Question 1

In a rhombus PQRS, each side has length 8 cm and the interior angle ∠PQR is 120°. Find the length (in centimetres) of the diagonal QS of the rhombus.

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Introduction:This question involves properties of a rhombus and the use of trigonometry to find the length of a diagonal. In a rhombus, all four sides are equal, but the diagonals are different and depend on the interior angles. The given information includes the side length and one interior angle, and we are asked to find one of the diagonals, specifically QS. This type of geometry question checks understanding of special quadrilaterals and the relationship between sides, angles, and diagonals. Given Data / Assumptions: • PQRS is a rhombus. • Each side of the rhombus is 8 cm. • ∠PQR = 120°. • We are asked to find the length of diagonal QS. • Diagonal QS is taken to be the shorter diagonal, associated with the acute angle of the rhombus. Concept / Approach:In a rhombus, opposite angles are equal and adjacent angles are supplementary. Given that ∠PQR = 120°, the adjacent angle is 60°. The diagonals of a rhombus bisect the angles and intersect at right angles. As a result, each diagonal can be related to the side length and the angles using trigonometric identities or using known formulas for diagonals in a rhombus. One useful fact is that for a rhombus of side a and one interior angle θ, the lengths of the diagonals d1 and d2 are connected to the sides and angle by formulas involving 2 ± 2 * cosθ. For the shorter diagonal, in this particular configuration, we obtain a simple relation that gives diagonal QS = side length. Step-by-Step Solution: Step 1: Understand the geometry. The rhombus has side 8 cm and angles of 120° and 60°. Diagonal QS is opposite an acute angle in the figure. Step 2: Use the diagonal formula for a rhombus. For a rhombus of side a and angle 120°, one diagonal corresponds to an expression involving 2 + 2 * cos(120°), and the other corresponds to 2 - 2 * cos(120°). Step 3: Recall that cos(120°) = -1/2. Then 2 + 2 * cos(120°) = 2 + 2 * (-1/2) = 2 - 1 = 1. Step 4: The shorter diagonal squared becomes a^2 * 1 = a^2. So, diagonal QS^2 = 8^2 = 64. Step 5: Take the square root. Diagonal QS = 8 cm. Verification / Alternative check:Alternatively, you can use vector or coordinate geometry to place the rhombus on a plane, for example by placing PQ along the x-axis and using the 120° angle at Q to locate point R. Coordinate methods again lead to the same diagonal length after computing distances. The consistency of both trigonometric and coordinate approaches supports that the shorter diagonal QS must be 8 cm for the given rhombus side and angle configuration. Why Other Options Are Wrong: Option 4√5 cm: This value is approximately 8.94 cm and does not match the specific diagonal length derived from the side and angle data. Option 6 cm: This is too small for a diagonal relative to a side length of 8 cm in this configuration. Option 12 cm: This is too large and would suggest a much more stretched shape than the given angle would allow for the shorter diagonal. Option 10 cm: This is again larger than the side and does not satisfy the derived formula based on the rhombus geometry and angle. Common Pitfalls:Students often misinterpret which diagonal is being asked for or assume that both diagonals must be longer than the side. Another mistake is to use the cosine rule incorrectly in a triangle formed by two sides and a diagonal without correctly identifying the included angle. Forgetting the exact value of cos(120°) or mixing up acute and obtuse angles can also lead to errors. Always map the geometry carefully and match it consistently with formulas. Final Answer:The length of diagonal QS of the rhombus is 8 cm.

Question 2

A square and a rhombus stand on the same base and lie between the same two parallel lines. What is the ratio of the area of the square to the area of the rhombus in this situation?

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Introduction:This is a conceptual geometry question that tests a fundamental idea: when two plane figures stand on the same base and lie between the same parallels, they have the same height. Since area is equal to base multiplied by height for many standard shapes, this property often leads to simple area relationships. Here, we compare a square and a rhombus, both resting on the same base and between the same parallels, and we are asked for the ratio of their areas. Given Data / Assumptions: • There is a square standing on a given base segment. • There is a rhombus standing on the same base segment. • Both figures lie between the same pair of parallel lines, meaning they have equal perpendicular height. • We are asked to find the ratio: area of square : area of rhombus. Concept / Approach:The area of any parallelogram-like figure, including a rectangle, square, rhombus, or general parallelogram, is given by base * height, where height is the perpendicular distance between the pair of opposite sides. If two such figures share the same base and have the same height (that is, they lie between the same parallels), then their areas are equal. The shape of the figure may change the side lengths or angles, but as long as both base and height are common, the area remains the same. This is a well-known geometric result often used to compare parallelograms and related figures. Step-by-Step Solution: Step 1: Let the common base length be b units for both the square and the rhombus. Step 2: Let the common perpendicular distance between the parallels (height) be h units. Step 3: Area of the square in this configuration. Even though a square normally has all sides equal, in this conceptual setup we only care that it stands on base b and has height h. Therefore, its area = base * height = b * h. Step 4: Area of the rhombus in this configuration. A rhombus also has area = base * height, where base is any side and height is the perpendicular distance to the opposite side. Here, base = b and height = h, so area of the rhombus = b * h. Step 5: Compute the ratio of areas. Area of square : area of rhombus = (b * h) : (b * h) = 1 : 1. Verification / Alternative check:We can take any specific numeric example to verify the concept. Suppose the base is 10 units and the common height is 6 units. Then area of the square standing on that base and height is 10 * 6 = 60 square units. The rhombus also has area 10 * 6 = 60 square units. The ratio is again 60 : 60 = 1 : 1. No matter what actual values of base and height are chosen (as long as they are the same for both figures), the ratio stays equal to 1 : 1. Why Other Options Are Wrong: Option 1 : 2: This would mean the rhombus always has double the area of the square, which is not true when they share both base and height. Option 1 : 3 and Option 1 : 4: These suggest that the square has much smaller area than the rhombus with the same base and height, which contradicts the base * height formula. Option 2 : 1: This would mean the square always has twice the area of the rhombus, which is also false under the given conditions. Common Pitfalls:Many learners mistakenly focus on the fact that a square has all angles equal to 90° and a rhombus generally has oblique angles, and therefore assume their areas must differ. However, area depends on base and perpendicular height, not on angles directly. Another misconception is to assume that the side length defines the area in all configurations without considering the effective height. Remember, when two figures share the same base and lie between the same parallels, their areas are equal regardless of their shape along the top edge. Final Answer:The ratio of the area of the square to the area of the rhombus is 1 : 1.

