Question 1

Circle — The area increases by 22 sq cm when the radius increases by 1 cm. Find the original radius.

Accepted Answer

Introduction / Context:The change in area of a circle due to a change in radius can be written directly. If radius goes from r to r + 1, the area change is π[(r + 1)^2 − r^2] = π(2r + 1). This gives a linear equation in r. Given Data / Assumptions: ΔA = 22 sq cm Radius increment = 1 cm Use π ≈ 22/7 (makes an exact integer solution here) Concept / Approach:Set π(2r + 1) = 22 and solve for r. Step-by-Step Solution:π(2r + 1) = 22 ⇒ (22/7)(2r + 1) = 222r + 1 = 22 * (7/22) = 7 ⇒ 2r = 6 ⇒ r = 3 cm Verification / Alternative check:Compute areas: A(r) = πr^2 = (22/7)*9 = 198/7; A(r + 1) = (22/7)*16 = 352/7; difference = 154/7 = 22 ✔ Why Other Options Are Wrong:3.5, 3.2, 4, and 6 cm do not satisfy π(2r + 1) = 22. Common Pitfalls:Using circumference difference formulas instead of area, or forgetting (a + b)^2 − a^2 = 2ab + b^2. Final Answer:3 cm

Question 2

Grazing rope — A cow must graze an area of 154 sq m. Find the required rope length (assume a circular patch).

Accepted Answer

Introduction / Context:When an animal is tethered by a rope of length r to a fixed point, the reachable grazing region is a circle of radius r. Thus, the grazing area equals πr^2, allowing us to solve for r given the area. Given Data / Assumptions: Area = 154 sq m π ≈ 22/7 (for neat integers) Area A = πr^2 Concept / Approach:Compute r from r^2 = A / π. With π = 22/7, the numbers are designed to give an integer r. Step-by-Step Solution:r^2 = 154 / (22/7) = 154 * 7 / 22 = 49 ⇒ r = 7 m Verification / Alternative check:Check: A = (22/7) * 49 = 154 sq m, matching the target area. Why Other Options Are Wrong:6, 8, 12, 13 m yield areas not equal to 154 sq m under A = πr^2. Common Pitfalls:Forgetting the relationship A = πr^2 or mixing up circumference with area. Final Answer:7 m

Question 3

Circular track — A cyclist travels one lap (circumference) at 14.4 km/h in 1 min 28 s. Find the field’s area.

Accepted Answer

Introduction / Context:One full lap around a circular field equals its circumference. Given speed and time, we find the distance per lap (circumference), then radius, and finally the area using A = πr^2. Given Data / Assumptions: Speed = 14.4 km/h = 4 m/s Time = 1 min 28 s = 88 s Circumference C = speed * time π ≈ 22/7 for exact result here Concept / Approach:Compute C, then r = C / (2π), and area A = πr^2. Numbers are chosen for a clean r. Step-by-Step Solution:C = 4 * 88 = 352 mr = 352 / (2 * 22/7) = 352 / (44/7) = 352 * 7 / 44 = 56 mA = πr^2 = (22/7) * 56^2 = (22/7) * 3136 = 22 * 448 = 9856 sq m Verification / Alternative check:With r = 56 m, circumference is 2πr = 2 * (22/7) * 56 = 352 m, which matches the distance covered. Why Other Options Are Wrong:7958 and 8842 sq m come from rounding or π ≈ 3.14 without keeping the exact nice fraction; “Cannot be determined” is false because data are sufficient. Common Pitfalls:Not converting km/h to m/s; mixing up diameter and radius in circumference calculations. Final Answer:9856 sq m

Question 4

Time difference on circle — It takes 30 s less along the diameter than around the boundary at speed 30 m/min. Find the circle radius (π = 22/7).

