Question 1

The area of a square is equal to the area of a rectangle. The rectangle’s length is 5 cm more than the side of the square, and the rectangle’s breadth is 3 cm less than the side of the square. Find the perimeter of the rectangle in cm, assuming both…

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Introduction / Context: This problem combines algebra with mensuration. You are told a square and a rectangle have equal area, but the rectangle’s sides depend on the square’s side. The correct method is to let the square side be s, write the rectangle dimensions as (s + 5) and (s - 3), equate areas, solve for s, then compute the rectangle perimeter 2 * (l + b). Given Data / Assumptions: Square side = s cm Rectangle length = s + 5 cm Rectangle breadth = s - 3 cm Areas are equal: s^2 = (s + 5)(s - 3) Perimeter of rectangle = 2 * (length + breadth) Concept / Approach: Form an equation using equal areas, solve for s, then substitute to find rectangle sides and perimeter. Ensure the breadth stays positive (s must be greater than 3). Step-by-Step Solution: Step 1: Set up area equality: s^2 = (s + 5)(s - 3) Step 2: Expand RHS: (s + 5)(s - 3) = s^2 + 2s - 15 Step 3: Equate: s^2 = s^2 + 2s - 15 Step 4: Cancel s^2: 0 = 2s - 15 => s = 7.5 Step 5: Rectangle length = 7.5 + 5 = 12.5 cm Step 6: Rectangle breadth = 7.5 - 3 = 4.5 cm Step 7: Perimeter = 2 * (12.5 + 4.5) = 2 * 17 = 34 cm Verification / Alternative check: Square area = 7.5^2 = 56.25. Rectangle area = 12.5 * 4.5 = 56.25. Areas match, so the perimeter computed from these rectangle sides is consistent. Why Other Options Are Wrong: 26 cm or 18 cm: imply smaller side sums that would not preserve the equal-area relationship. 15 cm: too small for a rectangle whose length must exceed the square side by 5. 38 cm: would require larger sides than (12.5, 4.5) while still matching the square area. Common Pitfalls: Students often forget to expand correctly, or incorrectly cancel terms. Another mistake is using perimeter of square instead of rectangle. Also, do not assume s is an integer; s can be fractional as seen here. Final Answer: 34 cm

Question 2

This is a data sufficiency question about a right-angled triangle. Question: What is the height (perpendicular) of the right-angled triangle? Statement A: The area of the triangle is 20 times its base. Statement B: The perimeter of the triangle is e…

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Introduction / Context: Data sufficiency questions test whether given statements provide enough information to answer a question, not whether you can compute many values. Here, you must decide if the height of a right-angled triangle can be uniquely determined. The right-triangle area formula involves base and height directly, while the perimeter statement relates to side lengths but may not uniquely fix the height without additional relationships. Given Data / Assumptions: We need the triangle’s height (perpendicular to the base) Right-angled triangle area formula: area = (1/2) * base * height Statement A: area is 20 times the base Statement B: triangle perimeter equals perimeter of a square of side 10 cm (so perimeter = 40 cm) Concept / Approach: Check each statement independently. A is sufficient if it directly yields height. B is sufficient only if perimeter alone uniquely determines height, which generally it does not for right triangles unless more constraints are provided. Step-by-Step Solution: Step 1: From statement A, area = 20 * base Step 2: Also, area = (1/2) * base * height Step 3: Set equal: (1/2) * base * height = 20 * base Step 4: Cancel base (base is non-zero for a triangle): (1/2) * height = 20 Step 5: height = 40 cm Step 6: From statement B, perimeter = 4 * 10 = 40 cm, but many different right triangles can have perimeter 40 cm, so height is not unique Verification / Alternative check: Statement A gives a single value height = 40 cm regardless of base, so it uniquely determines the height and is sufficient alone. Statement B gives only a perimeter total, which can correspond to multiple sets of sides (and thus multiple heights), so it is insufficient alone. Why Other Options Are Wrong: Only B: perimeter alone does not lock the height uniquely. Both A and B: A already determines height; B is unnecessary. Neither: A clearly gives height by direct area relationship. Either alone: only A works; B does not. Common Pitfalls: Many learners confuse data sufficiency with “can I compute something.” You must check uniqueness. Another mistake is forgetting that base cancels in statement A, making height independent of base value. For statement B, assuming a unique triangle from perimeter alone is incorrect. Final Answer: Only statement A is sufficient

Question 3

A rectangular room is “half as long again as it is broad”, meaning the length is 1.5 times the breadth. The cost of carpeting the floor at ₹5 per sq m is ₹270. The cost of papering the four walls at ₹10 per sq m is ₹1720. A door and 2 windows togeth…

