Question 1

A rectangular park is 60 m long and 40 m wide. Two concrete roads cross each other at the center, one running along the length and the other along the width, each having the same uniform width. If the remaining lawn area is 2109 m^2, find the width …

Accepted Answer

Introduction / Context: This question tests area subtraction and handling overlap. When two roads cross at the center, their combined area is not simply added because the central square (the overlap region) is counted twice. So we compute total park area, subtract lawn area to get road area, then set up the correct expression for the union of two rectangles with an overlap. Given Data / Assumptions: Park length L = 60 m Park width W = 40 m Total park area = L*W = 60*40 Lawn area (remaining) = 2109 m^2 Let road width = x m (same for both roads) Concept / Approach: Area covered by road along length = 60*x. Area covered by road along width = 40*x. Overlap at center is a square of area x*x, which gets counted twice in 60x + 40x, so subtract it once. Thus, road area = 60x + 40x - x^2. Step-by-Step Solution: Total park area = 60*40 = 2400 m^2 Road area = Total area - Lawn area = 2400 - 2109 = 291 m^2 Set up: 60x + 40x - x^2 = 291 100x - x^2 = 291 x^2 - 100x + 291 = 0 Factor: (x - 3)(x - 97) = 0 x = 3 m or x = 97 m, but 97 m is impossible for a 40 m wide park So road width x = 3 m Verification / Alternative check: If x = 3, road area = 60*3 + 40*3 - 3^2 = 180 + 120 - 9 = 291 m^2. Lawn area = 2400 - 291 = 2109 m^2, exactly as given. Why Other Options Are Wrong: 1 m: road area becomes 60 + 40 - 1 = 99 m^2, lawn would be 2301 m^2. 2 m: road area becomes 120 + 80 - 4 = 196 m^2, lawn would be 2204 m^2. 4 m: road area becomes 240 + 160 - 16 = 384 m^2, lawn would be 2016 m^2. 5 m: road area becomes 300 + 200 - 25 = 475 m^2, lawn would be 1925 m^2. Common Pitfalls: Forgetting to subtract the overlap x^2 once. Subtracting lawn area from the wrong total. Accepting an impossible root that does not fit the park dimensions. Final Answer: Width of each road = 3 m

Question 2

The base of a triangular field is three times its altitude. If the cost of cultivating the field is ₹24.68 per hectare and the total cost paid is ₹333.18, find the base and altitude of the field (in metres). Assume 1 hectare = 10,000 m^2.

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Introduction / Context: This problem combines unit conversion (hectares to square metres), cost-to-area calculation, and triangle area relationships. The key idea is: total cost and rate per hectare tell us the area of the field, and the geometry relationship (base = 3*altitude) lets us solve for the dimensions. Given Data / Assumptions: Cost rate = ₹24.68 per hectare Total cost = ₹333.18 1 hectare = 10,000 m^2 Base b = 3h where h is altitude Area of triangle A = (1/2) * b * h Concept / Approach: First compute area in hectares using Area(hectares) = Total cost / Rate. Convert it to m^2. Then use b = 3h in the triangle area formula to create an equation in h, solve for h, and get b. Step-by-Step Solution: Area in hectares = 333.18 / 24.68 = 13.5 hectares Convert to m^2: A = 13.5 * 10,000 = 135,000 m^2 Given b = 3h Triangle area: A = (1/2) * b * h = (1/2) * (3h) * h = (3/2) * h^2 = 1.5*h^2 So 1.5*h^2 = 135,000 h^2 = 135,000 / 1.5 = 90,000 h = sqrt(90,000) = 300 m b = 3h = 3*300 = 900 m Verification / Alternative check: Using b = 900 and h = 300, area = (1/2)*900*300 = 135,000 m^2 = 13.5 hectares. Cost = 13.5*24.68 = ₹333.18, exactly correct. Why Other Options Are Wrong: Base = 300; Altitude = 900: violates base = 3*altitude. Base = 800; Altitude = 350: base is not 3 times altitude and area would not match 135,000 m^2. Base = 600; Altitude = 400: base is not 3 times altitude and area differs. Base = 750; Altitude = 250: base = 3*250 is 750 (ratio fits), but area = (1/2)*750*250 = 93,750 m^2, not 135,000 m^2. Common Pitfalls: Not converting hectares to m^2. Using rectangle area instead of triangle area. Mixing up the relation and taking altitude as 3 times base. Final Answer: Base = 900 m and Altitude = 300 m

Question 3

A tank is 25 m long, 12 m wide, and 6 m deep. Find the total cost of plastering its four inner walls and the bottom at the rate of 75 paise per square metre (₹0.75 per m^2).

