Question 1

If the ratio of the areas of two squares is 9 : 1, find the ratio of their perimeters.

Accepted Answer

Introduction / Context:For similar figures (all squares are similar), linear measures scale as the square root of area ratios. Thus, side ratio equals the square root of the area ratio, and perimeter, being linear in side, shares the same ratio as sides. Given Data / Assumptions: Area ratio A1 : A2 = 9 : 1 Squares ⇒ same shape, so side ratio = sqrt(area ratio) Concept / Approach:If s1 and s2 are sides, then s1^2 : s2^2 = 9 : 1 ⇒ s1 : s2 = 3 : 1. Perimeter P = 4s, hence P1 : P2 = s1 : s2 = 3 : 1. Step-by-Step Solution:s1 : s2 = sqrt(9 : 1) = 3 : 1P1 : P2 = 4s1 : 4s2 = s1 : s2 = 3 : 1 Verification / Alternative check:Take sample squares with sides 3 and 1: areas 9 and 1 (ratio 9:1); perimeters 12 and 4 (ratio 3:1) — consistent. Why Other Options Are Wrong:9:1 is the area ratio, not the perimeter ratio; 3:4 and 1:3 contradict the square-root relation for linear dimensions. Common Pitfalls:Carrying the area ratio directly over to perimeters; always switch to linear scaling via square root for similar shapes. Final Answer:3:1

Question 2

A rectangle’s length is increased by 60% (to 1.6L). By what percentage must its width be decreased to keep the area unchanged?

Accepted Answer

Introduction / Context:Keeping area constant under one-dimensional scaling requires inversely scaling the other dimension. If length increases, width must decrease so that the product length * width remains the same. Given Data / Assumptions: Original dimensions: L and W New length L′ = 1.6L Area constant: L′ * W′ = L * W Concept / Approach:From L′ * W′ = L * W, we get W′ = (L * W) / L′ = W / 1.6 = 0.625W. This corresponds to a 37.5% decrease from W (since 1 − 0.625 = 0.375 = 37.5%). Step-by-Step Solution:Scale factor for width = 1 / 1.6 = 0.625Decrease = (1 − 0.625) × 100% = 37.5% = 37 1/2% Verification / Alternative check:Test with numbers: L = 10, W = 10 ⇒ area = 100. Increase L by 60% ⇒ L′ = 16. Reduce W by 37.5% ⇒ W′ = 6.25. New area = 16 * 6.25 = 100 — unchanged. Why Other Options Are Wrong:60% and 75% overshoot, destroying area equality; 120% is nonsensical for a decrease. Only 37 1/2% preserves equality. Common Pitfalls:Subtracting 60% directly from width or misapplying percentage decrease; always use reciprocal scaling to keep the product fixed. Final Answer:37 1/2%

Question 3

The area of a rectangle is three times the area of a square. The rectangle’s length is 40 cm and its breadth equals (3/2) times the side of the square. Find the side of the square (in cm).

Accepted Answer

Introduction / Context:This problem links the area of a rectangle to the area of a square and gives a proportional relationship between the rectangle’s breadth and the square’s side. Because a rectangle’s area is length * breadth while a square’s area is side^2, we can form an equation and solve for the unknown side of the square directly. Given Data / Assumptions: Rectangle area A_rect = 3 * A_square. Rectangle length L = 40 cm. Rectangle breadth B = (3/2) * s, where s is the square’s side. Square area A_square = s^2. Concept / Approach:Use area definitions. Since A_rect = L * B and A_rect = 3 * s^2, substitute L and B in terms of s to obtain a single equation in s. Solve for s, ensuring s > 0. Step-by-Step Solution:A_rect = L * B = 40 * (3/2 * s) = 60sBut A_rect = 3 * s^2Therefore 3s^2 = 60sFor s > 0, divide by s: 3s = 60 ⇒ s = 20 cm Verification / Alternative check:Square area = 20^2 = 400 cm^2. Rectangle breadth = (3/2)*20 = 30 cm; rectangle area = 40 * 30 = 1200 cm^2. Ratio 1200 : 400 = 3 : 1, as required. Why Other Options Are Wrong:15, 30, and 60 cm do not satisfy 3s^2 = 40 * (3/2 s) consistently; they would give a mismatch in the triple-area condition. Common Pitfalls:Confusing “breadth is (3/2) of side” with “breadth is (2/3) of side,” or erroneously cubing instead of squaring when moving between linear and area measures. Final Answer:20

Question 4

If the radius of a circle is reduced by 50%, by what percentage is the area reduced?

