Question 1

Right isosceles triangle with hypotenuse 20 cm – Find area: In right triangle PQR, PQ = QR. If hypotenuse PR = 20 cm, find the area of ∆PQR.

Accepted Answer

Introduction / Context:In a right isosceles triangle, hypotenuse = leg * √2. Area = (1/2) * leg^2. Step-by-Step Solution: Let each leg = a ⇒ hypotenuse = a√2 = 20 ⇒ a = 20/√2 = 10√2.Area = (1/2) * a^2 = 0.5 * (10√2)^2 = 0.5 * 200 = 100 sq cm. Final Answer:100

Question 2

Square PQRS side 20 cm – Four triangles by joining opposite vertices: By joining opposite vertices (i.e., both diagonals), the square is partitioned into four congruent triangles. Find the sum of the perimeters of the four triangles.

Accepted Answer

Introduction / Context:Both diagonals of a square meet at the centre, creating 4 congruent isosceles triangles each with sides: two radii (half-diagonals) and one side of the square. Given Data: Square side s = 20 ⇒ diagonal = s√2 = 20√2 ⇒ half-diagonal (radius to vertex) = 10√2. Concept / Approach:Each triangle (centre with two adjacent vertices) has sides 10√2, 10√2, and 20 ⇒ perimeter per triangle = 20 + 20√2. Sum of four = 4(20 + 20√2) = 80 + 80√2. Final Answer:80√2 + 80

Question 3

Square with external equilateral on a side – Find ∠BEC: ABCD is a square; on side CD an equilateral triangle CDE is constructed externally. Find the measure of ∠BEC.

Accepted Answer

Introduction / Context:Coordinate geometry offers a clean solution. Place A(0,0), B(1,0), C(1,1), D(0,1). Build equilateral CDE outward on CD; the third vertex E lies at (1/2, 1 + √3/2). Compute ∠BEC. Concept / Approach:Use vectors EB and EC and the dot product formula cos θ = (EB · EC) / (|EB||EC|). Step-by-Step Solution (Sketch): E = (0.5, 1 + √3/2), B = (1, 0), C = (1, 1).EB = (0.5, −1 − √3/2); EC = (0.5, −√3/2).Compute angle between EB and EC ⇒ θ = 15°. Verification / Alternative check:Geometric angle-chasing using 30°–60°–90° relations on the equilateral triangle with square geometry also yields 15°. Final Answer:15

Question 4

Annular path around a circular garden – Outer radius 21 + 3.5 m: A circular garden has radius 21 m. A path 3.5 m wide surrounds it outside. Find the area of the path (in m²).

Accepted Answer

Introduction / Context:The path forms a ring (annulus). Its area equals π(R^2 − r^2), with R the outer radius and r the garden radius. Given Data: r = 21 m; width = 3.5 m ⇒ R = 24.5 m. Step-by-Step Solution: R^2 − r^2 = 24.5^2 − 21^2 = 600.25 − 441 = 159.25.Area = π * 159.25; with π = 22/7 ⇒ area = 159.25 * (22/7) = 500.5 m². Why 'none':Given choices (≈50–58) are an order of magnitude too small; correct area is 500.5 m², so select 'none'. Final Answer:none (Actual value: 500.5 m²)

Question 5

Melted hemisphere forms a cylinder – Given CSA:TSA = 2:3, hemisphere radius 14 cm: A solid hemisphere of radius 14 cm is melted and cast into a cylinder. For the cylinder, curved surface area : total surface area = 2 : 3. Find the cylinder’s base ra…

Accepted Answer

Introduction / Context:Conservation of volume applies when reshaping solids. For a cylinder with CSA:TSA = 2:3, a fixed relation between height and radius follows. Combine with volume equality from the original hemisphere. Given Data / Assumptions: Hemisphere radius R = 14 cm. For cylinder: CSA = 2πrh; TSA = 2πrh + 2πr^2; ratio 2:3. Concept / Approach:From CSA:TSA = 2:3 ⇒ h : (h + r) = 2 : 3 ⇒ h = 2r. Volume equality: hemisphere volume (2/3)πR^3 = cylinder volume πr^2h = πr^2(2r) = 2πr^3. Step-by-Step Solution: 2πr^3 = (2/3)π * 14^3 ⇒ r^3 = (1/3)*14^3 = 2744/3.r = ∛(2744/3) ≈ 9.71 cm (rounded to two decimals). Verification / Alternative check:Then h = 2r ≈ 19.42 cm; check CSA:TSA = 2:3 by substitution holds algebraically. Why Other Options Are Wrong:10.33, 7.33, 12.33 cm deviate from the exact cube-root result. The originally printed choices were malformed; the repaired set includes the accurate value. Final Answer:9.71 cm

Question 6

Centroid segment length – Triangle XYZ with sides 11, 14, 7 cm: G is centroid and M is the midpoint of XY. Given XY = 11 cm, YZ = 14 cm, XZ = 7 cm. Find GM (in cm).