Question 3

A rectangular parking space is marked out by painting three of its sides. The remaining unpainted side has a length of 9 feet, and the sum of the lengths of the three painted sides is 37 feet. What is the area of this rectangular parking space in sq…

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Introduction:This is a rectangle perimeter puzzle where only partial information about the sides is given. Three sides are painted and their total length is known, while the fourth unpainted side's length is also given. From these details, we must reconstruct the dimensions of the rectangle and then find its area. The problem checks understanding of properties of rectangles (opposite sides equal) and ability to form and solve linear equations from word statements. Given Data / Assumptions: • The shape of the parking space is a rectangle. • One unpainted side has length 9 feet. • The sum of the lengths of the three painted sides is 37 feet. • Opposite sides of a rectangle are equal in length. • We are asked to find the area in square feet. Concept / Approach:Let the rectangle have sides of length L and B. Since one side is unpainted and has length 9 feet, that side must be either L or B. The opposite side, which has the same length, is part of the painted sides. The three painted sides therefore consist of one side of length 9 feet and the other two sides of the rectangle. Using the total length of painted sides, we can solve for the other side, and then compute the area L * B. Step-by-Step Solution: Step 1: Let us assume the unpainted side is of length 9 feet. Its opposite side is also 9 feet long and is painted. Step 2: Let the other dimension of the rectangle be x feet. Step 3: The painted sides are: one side of 9 feet (opposite the unpainted side) and the two sides of length x feet each. Step 4: Sum of painted sides is therefore 9 + x + x = 9 + 2x. Step 5: We are given that 9 + 2x = 37. Step 6: Solve for x: 2x = 37 - 9 = 28, so x = 14. Step 7: The dimensions of the rectangle are 9 feet and 14 feet. Step 8: Compute the area: area = length * breadth = 9 * 14 = 126 square feet. Verification / Alternative check:Check the perimeter and painted sides for consistency. The full perimeter of the rectangle is 2 * (9 + 14) = 2 * 23 = 46 feet. One side of 9 feet is unpainted, so the three painted sides must be 14 + 14 + 9 = 37 feet, which matches the given condition. The area computed from these dimensions is 126 sq ft, confirming that our reasoning and calculations are correct. Why Other Options Are Wrong: Option 120 sq ft: This would require dimensions such as 10 ft by 12 ft or some other pair that does not satisfy the painted sides condition. Option 130 sq ft: There is no integer pair of sides near 9 and 14 that would yield this exact area while fitting the painted side sum. Option 135 sq ft: This would correspond to something like 9 ft by 15 ft, whose perimeter and painted side totals do not match the problem statement. Option 144 sq ft: This might correspond to 12 ft by 12 ft, which would make it a square and would change both the perimeter and the sum of painted sides. Common Pitfalls:Some students mistakenly assume that the three painted sides form three equal sides or that the unpainted side must be the longer side. Others incorrectly treat 37 as the full perimeter instead of the sum of three sides. Understanding that opposite sides of a rectangle are equal and clearly identifying which sides are counted in the 37 feet is essential to avoid algebraic errors. Final Answer:The area of the parking space is 126 square feet.

Question 4

A cow is tethered by a rope so that it can graze exactly over an area of 2826 square metres. Assuming the grazing region is a perfect circle and the rope represents the radius, what is the length of the rope in metres?

