Question 1

Crossing a square field diagonally at 4 km/h: A person walks at 4 km/h and crosses a square field along its diagonal in 3 minutes (interpreting the time unit as minutes per standard usage). Find the area of the field in square metres.

Accepted Answer

Introduction / Context:We relate uniform speed to distance covered along the diagonal of a square, then recover the side and compute area. The original “3 meters” appears inconsistent with a speed in km/h; under Recovery-First, we repair it to “3 minutes,” a standard formulation. Given Data / Assumptions: Speed v = 4 km/h = 4000 m/60 min = 66.666… m/min. Time t = 3 min. Distance along diagonal d = v * t. Concept / Approach:For a square, area = (diagonal^2)/2. Compute diagonal from speed and time, then area. Step-by-Step Solution: 1) d = 66.666… * 3 = 200 m.2) Area = d^2 / 2 = 200^2 / 2 = 40000 / 2 = 20000 sq m. Verification / Alternative check:Side s = d/√2 ≈ 141.421 m; s^2 ≈ 20000 sq m as above. Why Other Options Are Wrong:10000, 30000, 40000 do not satisfy the diagonal–area relationship for the given speed and time. Common Pitfalls:Not converting km/h to m/min; treating the inconsistent “3 meters” literally, which breaks dimensional consistency. Final Answer:20000.

Question 2

Squares – area ratio to perimeter ratio: Given two squares whose areas are in the ratio 225 : 256, find the ratio of their perimeters.

Accepted Answer

Introduction / Context:For similar figures like squares, side lengths scale with the square root of area scale. Perimeter is linear in side length. Given Data / Assumptions: Area ratio A₁:A₂ = 225:256. Concept / Approach:Side ratio = √(area ratio) = 15:16. Perimeter ratio equals side ratio. Step-by-Step Solution: √225 = 15; √256 = 16.Perimeter ratio = 15 : 16. Verification / Alternative check:Pick side lengths 15 and 16; perimeters are 60 and 64 ⇒ 15:16. Why Other Options Are Wrong:25:16 confuses area with perimeter; 16:15 inverts; 17:16 is unrelated. Common Pitfalls:Forgetting that perimeter is linear, not quadratic. Final Answer:15:16.

Question 3

Square area 1024 sq cm → rectangle dimensions from square side: A square has area 1024 sq cm. Consider a rectangle whose length is twice the square’s side and whose breadth is 12 cm less than that side. Find the ratio length : breadth of the rectang…

Accepted Answer

Introduction / Context:Use the square’s area to get its side, then translate to the rectangle’s dimensions and simplify the ratio. Given Data / Assumptions: Square area = 1024 ⇒ side s = √1024 = 32 cm. Rectangle: length L = 2s = 64 cm; breadth B = s − 12 = 20 cm. Concept / Approach:Compute L:B and reduce to lowest terms. Step-by-Step Solution: L:B = 64:20 ⇒ divide by 4 ⇒ 16:5. Verification / Alternative check:Values are integers and consistent with the description. Why Other Options Are Wrong:14:9, 13:12, 17:13 do not match 64:20. Common Pitfalls:Mistaking “twice the side” as “side + 2”. Final Answer:16:5.

Question 4

Lawn 120 m × 60 m with 1.5 m inner border all around: A rectangular grassy plot 120 m by 60 m has a 1.5 m wide gravel path running all round on the inside. Find the cost of gravelling this path at Rs 0.65 per sq m.

Accepted Answer

Introduction / Context:Border (annulus-like) area equals outer rectangle minus the inner clear rectangle after shrinking each side by twice the border width. Given Data / Assumptions: Outer: 120 m × 60 m. Border width w = 1.5 m. Rate = Rs 0.65 per sq m. Concept / Approach:Inner dimensions = (120 − 2w) × (60 − 2w) = 117 × 57. Area(path) = 120×60 − 117×57. Step-by-Step Solution: 1) Outer area = 7200 sq m.2) Inner area = 117 × 57 = 6669 sq m.3) Path area = 7200 − 6669 = 531 sq m.4) Cost = 531 × 0.65 = Rs 345.15. Verification / Alternative check:Difference-of-areas method is standard for uniform borders. Why Other Options Are Wrong:They correspond to incorrect inner dimensions or rate multiplication. Common Pitfalls:Subtracting only once the width instead of twice (both sides). Final Answer:Rs. 345.15.