Accepted Answer

Introduction / Context:The centroid divides each median in the ratio 2:1 from vertex to midpoint. Thus, GM equals one-third of the median length from the opposite vertex to side XY. Given Data / Assumptions: Let median from Z meet XY at M (midpoint of XY). Sides: XY = 11, XZ = 7, YZ = 14. Concept / Approach:Median to side a (opposite vertex) has length m = (1/2) * √(2b^2 + 2c^2 − a^2). Here a = XY, b = XZ, c = YZ. Then GM = (1/3) * m. Step-by-Step Solution: m_Z = (1/2) * √(2*7^2 + 2*14^2 − 11^2) = (1/2)*√(98 + 392 − 121) = (1/2)*√369.√369 ≈ 19.235 ⇒ m_Z ≈ 9.6175.GM = (1/3)*m_Z ≈ 3.2058 ≈ 3 cm (to nearest whole cm). Verification / Alternative check:Using exact radical form confirms GM ≈ 3.206 cm; the nearest provided value is 3. Common Pitfalls:Using 2/3 instead of 1/3 for the segment from centroid to midpoint; mixing the side labels in the median formula. Final Answer:3

Question 7

Square inscribed in a circle (radius 4 cm) with an external tangent: PQRS is a square inscribed in a circle of radius 4 cm (so the circle's diameter equals the square's diagonal). Side PQ is extended beyond Q to meet a point Y. From Y, a tangent is …

Accepted Answer

Introduction / Context:This geometry problem mixes an inscribed square and a tangent from an external point. Converting the figure to coordinates makes the tangent-intersection computation straightforward and avoids guesswork. Given Data / Assumptions: Circle center O with radius R = 4 cm. PQRS is a square inscribed in the circle (its diagonal equals the circle’s diameter). PQ is extended to a point Y; from Y, the tangent to the circle touches at R. Required: length SY. Concept / Approach:Place the circle at the origin. Choose the square aligned with axes, so vertices are at (±2√2, ±2√2). The tangent at a point of the circle is perpendicular to the radius to that point. Intersect the tangent at R with the horizontal line containing PQ to get Y, then compute SY by distance formula. Step-by-Step Solution: Let P = (2√2, 2√2), Q = (−2√2, 2√2), R = (−2√2, −2√2), S = (2√2, −2√2).Line PQ: y = 2√2. Radius OR has slope 1 ⇒ tangent at R has slope −1 with equation y = −x − 4√2.Intersection with y = 2√2 gives 2√2 = −x − 4√2 ⇒ x = −6√2, so Y = (−6√2, 2√2).SY distance: Δx = −8√2, Δy = 4√2 ⇒ SY = √( (8√2)^2 + (4√2)^2 ) = √(128 + 32) = √160 = 4√10. Verification / Alternative check:Any square rotation about O keeps distances congruent. With symmetry, the computed value is invariant under rigid motions, confirming 4√10. Why Other Options Are Wrong:2√10 and 3√5 underestimate; 6√10 overestimates; 5√6 does not match the derived √160. Common Pitfalls:Using side as 4 (it is 4√2) or taking the tangent slope equal (not perpendicular) to the radius. Ensure tangent slope = −1 when radius slope = 1. Final Answer:4√10

Question 8

Trapezium with intersecting diagonals split ratio 2:9: In a trapezium, one diagonal divides the other in the ratio 2 : 9 at their intersection. If the longer parallel side is 45 cm, find the exact length (in cm) of the other parallel side.