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Introduction / Context: This is a combined mensuration problem using both floor area (for carpeting) and four-walls area (for papering). You must translate costs into areas using the given rates. Then, use the relationship between length and breadth to solve for floor dimensions. Finally, use wall papering area (after adding back openings) to solve for the height using 2 * (l + b) * H. Given Data / Assumptions: Length l = 1.5 * breadth b Carpet rate = ₹5 per sq m, carpet cost = ₹270 Paper rate = ₹10 per sq m, paper cost = ₹1720 Door + 2 windows area (unpapered) = 8 sq m Floor area = l * b Four walls area = 2 * (l + b) * H Concept / Approach: Convert cost to area. Use l = 1.5b in l*b to find b and l. Convert papering cost to papered area, add openings to get total wall area, then solve for height. Step-by-Step Solution: Step 1: Floor area = 270 / 5 = 54 sq m Step 2: l * b = 54 and l = 1.5b Step 3: 1.5b^2 = 54 => b^2 = 36 => b = 6 m Step 4: l = 1.5 * 6 = 9 m Step 5: Papered wall area = 1720 / 10 = 172 sq m Step 6: Total wall area = papered area + openings = 172 + 8 = 180 sq m Step 7: 2 * (l + b) * H = 180 => 2 * (9 + 6) * H = 180 Step 8: 30H = 180 => H = 6 m Verification / Alternative check: Check carpeting: 9 * 6 = 54 sq m, cost = 54 * 5 = ₹270. Check walls: 2*(9+6)*6 = 180 sq m total, minus openings 8 gives 172 sq m papered, cost = 172 * 10 = ₹1720. Everything matches perfectly. Why Other Options Are Wrong: Other dimension sets fail either the floor area 54 sq m or the wall-area-with-openings 180 sq m condition. Some options violate l = 1.5b or produce incorrect costs when recomputed. If any single check (carpet or paper) fails, the dimensions cannot be correct. The correct answer must satisfy all constraints simultaneously, not just one. Common Pitfalls: Common mistakes include interpreting “half as long again” incorrectly (it means 1.5 times, not 0.5 times). Another mistake is forgetting to add back the 8 sq m openings to compute total wall area for the 2(l+b)H formula. Some also confuse floor area with four walls area. Final Answer: b = 6 m, l = 9 m, H = 6 m

Question 4

A man runs around a circular field of radius 50 m at a speed of 12 km/h. Find the total time taken (in minutes) to complete 20 full rounds of the circular field, using pi = 22/7 for calculation consistency.

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Introduction / Context: This question combines circumference of a circle with speed-time-distance conversion. You must compute the distance for one round using circumference = 2*pi*r, multiply by 20 rounds, then divide by speed. The main trick is converting 12 km/h into metres per minute so the units match metres and minutes. Given Data / Assumptions: Radius r = 50 m Speed = 12 km/h Number of rounds = 20 Circumference per round = 2 * pi * r Use pi = 22/7 1 km = 1000 m and 1 hour = 60 minutes Concept / Approach: Total distance = (circumference) * (rounds). Time = distance / speed, with consistent units (metres and minutes). Convert speed from km/h to m/min for easy division. Step-by-Step Solution: Step 1: One round distance = 2 * pi * r = 2 * (22/7) * 50 = 2200/7 m Step 2: Distance for 20 rounds = 20 * (2200/7) = 44000/7 m Step 3: Convert speed: 12 km/h = 12*1000 m / 60 min = 12000/60 = 200 m/min Step 4: Time = (44000/7) / 200 = 44000 / (1400) = 220/7 minutes Verification / Alternative check: 220/7 ≈ 31.43 minutes. At 200 m/min, in about 31.43 minutes, distance covered ≈ 200 * 31.43 ≈ 6286 m, which matches 20 rounds of circumference about 314.16 m each (20 * 314.16 ≈ 6283 m). The slight difference is due to using pi = 22/7, so the result is consistent. Why Other Options Are Wrong: 110/7 min: corresponds roughly to only 10 rounds, not 20. 230/7 min and 200/7 min: come from incorrect speed conversion or circumference calculation. 90/7 min: far too small; would require a much higher speed than 12 km/h. Common Pitfalls: The most common error is forgetting to convert 12 km/h into m/min and directly dividing metres by km/h. Another mistake is using diameter instead of radius in circumference, or forgetting to multiply by 20 rounds. Always keep units consistent before dividing distance by speed. Final Answer: 220/7 min

Question 5

A rectangular lawn measures 55 m by 35 m. Two roads, each 4 m wide, run through the middle of the lawn: one road is parallel to the length and the other is parallel to the breadth. Find the total cost of gravelling the roads at the rate of 75 paise …

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Introduction / Context: This question tests area calculation of intersecting strips (roads) inside a rectangle and then converts area into cost. When two roads cross at the center, simply adding their areas double-counts the overlap region. The correct approach is: area of road along length + area of road along breadth - overlapping area. Then multiply by the gravelling rate, converting paise to rupees properly. Given Data / Assumptions: Lawn dimensions: 55 m by 35 m Two roads, each width = 4 m One road runs along length (55 m), other along breadth (35 m) Overlap area is a 4 m by 4 m square Rate = 75 paise per sq m = ₹0.75 per sq m Concept / Approach: Area of strip = (length of strip) * (width). Total road area = A1 + A2 - overlap. Cost = total area * rate. Step-by-Step Solution: Step 1: Area of road parallel to length = 55 * 4 = 220 sq m Step 2: Area of road parallel to breadth = 35 * 4 = 140 sq m Step 3: Overlap area counted twice = 4 * 4 = 16 sq m Step 4: Total road area = 220 + 140 - 16 = 344 sq m Step 5: Cost = 344 * 0.75 = ₹258 Verification / Alternative check: The combined road area must be less than 220 + 140 = 360 because the 16 sq m overlap should be removed once. 344 fits this logic. Multiplying 344 by ₹0.75 gives ₹258 exactly, which is consistent with a simple three-quarters multiplication. Why Other Options Are Wrong: ₹158 or ₹58: imply much smaller road areas and usually come from forgetting one road or wrong rate conversion. ₹358: often comes from forgetting to subtract the 16 sq m overlap. ₹278: does not match any correct road area at ₹0.75 per sq m. Common Pitfalls: Common mistakes include adding areas without subtracting overlap, or using 75 as rupees instead of paise (₹75) which inflates the cost massively. Another pitfall is using 55 + 35 as length for a single road, which is conceptually wrong because there are two separate strips. Final Answer: ₹258