Question 5

Isosceles triangle with perimeter 32 cm and equal sides 5/6 of base: Find the area (in cm²).

Accepted Answer

Introduction / Context:Use the perimeter condition to find the base and equal sides, then compute height via Pythagoras and the area. Given Data / Assumptions: Perimeter = 32 cm. Let base = b; each equal side = (5/6)b. Concept / Approach:Perimeter: b + 2*(5b/6) = 32 ⇒ solve b. Height comes from splitting the base and using right triangles. Step-by-Step Solution: 1) b + (10b/6) = 32 ⇒ (16b/6) = 32 ⇒ b = 12 cm.2) Equal side = (5/6)*12 = 10 cm.3) Height h = √(10^2 − (12/2)^2) = √(100 − 36) = 8 cm.4) Area = (1/2)*b*h = 0.5*12*8 = 48 cm². Verification / Alternative check:Perimeter check: 12 + 10 + 10 = 32. Why Other Options Are Wrong:39, 57, 64 arise from computation errors in b or h. Common Pitfalls:Not halving the base when using Pythagoras for an isosceles triangle. Final Answer:48.

Question 6

Rectangular field ratio 5:4 and breadth 20 m less than length: Find the perimeter of the field (metres).

Accepted Answer

Introduction / Context:Translate ratio and difference into algebra, solve dimensions, and compute the perimeter. Given Data / Assumptions: L : B = 5 : 4; B = L − 20. Concept / Approach:Let L = 5x, B = 4x; then 4x = 5x − 20 ⇒ x = 20 ⇒ L = 100, B = 80. Step-by-Step Solution: Perimeter P = 2(L + B) = 2(100 + 80) = 360 m. Verification / Alternative check:Breadth 80 is indeed 20 less than length 100. Why Other Options Are Wrong:260, 280, 320 do not equal 2(180). Common Pitfalls:Assigning the difference to the wrong side. Final Answer:360m.

Question 7

Motorcycle speed from wheel revolutions: A motorcycle wheel of diameter 70 cm makes 40 revolutions in 10 seconds. Find the speed of the motorcycle in km/h.

Accepted Answer

Introduction / Context:Distance traveled equals revolutions times circumference. Convert the resulting m/s to km/h. Given Data / Assumptions: Diameter d = 0.70 m ⇒ circumference C = πd ≈ 2.199 m. 40 rev in 10 s ⇒ 4 rev/s. Concept / Approach:Speed v = rev rate * circumference. Step-by-Step Solution: 1) v = 4 * π * 0.70 ≈ 8.796 m/s.2) Convert to km/h: 8.796 * 3.6 ≈ 31.6656 ≈ 31.68 km/h. Verification / Alternative check:Using π ≈ 22/7 gives the same rounded value. Why Other Options Are Wrong:They reflect incorrect circumference or conversion factors. Common Pitfalls:Forgetting to convert m/s to km/h by 3.6. Final Answer:31.68 km/h.

Question 8

Two triangles – one has base 15 cm and height 12 cm; find height of a second triangle with double the area and base 20 cm:

Accepted Answer

Introduction / Context:Relate triangle areas via A = (1/2) * base * height. If the second triangle has double the area of the first, scale accordingly. Given Data / Assumptions: Triangle 1: base = 15 cm, height = 12 cm ⇒ A₁ = (1/2)*15*12 = 90 cm². Triangle 2: A₂ = 2*A₁ = 180 cm²; base₂ = 20 cm; height₂ = ? Concept / Approach:height₂ = 2*A₂ / base₂ (rearranging A₂ = (1/2) * base₂ * height₂). Step-by-Step Solution: height₂ = (2*180)/20 = 360/20 = 18 cm. Verification / Alternative check:Check: (1/2)*20*18 = 180 cm², which is twice 90. Why Other Options Are Wrong:16, 17, 20 do not satisfy the area condition with base 20 cm. Common Pitfalls:Forgetting the 1/2 factor when solving for height. Final Answer:18.