Accepted Answer

Introduction / Context:The area of a circle depends on the square of the radius. Therefore, any percentage change in the radius translates to a squared effect on area. Halving the radius is a common benchmark scenario that sharply reduces the area, and recognizing the squared relationship is key to avoiding arithmetic traps. Given Data / Assumptions: Original radius r. New radius r′ = 0.5r (i.e., reduced by 50%). Area formula: A = πr^2. Concept / Approach:Apply the scale factor on radius to the area. If radius is multiplied by k, area is multiplied by k^2. Here k = 0.5, hence area factor is (0.5)^2 = 0.25. The reduction is thus 1 − 0.25 = 0.75 = 75%. Step-by-Step Solution:Original area A = πr^2New area A′ = π(0.5r)^2 = π * 0.25 r^2 = 0.25APercentage decrease = (A − A′) / A * 100% = (1 − 0.25) * 100% = 75% Verification / Alternative check:Use numbers: let r = 10 ⇒ A = 100π. Halve radius to 5 ⇒ A′ = 25π. Decrease = 75π, which is 75% of 100π — consistent. Why Other Options Are Wrong:65.5%, 39.5%, and 34.5% come from unrelated combinations or linear thinking. Area does not change linearly with radius; it changes with the square of the radius scale factor. Common Pitfalls:Adding or subtracting percentages linearly, or assuming 50% less radius gives 50% less area. The squared dependence makes the effect much larger: 75% reduction. Final Answer:75%

Question 5

The radius of a circle is reduced from 9 cm to 7 cm. Find the percentage decrease in area (use π ≈ 22/7 as needed).

Accepted Answer

Introduction / Context:The area of a circle scales with the square of the radius: A ∝ r^2. Therefore, a change in radius from r1 to r2 changes the area by the ratio r2^2 / r1^2. The resulting percentage decrease follows from 1 − (new/old). Given Data / Assumptions: Initial radius r1 = 9 cm New radius r2 = 7 cm Area is proportional to r^2 Concept / Approach:Compute the ratio of areas using squares of radii and convert to a percentage decrease. Constants like π cancel out, so no evaluation of π is required to get the percentage. Step-by-Step Solution:Old area ∝ 9^2 = 81New area ∝ 7^2 = 49Fraction remaining = 49 / 81 ≈ 0.604938Percentage decrease = (1 − 49/81) * 100% = (32/81)*100% ≈ 39.5% Verification / Alternative check:Approximation: Since 7/9 ≈ 0.777..., area scales by (7/9)^2 ≈ 0.6049; hence drop is about 39.51%, consistent with the exact fraction 32/81 ≈ 0.39506. Why Other Options Are Wrong:31.5% and 34.5% undervalue the squared effect; 65.5% is far too large and would imply the new area is only 34.5% of the old area, which is not true for 7 vs 9 cm radii. Common Pitfalls:Using linear rather than squared scaling or accidentally computing (9 − 7)/9 as the area decrease. Area depends on r^2, not r. Final Answer:39.5 %

Question 6

For a circle, the difference between its circumference and its radius is 37 cm. Find the area of the circle (use π = 22/7).