Question 6

The cost of fencing a square field at ₹20 per metre is ₹10,080. A 3 metre wide pavement is to be laid all along the inside of the fencing (uniform width on all four sides). Find the total cost of laying this pavement at ₹50 per sq m, assuming the pa…

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Introduction / Context: This question tests a two-step mensuration process. First, you use fencing cost to find the square’s perimeter and side. Second, you compute the area of an inside pavement strip of uniform width, which forms a “square ring”: area of outer square minus area of inner square. Finally, convert the pavement area into cost using the given rate per square metre. Given Data / Assumptions: Fencing rate = ₹20 per metre Total fencing cost = ₹10,080 Square perimeter = total cost / rate Pavement width inside = 3 m on each side Pavement rate = ₹50 per sq m Concept / Approach: Find side of outer square from perimeter. Inner square side = outer side - 2*width. Pavement area = outer side^2 - inner side^2. Cost = area * 50. Step-by-Step Solution: Step 1: Perimeter = 10080 / 20 = 504 m Step 2: Side of square = 504 / 4 = 126 m Step 3: Inner square side = 126 - 2*3 = 120 m Step 4: Pavement area = 126^2 - 120^2 = 15876 - 14400 = 1476 sq m Step 5: Cost = 1476 * 50 = ₹73,800 Verification / Alternative check: Width is 3 m on all sides, so subtracting 6 m from the side is correct. The pavement area should be noticeably smaller than the full field area 126^2, and 1476 sq m is a reasonable “ring” area. Multiplying by ₹50 gives ₹73,800, consistent in magnitude. Why Other Options Are Wrong: ₹83,800 or ₹78,800: would require a larger pavement area than 1476 sq m. ₹53,800 or ₹43,800: would require a smaller pavement area, often from using wrong side length or wrong width adjustment. Errors typically come from forgetting to divide perimeter by 4 or subtracting only 3 instead of 6. Common Pitfalls: Common mistakes include treating ₹10,080 as perimeter directly, forgetting the rate ₹20 per metre, computing inner side as 126 - 3 (instead of 126 - 6), or using 2*(126+120)*3 as area without understanding the ring method. The ring method (difference of squares) is clean and reliable here. Final Answer: ₹73,800

Question 7

A rectangular room has length 5.5 m and width 3.75 m. Find the total cost of paving (covering) the floor with slabs at the rate of ₹800 per sq m, assuming the entire floor area is covered and there is no wastage.

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Introduction / Context: This is a direct application of area and cost calculation. For a rectangular floor, area is simply length * width. Once area is found in square metres, multiply by the slab rate (₹ per sq m) to get the total paving cost. The main focus is clean multiplication and keeping units consistent. Given Data / Assumptions: Length = 5.5 m Width = 3.75 m Floor area = length * width Rate of paving = ₹800 per sq m No wastage or cutting loss assumed Concept / Approach: Compute rectangular area in sq m, then multiply by ₹800 per sq m to get total cost in rupees. Step-by-Step Solution: Step 1: Area = 5.5 * 3.75 = 20.625 sq m Step 2: Cost = 20.625 * 800 Step 3: 20.625 * 800 = 20.625 * (8 * 100) = (20.625 * 8) * 100 Step 4: 20.625 * 8 = 165 Step 5: Therefore cost = 165 * 100 = ₹16,500 Verification / Alternative check: Since 800 is a large rate, the cost should be in the tens of thousands. The area is about 20.6 sq m, so 20.6 * 800 is about 16,480, close to ₹16,500. The exact multiplication confirms ₹16,500. Why Other Options Are Wrong: ₹15,500 and ₹14,500: correspond to smaller areas than 20.625 sq m. ₹17,500 and ₹18,500: correspond to larger areas or incorrect multiplication of 800. None match the exact product of 20.625 and 800. Common Pitfalls: Common errors include multiplying 5.5 and 3.75 incorrectly, mixing centimetres with metres, or multiplying by 80 instead of 800. Also, some mistakenly add length and width instead of multiplying them when computing area. Final Answer: ₹16,500

Question 8

The difference between the length and breadth of a rectangle is 23 m. If the perimeter of the rectangle is 206 m, find the area of the rectangle in square metres, keeping the 23 m difference and 206 m perimeter conditions unchanged.