Accepted Answer

Introduction / Context:The walker compares two routes across a circle: the circumference 2πr and the straight diameter 2r. The difference in times at the same speed equals the difference in distances divided by that speed, allowing us to solve for r. Given Data / Assumptions: Speed v = 30 m/min Time difference Δt = 30 s = 0.5 min Distance difference Δd = 2πr − 2r = 2r(π − 1) π = 22/7 Concept / Approach:Use v = distance / time ⇒ Δd = v * Δt. Equate 2r(π − 1) with vΔt and solve for r. Step-by-Step Solution:Δd = 30 * 0.5 = 15 m2r(π − 1) = 15 ⇒ 2r((22/7) − 1) = 15 ⇒ 2r(15/7) = 15(30/7)r = 15 ⇒ r = 15 * 7 / 30 = 3.5 m Verification / Alternative check:Compute distances: circumference ≈ 2 * (22/7) * 3.5 = 22 m; diameter = 7 m; difference = 15 m. Time at 30 m/min: 0.5 min ✔ Why Other Options Are Wrong:All other radii give a different difference than 15 m when plugged into 2r(π − 1). Common Pitfalls:Using 30 s as 30 min or 0.3 min; forgetting that both routes are through the same circle center. Final Answer:3.5 m

Question 5

Rectangle — The area equals 15 times the sum of its length and breadth. If length = 40 m, find the breadth.

Accepted Answer

Introduction / Context:The statement “area = 15 times (length + breadth)” gives a linear equation in the unknown breadth when length is known. Solve it to find the missing side. Given Data / Assumptions: A = 15(L + B) L = 40 m Rectangle area A = L * B Concept / Approach:Equate L * B to 15(L + B) and solve for B with L fixed at 40. Step-by-Step Solution:L * B = 15(L + B) ⇒ 40B = 15(40 + B)40B = 600 + 15B ⇒ 25B = 600 ⇒ B = 24 m Verification / Alternative check:Check: A = 40 * 24 = 960; 15(L + B) = 15(40 + 24) = 15 * 64 = 960 ✔ Why Other Options Are Wrong:25, 28, 32, and 30 m do not satisfy 40B = 600 + 15B. Common Pitfalls:Distributing 15 incorrectly or mixing up A = L + B instead of A = L * B. Final Answer:24 m

Question 6

Rectangle — Difference of sides is 23 m and perimeter is 206 m. Find its area.

Accepted Answer

Introduction / Context:With the perimeter and the difference of sides known, one can reconstruct the two sides by solving simultaneous linear equations, then compute the area as the product of the two sides. Given Data / Assumptions: L − B = 23 Perimeter P = 2(L + B) = 206 ⇒ L + B = 103 Concept / Approach:Sum and difference determine the values: L = ( (L + B) + (L − B) ) / 2; B = ( (L + B) − (L − B) ) / 2. Step-by-Step Solution:L = (103 + 23) / 2 = 63B = (103 − 23) / 2 = 40Area = L * B = 63 * 40 = 2520 m2 Verification / Alternative check:Perimeter check: 2(63 + 40) = 206 ✔; difference 63 − 40 = 23 ✔ Why Other Options Are Wrong:2480 and 2420 come from arithmetic slips; “None” is incorrect because 2520 m2 fits perfectly. Common Pitfalls:Dividing perimeter by 2 twice or mixing up sum/difference formulas. Final Answer:2520 m2

Question 7

Rectangle — Area is 75 sq units and length is 5 more than twice the width. Find the perimeter.

Accepted Answer

Introduction / Context:This is a quadratic relation between width and length because area = L * W and L is a linear function of W. Solve for W, then compute L and the perimeter 2(L + W). Given Data / Assumptions: A = 75 L = 2W + 5 Concept / Approach:Set (2W + 5) * W = 75, solve the quadratic for a positive width, then compute L and the perimeter. Step-by-Step Solution:2W^2 + 5W − 75 = 0Discriminant = 25 + 600 = 625 ⇒ W = (−5 + 25)/4 = 5L = 2*5 + 5 = 15Perimeter = 2(L + W) = 2(15 + 5) = 40 unit Verification / Alternative check:A = 15 * 5 = 75 ✔ Why Other Options Are Wrong:30, 24, and 20 units are typical misreads; 50 unit overestimates. Common Pitfalls:Choosing the negative quadratic root or forgetting perimeter is twice the sum. Final Answer:40 unit

Question 8

Rectangle — Length 30 cm, area 720 sq cm. Increasing only length makes area 1.25 times the original. Find the new perimeter.