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Introduction:This is a straightforward application of the formula for the area of a circle. The situation describes a cow tied to a fixed point with a rope, and the cow can graze in a circular region around that point. The area grazed is given, and we are asked to determine the radius of the circle, which is the length of the rope. Such questions often appear in quantitative aptitude to test comfort with circle geometry, basic formula manipulation, and arithmetic. Given Data / Assumptions: • The grazing area is circular. • Area of the circle grazed by the cow = 2826 square metres. • The radius of the circle is equal to the length of the rope. • Use the standard formula: area of circle = pi * r^2. • Use pi = 3.14 for calculations. Concept / Approach:The area A of a circle is related to its radius r by A = pi * r^2. When A is known, we can rearrange the formula to solve for r: r^2 = A / pi and r = sqrt(A / pi). The problem becomes a simple substitution and square root calculation. Once the radius is obtained, we interpret it directly as the rope length in metres. Step-by-Step Solution: Step 1: Write the formula for the area of a circle. A = pi * r^2. Step 2: Substitute the given area A = 2826 sq m and pi = 3.14. 2826 = 3.14 * r^2. Step 3: Solve for r^2. r^2 = 2826 / 3.14. Step 4: Perform the division. 2826 / 3.14 = 900. Step 5: Take the square root of r^2. r = sqrt(900) = 30. Thus, the radius of the circular grazing area, and therefore the length of the rope, is 30 metres. Verification / Alternative check:Check by substitution: if r = 30 metres, area = pi * r^2 = 3.14 * 30 * 30 = 3.14 * 900 = 2826 square metres. This exactly matches the given grazing area, confirming that the radius value is correct. There is no ambiguity in the shape or formula used, so this verification is strong evidence that 30 metres is the correct rope length. Why Other Options Are Wrong: Option 15 m: Using this as radius gives area = 3.14 * 15 * 15 = 706.5 sq m, which is far smaller than 2826 sq m. Option 24 m: Area would be 3.14 * 24 * 24 = 3.14 * 576 = 1808.64 sq m, still much less than required. Option 36 m: Area would be 3.14 * 36 * 36 = 3.14 * 1296 ≈ 4069.44 sq m, which is significantly larger than the given area. Option 20 m: Area would be 3.14 * 20 * 20 = 1256 sq m, again smaller than the required 2826 sq m. Common Pitfalls:Students sometimes confuse radius and diameter and may accidentally set 2r^2 instead of r^2, or they may forget to take the square root after dividing the area by pi. Some also mistakenly plug the area into pi * d^2 / 4 directly without careful substitution, which complicates the calculation. The safest route is always to start from A = pi * r^2, rearrange clearly, and proceed with systematic substitution and simplification. Final Answer:The length of the rope is 30 metres.

Question 5

An order was placed for a carpet whose breadth was 6 metres and whose length was 1.44 times its breadth. What will be the cost of a second carpet whose length and breadth are 40% more and 25% more respectively than the first carpet, if the rate of t…

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Introduction:This question combines geometry of rectangles with percentage increase and cost calculation. The original carpet has given dimensions based on a relationship between length and breadth. The new carpet has both dimensions increased by certain percentages, and the cost is based on area at a fixed rate per square metre. The problem checks a student's ability to handle successive percentage increases, compute the area of a rectangle correctly, and then apply a unit rate to find the total cost. Given Data / Assumptions: • Breadth of the first carpet = 6 m. • Length of the first carpet = 1.44 times its breadth. • Length and breadth of the second carpet are increased by 40% and 25% respectively compared to the first carpet. • Rate of carpet = Rs. 45 per square metre. • We need the cost of the second carpet. Concept / Approach:First, compute the dimensions of the original carpet. Then apply the given percentage increases to find the new dimensions for the second carpet. Use the area formula for rectangles, area = length * breadth, to find the area of the second carpet. Once the area is known, multiply by the cost per square metre to find the total cost. Careful handling of percentages and decimal multiplication is essential. Step-by-Step Solution: Step 1: Compute length of the first carpet. Breadth of first carpet = 6 m. Length of first carpet = 1.44 * 6 = 8.64 m. Step 2: Compute area of the first carpet (for reference). Area first = 6 * 8.64 = 51.84 sq m. Step 3: Compute new length of second carpet. Increase in length = 40% of 8.64 = 0.40 * 8.64 = 3.456 m. New length = 8.64 + 3.456 = 12.096 m. Step 4: Compute new breadth of second carpet. Increase in breadth = 25% of 6 = 0.25 * 6 = 1.5 m. New breadth = 6 + 1.5 = 7.5 m. Step 5: Compute area of the second carpet. Area second = 12.096 * 7.5 = 90.72 sq m. Step 6: Compute cost at Rs. 45 per sq m. Cost = 90.72 * 45. Cost = 4082.4, that is Rs. 4082.40. Verification / Alternative check:We can also reason proportionally: The area of a rectangle scales by the product of the scale factors applied to each dimension. Here, length is multiplied by 1.40 and breadth is multiplied by 1.25. So area is multiplied by 1.40 * 1.25 = 1.75. The first carpet area is 51.84 sq m, so the second carpet area should be 51.84 * 1.75 = 90.72 sq m, matching our previous result. Multiplying by the rate 45 gives 51.84 * 1.75 * 45 = 4082.4, reinforcing the answer. Why Other Options Are Wrong: Option Rs. 1024.21: This is far too low and would correspond to a much smaller carpet area. Option Rs. 2810.60: This underestimates the scaled area and cost, as it corresponds to reducing one of the multipliers incorrectly. Option Rs. 3214.00: This might result from an error in applying only one percentage change correctly or miscalculating the rate. Option Rs. 3500.00: This is an approximate round number but does not come from correct dimensional increases or exact multiplication. Common Pitfalls:Common mistakes include applying the percentage increases to the area directly instead of to each dimension, or adding 40% and 25% and assuming a 65% increase in area, which is incorrect. Another error is misplacing decimals while multiplying, especially with non-integer dimensions like 8.64 and 12.096. A step-by-step approach with clear intermediate results avoids most calculation errors. Final Answer:The cost of the second carpet is Rs. 4082.40.

Question 6

In a rectangle, the length is three times its breadth. If the length and the breadth of the rectangle are increased by 30% and 100% respectively, by what percentage does the perimeter of the rectangle increase?