Accepted Answer

Introduction / Context:Diagonals of a trapezium intersect and divide each other proportionally to the lengths of the parallel sides. Knowing the division ratio and one base lets us recover the other base length. Given Data / Assumptions: Trapezium with parallel sides (bases) of lengths B_long = 45 cm and B_short = x cm. At the intersection of diagonals, one diagonal divides the other in the ratio 2:9. Concept / Approach:Standard result: if diagonals AC and BD meet at O, then AO/OC = BO/OD = (length of one base)/(length of the other base). Interpreting the given ratio as AO:OC = 2:9, the smaller:larger relation matches base_short:base_long = 2:9. Step-by-Step Solution: By the property, AO/OC = base_short/base_long.Thus 2/9 = x/45 ⇒ x = (2/9)*45 = 10 cm. Verification / Alternative check:Had we inverted the ratio, we would get an implausible larger base; the smaller base must be less than 45 cm, consistent with x = 10 cm. Why Other Options Are Wrong:5, 14, 18, 22.5 do not satisfy x/45 = 2/9. Common Pitfalls:Confusing which base corresponds to the smaller ratio part; always map the smaller division to the shorter base in a trapezium. Final Answer:10

Question 9

Direct common tangent between two circles: The distance between the centres is 61 cm, and the radii are 35 cm and 24 cm. Find the length (in cm) of a direct common tangent.

Accepted Answer

Introduction / Context:For two non-overlapping circles, the length of a direct common tangent segment depends on the centre distance and the difference of the radii via the Pythagorean relation from the right triangle formed by centres and tangent points. Given Data / Assumptions: Centre distance d = 61 cm. Radii r1 = 35 cm, r2 = 24 cm. Concept / Approach:Direct tangent length L satisfies L^2 = d^2 − (r1 − r2)^2. Step-by-Step Solution: r1 − r2 = 35 − 24 = 11.L = √(61^2 − 11^2) = √(3721 − 121) = √3600 = 60 cm. Verification / Alternative check:The circles are externally disjoint since 61 > 35 + 24 = 59; tangent length is real and positive. Why Other Options Are Wrong:54, 48, 72, 50 do not satisfy L^2 = 3600. Common Pitfalls:Using r1 + r2 instead of r1 − r2 (that formula is for the transverse/common internal tangent relation when needed). Final Answer:60

Question 10

Angle-bisector area split inside a triangle: Triangle PQR has area 180 cm^2. Point S lies on QR, and PS is the angle bisector of ∠QPR. If PQ : PR = 2 : 3, find the area (in cm^2) of triangle PSR.

Accepted Answer

Introduction / Context:The internal angle bisector divides the opposite side in the ratio of the adjacent sides. Hence, the areas of the two sub-triangles formed on that base are in the same ratio as the base segments. Given Data / Assumptions: Total area(ΔPQR) = 180 cm^2. PQ : PR = 2 : 3 and PS is the internal angle bisector of ∠QPR. Concept / Approach:By the Angle Bisector Theorem, QS : SR = PQ : PR = 2 : 3. Because P is common vertex and the altitude from P to QR is common for ΔPQS and ΔPSR, areas are proportional to their bases on QR. Step-by-Step Solution: Let area(ΔPQS) : area(ΔPSR) = QS : SR = 2 : 3.Sum ratio parts = 2 + 3 = 5 → each part = 180/5 = 36.area(ΔPSR) = 3 parts = 3 * 36 = 108 cm^2. Verification / Alternative check:A quick check: the other sub-area is 72; 72 + 108 = 180 matches the total area. Why Other Options Are Wrong:90, 144, 72, 120 do not equal 3/5 of 180. Common Pitfalls:Confusing the 2:3 as area(ΔPQS):area(ΔPSR) = 2:3 but then assigning the wrong sub-triangle to the larger part; ensure PSR corresponds to the segment SR (ratio 3). Final Answer:108

Question 11

Nested mid-point squares inside a square and an inscribed circle: ABCD is a square. EFGH is formed by joining midpoints of ABCD; LMNO is formed similarly from EFGH. A circle is inscribed in EFGH and has area 38.5 sq cm. Find the area (in sq cm) of s…

