : 1. Why Other Options Are Wrong: 25 and 28 miss the 2× sum step. 35 and 40 require different side lengths than given. Common Pitfalls: Treating the ratio as L − B rather than L : B. Final Answer: 30 cm.

Question 6

Rectangle — Area 144 cm^2 and Sides in Ratio 4 : 9: Find the perimeter of the rectangle.

Accepted Answer

Introduction / Context: When area and side ratio are known, scale factor k follows from (4k)*(9k) = 144. After finding sides, compute the perimeter. This avoids solving two unknowns independently by leveraging the ratio constraint. Given Data / Assumptions: Sides = 4k and 9k (cm) Area = 144 cm^2 Perimeter P = 2(4k + 9k) Concept / Approach: Compute k^2 from area and take the positive root. Then compute perimeter directly. Keep units in centimetres consistently. Step-by-Step Solution: (4k)(9k) = 36k^2 = 144 ⇒ k^2 = 4 ⇒ k = 2.Sides: 8 cm and 18 cm.Perimeter P = 2(8 + 18) = 52 cm. Verification / Alternative check: Area check: 8 * 18 = 144 cm^2 (matches). Why Other Options Are Wrong: 56, 60, 64 arise from incorrect k or arithmetic. 48 assumes equal sides (square), which is not given. Common Pitfalls: Using 4 + 9 = 13 as area; remember area uses multiplication. Final Answer: 52 cm.

Question 7

Rectangle — Area 3584 m^2 and Sides in Ratio 7 : 2: Find the perimeter of the rectangle (in metres).

Accepted Answer

Introduction / Context: This is a scale-factor problem: area fixes k^2 when sides are in a given ratio. After computing k, sides follow, and the perimeter is twice the sum. Carry exact integers to avoid rounding in the final perimeter value. Given Data / Assumptions: Sides = 7k and 2k (metres) Area = 3584 m^2 ⇒ (7k)(2k) = 14k^2 Perimeter P = 2(7k + 2k) Concept / Approach: Solve 14k^2 = 3584 ⇒ k^2 = 256 ⇒ k = 16. Then compute sides and the perimeter. All numbers are integral, giving an exact perimeter. Step-by-Step Solution: k^2 = 3584 / 14 = 256 ⇒ k = 16.Sides: 7k = 112 m; 2k = 32 m.Perimeter P = 2(112 + 32) = 2 * 144 = 288 m. Verification / Alternative check: Area check: 112 * 32 = 3584 m^2 (matches). Why Other Options Are Wrong: 246, 286, 292 are near misses from arithmetic errors. 274 is unsupported by the derived dimensions. Common Pitfalls: Treating 7 : 2 as a sum instead of a product when using area. Final Answer: 288 m.

Question 8

Rectangle — Given Ratio of Length to Perimeter 5 : 18: If L : P = 5 : 18 for a rectangle, what is the ratio L : B?

Accepted Answer

Introduction / Context: Relating a rectangle’s length directly to its perimeter allows back-solving for the breadth via P = 2(L + B). Converting the L : P ratio into an equation yields the desired L : B ratio without needing actual numbers. Given Data / Assumptions: L : P = 5 : 18 ⇒ P = (18/5)L Perimeter formula: P = 2(L + B) Concept / Approach: Set 2(L + B) = (18/5)L, solve for B in terms of L, then express L : B in simplest integer form by clearing denominators. Step-by-Step Solution: 2(L + B) = (18/5)L ⇒ L + B = (9/5)L.B = (9/5 − 1)L = (4/5)L.Therefore L : B = 1 : 4/5 = 5 : 4. Verification / Alternative check: Let L = 5 ⇒ P = 18; Then 2(L + B) = 18 ⇒ L + B = 9 ⇒ B = 4 ⇒ ratio 5 : 4. Why Other Options Are Wrong: 4 : 3 and 3 : 5 do not satisfy P = (18/5)L. 4 : 7 and 7 : 4 invert or distort the derived proportion. Common Pitfalls: Forgetting that perimeter is twice the semiperimeter leading to a factor-of-two error. Final Answer: 5 : 4.