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Introduction / Context: This problem tests solving rectangle dimensions using a system of two linear equations derived from perimeter and difference. Once length and breadth are obtained, area is computed using length * breadth. The core skill is translating words into equations correctly: perimeter gives l + b, and difference gives l - b. Given Data / Assumptions: Length - breadth = 23 m Perimeter = 206 m Perimeter formula: 2 * (l + b) = 206 Area formula: area = l * b Concept / Approach: From perimeter, compute l + b. Combine with l - b to solve for l and b. Then multiply to get area. Step-by-Step Solution: Step 1: From perimeter: 2 * (l + b) = 206 => l + b = 103 Step 2: Also given: l - b = 23 Step 3: Add equations: (l + b) + (l - b) = 103 + 23 => 2l = 126 Step 4: l = 63 m Step 5: Substitute into l + b = 103 => 63 + b = 103 => b = 40 m Step 6: Area = l * b = 63 * 40 = 2520 sq m Verification / Alternative check: Check difference: 63 - 40 = 23 m, correct. Check perimeter: 2*(63+40) = 2*103 = 206 m, correct. Therefore the derived area 63*40 is reliable and must be correct. Why Other Options Are Wrong: 2420, 2320: too low and would require smaller dimension products than 63*40. 2620, 2720: too high and would require larger products while still matching the fixed perimeter and difference, which is not possible. Only one unique rectangle matches both l+b and l-b, so only one area is possible. Common Pitfalls: Common mistakes include forgetting to divide by 2 when converting perimeter to l + b, reversing length and breadth in the difference, or subtracting equations incorrectly. Another error is using area = 2(l+b) which is actually perimeter, not area. Final Answer: 2520 sq m

Question 9

The inner circumference of a circular race track is 440 m. The track is uniformly 14 m wide (so the outer radius is 14 m more than the inner radius). Find the radius of the outer circle in metres, using pi = 22/7 for calculation consistency.

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Introduction / Context: This question tests circle circumference and the concept of a uniform-width track. The inner boundary is a circle whose circumference is given. From circumference = 2*pi*r, you find the inner radius. Then add the track width (14 m) to get the outer radius. The important point is that the track width increases the radius, not the diameter. Given Data / Assumptions: Inner circumference C_inner = 440 m Track width = 14 m Use pi = 22/7 Circumference formula: C = 2 * pi * r Outer radius = inner radius + 14 Concept / Approach: Compute inner radius from 440 = 2*pi*r. Then increase by 14 m to get outer radius. Keep units in metres throughout. Step-by-Step Solution: Step 1: 440 = 2 * (22/7) * r_inner Step 2: 2 * (22/7) = 44/7, so 440 = (44/7) * r_inner Step 3: r_inner = 440 * 7 / 44 = 10 * 7 = 70 m Step 4: r_outer = r_inner + 14 = 70 + 14 = 84 m Verification / Alternative check: The inner radius 70 m gives inner circumference 2*(22/7)*70 = 2*22*10 = 440 m, correct. Since the track is 14 m wide, moving outward increases the radius by 14 m, so 84 m is consistent and must be the outer radius. Why Other Options Are Wrong: 70 m: this is the inner radius, not the outer radius. 44 m or 22 m: arise from confusing circumference with radius or using pi incorrectly. 33 m: not consistent with 440 m circumference at all. Any value not equal to 70 + 14 cannot be correct if inner radius is computed correctly. Common Pitfalls: A common mistake is adding 14 to the diameter instead of the radius, or subtracting instead of adding. Another error is using C = pi*r instead of 2*pi*r. Always compute inner radius first from the given inner circumference and only then adjust by the track width. Final Answer: 84 m

Question 10

The length of a rectangle is twice its breadth. If the length is decreased by 5 cm and the breadth is increased by 5 cm, the area of the rectangle increases by 75 sq cm. Find the original length of the rectangle in cm, keeping the “twice” condition …

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Introduction / Context: This question tests algebraic modelling of rectangle area changes. You are given a relationship between length and breadth (length = 2 * breadth) and a change scenario that increases area. The right approach is to represent breadth as b and length as 2b, write original area and new area after modifications, then use the given area increase of 75 sq cm to form and solve an equation. Given Data / Assumptions: Let breadth = b cm Length = 2b cm New length = 2b - 5 New breadth = b + 5 New area - original area = 75 sq cm Concept / Approach: Original area = (2b)*b = 2b^2. New area = (2b - 5)(b + 5). Expand, subtract original, set equal to 75, solve for b, then compute length = 2b. Step-by-Step Solution: Step 1: Original area = 2b^2 Step 2: New area = (2b - 5)(b + 5) = 2b^2 + 10b - 5b - 25 = 2b^2 + 5b - 25 Step 3: Increase = (2b^2 + 5b - 25) - (2b^2) = 5b - 25 Step 4: Given increase is 75: 5b - 25 = 75 Step 5: 5b = 100 => b = 20 cm Step 6: Length = 2b = 40 cm Verification / Alternative check: Original: l = 40, b = 20, area = 800. New: l = 35, b = 25, area = 875. Increase = 875 - 800 = 75 sq cm, matches exactly, so 40 cm is correct. Why Other Options Are Wrong: 30 cm or 20 cm: would imply smaller breadth values that do not produce a 75 sq cm increase under the given ±5 changes. 10 cm: far too small and would not satisfy the increase condition. 50 cm: would produce a different area change, not 75. Common Pitfalls: Common errors include using (l+5)(b-5) instead of the given changes, expanding incorrectly, or forgetting that length is twice breadth from the start. Another pitfall is solving for length directly without first expressing everything in one variable b. Final Answer: 40 cm

Question 11

The area of a rectangular field is 460 sq m. The length of the rectangle is 15% more than its breadth. Find the breadth of the rectangle in metres, assuming length = 1.15 * breadth and area = length * breadth.

