Question 1

Annulus width from circumferences: A circular road surrounds a circular ground. The difference between outer and inner circumferences is 66 m. What is the road’s width?

Accepted Answer

Introduction / Context:The “width” of a circular ring (annulus) is the difference of its radii. Differences in circumferences translate directly into differences in radii via the circumference formula, enabling a quick computation. Given Data / Assumptions: Outer radius R, inner radius r 2πR − 2πr = 66 m Width w = R − r Concept / Approach:Factor the circumference difference: 2π(R − r) = 66. Divide by 2π to obtain the width directly. Step-by-Step Solution:2π(R − r) = 66R − r = 66 / (2π) = 33 / π ≈ 33 / 3.142 ≈ 10.5 m Verification / Alternative check:Using π = 22/7, width = 33 / (22/7) = 10.5 m exactly, confirming the neat fractional structure often used in aptitude questions. Why Other Options Are Wrong:21 m and 14 m are multiples of the exact width; 7 m and 5.25 m are fractional misreads of the same relation. Common Pitfalls:Attempting to find radii individually (unnecessary); mixing diameter and radius terms in the circumference formula. Final Answer:10.5 meters

Question 2

Outer radius from inner circumference and track width: A circular race track is 14 m wide. The inner circumference is 440 m. What is the radius of the outer circle?

Accepted Answer

Introduction / Context:Concentric circles model circular tracks: inner and outer edges are separated by the constant width. With an inner circumference given, we compute the inner radius and add the width to obtain the outer radius. Given Data / Assumptions: Track width w = 14 m Inner circumference C_in = 440 m Inner radius r_in = C_in / (2π) Outer radius r_out = r_in + w Concept / Approach:Use r = C / (2π) to get r_in, then add the width. Using π = 22/7 streamlines the arithmetic and yields an exact result often targeted by test setters. Step-by-Step Solution:r_in = 440 / (2π) = 220 / π ≈ 220 / (22/7) = 70 mr_out = 70 + 14 = 84 m Verification / Alternative check:Check C_out = 2π * 84 ≈ 528π/π? Using 22/7, C_out = 2 * (22/7) * 84 = 528 m, consistent with a 14 m width increment in radius (circumference increases by 2π * 14). Why Other Options Are Wrong:70 m is the inner radius; 56 m and 64 m are too small; 77 m ignores adding full width; only 84 m reflects inner radius plus width. Common Pitfalls:Subtracting rather than adding the width; using diameter in place of radius when applying circumference. Final Answer:84 m

Question 3

Area ratio from radius ratio: The radii of two circles are in the ratio 1 : 3. What is the ratio of their areas?

Accepted Answer

Introduction / Context:For similar figures like circles, linear measures scale by k, while areas scale by k^2. Translating a radius ratio into an area ratio is a fundamental proportional reasoning task. Given Data / Assumptions: r1 : r2 = 1 : 3 Area of a circle A = πr^2 Concept / Approach:If r2 = 3 * r1, then A2/A1 = (π * (3r1)^2) / (π * r1^2) = 9. Hence A1 : A2 = 1 : 9. Always square the scale factor for areas. Step-by-Step Solution:Let r1 = 1k, r2 = 3kA1 = π * (1k)^2 = πk^2A2 = π * (3k)^2 = 9πk^2A1 : A2 = 1 : 9 Verification / Alternative check:Pick numbers: r1 = 1, r2 = 3 → A1 = π, A2 = 9π → ratio 1:9, independent of π or k. Why Other Options Are Wrong:1:3 confuses linear with area scaling; 1:6 and 3:9 are inconsistent; “None” is incorrect since 1:9 is definitive. Common Pitfalls:Not squaring the ratio for areas; mixing up which circle is first in the ratio. Final Answer:1:9

Question 4

Circle on a square’s diagonal: A square has area 50 square units. A circle is drawn with the square’s diagonal as its diameter. Find the circle’s area.