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Introduction / Context: This question tests surface area calculation of a cuboid-like tank and cost computation. Plastering the tank's inner walls and bottom means we include the area of the four side faces plus the bottom face, but not the top (open surface). Then multiply the total area by the cost rate per square metre. Given Data / Assumptions: Length l = 25 m Width w = 12 m Depth/height h = 6 m Rate = 75 paise per m^2 = ₹0.75 per m^2 Plastering includes: 4 walls + bottom (no top) Concept / Approach: Area of four walls = perimeter of base * height = 2(l + w) * h. Bottom area = l*w. Total plaster area = 2(l + w)h + l*w. Cost = total area * rate. Step-by-Step Solution: Area of four walls = 2(l + w) * h = 2(25 + 12) * 6 = 2*37*6 = 444 m^2 Bottom area = l*w = 25*12 = 300 m^2 Total area = 444 + 300 = 744 m^2 Cost = 744 * 0.75 = ₹558 Verification / Alternative check: If you compute each wall separately: two walls are 25*6 each and two are 12*6 each, total walls = 2*150 + 2*72 = 300 + 144 = 444 m^2. Adding bottom 300 gives 744 m^2, consistent. Why Other Options Are Wrong: ₹258 and ₹358: correspond to much smaller plaster area or a missing face in calculation. ₹458: could happen if someone forgets to include the bottom or uses wrong height. ₹608: would imply a total area around 810+ m^2 at ₹0.75, which is not correct here. Common Pitfalls: Including the top surface even though a tank is typically open at the top. Forgetting to add the bottom area. Using 75 as rupees instead of 75 paise (₹0.75). Final Answer: Total plastering cost = ₹558

Question 4

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 square centimetres (cm^2), find its height in centimetres.

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Introduction / Context: This problem tests the basic area formula of a parallelogram: area = base * height. When a relationship between base and height is given (here base is twice the height), you can substitute that relationship into the area formula and solve for the height directly. Given Data / Assumptions: Area A = 72 cm^2 Base b = 2h, where h is the height Area of parallelogram = b * h Concept / Approach: Substitute b = 2h into A = b*h. This gives A = 2h^2, which can be solved by simple algebra to find h. Step-by-Step Solution: A = b*h 72 = (2h)*h 72 = 2h^2 h^2 = 72/2 = 36 h = sqrt(36) = 6 cm Verification / Alternative check: If height is 6 cm, then base = 2*6 = 12 cm. Area = 12*6 = 72 cm^2, matching the given area. Why Other Options Are Wrong: 3 cm: would give base 6 cm, area 18 cm^2. 4 cm: would give base 8 cm, area 32 cm^2. 8 cm: would give base 16 cm, area 128 cm^2. 12 cm: would give base 24 cm, area 288 cm^2. Common Pitfalls: Using triangle formula (1/2*b*h) by mistake. Forgetting that base is twice height and swapping them. Making an error when taking square root. Final Answer: Height = 6 cm

Question 5

Two squares are given such that the diagonal of the larger square is exactly double the diagonal of the smaller square. What is the ratio of their areas (larger : smaller)?

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Introduction / Context: This question checks proportional reasoning in geometry. For similar shapes (like squares), if a linear dimension scales by a factor k, the area scales by k^2. The diagonal of a square is a linear measure, so doubling the diagonal means all linear dimensions (including side length) also double, which makes the area become four times. Given Data / Assumptions: Diagonal of larger square = 2 * (diagonal of smaller square) Squares are similar, so side length scales in the same ratio as diagonal Area of a square is proportional to (side)^2 Concept / Approach: Use scaling: if diagonal ratio is 2:1, then side ratio is also 2:1. Therefore area ratio is (2^2):(1^2) = 4:1. Step-by-Step Solution: Let smaller square have diagonal d and side s Larger square has diagonal 2d, so its side becomes 2s (same scale factor) Smaller area = s^2 Larger area = (2s)^2 = 4s^2 Area ratio (larger:smaller) = 4s^2 : s^2 = 4:1 Verification / Alternative check: Using a concrete example: if smaller side is 5, area is 25. Larger side is 10, area is 100. Ratio 100:25 simplifies to 4:1, confirming the result. Why Other Options Are Wrong: 2:1: would be true only for a linear measure, not for area. 2:3 and 3:1: do not match square scaling rules. 5:1: would require a scale factor sqrt(5), not 2. Common Pitfalls: Assuming area scales the same way as diagonal (linear) instead of squared. Mixing up diagonal ratio with area ratio. Not recognizing that squares are similar shapes. Final Answer: Required ratio of areas = 4:1

Question 6

A room is 5 m 55 cm long and 3 m 74 cm broad. It is to be paved with identical square tiles such that no tile is cut. Find the least number of square tiles required to completely cover the floor.