Accepted Answer

Introduction / Context:When one side changes and the other remains fixed, area changes proportionally to the changing side. Use the target area to find the new length, then compute the new perimeter. Given Data / Assumptions: Original L = 30 cm Original A = 720 sq cm ⇒ B = 720 / 30 = 24 cm New area A' = 1.25 * 720 = 900 sq cm Only length changes Concept / Approach:New length L' = A' / B. Then perimeter P' = 2(L' + B). Step-by-Step Solution:B = 24 cmL' = 900 / 24 = 37.5 cmP' = 2(37.5 + 24) = 2 * 61.5 = 123 cm Verification / Alternative check:Area check: 37.5 * 24 = 900 sq cm, which is 1.25 × 720. Why Other Options Are Wrong:Other perimeters correspond to incorrect divisions or rounding errors. Common Pitfalls:Changing both sides or misreading 1 1/4 as 1.4 instead of 1.25. Final Answer:123 cm

Question 9

Trapezium — Area is 384 sq cm. Parallel sides are in ratio 3 : 5 and the distance between them (height) is 12 cm. Find the smaller parallel side.

Accepted Answer

Introduction / Context:The area of a trapezium (trapezoid) equals 1/2 * (sum of parallel sides) * height. If the parallel sides are in ratio 3 : 5, we can express them as 3x and 5x and solve for x from the area relation. Given Data / Assumptions: A = 384 sq cm Height h = 12 cm Parallel sides = 3x and 5x Concept / Approach:Use A = (1/2)(a + b)h = (1/2)(3x + 5x) * 12 = (1/2)(8x) * 12 = 48x. Step-by-Step Solution:48x = 384 ⇒ x = 8Smaller side = 3x = 3 * 8 = 24 cm Verification / Alternative check:Area check: (1/2)(24 + 40) * 12 = (1/2)(64) * 12 = 32 * 12 = 384 ✔ Why Other Options Are Wrong:20, 30, and 36 cm do not satisfy the exact ratio/area; 18 cm uses x = 6, not consistent with A = 384. Common Pitfalls:Mixing up which side is smaller; forgetting to halve the sum in the area formula. Final Answer:24 cm

Question 10

Square vs circle — A square has side 5 cm which is 13 cm less than the diameter of a circle. Find the approximate area of the circle.

Accepted Answer

Introduction / Context:Translate the relationship “square side is 13 cm less than the circle’s diameter” to find the circle’s diameter and radius, then compute the circle’s area. We round to the nearest of the provided options. Given Data / Assumptions: Square side = 5 cm Square side = diameter − 13 ⇒ diameter = 5 + 13 = 18 cm Radius r = 9 cm Concept / Approach:Area of a circle A = πr^2. With r = 9 cm, A ≈ π * 81. Using π ≈ 22/7 gives a value close to 255 sq cm. Step-by-Step Solution:A ≈ (22/7) * 81 ≈ 254.57 sq cmRounded to nearby option ≈ 255 sq cm Verification / Alternative check:Using π = 3.1416 gives A ≈ 254.47, which is again closest to 255 among the choices. Why Other Options Are Wrong:245, 235, 265, and 275 sq cm are farther from the true value; 255 is the best approximation. Common Pitfalls:Subtracting 13 from the radius instead of from the diameter; squaring 18 instead of 9 for the area calculation. Final Answer:255 sq cm

Question 11

Perimeter preserved — A square has area 64 sq cm. A circle with the same perimeter (circumference) is formed. Find its area.

Accepted Answer

Introduction / Context:We map a fixed perimeter from a square to a circle. The square perimeter becomes the circle circumference. From the circumference we get the circle radius and hence its area. Given Data / Assumptions: Square area = 64 ⇒ side s = 8 cm Square perimeter = 4s = 32 cm Circle circumference = 32 = 2πr Concept / Approach:From 2πr = 32, get r = 16/π. Then circle area A = πr^2 = π * (16/π)^2 = 256/π sq cm. Step-by-Step Solution:r = 16/πA = π * (256/π^2) = 256/π sq cm Verification / Alternative check:Units consistent; the transformation uses perimeter equality only, not area equality. Why Other Options Are Wrong:Other fractions do not follow from r = 16/π. The exact symbolic form 256/π is required. Common Pitfalls:Using square area as circle area or using diameter instead of circumference to compute r. Final Answer:256/π sq cm

Question 12

Incised square — A circle has circumference 25 cm. Find the side of a square inscribed inside it.