Accepted Answer

Introduction / Context:This mixes properties of squares and circles: from a square’s area, deduce its side and diagonal; that diagonal becomes the circle’s diameter. Then compute the circle’s area from its radius (half the diameter). Given Data / Assumptions: Square area = 50 Side s = √50 Diagonal d = s * √2 Circle diameter = d; radius r = d/2 Concept / Approach:Compute s from area. Use the square’s diagonal relation d = s√2. Halve d for the circle’s radius and apply A_circle = πr^2. Keep results exact in terms of π to avoid rounding errors. Step-by-Step Solution:s = √50d = √50 * √2 = √100 = 10r = d/2 = 10/2 = 5A_circle = π * 5^2 = 25π Verification / Alternative check:Back-check: A_square = 50, so s ≈ 7.071. Diagonal ≈ 10, radius ≈ 5 → area ≈ 78.54, which equals 25π numerically. Why Other Options Are Wrong:50π and 100π double-count; 12.5π halves incorrectly; “None” is false because 25π is exact. Common Pitfalls:Using side as diameter; forgetting the √2 factor for a square’s diagonal; squaring before or after halving incorrectly. Final Answer:25π sq. units

Question 5

Wire circle reshaped to rectangle: A circular wire has radius 42 cm. It is cut and bent into a rectangle whose sides are in the ratio 6 : 5. Find the smaller side of the rectangle.

Accepted Answer

Introduction / Context:Perimeter is conserved when a wire is reshaped. A circle turned into a rectangle retains its total length; with the side ratio given, we can recover the actual sides. Given Data / Assumptions: Circle radius r = 42 cm Rectangle sides ratio = 6 : 5 → sides 6x and 5x Perimeter(rectangle) = Circumference(circle) Concept / Approach:Circumference C = 2 * π * r. Rectangle perimeter P = 2(6x + 5x) = 22x. Equate C and P to solve x, then compute the smaller side 5x. Step-by-Step Solution:C = 2 * π * 42 = 84π cmWith π = 22/7, C = 84 * 22/7 = 264 cm22x = 264 → x = 12 cmSmaller side = 5x = 60 cm Verification / Alternative check:Larger side = 6x = 72 cm; perimeter = 2(72 + 60) = 264 cm which matches the original circumference, confirming correctness. Why Other Options Are Wrong:30 cm and 36 cm misuse the ratio; 72 cm is the larger side; 132 cm is the full half-perimeter, not a side. Common Pitfalls:Setting 6x + 5x equal to circumference instead of 2(6x + 5x); rounding π unnecessarily when a neat 22/7 result exists. Final Answer:60 cm

Question 6

Area of a circular plot from fencing length: A wire of length 88 m fences a circular plot (wire equals the circumference). Find the area of the plot.

Accepted Answer

Introduction / Context:Given the circumference, we can compute the radius and then the area. This is a direct application of circle formulae with careful substitution. Given Data / Assumptions: Circumference C = 88 m r = C / (2π) Area A = πr^2 Concept / Approach:Compute r first, preferably using π = 22/7 for clean arithmetic, then compute A exactly. Step-by-Step Solution:r = 88 / (2π) = 44 / π = 44 / (22/7) = 14 mA = π * 14^2 = π * 196 = 616 m2 (with π = 22/7) Verification / Alternative check:Using π ≈ 3.1416 gives A ≈ 615.75 m2, which rounds to 616 m2, matching the option. Why Other Options Are Wrong:526 and 556 m2 are off from the exact computation; “None” is false since 616 m2 is correct; 600 m2 is a rounded guess, not exact. Common Pitfalls:Forgetting to divide by 2π; squaring 44/π incorrectly; mixing units. Final Answer:616 m2

Question 7

Effect of increasing circumference by 50%: If a circle’s circumference increases by 50%, by what percentage does its area increase?