Question 9

Circular track 14 m wide; inner circumference 440 m: Find the radius of the outer circle.

Accepted Answer

Introduction / Context:A track has uniform width. Knowing the inner circumference gives the inner radius; the outer radius is inner radius plus width. Given Data / Assumptions: Width w = 14 m. Inner circumference = 440 m ⇒ 2πr_in = 440. Concept / Approach:r_in = 440 / (2π). Then r_out = r_in + 14. Step-by-Step Solution: 1) Using π ≈ 22/7: r_in = 440 / (2 * 22/7) = 440 * 7 / 44 = 70 m.2) r_out = 70 + 14 = 84 m. Verification / Alternative check:Outer circumference would then be 2π*84 ≈ 528π/ ? Not needed; radius result is direct. Why Other Options Are Wrong:24, 44, 54 do not equal 70 + 14. Common Pitfalls:Subtracting the width instead of adding for the outer radius. Final Answer:84 m.

Question 10

Two crossing roads inside an 80 m × 60 m lawn, each 10 m wide: Find the cost of gravelling the roads at Rs 3 per sq m.

Accepted Answer

Introduction / Context:Two strips cross at right angles. Sum their areas and subtract the central overlap counted twice. Given Data / Assumptions: Lawn: 80 m by 60 m. Road width w = 10 m. Rate = Rs 3 per sq m. Concept / Approach:Road area = (80×10) + (60×10) − (10×10). Step-by-Step Solution: 1) Lengthwise strip area = 800 sq m.2) Breadthwise strip area = 600 sq m.3) Overlap = 100 sq m.4) Total road area = 800 + 600 − 100 = 1300 sq m.5) Cost = 1300 * 3 = Rs 3900. Verification / Alternative check:Visualize two rectangles crossing; the overlap is a 10×10 square. Why Other Options Are Wrong:3600, 3700, 3800 result from miscounting the overlap or one strip. Common Pitfalls:Adding both strips without subtracting the overlapped square. Final Answer:Rs. 3900.

Question 11

Rectangle from perimeter and side difference; triangle on its diagonal: A rectangle has perimeter 84 m and its length is 6 m more than its breadth. Find the area of a triangle whose base equals the rectangle’s diagonal and height equals the rectangl…

Accepted Answer

Introduction / Context:First recover the rectangle’s dimensions, compute its diagonal, then evaluate the triangle area using base = diagonal and height = length. Given Data / Assumptions: Perimeter P = 84 ⇒ L + B = 42. L = B + 6. Triangle base = diagonal; height = L. Concept / Approach:Solve for L and B, compute diagonal d = √(L^2 + B^2), then triangle area A = (1/2)*d*L. Step-by-Step Solution: 1) (B + 6) + B = 42 ⇒ 2B = 36 ⇒ B = 18; L = 24.2) d = √(24^2 + 18^2) = √(576 + 324) = √900 = 30 m.3) Triangle area = 0.5 * 30 * 24 = 360 sq m. Verification / Alternative check:Dimensions satisfy perimeter 2(24 + 18) = 84. Why Other Options Are Wrong:380, 400 do not match the computed area; “360 metre” has wrong units for area. Common Pitfalls:Using 2(L+B) wrongly or forgetting to halve for triangle area. Final Answer:360 sq metre.

Question 12

Curved surface area of a hemisphere, diameter 28 cm: Find the curved surface area (in cm²).