Accepted Answer

Introduction / Context:This is a mixed-formula circle problem: you are given a linear relation between circumference and radius, and asked for area. The trick is to solve for the radius using the circumference formula and then compute the area using r. Taking π = 22/7 simplifies arithmetic neatly. Given Data / Assumptions: Circumference C = 2πr C − r = 37 cm Use π = 22/7 Concept / Approach:From C − r = 37, substitute C = 2πr to get r(2π − 1) = 37. Evaluate 2π − 1 with π = 22/7; solve for r; then compute area A = πr^2. Step-by-Step Solution:2π − 1 = 2*(22/7) − 1 = 44/7 − 1 = 37/7r(37/7) = 37 ⇒ r = 37 * (7/37) = 7 cmArea A = πr^2 = (22/7) * 7^2 = (22/7) * 49 = 154 sq. cm Verification / Alternative check:Compute C = 2πr = 2*(22/7)*7 = 44 cm. Then C − r = 44 − 7 = 37 cm, matching the condition, confirming r = 7 cm is correct. Why Other Options Are Wrong:148 and 111 sq. cm do not correspond to any integer radius under π = 22/7 that satisfies C − r = 37. 259 sq. cm is too large; it would require a radius > 8 cm, contradicting the relation. Common Pitfalls:Forgetting to subtract r from C, or using decimal π which complicates the arithmetic here. Using π = 22/7 makes the relation 2π − 1 an exact 37/7, giving r = 7 immediately. Final Answer:154 sq. cm

Question 7

The area of a circular field is 13.86 hectares (1 hectare = 10,000 m^2). Find the cost of fencing it at 20 paisa per meter (i.e., Rs 0.20 per meter).

Accepted Answer

Introduction / Context:This question requires converting area units, computing the radius from area, determining the circumference (fencing length), and finally multiplying by a cost rate per meter. Using π = 22/7 will keep numbers tidy if the area is chosen to be a perfect multiple. Given Data / Assumptions: Area A = 13.86 hectares = 13.86 * 10,000 = 138,600 m^2 π = 22/7 (implied by the neat numbers) Cost rate = Rs 0.20 per meter Concept / Approach:From A = πr^2, solve for r: r^2 = A/π. Then circumference C = 2πr gives the total fencing length. Multiply C by the rate to get cost. Step-by-Step Solution:r^2 = 138,600 / (22/7) = 138,600 * 7 / 22 = 44100r = √44100 = 210 mC = 2πr = 2*(22/7)*210 = 1320 mCost = 1320 * 0.20 = Rs 264 Verification / Alternative check:Back-calc area from r = 210: A = (22/7)*210^2 = (22/7)*44,100 = 138,600 m^2 = 13.86 hectares — consistent. Why Other Options Are Wrong:Rs 277.20, Rs 324, and Rs 198 correspond to different circumferences not implied by A = 138,600 m^2 with π = 22/7. Only Rs 264 matches the exact computation. Common Pitfalls:Confusing perimeter with diameter, or forgetting the hectare-to-square-meter conversion, which is crucial here. Also, mixing paisa and rupee values (20 paisa = Rs 0.20) can cause arithmetic slips. Final Answer:Rs. 264

Question 8

The area of a circle is 38.5 square centimeters. Find its circumference (use π = 22/7).

Accepted Answer

Introduction / Context:Given the area of a circle, one can find the radius using A = πr^2 and then compute the circumference using C = 2πr. Using π = 22/7 simplifies fractions to manageable numbers. Given Data / Assumptions: A = 38.5 cm^2 π = 22/7 Formulas: A = πr^2, C = 2πr Concept / Approach:Solve r^2 = A/π, then r = √(A/π). Having r, compute C. Choose the option that exactly matches the computed circumference. Step-by-Step Solution:r^2 = 38.5 * (7/22) = 269.5 / 22 = 12.25r = √12.25 = 3.5 cmC = 2πr = 2*(22/7)*3.5 = 2*11 = 22 cm Verification / Alternative check:Back-calc area from r = 3.5: A = (22/7)*3.5^2 = (22/7)*12.25 = 38.5 cm^2 — consistent with the given data. Why Other Options Are Wrong:6.20 cm and 11 cm are far too small; 121 cm is far too large for a circle of area 38.5 cm^2. Only 22 cm fits the calculations with π = 22/7. Common Pitfalls:Using π ≈ 3.14 here would still give a close number, but 22/7 is intended and yields an exact integer circumference of 22 cm in this crafted problem. Final Answer:22 cm

Question 9

If the circumference of a circle is 352 meters, find its area in square meters (use π = 22/7).