Accepted Answer

Introduction / Context:Area depends on radius squared, while circumference depends linearly on radius. A percentage change in circumference maps to the same percentage change in radius, and area responds quadratically. Given Data / Assumptions: C = 2πr; A = πr^2 New circumference C′ = 1.5 * C → r′ = 1.5 * r Concept / Approach:Since r scales by 1.5, area scales by (1.5)^2 = 2.25. The fractional increase is 2.25 − 1 = 1.25 → 125% increase. Step-by-Step Solution:Scale factor for r: k = 1.5Area scale factor: k^2 = 2.25Percentage increase: (2.25 − 1) * 100% = 125% Verification / Alternative check:Take r = 2 → C ≈ 12.566; increase 50% → r′ = 3; A from ~12.566 to ~28.274, a rise of ~125%. Why Other Options Are Wrong:50% confuses linear with area; 100% is doubling, not correct here; 225% is the final value vs. original, not the increase; 75% is arbitrary. Common Pitfalls:Not squaring the scale factor; mixing “increase to” vs. “increase by.” Final Answer:125%

Question 8

Perimeter of regular hexagon inscribed in a circle: If a regular hexagon is inscribed in a circle of radius r, what is its perimeter?

Accepted Answer

Introduction / Context:In a regular hexagon inscribed in a circle, each side equals the circle’s radius. This geometric fact allows a fast perimeter computation without trigonometry. Given Data / Assumptions: Regular hexagon with 6 equal sides Inscribed in a circle (circumradius = r) Concept / Approach:Each central angle is 360/6 = 60 degrees, making each chord equal to the radius. Therefore, side length s = r and perimeter P = 6s = 6r. Step-by-Step Solution:s = rP = 6 * s = 6r Verification / Alternative check:Construct equilateral triangles by joining the center to adjacent vertices; each has side r, confirming the chord length equals r. Why Other Options Are Wrong:3r, 9r, and 12r are incorrect multiples; 6πr confuses polygon perimeter with a circle’s circumference formula. Common Pitfalls:Assuming perimeter equals circumference; forgetting the chord property for regular hexagons. Final Answer:6r

Question 9

Ungrazed area with four tethered horses at a square’s corners: Four horses are tethered at the four corners of a square plot of side 63 m, each with a rope just long enough that adjacent horses cannot reach one another. Find the ungrazed area.

Accepted Answer

Introduction / Context:This classic geometry problem involves subtracting grazed circular sectors from the area of a square. “Just cannot reach one another” implies each rope length equals half the side, so the grazed regions touch but do not overlap across the square’s interior. Given Data / Assumptions: Square side a = 63 m Each rope length r = a/2 = 31.5 m Each horse grazes a quarter-circle sector inside the square Concept / Approach:Total grazed area = 4 * (1/4 * π * r^2) = π * r^2. Ungrazed area = area(square) − grazed area. Step-by-Step Solution:Area(square) = a^2 = 63^2 = 3969 m2r = 31.5 m → r^2 = 992.25Grazed = π * 992.25 = (22/7) * 992.25 = 3118.5 m2Ungrazed = 3969 − 3118.5 = 850.5 m2 Verification / Alternative check:Using π ≈ 3.1416 yields ~851.7 m2; standard test convention uses π = 22/7 giving the exact listed value 850.5 m2. Why Other Options Are Wrong:Other values do not match the precise subtraction using quarter-circle sectors with r = 31.5 m and a = 63 m. Common Pitfalls:Assuming r = 63 m instead of 31.5 m; forgetting there are four quarter-circles (which sum to one full circle). Final Answer:850.5 m2

Question 10

Difference between radii from circumferences: Two concentric circles have circumferences 176 m and 132 m. Find the difference between their radii.

Accepted Answer

Introduction / Context:For concentric circles, the difference in circumferences equals 2π times the difference in radii. This allows a one-step computation of the radial gap. Given Data / Assumptions: C1 = 176 m, C2 = 132 m ΔC = 44 m Δr = ΔC / (2π) Concept / Approach:Apply the circumference formula in difference form and divide by 2π. Step-by-Step Solution:Δr = 44 / (2π) = 22 / π = 22 / (22/7) = 7 m Verification / Alternative check:With π ≈ 3.1416, Δr ≈ 7.0 m; both approximations agree. Why Other Options Are Wrong:5 m, 6 m, and 8 m are off; 44 m confuses ΔC with Δr. Common Pitfalls:Dividing by π only (forgetting the factor 2); mixing units or misreading which is the larger circumference. Final Answer:7 meter

Question 11

Diameter 105 cm less than circumference: For a circle, the diameter is 105 cm less than its circumference. Find the diameter.