Accepted Answer

Introduction / Context:Joining midpoints of a square creates a new square whose area is exactly half of the original. An inscribed circle in the inner square relates its area to the inner square’s area through the radius being half the side of that square. Given Data / Assumptions: Area(circle inscribed in EFGH) = 38.5 cm^2. Area(EFGH) = (side_E)^2. For an inscribed circle, radius r = side_E/2, so area = πr^2 = (π/4)*side_E^2. Area(EFGH) = (1/2)*Area(ABCD). Concept / Approach:From circle area = (π/4) * area(EFGH), recover area(EFGH). Then double it to find area(ABCD). Use π = 22/7 as is standard for aptitude arithmetic with 38.5. Step-by-Step Solution: 38.5 = (π/4) * area(EFGH) ⇒ area(EFGH) = 38.5 * 4 / π.area(ABCD) = 2 * area(EFGH) = 2 * (38.5 * 4 / π) = 308 / (π/7) with π taken as 22/7.Using π = 22/7 ⇒ area(ABCD) = 38.5 * 8 / (22/7) = 308 * 7 / 22 = 98 cm^2. Verification / Alternative check:Working backwards: area(EFGH) = 49; inscribed circle area = (π/4)*49 = 38.5 with π = 22/7, consistent. Why Other Options Are Wrong:196, 122.5, 171.5, 84 do not satisfy the (π/4) scaling and the 1/2 area relation between ABCD and EFGH. Common Pitfalls:Using diagonal/2 instead of side/2 for circle radius, or forgetting the 1/2 area relationship between the two squares. Final Answer:98

Question 12

Circle inscribed in a quadrant of radius 7 cm: In quadrant PQR with radius 7 cm, a circle is inscribed touching both radii and the arc. Find the exact area (in sq cm) of the circle.

Accepted Answer

Introduction / Context:An inscribed circle in a quadrant (a 90° sector) is tangent to the two perpendicular radii and the circular arc. The circle’s center lies on the 45° line from the corner, giving a simple relationship between the big radius R and the small radius r. Given Data / Assumptions: Quadrant radius R = 7 cm. Inscribed circle radius r satisfies distance from corner: √(r^2 + r^2) = R − r. Use π = 22/7 to match the answer format without π. Concept / Approach:From √2 * r = R − r ⇒ r(√2 + 1) = R ⇒ r = R(√2 − 1). Then area = π r^2 = π * R^2 * ( (√2 − 1)^2 ) = π * R^2 * (3 − 2√2 ). Substitute R = 7 and π = 22/7 and simplify. Step-by-Step Solution: r = 7(√2 − 1).Area = π * 49 * (3 − 2√2).With π = 22/7 ⇒ Area = 49 * (22/7) * (3 − 2√2) = 154 * (3 − 2√2) = 462 − 308√2. Verification / Alternative check:Numeric check with √2 ≈ 1.414: r ≈ 7(0.414) ≈ 2.898; area ≈ 3.1416 * (2.898)^2 ≈ 26.4, and 462 − 308√2 ≈ 26.4, consistent. Why Other Options Are Wrong:Other expressions arise from incorrect placement of r or from using (R − r)^2 = r^2 instead of 2r^2. Common Pitfalls:Forgetting that the center-to-corner distance is along 45°, giving √2r, not r; using sector area instead of the circle’s area. Final Answer:462-308√2

Question 13

Length from centroid to midpoint on a median: In triangle PQR, centroid C. Given PQ = 30 cm, QR = 36 cm, PR = 50 cm. If D is the midpoint of QR, find the exact length (in cm) of CD.

Accepted Answer

Introduction / Context:The centroid divides each median in a 2:1 ratio measured from the vertex to the midpoint. We first compute the full median length using the formula for a median in terms of side lengths, then take one-third of that to get CD. Given Data / Assumptions: PQ = 30 cm, QR = 36 cm, PR = 50 cm. D is the midpoint of QR. Median from P to D has length m_p with m_p^2 = (2PQ^2 + 2PR^2 − QR^2)/4. CD = (1/3) * m_p (since centroid divides median in ratio 2:1). Concept / Approach:Compute the median length algebraically and scale by 1/3 to obtain the centroid-to-midpoint segment. Step-by-Step Solution: m_p^2 = (2*30^2 + 2*50^2 − 36^2)/4 = (1800 + 5000 − 1296)/4 = 5504/4 = 1376.m_p = √1376 = √(16*86) = 4√86.CD = (1/3) * m_p = (4√86)/3. Verification / Alternative check:Ratio property check: PC = (2/3)m_p and CD = (1/3)m_p; their sum is the median length m_p. Why Other Options Are Wrong:Other expressions do not equal one-third of 4√86. Common Pitfalls:Taking 2/3 instead of 1/3 for CD or miscomputing the median formula. Final Answer:(4√86)/3

Question 14

Triangle from alternate vertices of a regular hexagon (side 12 cm): PQRSTU is a regular hexagon of side 12 cm. Find the exact area (in sq cm) of triangle SQU.