Question 9

Carpeting a Rectangular Room — Cost from Width and Rate: Find the total cost of carpeting a room 8 m by 6 m with a carpet 0.75 m wide at ₹20 per metre (length charged by running metre).

Accepted Answer

Introduction / Context: When carpet is priced per running metre of a fixed width, the payable length equals floor area divided by carpet width. Multiplying this length by the per-metre rate yields the total cost. Unit consistency (metres) is crucial. Given Data / Assumptions: Room: 8 m × 6 m ⇒ area = 48 m^2 Carpet width = 0.75 m; priced at ₹20 per running metre No wastage assumed Concept / Approach: Length of carpet required = area / width. Cost = (required length) * (rate per metre). Choosing orientation to minimize seams does not affect total length because width is fixed and coverage is full. Step-by-Step Solution: Area = 8 * 6 = 48 m^2.Required length = 48 / 0.75 = 64 m.Cost = 64 * ₹20 = ₹1280. Verification / Alternative check: Rows of width strips: Total strip area = width * length; summing to area forces the same total running metres regardless of direction. Why Other Options Are Wrong: ₹1300, ₹1440, ₹1500, ₹1750 arise from rounding or using width 0.8 or 0.7 m mistakenly. Common Pitfalls: Mismatching metres and centimetres or pricing per square metre instead of per running metre. Final Answer: ₹ 1280.

Question 10

Rectangle — Length 20 cm, Area 200 cm^2; Area Increased by 1 1/5 Times by Increasing Length Only: Find the new perimeter (in cm).

Accepted Answer

Introduction / Context: The problem states the area is increased by “1 1/5 times” the original by changing length only. In standard wording, “increased by 1 1/5 times” means the increase equals 1.2 of the original, so the new area becomes 1 + 1.2 = 2.2 times. However, many aptitude items intend “becomes 1 1/5 times,” i.e., 1.2 times. We apply the Recovery-First policy and adopt the widely used interpretation: new area = 1.2 * original, since the answer options align with this reading. Given Data / Assumptions: Original: L = 20 cm, A = 200 cm^2 ⇒ B = A/L = 10 cm Only L changes; B fixed at 10 cm New area A' = 1.2 * 200 = 240 cm^2 (assumed per standard aptitude phrasing) Concept / Approach: With breadth fixed, new length L' = A' / B. Compute perimeter P' = 2(L' + B). This directly uses rectangle area and perimeter definitions under the clarified assumption. Step-by-Step Solution: B = 200/20 = 10 cm.A' = 1.2 * 200 = 240 cm^2.L' = 240 / 10 = 24 cm.Perimeter P' = 2(24 + 10) = 68 cm. Verification / Alternative check: If “increased by 1.2 times” were read as +120% (2.2×), perimeter would be 108 cm, absent from options—supporting our interpretation. Why Other Options Are Wrong: 60, 64, 72, 76 come from misreading the phrase or changing breadth too. Common Pitfalls: Linguistic ambiguity in “increased by x times.” Always reconcile with options. Final Answer: 68.

Question 11

Rectangular Field — Cost to Grass Bed ⇒ Area ⇒ Length: Breadth is 25 m. Total cost for grass at ₹15 per m^2 is ₹12375. Find the length of the field (in metres).

Accepted Answer

Introduction / Context: Unit pricing per square metre allows recovery of area from total cost. With one dimension known, the other follows from area = length * breadth. Keeping currency and unit conversions straight is essential. Given Data / Assumptions: Breadth B = 25 m Total cost = ₹12375; rate = ₹15 per m^2 Area A = cost / rate Concept / Approach: Compute area from cost and rate. Then length L = A / B. The rectangular field is assumed to be exactly covered (no wastage) and charged purely by area. Step-by-Step Solution: A = 12375 / 15 = 825 m^2.L = 825 / 25 = 33 m.Therefore, the length is 33 m. Verification / Alternative check: Area check: 33 * 25 = 825 m^2. Why Other Options Are Wrong: 27, 30, 32, 36 yield areas not equal to 825 m^2 with breadth 25 m. Common Pitfalls: Dividing by rate instead of multiplying, or using breadth as 20/30 in error. Final Answer: 33 m.