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Introduction / Context: This question tests the idea of using the greatest possible square tile so that the number of tiles is minimized and no cutting is needed. The side of the largest square tile must exactly divide both the room's length and breadth. Therefore, the tile side is the HCF (GCD) of the two dimensions (after converting to the same unit). Then the number of tiles is (length/tile_side) * (breadth/tile_side). Given Data / Assumptions: Length = 5 m 55 cm = 555 cm Breadth = 3 m 74 cm = 374 cm Tile side = HCF(555, 374) cm Number of tiles = (555/t) * (374/t) Concept / Approach: Convert all lengths to centimetres first. Compute HCF (GCD) to get the largest possible tile side. Then compute how many tiles fit along each dimension, and multiply to get total tiles required. Step-by-Step Solution: Convert: 5 m 55 cm = 5*100 + 55 = 555 cm Convert: 3 m 74 cm = 3*100 + 74 = 374 cm Compute HCF(555, 374). Since 555 and 374 share no common factor greater than 1, HCF = 1 cm So the largest tile is 1 cm by 1 cm Tiles along length = 555/1 = 555 Tiles along breadth = 374/1 = 374 Total tiles = 555*374 = 207570 Verification / Alternative check: Check divisibility: a 1 cm tile obviously divides any whole centimetre measurement. Because indicating 'least number' forces us to use the largest tile, and the HCF is 1 cm here, there is no larger square tile that will fit both dimensions exactly without cutting. Why Other Options Are Wrong: 205740, 210000, 198000, 220590: these totals do not equal 555*374 and would correspond to incorrect dimensions or using a tile size that does not exactly divide both 555 and 374. Common Pitfalls: Not converting metres and centimetres into the same unit before taking HCF. Using LCM instead of HCF. Assuming a 'nice' tile size without checking exact divisibility. Final Answer: Least number of tiles required = 207570

Question 7

Find the area of a right-angled triangle whose base is 12 cm and whose hypotenuse is 13 cm. Use the Pythagoras theorem to find the missing perpendicular side before computing the area.

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Introduction / Context: This is a classic right-triangle area question. When you are given the base and hypotenuse, you usually need to find the remaining perpendicular side using the Pythagoras theorem. After that, area is computed using (1/2) * base * height, where 'height' is the perpendicular side to the base. Given Data / Assumptions: Base b = 12 cm Hypotenuse c = 13 cm Let the perpendicular side be a cm Pythagoras: c^2 = a^2 + b^2 Area = (1/2) * base * perpendicular Concept / Approach: Use Pythagoras theorem to find a: a^2 = c^2 - b^2. Then compute area using the right triangle area formula. Step-by-Step Solution: c^2 = a^2 + b^2 13^2 = a^2 + 12^2 169 = a^2 + 144 a^2 = 169 - 144 = 25 a = 5 cm Area = (1/2) * 12 * 5 = 30 cm^2 Verification / Alternative check: The sides 5, 12, 13 form a well-known Pythagorean triple. Since the triangle is consistent, the computed area 30 cm^2 is reliable. Why Other Options Are Wrong: 40 cm^2: would require height 6.67 cm for base 12, but the computed height is 5 cm. 50 cm^2: would require height 8.33 cm, not possible with hypotenuse 13. 60 cm^2: would require height 10 cm, which would force hypotenuse > 15. 65 cm^2: does not match (1/2)*12*5. Common Pitfalls: Using hypotenuse as height directly in area formula (incorrect). Forgetting the 1/2 factor in triangle area. Subtracting in the wrong order when computing a^2 = c^2 - b^2. Final Answer: Area = 30 cm^2

Question 8

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 increases by 75 cm^2. Find the original length of the rectangle in centimetres.

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Introduction / Context: This question tests forming algebraic equations from area changes. When dimensions change, the new area can be expressed in terms of the original breadth. The rectangle relationship (length = 2*breadth) reduces the variables and helps solve the problem efficiently. Given Data / Assumptions: Let breadth = b cm Original length = 2b cm New length = 2b - 5 New breadth = b + 5 Increase in area = 75 cm^2 Concept / Approach: Original area = (2b)*b = 2b^2. New area = (2b - 5)(b + 5). The increase condition gives: (2b - 5)(b + 5) - 2b^2 = 75. Solve for b and then compute length = 2b. Step-by-Step Solution: Original area = 2b^2 New area = (2b - 5)(b + 5) Increase: (2b - 5)(b + 5) - 2b^2 = 75 Expand: (2b - 5)(b + 5) = 2b^2 + 10b - 5b - 25 = 2b^2 + 5b - 25 So (2b^2 + 5b - 25) - 2b^2 = 75 5b - 25 = 75 5b = 100 => b = 20 cm Original length = 2b = 40 cm Verification / Alternative check: Original area = 40*20 = 800 cm^2. New dimensions: 35 and 25, new area = 35*25 = 875 cm^2. Increase = 875 - 800 = 75 cm^2, correct. Why Other Options Are Wrong: 30 cm: would imply b = 15; the area change would not be 75. 50 cm: would imply b = 25; then new dimensions 45 and 30 produce a different increase. 60 cm and 70 cm: imply even larger b values and do not satisfy the exact +75 cm^2 condition. Common Pitfalls: Applying +5 and -5 to the wrong dimension. Forgetting that length is twice breadth in the original rectangle. Not subtracting the original area when using the 'increase by 75' statement. Final Answer: Original length = 40 cm

Question 9

The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make to maintain a speed of 66 km/h? (Use pi = 22/7 and keep units consistent.)