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Introduction / Context: This question tests translating percentage relationships into algebra and solving for dimensions using area. If length is 15% more than breadth, then length = breadth + 15% of breadth = 1.15 * breadth. Substitute this into the area formula area = length * breadth to get a quadratic in breadth, then solve and choose the physically meaningful positive value. Given Data / Assumptions: Area = 460 sq m Length is 15% more than breadth So length l = 1.15 * b Area formula: A = l * b Concept / Approach: Replace l in A = l*b with 1.15b. Then solve 1.15b^2 = 460 for b. Keep units in metres throughout. Step-by-Step Solution: Step 1: l = 1.15b Step 2: Area A = l*b = (1.15b)*b = 1.15b^2 Step 3: 1.15b^2 = 460 Step 4: b^2 = 460 / 1.15 Step 5: 1.15 = 115/100, so 460 / 1.15 = 460 * 100 / 115 = 400 Step 6: b = sqrt(400) = 20 m Verification / Alternative check: If breadth b = 20, then length l = 1.15 * 20 = 23 m. Area = 23 * 20 = 460 sq m, matching perfectly. Also, length is indeed 3 m more than breadth, which is 15% of 20. Why Other Options Are Wrong: 30 m or 40 m or 50 m: would make the area far larger than 460 when multiplied by 1.15 times the breadth. 25 m: would give l = 28.75 and area 718.75, not 460. Only b = 20 satisfies the exact equation 1.15b^2 = 460. Common Pitfalls: Common mistakes include interpreting “15% more” as “add 15 metres” instead of multiplying by 1.15, or using 0.15b instead of 1.15b for the length. Another error is mishandling 460/1.15; converting 1.15 into a fraction (115/100) makes division cleaner and avoids rounding errors. Final Answer: 20 m

Question 12

A circular piece of thin wire is bent and converted into a rhombus of side 11 cm. Assuming the wire length remains the same and there is no wastage, find the diameter of the original circular wire in cm, using pi = 22/7 for calculation consistency.

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Introduction / Context: This question tests the conservation of perimeter (or total wire length) when a wire is reshaped. The wire initially forms a circle, so its length is the circle’s circumference. After reshaping into a rhombus, the same wire length becomes the rhombus perimeter. So, equate circumference of the circle to the perimeter of the rhombus and solve for the circle’s diameter. Given Data / Assumptions: Rhombus side = 11 cm Perimeter of rhombus = 4 * side Circle wire length = circumference = pi * diameter No stretching or wastage, so lengths are equal Use pi = 22/7 Concept / Approach: Compute rhombus perimeter, set it equal to circle circumference, then solve for diameter = (perimeter)/pi. Step-by-Step Solution: Step 1: Perimeter of rhombus = 4 * 11 = 44 cm Step 2: This equals circle circumference: pi * d = 44 Step 3: d = 44 / pi Step 4: Using pi = 22/7, d = 44 / (22/7) = 44 * 7 / 22 Step 5: 44/22 = 2, so d = 2 * 7 = 14 cm Verification / Alternative check: If diameter is 14 cm, circumference = pi*d = (22/7)*14 = 44 cm, exactly matching the rhombus perimeter 44 cm. Since wire length is conserved, this confirms the answer. Why Other Options Are Wrong: 28 cm: would give circumference 88 cm, double the required wire length. 7 cm: would give circumference 22 cm, only half the wire length needed. 3.5 cm: far too small; circumference would be 11 cm. 11 cm: would give circumference about 34.57 cm, not 44 cm. Common Pitfalls: Common mistakes include equating circumference to 2*pi*r but then forgetting to convert to diameter, or using area formulas instead of perimeter/circumference. Another pitfall is assuming the rhombus side equals the circle radius or diameter, which is not implied. Always use the “same wire length” conservation idea. Final Answer: 14 cm

Question 13

For a circle of radius 5 units, compare its area and circumference numerically. The question asks: the area of the circle is numerically what percent of its circumference? (That is, compute (area / circumference) * 100 for a circle of radius 5.)