Accepted Answer

Introduction / Context:For a square inscribed in a circle, the square’s diagonal equals the circle’s diameter. If we know the circumference, we can find the radius/diameter and hence the square’s side. Given Data / Assumptions: Circumference C = 25 cm 2πr = 25 ⇒ r = 25/(2π) Diameter D = 2r = 25/π Inscribed square side s = D/√2 = (25/π)/√2 Concept / Approach:Use the relation between an inscribed square and its circumcircle (the given circle). Side is diagonal divided by √2. Step-by-Step Solution:s = (25/π)/√2 = 25/(π√2) cm Verification / Alternative check:Compute the square’s diagonal s√2 = 25/π, matching the circle’s diameter as required. Why Other Options Are Wrong:Other expressions use wrong radicals or constants; only 25/(π√2) follows from geometry. Common Pitfalls:Using the radius instead of the diameter for the square’s diagonal; missing the √2 relationship. Final Answer:25/(π√2) cm

Question 13

Equilateral triangle — Find the altitude when the side is 2√3 cm.

Accepted Answer

Introduction / Context:In an equilateral triangle of side a, the altitude (height) is a * (√3/2). This comes from splitting the triangle into two 30-60-90 right triangles, where the height is the longer leg. Given Data / Assumptions: a = 2√3 cm Altitude h = (√3/2) * a Concept / Approach:Apply the standard altitude formula for equilateral triangles directly. Step-by-Step Solution:h = (√3/2) * (2√3) = (2 * 3) / 2 = 3 cm Verification / Alternative check:Compute area via (√3/4)a^2 = (√3/4)*(12) = 3√3; also via (1/2) * base * height = (1/2)*(2√3)*3 = 3√3, consistent. Why Other Options Are Wrong:√3/2 cm and √3/4 are fragments of the formula; 1/2 cm is unrelated; 2 cm is not consistent with the 30-60-90 ratios. Common Pitfalls:Forgetting to multiply the side by √3/2; confusing altitude with median length without evaluating the factor correctly. Final Answer:3 cm

Question 14

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

Accepted Answer

Introduction / Context:As circumference C = 2πr is linear in r, a 5% increase in circumference means a 5% increase in radius. Area A = πr^2 then increases by the square of the radius factor. Given Data / Assumptions: C increases by 5% ⇒ r increases by 5% Area ∝ r^2 Concept / Approach:If r → 1.05r, then A → (1.05)^2 A. The exact percentage increase is (1.05^2 − 1) * 100%. Step-by-Step Solution:1.05^2 = 1.1025Increase = (1.1025 − 1) * 100% = 10.25% Verification / Alternative check:Small-change estimate gives ~10%; adding the square term (0.25%) yields the precise 10.25%. Why Other Options Are Wrong:10% and 10.5% are approximations; 8.75% and 9.5% are unrelated to the exact square factor. Common Pitfalls:Forgetting that area scales with r^2, not r. Final Answer:10.25 %

Question 15

A wall clock has a minute hand of length 7 cm. Find the area swept by the minute hand in 30 minutes (i.e., in a 180° rotation).

Accepted Answer

Introduction / Context:The tip of a clock’s minute hand traces a circle of radius equal to the hand’s length. The “area swept” in a given time is the sector area corresponding to the angular turn in that time. In 30 minutes, the minute hand turns through 180 degrees (half a revolution), so the swept area is simply half the area of the full circle formed by the minute hand’s rotation. Given Data / Assumptions: Radius r = 7 cm. Angular turn in 30 minutes = 180 degrees = π radians. Circle area formula: A = πr^2. Sector area (θ in radians): A_sector = (θ / (2π)) * πr^2 = (θ/2) * r^2. Concept / Approach:Because 30 minutes is exactly half a turn for the minute hand, the area swept is one-half of the full circle’s area. This avoids any need for degree-to-radian conversion beyond recognizing that 180 degrees is half a circle. Step-by-Step Solution:Full circle area = π * r^2 = π * 7^2 = 49πSwept fraction in 30 minutes = 1/2 of full revolutionArea swept = (1/2) * 49π = 24.5πUsing π = 22/7, area = 24.5 * 22/7 = 77 sq. cm Verification / Alternative check:Compute via sector formula directly with θ = π radians: A_sector = (θ/2) * r^2 = (π/2) * 49 = 24.5π ⇒ 77 with π = 22/7, agreeing with the above. Why Other Options Are Wrong:147 sq. cm is the full circle area (49π) rounded with π ≈ 3; 210 and 154 sq. cm correspond to mismatched multiples of π for different radii, not for 7 cm and a half-turn. Common Pitfalls:Forgetting that 30 minutes is half a rotation; using diameter instead of radius; or taking half the circumference instead of half the area. Final Answer:77 sq. cm