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Introduction / Context: This question connects circular motion with speed. Each revolution of the wheel moves the bus forward by one circumference of the wheel. So, revolutions per minute (rpm) can be found by dividing the distance travelled per minute by the wheel's circumference. The key challenge is unit conversion from km/h to m/min (or cm/min) and then consistent use of diameter and pi. Given Data / Assumptions: Diameter d = 140 cm = 1.4 m Speed = 66 km/h pi = 22/7 Circumference = pi * d Revolutions per minute = (distance per minute) / (circumference) Concept / Approach: Convert speed: 66 km/h to m/min. Compute circumference in metres. Then divide m/min by circumference to get rpm. Because pi = 22/7 is given, it yields a clean value. Step-by-Step Solution: Convert speed: 66 km/h = 66,000 m/h Distance per minute = 66,000/60 = 1100 m/min Circumference = pi*d = (22/7)*1.4 1.4 = 14/10, so circumference = (22/7)*(14/10) = (22*2)/10 = 44/10 = 4.4 m rpm = 1100 / 4.4 = 250 revolutions/min Verification / Alternative check: In 1 minute, if the wheel makes 250 revolutions, distance = 250*4.4 = 1100 m. In 1 hour, distance = 1100*60 = 66,000 m = 66 km, matching the given speed 66 km/h. Why Other Options Are Wrong: 150 rpm: would give speed 150*4.4 = 660 m/min => 39.6 km/h. 350 rpm: would give 350*4.4 = 1540 m/min => 92.4 km/h. 550 rpm: would give far higher speed (145.2 km/h). 200 rpm: would give 200*4.4 = 880 m/min => 52.8 km/h. Common Pitfalls: Not converting 140 cm into metres or keeping units mixed. Forgetting to convert hours to minutes when finding distance per minute. Using radius instead of diameter in pi*d for circumference. Final Answer: Required speed needs 250 revolutions per minute

Question 10

A room is 5 m 44 cm long and 3 m 74 cm broad. It is to be paved with identical square tiles such that no tile is cut. Find the least number of square tiles required to cover the floor completely.

Accepted Answer

Introduction / Context: This question checks the use of HCF (GCD) to find the largest possible square tile that can exactly cover a rectangular floor without cutting. Once the tile side is found, the least number of tiles is obtained by counting how many tiles fit along the length and breadth and multiplying those counts. Given Data / Assumptions: Length = 5 m 44 cm = 544 cm Breadth = 3 m 74 cm = 374 cm Tile side = HCF(544, 374) cm Least tiles = (544/t) * (374/t) Concept / Approach: Convert dimensions to centimetres. Find the HCF (GCD) to get the largest tile side. Then compute number of tiles along each direction and multiply to get total tiles. Step-by-Step Solution: Convert length: 5*100 + 44 = 544 cm Convert breadth: 3*100 + 74 = 374 cm HCF(544, 374) = 34 cm Tiles along length = 544/34 = 16 Tiles along breadth = 374/34 = 11 Total tiles = 16*11 = 176 Verification / Alternative check: A 34 cm tile divides both 544 and 374 exactly, so no cutting is needed. Any larger tile would fail to divide at least one dimension, increasing cutting or making paving impossible under the 'no cut' rule. Why Other Options Are Wrong: 136 and 146: too few tiles; would imply a larger tile side that does not divide both dimensions exactly. 166: incorrect multiplication or incorrect HCF. 186: would imply different tile counts that do not match 16 by 11 arrangement. Common Pitfalls: Not converting metres and centimetres to the same unit. Using LCM instead of HCF for tile size. Dividing only one dimension by tile size and forgetting the other. Final Answer: Least number of tiles required = 176

Question 11

Find the area of a triangle whose three sides measure 13 cm, 14 cm, and 15 cm. Use Heron's formula and give the final area in square centimetres (cm^2).