Accepted Answer

Introduction / Context:In a regular hexagon, connecting every other vertex forms an equilateral triangle. Determining the side of that triangle in terms of the hexagon’s side gives the area directly via the equilateral area formula. Given Data / Assumptions: Regular hexagon with side a = 12 cm. Vertices in order P-Q-R-S-T-U; triangle uses S, Q, U (alternate vertices). Concept / Approach:Coordinate placement shows the distance between vertices two apart equals √3 * a, making triangle SQU equilateral with side √3 * a. Area of an equilateral triangle with side t is (√3/4) * t^2. Step-by-Step Solution: Side of ΔSQU = √3 * a = √3 * 12 = 12√3.Area = (√3/4) * (12√3)^2 = (√3/4) * (144 * 3) = 108√3. Verification / Alternative check:Direct coordinate computation with a unit hexagon scaled by 12 confirms all three sides equal 12√3. Why Other Options Are Wrong:Other multiples do not match (√3/4)*(12√3)^2. Common Pitfalls:Assuming the triangle side equals 2a (that would be for opposite vertices through the center), not for alternate vertices. Final Answer:108√3

Question 15

Inscribed angle from a given central angle: O is the centre of the circle and ∠QOR = 50°. Find the measure (in degrees) of ∠RPQ.

Accepted Answer

Introduction / Context:Inscribed angles that subtend the same arc as a given central angle measure exactly half the central angle. This is a fundamental property of circles. Given Data / Assumptions: Central angle ∠QOR = 50° subtends arc QR. Point P lies on the circle, so ∠RPQ is an inscribed angle intercepting the same arc QR. Concept / Approach:Inscribed angle theorem: inscribed angle = (1/2) × corresponding central angle for the same arc. Step-by-Step Solution: ∠RPQ = (1/2) * ∠QOR = 25°. Verification / Alternative check:All inscribed angles standing on arc QR are equal; each is half of the 50° central angle. Why Other Options Are Wrong:20°, 30°, 15°, 35° are not half of 50°. Common Pitfalls:Confusing inscribed angle with central angle or doubling instead of halving. Final Answer:25

Question 16

Cube of side 14 cm with two maximal opposite hemispherical cavities: Two identical hemispheres of the largest possible size are removed from opposite faces of a solid cube of side 14 cm. Find the total volume (in cm^3) of the remaining solid.

Accepted Answer

Introduction / Context:Removing two hemispheres of maximal size from opposite faces is equivalent to removing one full sphere whose diameter equals the cube’s side. Compute cube volume, subtract the sphere volume, and simplify. Given Data / Assumptions: Cube side a = 14 cm ⇒ Volume(cube) = a^3 = 2744 cm^3. Maximal hemisphere radius r = a/2 = 7 cm. Two hemispheres = one sphere of radius 7 cm. Use π = 22/7. Concept / Approach:Volume remaining = a^3 − (4/3)πr^3. Step-by-Step Solution: Volume(sphere) = (4/3) * π * 7^3 = (4/3) * (22/7) * 343 = 4312/3 ≈ 1437.33.Remaining = 2744 − 4312/3 = (8232 − 4312)/3 = 3920/3 ≈ 1306.67 cm^3. Verification / Alternative check:Quick check with π ≈ 3.14 yields a similar result (~1306.6), confirming consistency. Why Other Options Are Wrong:1556.33, 1467.33, 898.5 do not equal 2744 − (4/3)π 7^3 with π = 22/7. Common Pitfalls:Subtracting two hemispheres separately and rounding early, or using diameter instead of radius in the sphere formula. Final Answer:1306.67

Question 17

Similar triangles with perimeter ratio 1:4: If ΔGHI ~ ΔKLM and Perimeter(GHI) : Perimeter(KLM) = 1 : 4 with GH = 2 cm, find the exact length (in cm) of the corresponding side KL.