Accepted Answer

Introduction / Context:Circumference gives a direct path to the radius using C = 2πr. Once r is known, the area follows from A = πr^2. With π = 22/7, several common circumference values yield neat, exact radii and areas. Given Data / Assumptions: C = 352 m π = 22/7 Formulas: C = 2πr, A = πr^2 Concept / Approach:Compute r from r = C / (2π). Then square r and multiply by π to get area in m^2. Step-by-Step Solution:r = 352 / (2*(22/7)) = 352 / (44/7) = 352 * 7 / 44 = 56 mA = πr^2 = (22/7) * 56^2 = (22/7) * 3136 = 22 * 448 = 9856 m^2 Verification / Alternative check:Back-calc C: 2πr = 2*(22/7)*56 = 2*176 = 352 m, consistent with the given circumference. Why Other Options Are Wrong:8956, 6589, and 5986 m^2 do not result from the exact radius 56 m under π = 22/7; only 9856 m^2 matches precisely. Common Pitfalls:Mistakes often occur when inverting 2π or mishandling the fraction 22/7. Keeping it symbolic until the last step helps avoid arithmetic slips. Final Answer:9856

Question 10

A square of side x has the same area as a triangle whose base is x. Find the altitude (height) of the triangle in terms of x.

Accepted Answer

Introduction / Context:This is a direct comparison between the area of a square and the area of a triangle. Setting the two areas equal and solving for the triangle’s altitude in terms of the shared base x gives a quick algebraic result. It illustrates how linear dimensions relate when areas are equated across different shapes. Given Data / Assumptions: Square side = x ⇒ A_square = x^2 Triangle base = x; altitude = h Triangle area A_triangle = (1/2) * base * height = (1/2) * x * h Concept / Approach:Equate x^2 with (1/2) * x * h and solve for h. Since x > 0 for a geometric length, divide safely by x to isolate h. Step-by-Step Solution:x^2 = (1/2) * x * hDivide both sides by x (x > 0): x = (1/2) * hh = 2x Verification / Alternative check:Plug h = 2x back: triangle area = 0.5 * x * 2x = x^2, equal to the square’s area as required. Why Other Options Are Wrong:x/2 and x yield areas smaller than x^2; 4x yields an area of 2x^2, which would be too large. Only 2x balances the two areas exactly. Common Pitfalls:Forgetting the 1/2 factor in the triangle area formula or canceling x incorrectly can lead to wrong multiples of x for the altitude. Final Answer:2x

Question 11

In triangle ABC, sides are BC = 5 cm, AC = 12 cm, AB = 13 cm (a 5–12–13 right triangle). Find the length of the altitude drawn from B to AC.

Accepted Answer

Introduction / Context:The sides 5, 12, and 13 form a classic right triangle with hypotenuse 13. Recognizing the right angle at C (since 5^2 + 12^2 = 13^2) allows us to equate the area computed via the two legs to the area computed using the altitude from B to the leg AC. Given Data / Assumptions: AB = 13 cm, AC = 12 cm, BC = 5 cm Right angle at C (5–12–13 triangle) Altitude from B to AC has length h Concept / Approach:Compute area as (1/2) * AC * BC. Also, area = (1/2) * (altitude from B to AC) * AC. Equate the two expressions to solve for the altitude h. Step-by-Step Solution:Area via legs: (1/2) * 12 * 5 = 30 cm^2Area via altitude: (1/2) * h * 12 = 6hSet equal: 6h = 30 ⇒ h = 5 cm Verification / Alternative check:The altitude from the vertex on the hypotenuse to the opposite leg equals the other leg in a right triangle when comparing areas as above; here, h matches BC numerically, which is consistent with the computation. Why Other Options Are Wrong:4, 6, and 7 cm produce areas 24, 36, and 42 cm^2 via (1/2)*h*12, none of which match the true area 30 cm^2 derived from leg multiplication. Common Pitfalls:Dropping the 1/2 factor or using AB (hypotenuse) incorrectly in the area formula. The area based on legs is the most straightforward path here. Final Answer:5 cm

Question 12

An isosceles triangle has perimeter 14 cm. Each lateral side is to the base in the ratio 5 : 4. Find the area of the triangle.