Accepted Answer

Introduction / Context:The curved surface area (CSA) of a hemisphere depends only on the radius via CSA = 2πr^2. Given Data / Assumptions: Diameter = 28 cm ⇒ r = 14 cm. Concept / Approach:Apply CSA = 2πr^2 directly and evaluate numerically. Step-by-Step Solution: 1) CSA = 2 * π * 14^2 = 2 * π * 196 = 392π.2) Using π ≈ 22/7 ⇒ CSA = 392 * 22/7 = 56 * 22 = 1232 cm². Verification / Alternative check:Using π ≈ 3.1416 gives ≈ 1231.5, which rounds to 1232. Why Other Options Are Wrong:1152, 1024, 956 do not equal 392π. Common Pitfalls:Using sphere area 4πr^2 or including the base area; only curved area is asked. Final Answer:1232.

Question 13

Rectangle with sides in the ratio 3:1 and perimeter 96 m: The length and breadth of a rectangle are in the ratio 3 : 1. If its total perimeter is 96 m, compute the exact length (in metres) of the rectangle.

Accepted Answer

Introduction / Context:This is a standard perimeter-to-dimensions problem for rectangles using a given side ratio. We translate perimeter information into algebraic equations and then identify the rectangle’s length. Given Data / Assumptions: Length : Breadth = 3 : 1 → Let length = 3k and breadth = 1k. Perimeter P = 96 m. Perimeter formula: P = 2 * (L + B). Concept / Approach:Substitute the ratio-based expressions into the perimeter formula to solve for k, then compute the length 3k. Step-by-Step Solution: Let L = 3k and B = k.P = 2(L + B) = 2(3k + k) = 2(4k) = 8k.8k = 96 → k = 12.Therefore, length L = 3k = 3 * 12 = 36 m. Verification / Alternative check:With L = 36 m and B = 12 m, P = 2(36 + 12) = 2 * 48 = 96 m, which matches the given perimeter. Why Other Options Are Wrong:24m and 12m are the breadth or its multiples; 48m and 40m do not satisfy the ratio and perimeter simultaneously. Common Pitfalls:Forgetting to double the sum in the perimeter formula or mixing up length and breadth magnitudes (3k vs k). Final Answer:36m

Question 14

Right triangle transformation and new hypotenuse: A right-angled triangle has hypotenuse 10 cm and area 24 cm^2. If the shorter leg is halved and the longer leg is doubled, find the length of the new hypotenuse.

Accepted Answer

Introduction / Context:This problem uses two invariants of a right triangle—Pythagoras and area—to recover the legs, then applies specified changes to the legs and recomputes the hypotenuse. Given Data / Assumptions: Right triangle with legs a and b, hypotenuse 10 cm. Area = 24 cm^2 → (1/2) * a * b = 24 → ab = 48. Pythagoras: a^2 + b^2 = 10^2 = 100. Shorter leg is halved; longer leg is doubled. Concept / Approach:From ab and a^2 + b^2, obtain (a + b)^2 = a^2 + b^2 + 2ab = 100 + 96 = 196 → a + b = 14. Solve a and b from t^2 − 14t + 48 = 0. Step-by-Step Solution: t^2 − 14t + 48 = 0 → (t − 6)(t − 8) = 0 → {a, b} = {6, 8}.Shorter = 6, longer = 8.New legs: 6/2 = 3 and 8*2 = 16.New hypotenuse = √(3^2 + 16^2) = √(9 + 256) = √265. Verification / Alternative check:Original: ab = 48 and a^2 + b^2 = 100 hold for 6 and 8. Transform as per instruction; recomputation matches √265 exactly. Why Other Options Are Wrong:√245, √255, √275, √225 are not equal to √(3^2 + 16^2). Common Pitfalls:Halving/doubling the wrong legs or attempting to scale the hypotenuse directly rather than recomputing with Pythagoras after the change. Final Answer:√265cm

Question 15

Circle from circumference 26.4 cm: What is the exact area (in square centimetres) of a circle whose circumference is 26.4 cm?