Question 16

A circle is inscribed in an equilateral triangle. The circle’s area is 462 sq. cm (use π = 22/7). Find the perimeter of the equilateral triangle.

Accepted Answer

Introduction / Context:An equilateral triangle has a well-known relation between its side length and the inradius (radius of the inscribed circle). If the inradius is known via the circle’s area, we can back-compute the side length and hence the perimeter of the triangle. The algebra is clean when π = 22/7 and the numbers are crafted to be exact. Given Data / Assumptions: Area of inscribed circle: A_circ = 462 sq. cm. π = 22/7. For equilateral triangle of side a: inradius r = (a√3) / 6. Perimeter P = 3a. Concept / Approach:From A_circ = πr^2, compute r. Then use r = (a√3)/6 to find a. Finally compute P = 3a. The numbers are chosen so that r comes out as a neat radical, and a, P are integers. Step-by-Step Solution:r^2 = A_circ / π = 462 * (7/22) = 147r = √147 = 7√3r = (a√3)/6 ⇒ a = 6r/√3 = 6 * (7√3) / √3 = 42Perimeter P = 3a = 3 * 42 = 126 cm Verification / Alternative check:Compute area from a = 42: inradius r = a√3/6 = 7√3; circle area = πr^2 = (22/7) * 147 = 462 sq. cm, consistent. Why Other Options Are Wrong:42√3 cms (option A) is the triangle’s side expressed differently (not the perimeter). 72.6 and 168 cms do not match the consistent inradius–side relationships. Common Pitfalls:Mixing up inradius with circumradius, or using the area formula of the triangle instead of the inscribed circle when working backward; also forgetting that perimeter is 3a. Final Answer:126 cms

Question 17

A square park has side 100 m (area = 10,000 m^2). The area outside a central circular lawn but inside the square is 8,614 m^2. Find the radius of the circular lawn.

Accepted Answer

Introduction / Context:The problem gives the square’s total area and the square area remaining after removing a centrally placed circular lawn. The difference is the circle’s area. From that, compute the radius by applying the circle area formula. The values are chosen so π = 22/7 yields perfect squares for r^2. Given Data / Assumptions: Square side s = 100 m ⇒ area A_square = 10,000 m^2. Area outside circle but inside square = 8,614 m^2. Hence circle area A_circ = 10,000 − 8,614 = 1,386 m^2. Use π = 22/7 unless noted otherwise. Concept / Approach:Area of a circle is A_circ = πr^2. Thus r^2 = A_circ / π. With π = 22/7, division becomes multiplication by 7/22, often leading to a neat square for r^2 if the problem is crafted cleanly. Step-by-Step Solution:A_circ = 1,386 m^2r^2 = 1,386 * (7/22) = 441r = √441 = 21 m Verification / Alternative check:Back-calc: πr^2 = (22/7)*441 = 22*63 = 1,386 m^2, consistent. Subtracting from 10,000 gives 8,614 m^2 as stated. Why Other Options Are Wrong:31 m and 41 m produce circle areas that do not equal 1,386 m^2 with π = 22/7; only 21 m satisfies the equality. Common Pitfalls:Forgetting to convert the complement area to the circle area, or using π ≈ 3.14 and rounding prematurely. With π = 22/7 the arithmetic is exact. Final Answer:21 m

Question 18

A room measures 5.44 m by 3.75 m and is to be paved with identical square tiles (no cutting). Find the least number of tiles required to exactly cover the floor.