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Introduction:This question is about how the perimeter of a rectangle changes when its length and breadth are increased by different percentages. Such problems test a student's understanding of ratios, percentage changes, and the formula for the perimeter of a rectangle. It also checks whether students can correctly track changes through algebra rather than relying on guesswork. Given Data / Assumptions: • Let the original breadth be B. • Original length = 3B (because length is three times breadth). • Length is increased by 30%, so new length = 130% of original length. • Breadth is increased by 100%, so new breadth = 200% of original breadth. • We must find the percentage increase in the perimeter. Concept / Approach:The perimeter P of a rectangle is given by P = 2 * (length + breadth). We can express the original perimeter in terms of B, then express the new perimeter after the percentage increases, again in terms of B, and finally compute the percentage increase using the formula: percentage increase = ((new - old) / old) * 100. Using algebra keeps the solution general and avoids choosing specific numeric values unnecessarily. Step-by-Step Solution: Step 1: Express original dimensions. Original breadth = B. Original length = 3B. Step 2: Compute original perimeter. Original perimeter P1 = 2 * (length + breadth) = 2 * (3B + B) = 2 * 4B = 8B. Step 3: Compute new dimensions after increase. Length increased by 30%: new length L2 = 1.30 * 3B = 3.9B. Breadth increased by 100%: new breadth B2 = 2 * B = 2B. Step 4: Compute new perimeter. New perimeter P2 = 2 * (L2 + B2) = 2 * (3.9B + 2B) = 2 * 5.9B = 11.8B. Step 5: Compute percentage increase in perimeter. Increase in perimeter = P2 - P1 = 11.8B - 8B = 3.8B. Percentage increase = (increase / original) * 100 = (3.8B / 8B) * 100. Simplify: (3.8 / 8) * 100 = 0.475 * 100 = 47.5%. Verification / Alternative check:Pick a specific value, for example B = 10 units. Then original length = 30 units and original perimeter P1 = 2 * (30 + 10) = 80 units. New length = 1.3 * 30 = 39 units and new breadth = 2 * 10 = 20 units, so new perimeter P2 = 2 * (39 + 20) = 2 * 59 = 118 units. The increase is 118 - 80 = 38 units. Percentage increase = (38 / 80) * 100 = 47.5%, matching our algebraic result. Why Other Options Are Wrong: Option 40%: Underestimates the effect of doubling the breadth and is not supported by the calculations. Option 45%: Close but not exact; it may result from rounding or partial computation errors. Option 50%: Assumes a simple half increase in perimeter, which is not consistent with the different percentage changes in length and breadth. Option 55%: Overestimates the change and does not match either numerical or algebraic checks. Common Pitfalls:Some students incorrectly assume that the percentage change in perimeter is simply the average or sum of the percentage changes of length and breadth. Others may recompute side changes but forget that perimeter uses the sum of both sides before doubling. Careless handling of decimal multipliers (such as 1.30 and 2.00) and incorrect arithmetic when simplifying 3.8 / 8 also cause mistakes. Writing each step and using a test value helps avoid these errors. Final Answer:The perimeter of the rectangle increases by 47.5%.

Question 7

A rectangular grassy plot measures 110 metres by 65 metres. There is a gravel path of uniform width 2.5 metres all around the inside of the plot. Find the total cost of gravelling this path at the rate of 80 paise (Rs. 0.80) per square metre.

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Introduction:This problem is a classic example of finding the area of a border or path around a rectangle. The rectangular grassy plot has a gravel path of uniform width running all around the inside. We must compute the area of this path and then multiply by the cost per square metre. The question tests understanding of inner and outer dimensions, subtraction of areas, and simple cost calculation based on unit area. Given Data / Assumptions: • Outer dimensions of the rectangular plot: 110 m by 65 m. • Gravel path runs all around the inside of the plot. • Width of the path = 2.5 m (uniform on all sides). • Cost of gravelling = 80 paise per square metre = Rs. 0.80 per sq m. • We must find the total cost of gravelling the path only. Concept / Approach:The area of the gravel path is equal to the area of the outer rectangle minus the area of the inner rectangle that remains grassy. When a border of width w is taken all around inside a rectangle of dimensions L and B, the inner rectangle has dimensions (L - 2w) and (B - 2w). Once we compute both areas, we subtract to get the area of the path. Finally, we multiply by the cost per square metre to get the total cost. Step-by-Step Solution: Step 1: Compute the area of the outer rectangle. Outer length L = 110 m. Outer breadth B = 65 m. Area_outer = L * B = 110 * 65 = 7150 sq m. Step 2: Compute dimensions of the inner rectangle (inside the path). Width of path w = 2.5 m on all sides. Inner length = L - 2w = 110 - 2 * 2.5 = 110 - 5 = 105 m. Inner breadth = B - 2w = 65 - 5 = 60 m. Step 3: Compute the area of the inner rectangle. Area_inner = 105 * 60 = 6300 sq m. Step 4: Compute the area of the gravel path. Area_path = Area_outer - Area_inner = 7150 - 6300 = 850 sq m. Step 5: Compute the cost of gravelling the path. Cost per sq m = Rs. 0.80. Total cost = Area_path * cost per sq m = 850 * 0.80 = Rs. 680. Verification / Alternative check:We can also reason by expansion: The path occupies a border of width 2.5 m along the entire boundary. Checking consistency, 2.5 m is reasonable relative to dimensions 110 m and 65 m, and the remaining inner dimensions 105 m and 60 m are positive and realistic. The area difference of 850 sq m is a moderate fraction of the total 7150 sq m, which is plausible for a relatively narrow border. Multiplying 850 by 0.80 gives 680, a clean and consistent number matching one of the options. Why Other Options Are Wrong: Option Rs. 480: This would correspond to an area of only 600 sq m at Rs. 0.80 per sq m, which is smaller than the actual path area. Option Rs. 340: This implies an area of 425 sq m, which is too small for the given path width and rectangle size. Option Rs. 320: This corresponds to 400 sq m of path, which is clearly less than the actual border area. Option Rs. 540: This corresponds to an area of 675 sq m, again not matching the area difference of 850 sq m. Common Pitfalls:Some students misinterpret the width as being added outside the rectangle instead of inside, leading to wrong inner dimensions. Others subtract w only once from each side instead of 2w, forgetting that the path runs on both ends of each dimension. Another frequent error is to compute the area of the inner rectangle incorrectly or to multiply by 80 instead of 0.80 when converting paise to rupees. Careful reading of units and consistent application of the 2w adjustment prevent most of these mistakes. Final Answer:The total cost of gravelling the path is Rs. 680.