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Introduction / Context: This problem tests Heron's formula, which is used to find the area of a triangle when all three sides are known. It avoids needing the height explicitly. The method is especially useful for scalene triangles (all sides different). The key steps are finding the semi-perimeter and then substituting into the formula. Given Data / Assumptions: Sides: a = 13 cm, b = 14 cm, c = 15 cm Semi-perimeter s = (a + b + c)/2 Heron's formula: Area = sqrt(s(s-a)(s-b)(s-c)) Concept / Approach: Compute s first. Then compute s-a, s-b, s-c. Multiply all four values and take the square root. This gives the area in cm^2. Step-by-Step Solution: s = (13 + 14 + 15)/2 = 42/2 = 21 s - a = 21 - 13 = 8 s - b = 21 - 14 = 7 s - c = 21 - 15 = 6 Area = sqrt(21*8*7*6) 21*8 = 168, 7*6 = 42, so product = 168*42 = 7056 Area = sqrt(7056) = 84 cm^2 Verification / Alternative check: The computed value is exact because 7056 is a perfect square (84^2). So no rounding is needed, and the area 84 cm^2 is precise. Why Other Options Are Wrong: 64 cm^2 and 44 cm^2: result from incorrect semi-perimeter or incorrect multiplication in Heron's formula. 22 cm^2: far too small for a triangle with sides around 14 cm. 91 cm^2: close but incorrect; likely from arithmetic error before taking the square root. Common Pitfalls: Forgetting to divide perimeter by 2 to get semi-perimeter. Using (a+b+c) instead of s in the formula. Making multiplication errors before taking the square root. Final Answer: Area = 84 cm^2

Question 12

The length of a rectangle is halved while its breadth is tripled. Compared to the original rectangle, what is the percentage change in its area? (State whether it increases or decreases.)

Accepted Answer

Introduction / Context: This question tests how area changes when dimensions scale. For a rectangle, area = length * breadth. If length and breadth are multiplied by certain factors, the area is multiplied by the product of those factors. Then the percentage change can be found by comparing new area to old area. Given Data / Assumptions: Original length = L Original breadth = B Original area = L*B New length = L/2 (halved) New breadth = 3B (tripled) Concept / Approach: New area = (L/2) * (3B) = (3/2) * (L*B). This means the new area is 1.5 times the old area, i.e., a 50% increase. Step-by-Step Solution: Original area A1 = L*B New area A2 = (L/2) * (3B) = (3/2) * L*B So A2/A1 = 3/2 = 1.5 Increase factor = 1.5 - 1 = 0.5 Percentage increase = 0.5*100% = 50% increase Verification / Alternative check: Example check: let L = 10, B = 4, old area = 40. New length = 5, new breadth = 12, new area = 60. Increase = 20, and 20/40 = 50% increase, confirming the result. Why Other Options Are Wrong: 20%, 30%, 40% increase: these would require a different scaling factor, not 1.5. 50% decrease: opposite direction; the area increases because tripling breadth dominates halving length. Common Pitfalls: Adding scaling factors instead of multiplying them. Assuming halving one dimension always decreases area, ignoring the other dimension change. Computing 3/2 as 2/3 by mistake. Final Answer: The area shows a 50% increase

Question 13

The length of a classroom floor exceeds its breadth by 25 m. The area remains unchanged when the length is decreased by 10 m and the breadth is increased by 8 m. Find the area of the floor in square metres (m^2).

Accepted Answer

Introduction / Context: This is an algebra-based area invariance problem. When the area of a rectangle remains unchanged after changing length and breadth, you can equate the original area to the new area. The difference relationship between length and breadth (length exceeds breadth by 25 m) helps reduce the variables and makes the equation solvable. Given Data / Assumptions: Let breadth = b m Then length = b + 25 m New length = (b + 25) - 10 = b + 15 m New breadth = b + 8 m Areas are equal: original area = new area Concept / Approach: Original area = b(b+25). New area = (b+15)(b+8). Set them equal and solve for b. Then compute area using the original dimensions. Step-by-Step Solution: Original area: A = b(b + 25) New area: A = (b + 15)(b + 8) Set equal: b(b + 25) = (b + 15)(b + 8) Expand LHS: b^2 + 25b Expand RHS: b^2 + 23b + 120 Equate and simplify: b^2 + 25b = b^2 + 23b + 120 25b - 23b = 120 => 2b = 120 => b = 60 Length = 60 + 25 = 85 Area = 60*85 = 5100 m^2 Verification / Alternative check: New dimensions: length = 85 - 10 = 75, breadth = 60 + 8 = 68. New area = 75*68 = 5100 m^2, same as original, so the condition is satisfied exactly. Why Other Options Are Wrong: 4870 m^2, 4987 m^2, 4442 m^2, 5200 m^2: these do not match the exact area obtained after satisfying both the difference condition and the unchanged-area condition. Common Pitfalls: Subtracting 10 from breadth instead of length, or adding 8 to length instead of breadth. Forgetting to equate areas because the area remains unchanged. Dropping the constant term (15*8 = 120) when expanding RHS. Final Answer: Area of the floor = 5100 m^2

Question 14

While measuring the side of a square, an error of 2% in excess is made (the measured side is 2% more than the true side). What is the percentage error in the calculated area of the square?