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Introduction / Context: This question checks understanding of circle formulas and the meaning of “numerically what percent.” You are not comparing units (area is square units and circumference is units), but comparing the numerical values produced by the formulas. The required calculation is (area / circumference) * 100 for radius 5. The pi factor cancels, making the computation simple. Given Data / Assumptions: Radius r = 5 Area A = pi * r^2 Circumference C = 2 * pi * r Percent asked = (A / C) * 100 Concept / Approach: Compute A/C symbolically so pi cancels. Then substitute r = 5 and convert to percent by multiplying by 100. Step-by-Step Solution: Step 1: A = pi * r^2 = pi * 5^2 = 25pi Step 2: C = 2 * pi * r = 2 * pi * 5 = 10pi Step 3: A/C = (25pi) / (10pi) = 25/10 = 2.5 Step 4: Convert to percent: 2.5 * 100 = 250% Verification / Alternative check: Since A/C simplifies to r/2 (because (pi*r^2)/(2*pi*r) = r/2), for r = 5 we get 5/2 = 2.5, which is 250%. This confirms the same result without computing pi explicitly. Why Other Options Are Wrong: 150% and 200%: would correspond to A/C values 1.5 and 2.0, not 2.5. 350% and 450%: too large and would require much larger r or wrong formulas. Only 250% matches the exact ratio r/2 converted to percent for r = 5. Common Pitfalls: Many students confuse area and circumference formulas or forget that percent means multiplying by 100. Another mistake is canceling incorrectly and ending up with 5/10 instead of 25/10. Also, some interpret “percent” as “difference percent” rather than the ratio-based definition given here. Final Answer: 250%

Question 14

A cow is tethered with a rope so that it can graze in a circular region. If the cow must be able to graze an area of 9856 sq m, find the required length of the rope in metres (assume the grazed region is a perfect circle and use pi = 22/7).

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Introduction / Context: This question connects real-life interpretation (rope length) with circle geometry. If the cow is tied to a fixed point, the rope length acts as the radius of the circular grazing area. The area is given, so you use the circle area formula A = pi * r^2 to solve for r. The value of pi is specified (22/7) to avoid approximation differences. Given Data / Assumptions: Grazing area A = 9856 sq m Circle area formula: A = pi * r^2 Rope length = radius r Use pi = 22/7 Concept / Approach: Rearrange A = pi*r^2 to r^2 = A/pi, then compute r by taking square root. The rope length equals the computed radius. Step-by-Step Solution: Step 1: A = pi * r^2 => r^2 = A / pi Step 2: r^2 = 9856 / (22/7) = 9856 * 7 / 22 Step 3: 9856 / 22 = 448, so r^2 = 448 * 7 = 3136 Step 4: r = sqrt(3136) = 56 m Verification / Alternative check: Check by forward calculation: area = (22/7) * 56^2 = (22/7) * 3136. Since 3136/7 = 448, area = 22 * 448 = 9856 sq m, matching exactly. So rope length must be 56 m. Why Other Options Are Wrong: 28 m: would give area about 2464 sq m, too small. 76 m: would give area far larger than 9856 sq m. 16 m and 14 m: produce very small areas and cannot match 9856 sq m. Only 56 m yields exactly 9856 sq m using pi = 22/7. Common Pitfalls: Common errors include using circumference 2*pi*r instead of area, forgetting to divide by pi, or making mistakes in fraction division by (22/7). Also, some compute r^2 correctly but forget to take the square root at the end. Always interpret rope length as the radius of the grazed circle. Final Answer: 56 m

Question 15

An office floor is shaped like a rectangle that measures 13 units by 9 units. Find the area of the office in square units, assuming the given 13 and 9 are the rectangle’s length and breadth.

Accepted Answer

Introduction / Context: This is a direct rectangle area question. When a floor or region is rectangular, its area is found by multiplying its length and breadth. The phrase “square units” means the numeric answer should be expressed as a count of unit squares that can cover the rectangle, so the multiplication result is the area. Given Data / Assumptions: Length = 13 units Breadth = 9 units Rectangle area formula: area = length * breadth Units are generic, so the result is in square units Concept / Approach: Use area = 13 * 9. This is a single-step multiplication problem with no unit conversion needed. Step-by-Step Solution: Step 1: Area = length * breadth Step 2: Area = 13 * 9 Step 3: 13 * 9 = (10 * 9) + (3 * 9) = 90 + 27 = 117 square units Verification / Alternative check: You can verify by reversing: if area is 117 and one side is 13, the other side must be 117/13 = 9, which matches the given breadth. That consistency confirms the multiplication is correct. Why Other Options Are Wrong: 97, 107, 127: are not products of 13 and 9, and do not represent the correct rectangle area. 135: would be 15 * 9 or 13 * 10.38, not 13 * 9. Only 117 is the exact area for a 13-by-9 rectangle. Common Pitfalls: Some learners mistakenly add 13 + 9 instead of multiplying, which gives 22, not an area. Another mistake is confusing area with perimeter (2*(13+9)=44). Always choose multiplication for area of rectangles, and addition-based formula for perimeter. Final Answer: 117

Question 16

A trapezium has two parallel sides of lengths 12 cm and 10 cm. The perpendicular distance (height) between these parallel sides is 14 cm. Find the area of the trapezium in sq cm using the standard trapezium area formula.