Accepted Answer

Introduction / Context:This linear relation between circumference and diameter can be solved algebraically using C = πd. Rearranging yields the diameter directly. Given Data / Assumptions: d = diameter C = circumference = πd d = C − 105 Concept / Approach:Substitute C = πd into d = C − 105 to get d = πd − 105 → (π − 1)d = 105 → d = 105 / (π − 1). Use π = 22/7 for an exact result. Step-by-Step Solution:π − 1 = 22/7 − 1 = 15/7d = 105 / (15/7) = 105 * 7 / 15 = 49 cm Verification / Alternative check:C ≈ π * 49 ≈ 153.94 cm; C − d ≈ 104.94 cm, which equals 105 cm with π = 22/7 exactly. Why Other Options Are Wrong:44–52 cm values do not satisfy d = πd − 105 when substituted back. Common Pitfalls:Using radius instead of diameter; subtracting 105 from π instead of from the circumference; numerical rounding too early. Final Answer:49 cm

Question 12

Sector area from radius and central angle: Find the area of a sector of a circle with radius 12 m and central angle 42°.

Accepted Answer

Introduction / Context:Sector area scales with the fraction of the full angle: (θ/360) of the full circle’s area. Plug the radius and given angle directly. Given Data / Assumptions: r = 12 m θ = 42° Area(circle) = πr^2 Concept / Approach:A_sector = (θ/360) * π * r^2. Compute numerically without rounding too early. Step-by-Step Solution:A_sector = (42/360) * π * 12^2 = (7/60) * π * 144= (1008/60) * π = 16.8π ≈ 52.8 sq. meters Verification / Alternative check:Using π = 22/7: 16.8 * 22/7 = 52.8 exactly, matching the option. Why Other Options Are Wrong:26.4 and 39.6 undercount; 79.2 overcounts; 63.0 does not correspond to this angle fraction. Common Pitfalls:Using degrees directly as radians; forgetting to square the radius; mixing up 42 with 72 in mental math. Final Answer:52.8 sq. meters

Question 13

Arc length from radius and angle: In a circle of radius 21 cm, an arc subtends a central angle of 72°. Find the length of this arc.

Accepted Answer

Introduction / Context:Arc length is the same fraction of the full circumference as the angle is of 360°. With radius and angle known, the computation is straightforward. Given Data / Assumptions: r = 21 cm θ = 72° Full circumference = 2πr Concept / Approach:L = (θ/360) * 2πr. Substitute values and simplify. Step-by-Step Solution:L = (72/360) * 2π * 21 = (1/5) * 42π = 8.4π cmWith π = 22/7, L = 8.4 * 22/7 = 26.4 cm Verification / Alternative check:Using π ≈ 3.1416 gives L ≈ 26.39 cm, which rounds to 26.4 cm, matching the option. Why Other Options Are Wrong:Other lengths correspond to different angles or radii; they do not match the 72° fraction of the full circumference. Common Pitfalls:Using degree measure incorrectly (forgetting the /360 factor); using diameter instead of radius. Final Answer:26.4 cm

Question 14

Sector area from arc length and radius: A circle of radius 5 cm subtends an arc of length 3.5 cm. Find the area of the corresponding sector.

Accepted Answer

Introduction / Context:When arc length and radius are known, the sector area can be found without the angle by using the identity L = r * θ (θ in radians) and A = (1/2) * r^2 * θ, which combine to A = (1/2) * r * L. Given Data / Assumptions: r = 5 cm Arc length L = 3.5 cm No need to convert to degrees; work directly with the relation between L and A Concept / Approach:Use A_sector = (1/2) * r * L. This avoids computing the central angle explicitly. Step-by-Step Solution:A = (1/2) * r * L = 0.5 * 5 * 3.5 = 8.75 sq.cms Verification / Alternative check:Compute θ = L / r = 3.5 / 5 = 0.7 rad. Then A = (1/2) * r^2 * θ = 0.5 * 25 * 0.7 = 8.75 sq.cms, confirming the same result. Why Other Options Are Wrong:35 and 17.5 use r*L or r^2*θ incorrectly; 55 is unrelated; 7.00 undercounts by taking 0.4 * r * L instead of 0.5 * r * L. Common Pitfalls:Using degrees instead of radians with L = rθ; forgetting the 1/2 factor in the sector area formula. Final Answer:8.75 sq.cms

Question 15

Rectangle from area and perimeter — find the diagonal: A rectangular carpet has an area of 120 sq m and a perimeter of 46 m. Determine the length of its diagonal (in meters).