Question 12

Square — Perimeter 68 cm: Find the area (in square centimetres).

Accepted Answer

Introduction / Context: Given a square’s perimeter, the side is perimeter/4. Area is side^2. This item is a quick application of square properties and arithmetic squaring of an integer side length. Given Data / Assumptions: Perimeter P = 68 cm Side s = P/4 Area A = s^2 Concept / Approach: Compute the side, then square it to get the area. Keep units in centimetres squared for the final answer to match options exactly. Step-by-Step Solution: s = 68 / 4 = 17 cm.A = 17^2 = 289 cm^2. Verification / Alternative check: Reverse: A = 289 ⇒ s = 17 ⇒ P = 4 * 17 = 68 cm (consistent). Why Other Options Are Wrong: 361 and 324 are 19^2 and 18^2, not 17^2. 284 and 269 are near misses from arithmetic errors. Common Pitfalls: Dividing by 2 instead of 4 to find the side. Final Answer: 289 sq cm.

Question 13

Squares — Third Square from Difference of Areas (Recovery Applied): The perimeters of two squares are 68 cm and 32 cm. Find the perimeter of a third square whose area equals the difference of their areas.

Accepted Answer

Introduction / Context: The original stem was incomplete (“The perimeter of two squares are 68 cm.”) lacking the second perimeter. By the Recovery-First Policy, we minimally repair it to a standard, solvable form by specifying the second perimeter as 32 cm so the item remains meaningful and aligned with common exam patterns and the given options. Given Data / Assumptions: Square 1: P1 = 68 cm ⇒ s1 = 17 cm ⇒ A1 = 289 cm^2 Square 2: P2 = 32 cm ⇒ s2 = 8 cm ⇒ A2 = 64 cm^2 Third square area A3 = A1 − A2 = 225 cm^2 Concept / Approach: Compute areas from perimeters, take the difference, then find the square whose area equals that difference. From area, obtain side and then perimeter. This ties together square perimeter–area conversions consistently. Step-by-Step Solution: A3 = 289 − 64 = 225 cm^2.Side s3 = √225 = 15 cm.Perimeter P3 = 4 * 15 = 60 cm. Verification / Alternative check: Reverse: P3 = 60 ⇒ s3 = 15 ⇒ A3 = 225 = 289 − 64 (checks out). Why Other Options Are Wrong: 64 cm and 32 cm correspond to perimeters of the given squares, not the third. 8 cm and 40 cm do not match the difference area 225 cm^2. Common Pitfalls: Treating “difference” as “sum” or mixing perimeters with areas. Final Answer: 60 cm.

Question 14

Wheel Revolutions to Radius — Distance 44 km in 4000 Revolutions (Recovery Applied): A wheel makes 4000 revolutions while moving 44 km. Find the radius of the wheel.

Accepted Answer

Introduction / Context: The original options listed very large radii in metres that did not align with the computation, suggesting inconsistency. By the Recovery-First Policy, we keep the given distance and revolutions and reconstruct coherent options around the correct result. The problem uses distance = revolutions * circumference and circumference = 2 * π * r. Given Data / Assumptions: Total distance D = 44 km = 44000 m Number of revolutions N = 4000 Circumference C = D / N C = 2 * π * r Concept / Approach: Each revolution covers one circumference. Compute C, then solve for r = C / (2 * π). Keep metres throughout and round sensibly only at the end to match practical answer choices. Step-by-Step Solution: C = 44000 / 4000 = 11 m.r = C / (2 * π) = 11 / (2 * π) ≈ 11 / 6.28318 ≈ 1.75 m.Thus, radius ≈ 1.75 m. Verification / Alternative check: Check: 2 * π * 1.75 ≈ 10.9956 m; 4000 * 10.9956 ≈ 43982 m ≈ 44 km (rounding error acceptable). Why Other Options Are Wrong: 1.25 m and 1.50 m yield too small circumferences. 2.00 m and 2.25 m yield distances exceeding 44 km. Common Pitfalls: Mistaking diameter for radius; using C = π * d means d = 11/π and r = d/2. Final Answer: 1.75 m.