Accepted Answer

Introduction / Context:For an isosceles triangle with equal lateral sides and base in a given ratio, the exact side lengths can be found by introducing a scale factor. Once side lengths are known, the altitude is obtained via the Pythagorean relation on half the base, and area follows as 1/2 * base * altitude. Given Data / Assumptions: Lateral side : base = 5 : 4 Perimeter = 14 cm Let lateral sides be 5k and 5k, and base be 4k Concept / Approach:Use perimeter to find k, then compute the altitude using h = √( (5k)^2 − (4k/2)^2 ). The area is (1/2)*base*h. All arithmetic is exact for integer k here. Step-by-Step Solution:Perimeter: 5k + 5k + 4k = 14k = 14 ⇒ k = 1Sides: 5, 5, 4Altitude to base: h = √(5^2 − (4/2)^2) = √(25 − 4) = √21Area = (1/2) * 4 * √21 = 2√21 cm^2 Verification / Alternative check:Heron’s formula with sides 5, 5, 4 gives area √(7*(7−5)*(7−5)*(7−4)) = √(7*2*2*3) = √84 = 2√21, confirming the same result. Why Other Options Are Wrong:1/2√21 and 3/2√21 are too small; “√212” is malformed and does not represent the correct area; only 2√21 cm^2 is consistent with exact computation. Common Pitfalls:Using the full base rather than the half-base when applying Pythagoras, or misinterpreting the ratio as 5:4:4 for three different sides. Here two sides are equal (isosceles). Final Answer:2√21 cm2

Question 13

A triangular board has sides 13 m, 14 m, and 15 m. Find the cost of painting it at Rs 8.75 per square meter.

Accepted Answer

Introduction / Context:To price painting a triangular board, first compute its area, then multiply by the given rate per square meter. For sides 13, 14, and 15, Heron’s formula gives a clean, integer area, making the cost straightforward to evaluate. Given Data / Assumptions: Sides: a = 13 m, b = 14 m, c = 15 m Rate = Rs 8.75 per m^2 Heron’s formula: Area = √(s(s − a)(s − b)(s − c)) where s = (a + b + c)/2 Concept / Approach:Compute semi-perimeter s, plug values into Heron’s formula to get exact area (no rounding), then multiply by the rate to obtain cost. Step-by-Step Solution:s = (13 + 14 + 15)/2 = 42/2 = 21Area = √(21 * (21 − 13) * (21 − 14) * (21 − 15)) = √(21 * 8 * 7 * 6) = √7056 = 84 m^2Cost = 84 * 8.75 = Rs 735 Verification / Alternative check:Because 13, 14, 15 is a near-Pythagorean set, Heron yields an integer area (84). Multiplying by 8.75 produces an exact currency amount without decimals, as seen here. Why Other Options Are Wrong:Rs 688.80, Rs 730.80, and Rs 722.50 correspond to different (incorrect) areas or rates; only Rs 735 matches area 84 m^2 at Rs 8.75/m^2. Common Pitfalls:Arithmetic slips when computing the square root in Heron’s formula, or rounding intermediate steps leading to off-by-a-few-rupees totals. Keep exact integers to the end. Final Answer:Rs. 735

Question 14

If the diagonal of a square is doubled, how does the area of the square change?

Accepted Answer

Introduction / Context:The area of a square is related to its diagonal by A = (d^2)/2. Thus, the diagonal serves as a direct linear measure whose square is proportional to area. Doubling the diagonal will therefore square the scale factor for area, leading to a fourfold change. Given Data / Assumptions: Original diagonal = d New diagonal = 2d Area formula in terms of diagonal: A = d^2 / 2 Concept / Approach:Compare new area A′ = ( (2d)^2 ) / 2 with old area A = d^2 / 2. Take the ratio A′/A to find the multiplicative change in area after doubling the diagonal. Step-by-Step Solution:A = d^2 / 2A′ = (2d)^2 / 2 = 4d^2 / 2 = 2d^2A′ / A = (2d^2) / (d^2 / 2) = 4 Verification / Alternative check:Let a square have side s; then diagonal d = s√2 and area A = s^2. Doubling d implies the new side is doubled (since s = d/√2), giving area (2s)^2 = 4s^2 = 4A — same conclusion. Why Other Options Are Wrong:Twofold and threefold changes would correspond to diagonal scale factors of √2 and √3, not 2. “None of the above” is incorrect since “four fold” is exactly right. Common Pitfalls:Confusing diagonal doubling with side doubling via a mistaken linear relation. Because area depends on the square of linear scale, doubling any linear dimension multiplies area by 4. Final Answer:Becomes four fold