Accepted Answer

Introduction / Context:When covering a rectangular floor with identical square tiles without cutting, the largest possible tile side that fits both dimensions exactly must be used to minimize the number of tiles. This side length equals the greatest common divisor (GCD) of the two side lengths, expressed in the same unit with integers (usually centimeters). Then, the number of tiles equals the area of the floor divided by the area of one tile. Given Data / Assumptions: Room: 5.44 m by 3.75 m. Convert to centimeters: 544 cm by 375 cm. Square tiles must tile both sides exactly (no cutting). Concept / Approach:Find tile side = GCD(544, 375) (in cm). Because the two lengths are coprime (no common factor > 1), the GCD is 1 cm. Thus, the largest square tile that exactly fits both dimensions has side 1 cm. The least number of tiles then equals floor area in cm^2. Step-by-Step Solution:Compute GCD(544, 375): successive remainders → … → 1 ⇒ GCD = 1 cmTile side = 1 cm ⇒ tile area = 1 cm^2Floor area = 544 * 375 = 204,000 cm^2Number of tiles = 204,000 / 1 = 204,000 Verification / Alternative check:Any larger tile side (e.g., 2 cm) fails to divide at least one dimension (375). Therefore, 1 cm is the largest exact-fitting square side, which gives the minimum count. Why Other Options Are Wrong:203,500, 205,000, and 192,000 do not equal the exact area 204,000 cm^2 obtainable only with the maximal exact tile side 1 cm. Common Pitfalls:Confusing “least number of tiles” with “smallest tile side”; here, using the GCD produces the largest allowable tile side that still fits, which minimizes the count. Final Answer:204,000

Question 19

Two similar triangles have a ratio of corresponding sides 3 : 4. Find the ratio of their areas.

Accepted Answer

Introduction / Context:For similar figures, linear dimensions scale by a factor k, areas scale by k^2, and volumes (if any) scale by k^3. For triangles in particular, if the ratio of corresponding sides is known, the area ratio is simply the square of that linear ratio. This property follows from area formulas whose units are squared lengths. Given Data / Assumptions: Triangles are similar; side ratio (triangle 1 : triangle 2) = 3 : 4. Thus the linear scale factor k = 3/4 from the smaller to the larger. Concept / Approach:Area ratio for similar figures is (side ratio)^2. Hence, the area ratio of the first to the second triangle equals (3/4)^2 = 9/16. Step-by-Step Solution:Let corresponding sides be 3x and 4x.Then areas are proportional to (3x)^2 and (4x)^2.Area ratio = 9x^2 : 16x^2 = 9 : 16. Verification / Alternative check:Pick concrete numbers: if sides are 3 and 4 with included equal angle, areas scale by base*height; both base and height scale by the same factor, giving a squared effect and confirming 9 : 16. Why Other Options Are Wrong:4 : 3 and 3 : 4 are linear ratios, not area ratios; √3 : 2 is irrelevant. Common Pitfalls:Confusing linear scaling (k) with area scaling (k^2). Always square the side ratio to get the area ratio for similar plane figures. Final Answer:9 : 16

Question 20

A polygon has 44 diagonals. Find the number of its sides.

Accepted Answer

Introduction / Context:The number of diagonals in a convex n-gon is given by the standard combinatorial formula n(n − 3)/2. This counts all ways to join non-adjacent vertices by a segment, avoiding sides. Given the number of diagonals, invert the formula to find n. Given Data / Assumptions: Diagonals D = 44. Formula: D = n(n − 3)/2 for integer n ≥ 3. Concept / Approach:Set n(n − 3)/2 = 44 ⇒ n(n − 3) = 88. Solve the quadratic equation n^2 − 3n − 88 = 0 and take the positive root. Step-by-Step Solution:n^2 − 3n − 88 = 0Factor: (n − 11)(n + 8) = 0n = 11 (discard n = −8) Verification / Alternative check:Plug back: D = 11(11 − 3)/2 = 11*8/2 = 44, consistent. Why Other Options Are Wrong:n = 9, 10, 12 yield diagonals 27, 35, and 54 respectively, not 44. Common Pitfalls:Using n(n − 1)/2 (which counts all vertex pairs) without subtracting edges; the correct formula removes the n polygon sides, giving n(n − 3)/2. Final Answer:11

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