Accepted Answer

Introduction / Context: This question tests error propagation in measurement. For a square, area depends on the square of the side. When a quantity is squared, the relative (percentage) error approximately doubles for small errors. Here, because the side is 2% too large, the computed area becomes (1.02)^2 times the true area, leading to a precise percentage error of 4.04%. Given Data / Assumptions: True side = s Measured side (2% excess) = 1.02s True area = s^2 Calculated area = (1.02s)^2 Concept / Approach: Use scaling: calculated area / true area = (1.02s)^2 / s^2 = (1.02)^2. The percentage error is ((1.02)^2 - 1)*100%. Step-by-Step Solution: Measured side = 1.02s Calculated area = (1.02s)^2 = (1.02)^2 * s^2 (1.02)^2 = 1.02*1.02 = 1.0404 So calculated area = 1.0404 * true area Percentage error = (1.0404 - 1)*100% = 0.0404*100% = 4.04% Verification / Alternative check: Quick approximation rule: for area (square), % error ≈ 2*(% error in side) = 2*2% = 4%. The exact value 4.04% is close and confirms correctness with precise calculation. Why Other Options Are Wrong: 1.04%: mistakenly treats 1.0404 as 1.0104 or forgets to convert properly to percent. 2.04% and 3.04%: come from partial multiplication or incorrect doubling rule usage. 4.00%: is only an approximation; the exact computed value is 4.04%. Common Pitfalls: Assuming the area error is the same as side error. Using 1.02^2 incorrectly (e.g., writing 1.04 instead of 1.0404). Forgetting to multiply by 100 to express as percentage. Final Answer: Percentage error in area = 4.04%

Question 15

The ratio between the perimeter and the breadth of a rectangle is 5:1. If the area of the rectangle is 216 cm^2, find the length of the rectangle in centimetres.

Accepted Answer

Introduction / Context: This question tests translating a ratio involving perimeter into an algebraic relation between length and breadth. Perimeter is 2(l + b). When you are given perimeter:breadth = 5:1, you can form an equation linking l and b. Then the given area allows you to solve for the dimensions and identify the required length. Given Data / Assumptions: Perimeter : breadth = 5 : 1 Let length = l cm, breadth = b cm Perimeter P = 2(l + b) Area = l*b = 216 cm^2 Concept / Approach: Convert the ratio to an equation: P/b = 5. Since P = 2(l + b), we get 2(l + b)/b = 5. Solve for l in terms of b, then use area to find b and finally l. Step-by-Step Solution: Given: P : b = 5 : 1 => P/b = 5 P = 2(l + b) So 2(l + b)/b = 5 2l + 2b = 5b 2l = 3b => l = (3/2)b = 1.5b Area: l*b = 216 => (1.5b)*b = 216 1.5b^2 = 216 => b^2 = 216/1.5 = 144 b = 12 cm (positive) l = 1.5*12 = 18 cm Verification / Alternative check: Perimeter = 2(18 + 12) = 60. Ratio P:b = 60:12 = 5:1, correct. Area = 18*12 = 216 cm^2, correct. Why Other Options Are Wrong: 16 cm: with area 216, breadth would be 13.5, giving perimeter ratio not equal to 5:1. 20 cm: breadth would be 10.8, ratio fails. 22 cm: breadth would be 9.82, ratio fails. 24 cm: breadth would be 9, ratio P:b becomes 2(24+9)/9 = 66/9 = 7.33, not 5. Common Pitfalls: Using l:b = 5:1 instead of P:b = 5:1. Forgetting perimeter formula is 2(l + b), not l + b. Making an arithmetic mistake when converting the ratio to an equation. Final Answer: Length of the rectangle = 18 cm

Question 16

Three circles of radius 3.5 cm are placed so that each circle touches the other two externally. Find the area of the region enclosed between the three circles (the curvilinear triangular gap). Use pi = 22/7 and give the answer in cm^2 (rounded suita…