Accepted Answer

Introduction / Context:Convert circumference to radius using C = 2πr, then compute the area πr^2. This is a direct formula application with clean decimals. Given Data / Assumptions: C = 26.4 cm. Use π = 22/7 for exact fractional handling (common in aptitude questions). Concept / Approach:r = C / (2π). Then A = πr^2. Carefully handle decimals to avoid rounding errors. Step-by-Step Solution: r = 26.4 / (2 * 22/7) = 26.4 / (44/7) = 26.4 * 7 / 44 = 0.6 * 7 = 4.2 cm.A = πr^2 = (22/7) * (4.2)^2 = (22/7) * 17.64 = 22 * 2.52 = 55.44 cm^2. Verification / Alternative check:Compute with C = 2πr → 2 * (22/7) * 4.2 = (44/7) * 4.2 = 26.4, confirming r = 4.2 cm is consistent. Why Other Options Are Wrong:44.55, 33.44, 44.33, 50.24 do not equal π*(4.2)^2 under the given circumference constraint. Common Pitfalls:Using C = πd but forgetting d = 2r, or squaring 26.4 accidentally instead of 4.2 when computing area. Final Answer:55.44

Question 16

Similar triangles and area removal: In ΔABC, points D on AB and E on AC are such that DE ∥ BC. Given AD : DB = 2 : 5 and area(ΔABC) = 98 cm^2, find the area (in cm^2) of quadrilateral BDEC.

Accepted Answer

Introduction / Context:When a segment joins the sides of a triangle parallel to the base, it forms a smaller triangle similar to the original. Areas scale by the square of the similarity ratio. We subtract to obtain the quadrilateral’s area. Given Data / Assumptions: DE ∥ BC with D on AB and E on AC. AD : DB = 2 : 5 → AD/AB = 2/(2+5) = 2/7. Area(ΔABC) = 98 cm^2. Concept / Approach:ΔADE ~ ΔABC with linear ratio 2/7. Hence area ratio is (2/7)^2 = 4/49. Compute area(ΔADE), then subtract from 98 to get area(BDEC). Step-by-Step Solution: Area(ΔADE) = (4/49) * 98 = 4 * 2 = 8 cm^2.Area(BDEC) = Area(ΔABC) − Area(ΔADE) = 98 − 8 = 90 cm^2. Verification / Alternative check:Consistency: If linear scale is 2/7, then the removed top triangle is small (8 cm^2), leaving most of the 98 cm^2 for the lower trapezoid-like region BDEC, i.e., 90 cm^2. Why Other Options Are Wrong:98 is the whole triangle; 94, 86, 88 do not equal 98 − 8 given the similarity ratio calculation. Common Pitfalls:Using 2/7 for area directly (forgetting to square the linear ratio), or mixing AD:DB with AD:AB. Final Answer:90

Question 17

Recover circumference from area: If the area of a circle is 616 cm^2, compute its circumference (in centimetres).

Accepted Answer

Introduction / Context:We invert area = πr^2 to find r, then compute circumference = 2πr. Integer-friendly values arise with π = 22/7. Given Data / Assumptions: Area A = 616 cm^2. Use π = 22/7. Concept / Approach:Find r from r^2 = A/π, then compute C = 2πr. Step-by-Step Solution: r^2 = 616 / (22/7) = 616 * 7 / 22 = 28 * 7 = 196.r = 14 cm.C = 2 * (22/7) * 14 = 88 cm. Verification / Alternative check:Area from r = 14 is (22/7)*14*14 = 22*28 = 616, consistent with the input. Why Other Options Are Wrong:44 cm and 22 cm are too small; 176 cm is twice the correct value; 110 cm is unrelated to 2πr here. Common Pitfalls:Forgetting to take the positive square root for r, or mixing up diameter with radius in circumference. Final Answer:88 cm

Question 18

Largest circle inside a square of side 18 cm (repaired stem): A wire is used to draw the largest possible circle that fits inside a square of side 18 cm. What is the area (in cm^2) of this circle?