Question 8

The ratio of the bases of two triangles is x : y and the ratio of their areas is a : b. What is the ratio of their corresponding altitudes?

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Introduction:This is a conceptual question about the relationship between the areas, bases, and altitudes (heights) of two triangles. The problem gives the ratio of the bases and the ratio of the areas and then asks for the ratio of the corresponding altitudes. This type of question tests whether students are comfortable manipulating proportional relationships and applying the area formula for a triangle, which involves both base and height. Given Data / Assumptions: • We have two triangles, say triangle 1 and triangle 2. • Ratio of their bases: base1 : base2 = x : y. • Ratio of their areas: area1 : area2 = a : b. span style="display:block;">• Let h1 and h2 be the corresponding altitudes (heights) for bases base1 and base2 respectively. • We need to find h1 : h2 in terms of a, b, x, and y. Concept / Approach:The area of a triangle is given by area = (1/2) * base * height. For each triangle, area1 = (1/2) * base1 * h1 and area2 = (1/2) * base2 * h2. If we divide area1 by area2, the factor (1/2) cancels out, and the ratio of areas becomes a ratio involving base1, base2, h1, and h2. Using the given ratios area1 : area2 = a : b and base1 : base2 = x : y, we can manipulate this relationship to solve for the ratio h1 : h2. Step-by-Step Solution: Step 1: Write expressions for areas of the two triangles. Area1 = (1/2) * base1 * h1. Area2 = (1/2) * base2 * h2. Step 2: Consider the ratio of the areas. Area1 / Area2 = [(1/2) * base1 * h1] / [(1/2) * base2 * h2] = (base1 * h1) / (base2 * h2). Step 3: Use the given area ratio. Area1 / Area2 = a / b. So, (base1 * h1) / (base2 * h2) = a / b. Step 4: Insert the base ratio base1 / base2 = x / y. (base1 / base2) * (h1 / h2) = a / b. So, (x / y) * (h1 / h2) = a / b. Step 5: Solve for h1 / h2. h1 / h2 = (a / b) * (y / x) = (a * y) / (b * x). Thus, h1 : h2 = ay : bx. Verification / Alternative check:To verify, take a concrete example. Let x = 2, y = 3, a = 4, and b = 6. Then base1 : base2 = 2 : 3 and area1 : area2 = 4 : 6 = 2 : 3. From the formula, h1 : h2 should be (a * y) : (b * x) = 4 * 3 : 6 * 2 = 12 : 12 = 1 : 1. If base1 = 2 units and base2 = 3 units, and h1 = h2, the area ratio is (1/2 * 2 * h1) : (1/2 * 3 * h1) = 2h1 : 3h1 = 2 : 3, consistent with the given ratio. This supports our derived result. Why Other Options Are Wrong: Option a : b: This ignores the effect of the base ratio and would be correct only if the bases were equal. Option x : y: This assumes that area ratio has no effect on heights, which contradicts the formula for area. Option bx : ay: This is simply the inverse of the correct ratio and would be obtained by mistakenly rearranging the equation. Option x * a : y * b: While superficially similar, this does not reduce to the derived ratio and misplaces which factors multiply together. Common Pitfalls:Students sometimes incorrectly cancel the base and height terms or forget that area of triangle depends on both base and height, not just one dimension. Another common error is to directly equate area ratio with height ratio, overlooking the base ratio. Careful algebraic manipulation of the area formula and stepwise reasoning prevents such mistakes. Final Answer:The ratio of the corresponding altitudes is ay : bx.

Question 9

Each side of a square is increased by 25% of its original length. By what percentage does the area of the square change as a result of this increase?