Question 15

Semicircle — If the area of a semicircle is 77 sq m, find its perimeter (use perimeter = πr + 2r for a semicircle).

Accepted Answer

Introduction / Context: This problem tests formulas for a circle and a semicircle. For a semicircle of radius r, the area is (1/2) * π * r^2 and the perimeter (often called the boundary length) equals πr + 2r because it includes the curved arc (half the circumference) plus the diameter as the straight edge. Given Data / Assumptions: Area of semicircle = 77 sq m Use π ≈ 22/7 for exact nice arithmetic Perimeter of semicircle P = πr + 2r Concept / Approach: First recover the radius from the given area. Then plug r into the perimeter expression πr + 2r. With π = 22/7 here, the arithmetic becomes exact. Step-by-Step Solution: (1/2) * π * r^2 = 77(1/2) * (22/7) * r^2 = 77 ⇒ r^2 = 154 / (22/7) = 49 ⇒ r = 7 mPerimeter P = πr + 2r = (22/7)*7 + 14 = 22 + 14 = 36 m Verification / Alternative check: Back-substitute r = 7 in area formula for a semicircle to reconfirm 77 sq m, which matches the prompt exactly. Why Other Options Are Wrong: 42 m, 48 m, and 54 m arise from misusing full circumference or omitting the diameter term; 32 m comes from mixing up half/whole factors. Common Pitfalls: Using full-circle perimeter 2πr or forgetting to add the diameter when computing the boundary length of a semicircle. Final Answer: 36 m

Question 16

Semicircular park — A railing of 288 m is used to fence a semicircular park (π = 22/7). Find the area of the park.

Accepted Answer

Introduction / Context: The fence length around a semicircle equals the semicircle perimeter P = πr + 2r. From this we can find the radius r and then compute the area of the semicircle, A = (1/2) * π * r^2. Given Data / Assumptions: Total railing (perimeter) P = 288 m π = 22/7 Semicircle perimeter P = πr + 2r Semicircle area A = (1/2) * π * r^2 Concept / Approach: Solve r from 288 = r(π + 2). Then compute A using the semicircle area formula with r obtained and π = 22/7. Step-by-Step Solution: π + 2 = 22/7 + 2 = 22/7 + 14/7 = 36/7r = 288 / (36/7) = 288 * 7 / 36 = 56 mA = (1/2) * π * r^2 = (1/2) * (22/7) * 56^2 = (11/7) * 3136 = 448 * 11 = 4928 m2 Verification / Alternative check: Check perimeter with r = 56: P = (22/7)*56 + 112 = 176 + 112 = 288 m ✔ Why Other Options Are Wrong: 9865, 8956, and 9856 m2 are not consistent with r = 56; they reflect arithmetic slips; 4480 m2 forgets the exact factor (11/7) * r^2. Common Pitfalls: Using full-circle perimeter 2πr or omitting the diameter when computing a semicircle’s perimeter. Final Answer: 4928 m2

Question 17

Circle scaling — If the radius of a circle is increased by 6%, find the percentage increase in its area.

Accepted Answer

Introduction / Context: This is a percentage change under geometric scaling. Area of a circle is proportional to r^2, so a change in radius scales the area by the square of the same linear factor. Given Data / Assumptions: Radius increases by 6% ⇒ new radius factor = 1.06 Area ∝ r^2 Concept / Approach: If r → 1.06r, then area A → (1.06)^2 * A. The percentage increase is (1.06^2 − 1) * 100%. Step-by-Step Solution: 1.06^2 = 1.1236Increase = (1.1236 − 1) * 100% = 0.1236 * 100% = 12.36% Verification / Alternative check: Use small-change approximation: for small x, (1 + x)^2 ≈ 1 + 2x = 1 + 12%. Exact result adds x^2 = 0.36%, giving 12.36%. Why Other Options Are Wrong: 10.25% corresponds to a 5% linear increase, not 6%; 15% and 17% are overestimates; 8.39% is unrelated. Common Pitfalls: Doubling the 6% directly (i.e., 12%) without the square term; here, the extra 0.36% comes from 0.06^2. Final Answer: 12.36%

Question 18

Wheel and distance — An engine wheel turns 350 revolutions to cover 1.76 km. Find the wheel’s diameter (in meters).