Accepted Answer

Introduction / Context: This question tests the standard area formula of a trapezium (also called a trapezoid in some regions). A trapezium has exactly one pair of parallel sides. The area is the average of the lengths of the parallel sides multiplied by the perpendicular distance between them. Since the sides and height are given directly, it is a straightforward substitution problem. Given Data / Assumptions: Parallel side a = 12 cm Parallel side b = 10 cm Height h (distance between parallel sides) = 14 cm Area formula: area = (1/2) * (a + b) * h Concept / Approach: Add the two parallel sides, take half of that sum (which represents the average base length), then multiply by the height. Keep everything in cm so the final unit is sq cm. Step-by-Step Solution: Step 1: a + b = 12 + 10 = 22 Step 2: (1/2) * (a + b) = 22/2 = 11 Step 3: Area = 11 * 14 = 154 sq cm Verification / Alternative check: The average of 12 and 10 is 11, so area should be close to 11 * 14 = 154. This is also between the areas of rectangles 10*14=140 and 12*14=168, which makes sense because a trapezium with bases 10 and 12 must have area between those two rectangle areas. Hence 154 is consistent. Why Other Options Are Wrong: 164 and 186: too large and would exceed the rectangle area with base 12 (168) in this configuration. 121: too small and would require either much smaller height or bases. 144: equals 12*12, a common incorrect multiplication not based on trapezium formula. Common Pitfalls: Common mistakes include forgetting the 1/2 factor, multiplying the two bases directly, or using non-perpendicular distance as height. Another frequent error is treating the trapezium like a triangle with base 12 and height 14, which ignores the second parallel side of length 10. Final Answer: 154 sq cm

Question 17

The area of a circle is 220 sq m.
A square is inscribed in this circle (all 4 vertices of the square lie on the circle).
What will be the area of the inscribed square in sq m?

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Introduction: This question tests the relationship between a circle and an inscribed square. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Using this key geometric fact, we can connect the given area of the circle to the area of the square. The problem is not about guessing values; it is about converting circle area into radius, then using the diagonal-to-side relationship of a square to compute the square’s area. This also checks whether you can move between area formulas and Pythagoras-style relationships without using any advanced geometry. Given Data / Assumptions: Area of circle = 220 sq m Circle area formula: A = pi * r^2 For an inscribed square: diagonal of square = diameter of circle = 2r For a square: diagonal = side * sqrt(2) Concept / Approach: First find r^2 from the circle area. Then use the inscribed-square relation: side = (2r)/sqrt(2) = r*sqrt(2). Finally, square area = side^2 = (r*sqrt(2))^2 = 2*r^2. This avoids unnecessary steps and directly links square area to r^2. Step-by-Step Solution: pi * r^2 = 220r^2 = 220 / piArea of inscribed square = 2 * r^2So square area = 2 * (220 / pi) = 440 / piUsing pi = 22/7: square area = 440 * (7/22) = 140 Verification / Alternative Check: Since the square’s diagonal is the circle’s diameter, the square is “as large as possible” inside the circle. That means its area must be a fixed fraction of the circle’s r^2 value. The derived relation square area = 2*r^2 is standard and confirms consistency. Plugging r^2 = 220/pi gives 440/pi, and with pi = 22/7 it becomes exactly 140, matching one option cleanly. Why Other Options Are Wrong: 150 sq m: would require a slightly larger r^2 than given by the circle area.70 sq m: is half of the correct value, usually from missing the factor 2.49 sq m: unrelated to the geometry; typically a random guess.110 sq m: comes from incorrect diagonal-side conversion. Common Pitfalls: Assuming square side equals circle radius (incorrect).Forgetting diagonal of square equals diameter of circle, not radius.Not squaring correctly when converting side to area.Mixing up pi approximations inconsistently. Final Answer: 140 sq m

Question 18

A right-angled triangle has side lengths 12 cm, 16 cm, and 20 cm.
A new triangle is formed by joining the midpoints of all three sides (the medial triangle). The same midpoint-joining process is repeated on the newly formed triangle, and this conti…

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Introduction: This question tests two ideas together: (1) area of a right triangle using base and height, and (2) how joining midpoints of a triangle creates a similar triangle with a predictable area ratio. The triangle formed by joining midpoints is called the medial triangle. It is similar to the original triangle, with each side half the original length. Since area scales as the square of the scale factor, the medial triangle’s area becomes one-fourth of the previous triangle’s area. Repeating the process creates an infinite geometric series of areas. The goal is to compute the total sum of that infinite series in a controlled, step-by-step way. Given Data / Assumptions: Right triangle sides: 12 cm, 16 cm, 20 cm Right angle is between 12 cm and 16 cm Area of right triangle = (1/2) * base * height Midpoint-joining creates a similar triangle with scale factor 1/2 Area scale factor = (1/2)^2 = 1/4 per step Concept / Approach: Compute the original area first. Then treat the repeated areas as a geometric series: A + A/4 + A/16 + ... with common ratio r = 1/4. Infinite sum = A / (1 - r). This is valid because |r| < 1 so the series converges. Step-by-Step Solution: Original area A0 = (1/2) * 12 * 16 = 96 sq cmEach medial triangle area is 1/4 of the previous: A1 = 96/4 = 24Next: A2 = 24/4 = 6, and so onSo total area sum = 96 + 24 + 6 + ...This is a geometric series with first term a = 96 and common ratio r = 1/4Sum to infinity S = a / (1 - r) = 96 / (1 - 1/4)S = 96 / (3/4) = 96 * (4/3) = 128 sq cm Verification / Alternative Check: Partial sums approach 128: 96 + 24 = 120, plus 6 = 126, plus 1.5 = 127.5, plus 0.375 = 127.875, clearly converging to 128. This confirms the infinite series result is sensible and consistent with repeated quartering of area. Why Other Options Are Wrong: 312 or 412: far too large; would require areas increasing, not shrinking.246 or 192: typically come from adding perimeters or using wrong ratio (like 1/2 instead of 1/4 for area). Common Pitfalls: Using scale factor 1/2 for area instead of 1/4.Forgetting to include the original triangle area in the sum.Using the hypotenuse (20 cm) as base/height in right-triangle area.Applying finite-series formula incorrectly to an infinite series. Final Answer: 128 sq cm