Accepted Answer

Introduction:This is a classic percentage increase problem involving area. The question describes a square whose sides are increased by a certain percentage and asks how much the area changes. Because area depends on the square of the side length, a percentage increase in side does not translate to the same percentage increase in area. This concept is very important in aptitude exams and geometry-based word problems. Given Data / Assumptions: • We start with a square of side length s. • Each side is increased by 25%. • That means the new side length is 125% of the original. • We are asked for the percentage change in the area of the square. Concept / Approach:The area A of a square with side s is s^2. If the side is increased by 25%, the new side becomes 1.25s. The new area is (1.25s)^2. The percentage change in area is computed as (new area - old area) / old area * 100. Because both dimensions (in this case, the same dimension counted twice) are scaled by the same factor, the area scales by the square of that factor, not by the factor itself. Step-by-Step Solution: Step 1: Let the original side be s. Original area A1 = s^2. Step 2: Compute the new side length after a 25% increase. New side length = s + 25% of s = s + 0.25s = 1.25s. Step 3: Compute the new area A2. A2 = (1.25s)^2 = 1.25^2 * s^2. Step 4: Compute 1.25^2. 1.25^2 = 1.25 * 1.25 = 1.5625. So, A2 = 1.5625 * s^2. Step 5: Compute the increase in area. Increase = A2 - A1 = (1.5625 * s^2) - s^2 = 0.5625 * s^2. Step 6: Compute percentage increase in area. Percentage increase = (increase / A1) * 100 = (0.5625 * s^2 / s^2) * 100 = 0.5625 * 100 = 56.25%. Verification / Alternative check:Take an easy numeric example. Suppose the original side s = 4 units. Then original area A1 = 4^2 = 16 square units. The new side is 1.25 * 4 = 5 units, so new area A2 = 5^2 = 25 square units. The increase in area is 25 - 16 = 9 square units. Percentage increase = (9 / 16) * 100 = 56.25%, exactly matching our earlier calculation. This confirms the formula-based result. Why Other Options Are Wrong: Option 36.25%: This likely comes from incorrectly squaring or partially applying the percentage change. Option 16.25%: This is far too small and fails to account for the fact that both dimensions are increased. Option 12.25%: This might come from mistakenly squaring 3.5% or some other wrong value, but it does not match the actual transformation. Option 25%: This would be correct only if area increased linearly with side length, which is not the case, since area depends on side squared. Common Pitfalls:One of the most common mistakes is to assume that increasing each side by 25% increases the area by the same 25%. This ignores the fact that area is a product of two dimensions. Another frequent error is miscomputing 1.25^2, sometimes writing it as 1.25 or 1.5 instead of 1.5625. To avoid errors, it is useful to set up the ratio new area / old area as (1.25s)^2 / s^2 = 1.25^2 and then calculate carefully. Final Answer:The area of the square increases by 56.25%.

Question 10

A square and a rectangle have the same perimeter. Let a be the area of the square and b be the area of the rectangle. Which of the following relations between a and b is true?

Accepted Answer

Introduction:This is a conceptual question involving optimization of area for a given perimeter. Among all rectangles with a fixed perimeter, the one with maximum area is a square. Therefore, when a square and a rectangle share the same perimeter, the area of the square is greater than or equal to the area of the rectangle, and it is strictly greater unless the rectangle is also a square. This problem tests whether students understand this classical result and can apply it symbolically using the variables a and b for areas. Given Data / Assumptions: • There is a square with area a. • There is a rectangle with area b. • The perimeter of the square equals the perimeter of the rectangle. • The shapes are general; the rectangle need not be a square. • We must determine the correct relation between a and b. Concept / Approach:Let the side of the square be s, so its perimeter is 4s. If the rectangle has sides L and W, its perimeter is 2(L + W). Given that the perimeters are equal, we have 4s = 2(L + W), or L + W = 2s. The area of the square is s^2, whereas the area of the rectangle is L * W. For a given sum L + W, the product L * W is maximized when L = W, that is, when the rectangle becomes a square. Thus, s^2 is at least as large as L * W, and strictly larger for any non-square rectangle. So a > b in general. Step-by-Step Solution: Step 1: Let the square have side length s. Then its perimeter = 4s and its area a = s^2. Step 2: Let the rectangle have length L and breadth W. Its perimeter = 2(L + W). Step 3: Use the condition of equal perimeters. 4s = 2(L + W) ⇒ L + W = 2s. Step 4: Express the rectangle’s area in terms of L and W. Rectangle area b = L * W. Step 5: Use the inequality for a fixed sum. For a fixed sum S = L + W, the product L * W is maximized when L = W = S / 2. Here S = 2s, so maximum L * W occurs when L = W = s, giving b_max = s^2. Step 6: Conclude relation between a and b. The maximum possible area for the rectangle is s^2, which equals the square’s area, and any other rectangle with unequal sides has a smaller area than s^2. Therefore, in general, a ≥ b, and for a non-square rectangle, a > b. Since typical exam questions consider a general rectangle, we select a > b. Verification / Alternative check:Consider a concrete example. Let the common perimeter be 40 units. Then square side s = 40 / 4 = 10, so square area a = 10^2 = 100. For a rectangle with the same perimeter 40, choose L = 12 and W = 8. Then L + W = 20, perimeter = 40, and area b = 12 * 8 = 96. Here a = 100 and b = 96, so a > b. Trying other combinations like 11 and 9 gives area 99, still less than 100. Only when L = W = 10 does the rectangle become a square and match the area of the square, not exceed it. Why Other Options Are Wrong: Option a < b: This would mean that a rectangle with given perimeter could always exceed the square’s area, which contradicts the maximum area property of the square. Option a = b: This holds only when the rectangle itself is a square. The question, however, refers to a general rectangle. Option none of these: This is incorrect because we have a definite relationship a > b under the usual interpretation. Option a + b is constant: This has no basis in the problem statement and mixes two unrelated areas. Common Pitfalls:Students may incorrectly assume that if two shapes have the same perimeter, their areas must also be equal. Others may know the result but misinterpret which shape has the larger area. It is important to remember the key fact: for a given perimeter, the square has the maximum area among all rectangles. Careful use of this principle avoids confusion. Final Answer:The correct relation is a > b, that is, the square’s area is greater than the rectangle’s area for the same perimeter.