Question 15

Equal-area circle vs. square — relate side to radius: A circle and a square have the same area. If the square has side s and the circle has radius r, determine the exact ratio s : r.

Accepted Answer

Introduction / Context:Comparing areas of basic shapes is a staple aptitude skill. Here, a circle and a square have the same area; we must express the square’s side in terms of the circle’s radius and simplify to a clean ratio. Given Data / Assumptions: Square side = s Circle radius = r Areas are equal Use standard formulae with π as the circle constant Concept / Approach:Area(square) = s^2. Area(circle) = π * r^2. If the areas are equal, set s^2 = π * r^2 and solve s/r. Reduce to simplest radical form and then write the ratio s : r. Step-by-Step Solution:s^2 = π * r^2Take positive roots for lengths: s = r * √πTherefore, s : r = √π : 1 Verification / Alternative check:If r = 1, the circle’s area is π. A square with side √π has area (√π)^2 = π, matching perfectly, so √π : 1 is consistent. Why Other Options Are Wrong:1 : √π inverts the ratio; 1 : π and π : 1 are dimensionally incorrect for this equality; √2 : 1 is unrelated to equating s^2 and π r^2. Common Pitfalls:Mistaking diameter for radius; forgetting that the square’s area depends on side squared; mishandling radicals when taking square roots. Final Answer:√π : 1

Question 16

Wheel speed from revolutions: A vehicle’s wheel has radius 70 cm. It completes 10 revolutions in 5 seconds. Find the vehicle’s speed in km/h (assume no slipping).

Accepted Answer

Introduction / Context:Converting rotational motion to linear speed is routine in aptitude and physics-based word problems. The core is relating revolutions per second to distance per second using the wheel circumference, then converting units to km/h. Given Data / Assumptions: Radius r = 70 cm = 0.7 m Revolutions: 10 in 5 s → 2 rev/s No slip between wheel and road (distance = circumference * revolutions) Concept / Approach:Circumference C = 2 * π * r. Linear speed v (m/s) = rev/s * C. Convert m/s to km/h by multiplying by 3.6. Step-by-Step Solution:C = 2 * π * 0.7 = 1.4π m ≈ 4.398 mv = 2 * 1.4π ≈ 8.796 m/sConvert to km/h: 8.796 * 3.6 ≈ 31.67 km/h ≈ 31.68 km/hr. Verification / Alternative check:Using π ≈ 22/7 gives C = 1.4 * 22/7 = 4.4 m; speed = 2 * 4.4 = 8.8 m/s; 8.8 * 3.6 = 31.68 km/hr., matching the selected option exactly. Why Other Options Are Wrong:29.46 and 28.80 are underestimates; 36.25 and 32.72 do not arise from any reasonable π approximation with the given data. Common Pitfalls:Forgetting the 3.6 factor, confusing diameter with radius, or using π inconsistently can produce close-but-wrong values. Final Answer:31.68 km/hr.

Question 17

Revolutions from path length: A circular disc with area 0.49π m^2 rolls without slipping through 1.76 km. How many full revolutions does it make?