Accepted Answer

Introduction / Context:This problem combines two standard rectangle formulas—area and perimeter—to recover the side lengths and then uses the Pythagoras relation to get the diagonal. These are core skills for geometry aptitude questions. Given Data / Assumptions: Area = 120 sq m Perimeter = 46 m → sum of adjacent sides l + w = 23 Rectangle right angle allows diagonal via l^2 + w^2 Concept / Approach:If l + w = 23 and l * w = 120, then l and w are the roots of t^2 − 23t + 120 = 0. After finding l and w, compute diagonal d using d^2 = l^2 + w^2. Step-by-Step Solution:Quadratic: t^2 − 23t + 120 = 0Discriminant = 23^2 − 4*120 = 529 − 480 = 49Roots: (23 ± 7)/2 → 15 and 8 → (l, w) = (15, 8)Diagonal d = sqrt(15^2 + 8^2) = sqrt(225 + 64) = sqrt(289) = 17 m Verification / Alternative check:Perimeter: 2*(15 + 8) = 46 ✓; Area: 15*8 = 120 ✓; Pythagorean triple (8, 15, 17) confirms. Why Other Options Are Wrong:15 m and 16 m are smaller than the longer side; 20 m exceeds required hypotenuse length for these sides. Common Pitfalls:Using 23^2 = l^2 + w^2 by mistake; Pythagoras applies to sides, not their sum. Final Answer:17 m

Question 16

Perimeter from speed and time — rectangular park area: Radhika runs along the boundary of a rectangular park at 12 km/h and completes one full round in 15 minutes. If the park’s length is four times its breadth, find the area of the park in square m…

Accepted Answer

Introduction / Context:Speed and time give distance, which here equals the perimeter of a rectangle. With a length-to-breadth ratio, we can back out the sides and area. Given Data / Assumptions: Speed = 12 km/h Time = 15 minutes = 0.25 h → distance (perimeter) = 12 * 0.25 = 3 km = 3000 m Length = 4 * breadth Concept / Approach:Perimeter P = 2(L + B). With L = 4B → P = 2(5B) = 10B. Solve B from P, then find L and area A = L * B. Step-by-Step Solution:10B = 3000 → B = 300 mL = 4B = 1200 mArea A = 1200 * 300 = 360000 m2 Verification / Alternative check:Perimeter check: 2(1200 + 300) = 3000 m, which matches the lap distance. Why Other Options Are Wrong:360002 is dimensionally incorrect; 3600 m2 mis-scales by a factor of 100. Common Pitfalls:Converting 15 minutes incorrectly or mixing km and meters. Final Answer:360000 m2

Question 17

Midpoint (medial) triangle area fraction: In triangle ABC, points D, E, F are midpoints of BC, CA, and AB, respectively. If area(ΔABC) = 36 sq m, find area(ΔDEF).

Accepted Answer

Introduction / Context:The triangle formed by joining the midpoints of the sides of a triangle is called the medial triangle. It is similar to the original triangle, and its area is a fixed fraction of the original. Given Data / Assumptions: D, E, F are midpoints of BC, CA, AB area(ΔABC) = 36 sq m Standard midpoint (mid-segment) theorem applies Concept / Approach:The medial triangle is similar to the original triangle with linear scale factor 1/2 (each side is half). Area scales as the square of the linear factor → (1/2)^2 = 1/4. Step-by-Step Solution:Scale factor (length) = 1/2Area factor = (1/2)^2 = 1/4area(ΔDEF) = (1/4) * area(ΔABC) = (1/4) * 36 = 9 sq m Verification / Alternative check:Coordinate geometry check: place ABC conveniently and compute midpoints; the determinant-based area confirms the 1/4 rule. Why Other Options Are Wrong:12 sq m equals 1/3, not the correct 1/4; 18 sq m is half; 19 sq m has no basis in the midpoint theorem. Common Pitfalls:Confusing side-halving with area-halving; area reduces by the square of the scale factor. Final Answer:9 sq m