Accepted Answer

Introduction / Context: Each revolution advances the wheel by its circumference. Total distance equals revolutions times circumference. Knowing revolutions and distance, we solve circumference, then diameter via C = πD. Given Data / Assumptions: Total distance = 1.76 km = 1760 m Revolutions = 350 Circumference C = distance / revs C = πD, with π ≈ 22/7 for neat arithmetic Concept / Approach: Compute C, then divide by π to get D. Units must be consistent (meters throughout). Step-by-Step Solution: C = 1760 / 350 = 176 / 35 ≈ 5.028571… mD = C / π ≈ 5.028571… / (22/7) = 5.028571… * 7 / 22 ≈ 1.6 m Verification / Alternative check: Use π ≈ 3.1416: D ≈ 5.0286 / 3.1416 ≈ 1.60 m, consistent with the above. Why Other Options Are Wrong: 1.4, 1.2, 1.8, and 2.0 m do not satisfy πD ≈ 5.0286 m. Options in centimeters sometimes appear in misprints; correct scale is meters. Common Pitfalls: Mixing units (km, m); forgetting that 1 revolution equals one full circumference. Final Answer: 1.6 m

Question 19

Square — What is the ratio of the areas of the circumcircle and the incircle of a square?

Accepted Answer

Introduction / Context: For a square of side a, the incircle fits inside touching the sides, and the circumcircle passes through all four vertices. Their radii relate directly to a: r_in = a/2 and r_out = a/√2. Areas scale with the square of the radius. Given Data / Assumptions: Incircle radius r_in = a/2 Circumcircle radius r_out = a/√2 Area of a circle ∝ r^2 Concept / Approach: The required ratio is (π r_out^2) : (π r_in^2) = r_out^2 : r_in^2. Step-by-Step Solution: r_out^2 = (a/√2)^2 = a^2/2r_in^2 = (a/2)^2 = a^2/4Ratio = (a^2/2) : (a^2/4) = (1/2) : (1/4) = 2 : 1 Verification / Alternative check: Choose a = 2: r_out = √2, r_in = 1 ⇒ areas π(2) and π(1) ⇒ ratio 2:1. Why Other Options Are Wrong: 1:2 inverts the relation; √2:1 and 1:√2 compare radii, not areas; 4:1 overstates the area ratio. Common Pitfalls: Confusing the ratio of radii (which is √2:1) with the ratio of areas (square of that), which is 2:1. Final Answer: 2 : 1

Question 20

Circle — The radius is increased so that the circumference increases by 5%. By what percentage does the area increase?

Accepted Answer

Introduction / Context: Circumference C = 2πr scales linearly with r. If the circumference increases by 5%, then r also increases by 5%. Area then scales with the square of the linear factor. Given Data / Assumptions: C increases by 5% ⇒ r increases by 5% Area ∝ r^2 Concept / Approach: If r → 1.05r, then area A → (1.05)^2 * A. The increase is (1.05^2 − 1) * 100%. Step-by-Step Solution: 1.05^2 = 1.1025Percentage increase = (1.1025 − 1) * 100% = 10.25% Verification / Alternative check: Small-change rule: 2 * 5% = 10% plus the square term 0.25% gives the precise 10.25%. Why Other Options Are Wrong: 12.5% corresponds to 1.118… factor; 10.5% and 11.25% are off; 9.75% is an underestimate. Common Pitfalls: Assuming a simple 10% (doubling 5%) without adding the 0.25% correction from the square. Final Answer: 10.25%

Area Questions