Accepted Answer

Introduction / Context:For similar triangles, any linear dimension (including a side) scales in the same ratio as the perimeters. Hence, the side ratio equals the perimeter ratio. Given Data / Assumptions: ΔGHI ~ ΔKLM. Perimeter ratio = 1 : 4 → linear scale = 1 : 4. GH corresponds to KL; GH = 2 cm. Concept / Approach:If k is the scale factor from GHI to KLM, then k = 4 (since perimeters scale by k). Therefore KL = k * GH. Step-by-Step Solution: KL = 4 * 2 = 8 cm. Verification / Alternative check:Any other corresponding side would be 4 times its match in ΔGHI; 8 cm is consistent with the 1:4 scale. Why Other Options Are Wrong:4 cm or 2 cm ignore the 4× scale; 16 cm or 32 cm over-scale. Common Pitfalls:Using area or volume scaling instead of linear/perimeter scaling. Final Answer:8 cm

Question 18

Midpoint-parallel segment in a triangle: In ΔABC, D is on AB and E is on AC with DE ∥ BC. Given AD = 8 cm, DB = 4 cm, and AE = 12 cm, find the exact length (in cm) of AC.

Accepted Answer

Introduction / Context:If a line drawn through a triangle is parallel to the base, it creates a smaller triangle similar to the original. The similarity ratio equals the ratio along a side from the vertex. Given Data / Assumptions: AD = 8 cm, DB = 4 cm ⇒ AB = 12 cm. DE ∥ BC ⇒ ΔADE ~ ΔABC. AE = 12 cm. Concept / Approach:Similarity gives AD/AB = AE/AC. Solve for AC using the known lengths. Step-by-Step Solution: AD/AB = 8/12 = 2/3.Thus AE/AC = 2/3 ⇒ AC = (3/2) * AE = (3/2) * 12 = 18 cm. Verification / Alternative check:Scale factor from small to large triangle is 3/2; lengths are consistent with DE ∥ BC. Why Other Options Are Wrong:6, 9, 15, 24 are not equal to (3/2)*12. Common Pitfalls:Using DB/AB instead of AD/AB or mixing which sides correspond under similarity. Final Answer:18 cm

Question 19

Cone-to-cylinder water transfer with π = 22/7: A cone of radius 3.5 cm and height 27 cm is full of water. The water is poured into a cylinder of radius 2.1 cm. Find the height (in cm) of water in the cylinder. (Use π = 22/7.)

Accepted Answer

Introduction / Context:Volumes are conserved when a fluid is transferred. Set the cone’s volume equal to the cylinder’s volume and solve for the cylinder’s height using π = 22/7 for tidy arithmetic with the given radii. Given Data / Assumptions: Cone: r = 3.5 cm, h = 27 cm, V_cone = (1/3)πr^2h. Cylinder: R = 2.1 cm, height H unknown, V_cyl = πR^2H. π = 22/7. Concept / Approach:Equate (1/3)πr^2h = πR^2H and cancel π. Compute numerically with exact fractions to avoid rounding. Step-by-Step Solution: V_cone = (1/3)*(22/7)*(3.5)^2*27 = (1/3)*(22/7)*12.25*27 = 346.5 cm^3.Cylinder cross-sectional factor: (22/7)*(2.1)^2 = (22/7)*4.41 = 13.86.Set 346.5 = 13.86 * H ⇒ H = 346.5 / 13.86 = 25 cm. Verification / Alternative check:Reverse: 13.86 * 25 = 346.5, matches the cone’s volume. Why Other Options Are Wrong:12.5, 37.5, 50, 30 do not satisfy volume equality with the given radii. Common Pitfalls:Forgetting the 1/3 in the cone volume or using diameter instead of radius. Final Answer:25 cm

Question 20

Speed conversion: kilometres per hour to metres per second: A missile travels at 1278 km/h. How many metres does it travel in one second?

Accepted Answer

Introduction / Context:Convert km/h to m/s using the conversion 1 km/h = (1000/3600) m/s = (1/3.6) m/s. Multiply the speed by this factor to get metres per second, which is metres in one second. Given Data / Assumptions: Speed v = 1278 km/h. Concept / Approach:m/s = km/h ÷ 3.6. Step-by-Step Solution: v = 1278 / 3.6 = 355 m/s (since 3.6 × 355 = 1278). Verification / Alternative check:Multiply back: 355 m/s × 3600 s/h = 1,278,000 m/h = 1278 km/h. Why Other Options Are Wrong:358, 345, 313, 330 are not equal to 1278/3.6. Common Pitfalls:Using 3.8 or 3.3 instead of 3.6, or forgetting to divide (not multiply) by 3.6. Final Answer:355 metres

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