Question 11

The area of a square field is 69696 square centimetres. What is the length of its diagonal, in centimetres?

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Introduction:This geometry question involves the relationship between the area, side length, and diagonal of a square. The area of the square is given, and we are asked to find the diagonal. This requires first finding the side length from the area and then using the well-known relationship between the side and diagonal of a square. The problem tests comfort with square roots and the Pythagorean-type relation for squares. Given Data / Assumptions: • Area of the square field = 69696 sq cm. • Let the side length of the square be s cm. • Area of a square = s^2. • Diagonal d of a square with side s = s * √2. Concept / Approach:Since the area of a square is s^2, we can find s by taking the square root of the given area. Once we know s, we use the fact that the diagonal of a square forms the hypotenuse of a right-angled triangle with both legs equal to s. By the Pythagorean theorem, diagonal d satisfies d^2 = s^2 + s^2 = 2s^2, so d = s * √2. After computing s, we multiply by an approximate value of √2 (or keep in mind approximate products) to find the diagonal. Step-by-Step Solution: Step 1: Use the area formula to find the side length. Area = s^2 = 69696 sq cm. Step 2: Take the square root of 69696. s = √69696. Step 3: Compute √69696. 69696 = 264^2, so s = 264 cm. Step 4: Use the diagonal formula for a square. d = s * √2 = 264 * √2. Step 5: Approximate √2 ≈ 1.414. d ≈ 264 * 1.414. Step 6: Multiply 264 by 1.414. 264 * 1.414 ≈ 373.35 cm (approximately). Step 7: Choose the nearest option. The nearest option to 373.35 cm is 373 cm. Verification / Alternative check:We can verify by squaring 373 to see if it is close to the theoretical diagonal squared. If d ≈ 373.35 cm, then d^2 ≈ 139,384. For 373^2, we get 139,129, which is close enough to the computed 2 * 69696 = 139,392, given rounding of √2. This confirms that 373 cm is a very good approximation for the diagonal and matches the nearest option provided. Why Other Options Are Wrong: Option 473 cm: This corresponds to a much larger diagonal that would produce a side longer than 264 cm, contradicting the given area. Option 573 cm and 673 cm: These values are even larger and would lead to side lengths that make the area far greater than 69696 sq cm. Option 273 cm: This is too small for the diagonal of a square with side 264 cm, because the diagonal must be longer than any side. Common Pitfalls:Some students may mistakenly take the square root of the area to be the diagonal instead of the side, or they may forget to multiply the side by √2. Others may approximate √2 too coarsely or round intermediate results incorrectly. It is also possible to make arithmetic mistakes while finding the exact square root of 69696. Writing each step clearly and using known squares (like 264^2) helps avoid confusion. Final Answer:The length of the diagonal of the square field is approximately 373 cm.

Question 12

What is the least number of square tiles required to completely pave the floor of a rectangular room that is 15 metres 17 centimetres long and 9 metres 2 centimetres broad, assuming all tiles are identical and square?

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Introduction:This question involves the concept of tiling a floor with square tiles and using the highest possible tile size so that no cutting is needed. The least number of tiles is achieved when each tile is as large as possible, and its side length is the greatest common divisor (GCD) of the room’s dimensions expressed in the same units. This problem combines knowledge of unit conversion, greatest common divisor, and area-based counting. Given Data / Assumptions: • Length of the room = 15 metres 17 centimetres. • Breadth of the room = 9 metres 2 centimetres. • Tiles are square and identical in size. • We want the least number of tiles, so each tile should be as large as possible. Concept / Approach:To find the maximum possible side length of the square tile that fits exactly along both dimensions, we convert both dimensions to the same unit (for example, centimetres) and then compute their greatest common divisor (GCD). This GCD is the largest possible tile side length in centimetres. The number of tiles is then the area of the floor divided by the area of one tile, or equivalently the product of (length / tile_side) and (breadth / tile_side). Step-by-Step Solution: Step 1: Convert dimensions to centimetres. 1 metre = 100 cm. Length = 15 m 17 cm = 15 * 100 + 17 = 1517 cm. Breadth = 9 m 2 cm = 9 * 100 + 2 = 902 cm. Step 2: Find the greatest common divisor (GCD) of 1517 and 902. Using the GCD concept, gcd(1517, 902) = 41. Step 3: The side of each square tile = 41 cm. Step 4: Find how many tiles fit along each dimension. Number of tiles along length = 1517 / 41 = 37. Number of tiles along breadth = 902 / 41 = 22. Step 5: Compute total number of tiles. Total tiles = 37 * 22 = 814. Verification / Alternative check:We can check whether 41 cm really divides both dimensions exactly. 41 * 37 = 1517 and 41 * 22 = 902, so the tile side length fits perfectly into both the length and breadth without leaving any remainder. Thus no cutting of tiles is required, and we are using the largest possible tile size. Therefore, any smaller tile size would increase the total number of tiles, making 814 the minimum possible number. Why Other Options Are Wrong: Option 844: This corresponds to a smaller tile size, which would not be the largest possible divisor of both dimensions. Option 840 and 820: These also assume different tile sizes that do not exploit the maximum GCD, leading to more tiles than necessary. Option 780: This is even smaller and cannot be achieved with the maximum tile side, given the exact room dimensions. Common Pitfalls:A common mistake is to convert only one dimension to centimetres or to find the least common multiple (LCM) instead of the greatest common divisor. Another pitfall is to guess a tile side that divides one dimension but not the other, which would require cutting and would contradict the requirement for identical, uncut tiles. Always remember to convert both dimensions to the same unit and to use the GCD for such tiling problems. Final Answer:The least number of square tiles required is 814.