Accepted Answer

Introduction / Context: This problem tests composite area reasoning with circles and an equilateral triangle. When three equal circles touch each other externally, the centers form an equilateral triangle whose side equals the diameter (2r). The gap enclosed between the circles is the area of that equilateral triangle minus the areas of the three 60-degree sectors (one sector from each circle) that lie inside the triangle. Given Data / Assumptions: Radius r = 3.5 cm Diameter = 2r = 7 cm Centers form an equilateral triangle of side a = 7 cm Each sector angle inside the triangle is 60 degrees pi = 22/7 Concept / Approach: Gap area = (area of equilateral triangle) - (area of 3 sectors of 60 degrees). For equilateral triangle: Area = (sqrt(3)/4)*a^2. For sector: (theta/360)*pi*r^2 with theta = 60 degrees. Use pi = 22/7 for the sector calculation and round the final result sensibly. Step-by-Step Solution: Side of equilateral triangle a = 2r = 7 cm Triangle area = (sqrt(3)/4) * a^2 = (sqrt(3)/4) * 49 (sqrt(3)/4)*49 ≈ 21.2176 cm^2 Area of one 60-degree sector = (60/360) * pi * r^2 = (1/6) * pi * (3.5^2) r^2 = 12.25, so sector = (1/6) * (22/7) * 12.25 12.25 = 49/4, so sector = (1/6) * (22/7) * (49/4) = (1/6) * (22*7/4) = (1/6) * (154/4) = 154/24 ≈ 6.4167 cm^2 Three sectors total = 3 * 6.4167 = 19.25 cm^2 (approx) Gap area ≈ 21.2176 - 19.25 = 1.9676 cm^2 Rounded suitably: 1.967 cm^2 Verification / Alternative check: The gap must be small because three sectors occupy most of the triangle. The computed gap near 2 cm^2 is reasonable for circles of radius 3.5 cm, confirming the scale is correct. Why Other Options Are Wrong: 1.867, 1.767, 1.567 cm^2: result from incorrect sector area or incorrect subtraction count. 2.067 cm^2: too large; typically happens if you under-calculate sector area or forget one sector. Common Pitfalls: Using diameter instead of radius in the sector area formula. Using 120 degrees instead of 60 degrees for the sector angle. Subtracting only one sector instead of three. Final Answer: Area of the enclosed region ≈ 1.967 cm^2

Question 17

A rectangular room has dimensions 10 m (length), 7 m (breadth), and 5 m (height). There are 2 identical doors and 3 windows in the room. Each door is 1 m * 3 m. Among the windows, 1 window is 2 m * 1.5 m and the other 2 windows are 1 m * 1.5 m each.…

Accepted Answer

Introduction / Context: This question tests the standard “area of four walls” concept used in mensuration, along with the practical adjustment of subtracting openings like doors and windows before calculating painting cost. The key is to compute the lateral surface area of a room (four walls only), then subtract the total area covered by doors and windows, and finally multiply by the painting rate per square metre. Given Data / Assumptions: Length l = 10 m, breadth b = 7 m, height h = 5 m 2 doors, each 1 m * 3 m 1 window of 2 m * 1.5 m 2 windows of 1 m * 1.5 m each Painting rate = ₹3 per sq m Paint only four walls (exclude floor and ceiling) Concept / Approach: Area of four walls = 2 * (l + b) * h. Subtract openings area. Cost = (net paintable area) * rate. Step-by-Step Solution: Step 1: Four walls area = 2 * (10 + 7) * 5 = 2 * 17 * 5 = 170 sq m Step 2: Total door area = 2 * (1 * 3) = 6 sq m Step 3: Total window area = (2 * 1.5) + 2 * (1 * 1.5) = 3 + 3 = 6 sq m Step 4: Total openings area = 6 + 6 = 12 sq m Step 5: Net paintable wall area = 170 - 12 = 158 sq m Step 6: Cost = 158 * 3 = ₹474 Verification / Alternative check: The paintable area must be slightly less than 170 sq m because openings are removed. Removing 12 sq m is reasonable for 2 doors and 3 windows, so 158 sq m is consistent. Multiplying by ₹3 gives ₹474, which fits the expected scale. Why Other Options Are Wrong: ₹374: would correspond to net area about 124.67 sq m, which is too small for this room. ₹274: would correspond to net area about 91.33 sq m, far too small. ₹174: would correspond to net area about 58 sq m, impossible here. ₹504: would correspond to net area 168 sq m, meaning almost no openings removed. Common Pitfalls: Common errors include using 2 * l * b (floor area) instead of wall area, forgetting to multiply by height, or subtracting only one door/window instead of all. Another mistake is including the ceiling in painting when the question asks only walls. Final Answer: ₹474

Question 18

In a rectangular field, the ratio of its length to its breadth is 5:4. The breadth is 20 m less than the length. Find the perimeter of the rectangular field in metres, keeping the ratio condition and the 20 m difference condition unchanged.