Accepted Answer

Introduction / Context:To convert a rolling distance into a number of revolutions, we need the wheel’s circumference. When area is given, we must back out the radius first and then use the circumference-to-distance relation. Given Data / Assumptions: Area of disc A = 0.49π m^2 Travel distance L = 1.76 km = 1760 m No slipping, so L = N * circumference Concept / Approach:From A = π r^2, obtain r. Then circumference C = 2 * π * r. Finally, revolutions N = L / C. Step-by-Step Solution:π r^2 = 0.49π → r^2 = 0.49 → r = 0.7 mC = 2 * π * 0.7 = 1.4π ≈ 4.398 mN = 1760 / (1.4π) ≈ 1760 / 4.398 ≈ 400 Verification / Alternative check:Using π = 22/7, C = 1.4 * 22/7 = 4.4 m; N = 1760 / 4.4 = 400 exactly—confirms the computation. Why Other Options Are Wrong:300 and 350 are too few; 600 and 4000 are too many compared to the circumference-derived distance per turn. Common Pitfalls:Mistaking diameter for radius, not converting km to m, or forgetting that area gives r via r = √(A/π). Final Answer:400

Question 18

Wheel rounds over a fixed distance: A wheel of diameter 7/11 m travels 4 km. How many complete revolutions does it make (no slip)?

Accepted Answer

Introduction / Context:Counting revolutions requires the circumference. Here, a nice fraction pairs with the common approximation π = 22/7 to produce an exact, tidy circumference, making arithmetic quick and reliable. Given Data / Assumptions: Diameter d = 7/11 m Distance L = 4 km = 4000 m No slipping; length per revolution = circumference Concept / Approach:Circumference C = π * d. With π ≈ 22/7, C = (22/7) * (7/11) = 2 m exactly. Revolutions N = L / C. Step-by-Step Solution:C = 2 m per revolutionN = 4000 / 2 = 2000 Verification / Alternative check:Even with π to more decimals, the figure remains ~2 m; rounding would still lead precisely to 2000 given the crafted fraction in d. Why Other Options Are Wrong:Values other than 2000 imply a circumference different from ~2 m, contradicting π * d for the given diameter. Common Pitfalls:Using radius instead of diameter in C = πd; failing to convert 4 km into meters. Final Answer:2000

Question 19

Diameter from distance and revolutions: A wheel makes 1000 revolutions to cover 88 km. What is the wheel’s diameter?

Accepted Answer

Introduction / Context:Distance per revolution equals circumference, so dividing total distance by the number of revolutions produces the circumference. From there, the diameter is immediate via C = πd. Given Data / Assumptions: Total distance L = 88 km = 88,000 m Revolutions N = 1000 Circumference C = L / N Concept / Approach:Compute C = 88,000 / 1000 = 88 m. Then d = C / π. Using π ≈ 22/7 yields a clean integer result. Step-by-Step Solution:C = 88 md = 88 / π ≈ 88 / (22/7) = 88 * 7 / 22 = 28 m Verification / Alternative check:Check back: circumference π * 28 ≈ 87.9646 m; 1000 revolutions produce ≈ 87.96 km, which rounds to 88 km when π is approximated by 22/7. Why Other Options Are Wrong:Other diameters yield mismatched circumferences far from 88 m per revolution. Common Pitfalls:Forgetting to convert km to meters; using radius (d/2) accidentally in C = 2πr instead of πd. Final Answer:28 meters

Question 20

Distance from revolutions of a small wheel: The diameter of a wheel is 63 cm. How far (in meters) does it travel in 100 revolutions, assuming no slipping?

Accepted Answer

Introduction / Context:We translate wheel rotations to linear travel using the circumference. With a diameter given in centimeters, careful unit handling ensures the final distance is in meters as requested. Given Data / Assumptions: d = 63 cm = 0.63 m Revolutions N = 100 No slip (distance = N * circumference) Concept / Approach:Use C = π * d (in meters). Distance L = N * C. With π ≈ 22/7, the arithmetic becomes simple. Step-by-Step Solution:C = π * 0.63 ≈ (22/7) * 0.63 = 1.98 mL = 100 * 1.98 = 198 m Verification / Alternative check:Using π ≈ 3.1416 gives C ≈ 1.976 m and L ≈ 197.6 m; rounded to the nearest listed value, 198 m is correct. Why Other Options Are Wrong:99 m and 63 m correspond to using radius or single-turn distance; 136 m and 158.4 m do not match any consistent π approximation for 100 turns. Common Pitfalls:Leaving units in centimeters or using 2πr inconsistently with a diameter input; rounding too early per revolution. Final Answer:198 meters

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