Question 19

In a trapezium, the difference between the lengths of the two parallel sides is 4 cm.
The perpendicular distance (height) between these parallel sides is 19 cm.
If the area of the trapezium is 475 sq cm, find the lengths (in cm) of the two paralle…

Accepted Answer

Introduction: This question tests the area formula of a trapezium and solving a pair of linear equations. A trapezium (trapezoid) has one pair of parallel sides, and its area is based on the average of those parallel sides multiplied by the perpendicular distance between them. Here, the area and height are known, and the difference between the parallel sides is known. That naturally produces two equations: one for the sum (from area) and one for the difference (given directly). Solving those two equations gives the two side lengths uniquely. Given Data / Assumptions: Area of trapezium = 475 sq cm Height (perpendicular distance) = 19 cm Difference between parallel sides = 4 cm Let parallel sides be a and b, where a > b Trapezium area formula: Area = (1/2) * (a + b) * h Concept / Approach: Use the area formula to derive (a + b). Use the given difference to derive (a - b). Then solve by adding and subtracting the equations to isolate a and b. This is a standard “sum and difference” method and avoids trial-and-error. Step-by-Step Solution: 475 = (1/2) * (a + b) * 19(a + b) = (2 * 475) / 19 = 950 / 19 = 50Given: a - b = 4Add equations: (a + b) + (a - b) = 50 + 42a = 54 => a = 27Then b = 50 - 27 = 23 Verification / Alternative Check: Check area using a = 27 and b = 23:
Average of parallel sides = (27 + 23)/2 = 25.
Area = 25 * 19 = 475 sq cm, exactly matches the given area. The difference is also 27 - 23 = 4 cm, matching the condition. So the solution is consistent in both constraints. Why Other Options Are Wrong: 25 and 23 / 24 and 23 / 22 and 23: these do not satisfy (a + b) = 50.28 and 24: difference becomes 4 but sum becomes 52, so area changes. Common Pitfalls: Using wrong area formula (like base*height) instead of trapezium average-base formula.Forgetting the 1/2 factor in trapezium area.Mixing up which side is larger when using a - b = 4.Arithmetic mistake in dividing 950 by 19. Final Answer: 27 cm and 23 cm

Question 20

The length of a rectangular plot is 20 m more than its breadth.
The plot is fenced completely around its boundary. The fencing cost is ₹26.50 per metre, and the total fencing cost is ₹5300.
What is the length of the plot in metres?

Accepted Answer

Introduction: This problem connects perimeter of a rectangle with a real-life cost scenario (fencing). The fencing cost per metre multiplied by the total perimeter gives the total cost. Once we compute the perimeter, we form an equation using the relationship between length and breadth. Because the length is expressed in terms of breadth (length = breadth + 20), the rectangle has only one unknown, so we can solve using basic algebra. This is a common aptitude question that checks whether you correctly translate a word problem into perimeter equations and handle units and cost conversion accurately. Given Data / Assumptions: Fencing rate = ₹26.50 per metre Total fencing cost = ₹5300 Rectangle perimeter P = total fence length (metres) Length L = breadth B + 20 Perimeter formula: P = 2(L + B) Concept / Approach: First compute the perimeter from cost: P = total cost / rate. Then substitute L = B + 20 into P = 2(L + B) to solve for B, and then find L. This is a two-step process: cost-to-perimeter, then perimeter-to-dimensions. Step-by-Step Solution: Perimeter P = 5300 / 26.50 = 200 metresSo 2(L + B) = 200 => L + B = 100Given L = B + 20Substitute: (B + 20) + B = 1002B + 20 = 1002B = 80 => B = 40L = B + 20 = 40 + 20 = 60 Verification / Alternative Check: Check perimeter with L = 60 and B = 40:
P = 2(60 + 40) = 200 m. Cost = 200 * 26.50 = ₹5300, matches perfectly. The difference L - B = 20 m also matches the statement. This confirms the computed length is correct without any ambiguity. Why Other Options Are Wrong: 50 m: would imply breadth 30 m and perimeter 160 m, not matching cost.40 m: would imply breadth 20 m and perimeter 120 m, too small.70 m or 80 m: would create perimeter larger than 200 m, exceeding the given cost. Common Pitfalls: Multiplying cost instead of dividing to get perimeter.Forgetting perimeter uses 2(L + B), not L + B.Using ₹26.50 incorrectly as 26.05 or 265.Mistake in algebra when substituting L = B + 20. Final Answer: 60 m

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