Accepted Answer

Introduction / Context:(Recovery–First applied) The original stem mentioned “sphere” with area options in terms of π. A sphere’s surface area would not match these choices for side 18 cm. The natural, minimal-repair interpretation is the largest circle that can be inscribed in the square. We therefore correct the stem to “largest circle inside a square.” Given Data / Assumptions: Square side s = 18 cm. Inscribed circle has diameter = s, hence radius r = s/2 = 9 cm. Concept / Approach:An inscribed circle in a square touches all sides; its diameter equals the square’s side. Area of the circle is A = πr^2. Step-by-Step Solution: r = 18 / 2 = 9 cm.A = π * 9^2 = 81π cm^2. Verification / Alternative check:If one attempted a sphere-in-cube interpretation, the correct surface area would be 4πr^2 with r = 9 → 324π, which is not offered. The circle interpretation gives 81π, which is present and consistent with “inside a square.” Why Other Options Are Wrong:972π, 36π, 288π, 100π do not equal π * 9^2 for the inscribed circle in a side-18 square. Common Pitfalls:Confusing inscribed circle’s radius with the square’s diagonal/2; here diameter equals side, not the diagonal. Final Answer:81π

Question 19

Quadrilateral sides in ratio 2:3:4:5 with perimeter 280 m: Find the exact length (in metres) of the longest side.

Accepted Answer

Introduction / Context:When the sides of a polygon are in a given ratio and the perimeter is known, multiply the sum of the ratio parts by a common scale factor to match the perimeter and then identify requested side lengths. Given Data / Assumptions: Side ratio: 2 : 3 : 4 : 5. Perimeter = 280 m. Concept / Approach:Let the common factor be k. Then the sides are 2k, 3k, 4k, 5k and perimeter is (2+3+4+5)k = 14k. Compute k and find the longest side 5k. Step-by-Step Solution: 14k = 280 → k = 20.Longest side = 5k = 5 * 20 = 100 m. Verification / Alternative check:All sides: 40, 60, 80, 100 sum to 280 → matches the given perimeter. Why Other Options Are Wrong:150m, 175m, 180m, 120m do not equal 5k when k = 20 from the exact perimeter constraint. Common Pitfalls:Using average or guessing without computing k; or mis-summing the ratio parts (they total 14). Final Answer:100m

Question 20

Angle chase with isosceles condition and external division: In ΔABC where AC = BC and ∠ABC = 50°, extend BC to point D so that BC = CD. Find the value of ∠BAD.

Accepted Answer

Introduction / Context:This is a classic Euclidean angle-chasing configuration combining an isosceles triangle (AC = BC) with an external point D chosen so that BC = CD. The task is to find the angle at A between AB and AD. Given Data / Assumptions: AC = BC (isosceles with apex at C). ∠ABC = 50° (angle at B). BC is produced to D with BC = CD. Concept / Approach:One method is coordinate/analytic geometry: set B = (0,0), C = (1,0), so D = (2,0). Choose A so that AC = 1 and ∠ABC = 50°. Alternatively, pure geometry using isosceles properties and external point reflection arguments also leads to a right angle at ∠BAD. Step-by-Step Solution (analytic sketch): Place B = (0,0), C = (1,0); let direction of BA make 50° with BC. Then A = (2cos50° * cos50°, 2cos50° * sin50°).Compute vectors AB and AD; their dot product evaluates to approximately 0, implying orthogonality.Hence ∠BAD = 90°. Verification / Alternative check:Known result: with AC = BC and external point D such that BC = CD, the locus creates a right angle at A between AB and AD due to reflective symmetry about the perpendicular at the midpoint of BD. Why Other Options Are Wrong:80°, 40°, 50°, 70° do not satisfy the orthogonality deduction; only 90° is consistent with both construction and computations. Common Pitfalls:Assuming interior angle sums directly determine ∠BAD without constructing D carefully, or overlooking the equality BC = CD which enforces a key symmetry. Final Answer:90°

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