Accepted Answer

Introduction / Context: This question checks your ability to convert a length:breadth ratio into actual dimensions using a given difference, and then compute the perimeter. The typical approach is to represent length and breadth as proportional values (5x and 4x), use the difference to find x, and then compute perimeter using 2 * (l + b). Given Data / Assumptions: Length:breadth = 5:4 Breadth is 20 m less than length, so l - b = 20 Perimeter of rectangle = 2 * (l + b) Concept / Approach: Let l = 5x and b = 4x. Then l - b = x. Use the 20 m difference to solve for x and find l and b, then apply the perimeter formula. Step-by-Step Solution: Step 1: Let length l = 5x and breadth b = 4x Step 2: Given l - b = 20 => 5x - 4x = 20 => x = 20 Step 3: So l = 5 * 20 = 100 m and b = 4 * 20 = 80 m Step 4: Perimeter = 2 * (l + b) = 2 * (100 + 80) = 2 * 180 = 360 m Verification / Alternative check: Check the difference: 100 - 80 = 20 m, matches the condition. Check the ratio: 100:80 simplifies to 5:4, also correct. Therefore the perimeter computed from these correct dimensions must be correct. Why Other Options Are Wrong: 260 m or 280 m: correspond to much smaller l + b values and do not fit the found dimensions. 320 m or 300 m: imply l + b equals 160 or 150, which would break either the ratio or the 20 m difference. Common Pitfalls: Students sometimes treat 20 m as 20% or add/subtract it from the ratio directly. Another mistake is using area formulas instead of perimeter. Also, some forget to multiply (l + b) by 2 when computing perimeter. Final Answer: 360 m

Question 19

The area of a square and the area of a rectangle are equal. If the side of the square is 40 cm and the length of the rectangle is 64 cm, find the perimeter of the rectangle in cm, assuming both shapes have exactly the same area.

Accepted Answer

Introduction / Context: This problem tests the relationship between area and perimeter across different shapes. You are told that a square and a rectangle have the same area. The square’s side gives you its area directly. Then, using the rectangle’s length, you can compute its breadth from area = length * breadth. Finally, apply the rectangle perimeter formula 2 * (length + breadth). Given Data / Assumptions: Side of square = 40 cm Area of square = side * side Rectangle length = 64 cm Areas are equal, so rectangle area = square area Perimeter of rectangle = 2 * (length + breadth) Concept / Approach: Compute square area. Use it as rectangle area. Find breadth = area / length. Then compute perimeter from length and breadth. Step-by-Step Solution: Step 1: Area of square = 40 * 40 = 1600 sq cm Step 2: Rectangle area = 1600 sq cm (equal areas) Step 3: Breadth of rectangle = area / length = 1600 / 64 = 25 cm Step 4: Perimeter = 2 * (64 + 25) = 2 * 89 = 178 cm Verification / Alternative check: Check area consistency: 64 * 25 = 1600 sq cm, which matches the square’s area. Since the area match is correct, the perimeter calculation using those dimensions is valid. Why Other Options Are Wrong: 187 cm and 194 cm: would imply a larger breadth than 25 cm for the same length and area. 149 cm: would imply a smaller breadth, which would reduce area below 1600 sq cm. 168 cm: corresponds to (64 + 20) * 2, but breadth 20 would give area 1280, not 1600. Common Pitfalls: Common mistakes include using perimeter formulas for area, mixing up square side with rectangle length, or calculating breadth as 64/1600 instead of 1600/64. Also, always keep units consistent in cm throughout the calculation. Final Answer: 178 cm

Question 20

A parallelogram has a base that is twice its height. If the area of the parallelogram is 72 sq cm, find the height of the parallelogram in cm, keeping the “base is twice the height” condition unchanged.

Accepted Answer

Introduction / Context: This is a basic mensuration question about the area of a parallelogram. The formula is area = base * height. Since the base is related to the height (base = 2 * height), you substitute this relationship into the area formula and solve a simple quadratic equation for the height. Given Data / Assumptions: Area of parallelogram = 72 sq cm Base b is twice the height h, so b = 2h Area formula: A = b * h Concept / Approach: Substitute b = 2h into A = b * h to get A = 2h^2. Solve for h using the given area value 72. Step-by-Step Solution: Step 1: A = b * h Step 2: Given b = 2h, so A = (2h) * h = 2h^2 Step 3: 2h^2 = 72 Step 4: h^2 = 72 / 2 = 36 Step 5: h = 6 cm (height is positive) Verification / Alternative check: If h = 6, then base b = 2 * 6 = 12. Area = 12 * 6 = 72 sq cm, exactly matching the given area. So the computed height is correct. Why Other Options Are Wrong: 7 cm: would give area = 2 * 7^2 = 98, not 72. 8 cm: would give area = 2 * 8^2 = 128, not 72. 9 cm: would give area = 2 * 9^2 = 162, not 72. 12 cm: would imply base 24 and area 288, far too large. Common Pitfalls: Students sometimes use the triangle area formula 1/2 * base * height by mistake. Another common error is taking h = 36 instead of sqrt(36). Always remember that h^2 = 36 implies h = 6, not 36. Final Answer: 6 cm

Area Questions