Question 18

Rectangle with area 60 — diagonal plus longer side equals 5 times shorter side: A rectangular carpet has an area of 60 sq m. The diagonal and the longer side together equal 5 times the shorter side. Find the length of the carpet (the longer side).

Accepted Answer

Introduction / Context:This problem blends area, Pythagoras, and a linear relation between diagonal and sides. It is a classic rectangle system intended to test simultaneous reasoning steps. Given Data / Assumptions: Let shorter side = a, longer side = b Area: a * b = 60 Diagonal: d = sqrt(a^2 + b^2) Relation: d + b = 5a Concept / Approach:Use b = 60 / a from area, then substitute into d + b = 5a with d = sqrt(a^2 + (60/a)^2). Solve for a, then compute b. Favor factorable integer triples if they appear. Step-by-Step Solution:Try a = 5 → b = 60/5 = 12d = sqrt(5^2 + 12^2) = sqrt(25 + 144) = 13Check relation: d + b = 13 + 12 = 25 = 5a (with a = 5) ✓Hence longer side = b = 12 m Verification / Alternative check:Area 5 * 12 = 60 ✓; Pythagorean triple (5, 12, 13) matches and satisfies the linear condition. Why Other Options Are Wrong:5 m is the shorter side; 13 m is the diagonal; 14.5 m does not satisfy area and diagonal constraints simultaneously. Common Pitfalls:Mixing up which side is longer or treating d + a = 5b by mistake. Final Answer:12 m

Question 19

Equal-area relation gives altitude on a different side: In triangle ABC, side BC = 10 cm and the altitude from A to BC (AD) is 4.4 cm. If AC = 11 cm, find the altitude from B to AC (i.e., BE).

Accepted Answer

Introduction / Context:The area of a triangle can be expressed using any side as base and its corresponding altitude. Equating these expressions lets us transfer from one altitude to another. Given Data / Assumptions: |BC| = 10 cm Altitude AD to BC = 4.4 cm |AC| = 11 cm Concept / Approach:Area via base BC: A = (1/2) * BC * AD. Also A = (1/2) * AC * BE. Equate and solve BE = (BC * AD) / AC. Step-by-Step Solution:A = (1/2) * 10 * 4.4 = 22 sq cm22 = (1/2) * 11 * BE ⇒ BE = (44) / 11 = 4 cm Verification / Alternative check:Units consistent (cm for lengths, sq cm for area). Both formulas return the same area. Why Other Options Are Wrong:5 cm, 5.5 cm, and 5.6 cm do not satisfy the area equality with given sides. Common Pitfalls:Forgetting that area remains constant regardless of base/altitude choice, or mixing which altitude corresponds to which base. Final Answer:4 cm

Question 20

Trapezium cross-section — solve for depth from area and widths: A canal cross-section is a trapezium. The top width is 10 m, the bottom width is 6 m, and the cross-sectional area is 640 sq m. Find the depth (perpendicular distance between the parall…

Accepted Answer

Introduction / Context:For a trapezium, area = (1/2) * (sum of parallel sides) * height. Here height is the canal depth. Given Data / Assumptions: a = 10 m (top width) b = 6 m (bottom width) Area = 640 sq m Concept / Approach:Let depth = h. Then 640 = (1/2) * (10 + 6) * h = 8h. Solve for h. Step-by-Step Solution:8h = 640 ⇒ h = 80 m Verification / Alternative check:Back-substitute: (1/2)*(16)*80 = 640 ✓ Why Other Options Are Wrong:40 m gives half the area; 160 m doubles the area; 384 m is unrelated. Common Pitfalls:Using difference of widths instead of sum; forgetting the 1/2 factor. Final Answer:80 meters

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