Question 13

In a parallelogram PQRS, the lengths of the sides are 8 cm and 12 cm, and one of its diagonals has length 10 cm. What is the approximate length (in centimetres) of the other diagonal?

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Introduction:This question is about a parallelogram and the relationship between its sides and diagonals. For any parallelogram with sides of lengths a and b and diagonals of lengths d1 and d2, there is an important formula connecting them. Here we are given the side lengths and one diagonal, and we are asked to find the length of the other diagonal. This is a typical application of the parallelogram diagonal identity. Given Data / Assumptions: • Parallelogram PQRS has adjacent sides of lengths 8 cm and 12 cm. • One diagonal (say PR) has length 10 cm. • Let the other diagonal (QS) have length d cm. • We need to find the approximate value of d. Concept / Approach:For a parallelogram with side lengths a and b and diagonals d1 and d2, there is a standard relation: d1^2 + d2^2 = 2(a^2 + b^2). This formula is derived from vector geometry or from splitting the parallelogram into congruent triangles. We can use this identity by taking a = 8, b = 12, d1 = 10, and then solving for d2. Finally, we take the square root to obtain the length of the second diagonal, and then choose the closest option. Step-by-Step Solution: Step 1: Let a = 8 cm, b = 12 cm, d1 = 10 cm, and d2 = d cm. Step 2: Use the parallelogram diagonal relation. d1^2 + d2^2 = 2(a^2 + b^2). Step 3: Compute the right-hand side. a^2 = 8^2 = 64. b^2 = 12^2 = 144. a^2 + b^2 = 64 + 144 = 208. 2(a^2 + b^2) = 2 * 208 = 416. Step 4: Substitute into the formula. 10^2 + d^2 = 416. So, 100 + d^2 = 416. Step 5: Solve for d^2. d^2 = 416 - 100 = 316. Step 6: Find d by taking the square root. d = √316. Step 7: Approximate √316. Since 17^2 = 289 and 18^2 = 324, √316 is between 17 and 18, closer to 18. A more accurate value is d ≈ 17.78 cm. Verification / Alternative check:We can check the reasonableness: for sides 8 cm and 12 cm, the diagonals should both be larger than the shorter side but smaller than the sum of the two sides. The computed value d ≈ 17.78 cm is indeed greater than 12 and less than 20, which is consistent. Also, if we square 17.8, we get approximately 316.84, very close to 316, which confirms that 17.8 cm is a good approximation. Why Other Options Are Wrong: Option 17.5 cm: Squaring 17.5 gives 306.25, which is significantly less than 316. Option 17 cm: 17^2 = 289, which is far from 316. Option 18 cm: 18^2 = 324, which is slightly more than 316, but 17.8 cm is a closer match to √316. Option 16 cm: 16^2 = 256, which is much too small compared to the required 316. Common Pitfalls:Common errors include misremembering or misapplying the diagonal formula, mixing it up with a formula for rectangles, or trying to use the cosine rule without clearly identifying the included angle. Another mistake is approximating √316 too roughly as 18 without checking how far off that is. Knowing simple square numbers and interpolation between them helps in making a better approximation. Final Answer:The approximate length of the other diagonal is 17.8 cm.

Question 14

The area of a field in the shape of a trapezium is 1440 square metres. The perpendicular distance between the two parallel sides is 24 metres, and the ratio of the lengths of the parallel sides is 5 : 3. What is the length of the longer parallel sid…

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Introduction:This question is about the area of a trapezium, also known as a trapezoid in some curricula. We are given the total area, the height (perpendicular distance between the parallel sides), and the ratio in which the two parallel sides are related. From this information, we need to find the length of the longer parallel side. The problem checks whether students can correctly use the area formula for a trapezium and work with ratios and simple algebra. Given Data / Assumptions: • Area of the trapezium = 1440 sq m. • Perpendicular distance between the parallel sides (height) = 24 m. • Ratio of the lengths of the two parallel sides = 5 : 3. • Let the parallel sides be 5x and 3x metres. • We are asked to find the length of the longer parallel side, that is 5x. Concept / Approach:The area A of a trapezium with parallel sides a and b and height h is given by A = (1/2) * (a + b) * h. Here, a and b are the lengths of the two parallel sides. Using the given ratio, we can express the sides as 5x and 3x. Then we substitute A = 1440 and h = 24 into the formula and solve for x. Once x is known, the longer side 5x can be found by simple multiplication. Step-by-Step Solution: Step 1: Represent the parallel sides using the ratio. Let the longer side be 5x and the shorter side be 3x. Step 2: Write the area formula for a trapezium. Area = (1/2) * (sum of parallel sides) * height. So, 1440 = (1/2) * (5x + 3x) * 24. Step 3: Simplify inside the brackets. 5x + 3x = 8x. Step 4: Substitute back. 1440 = (1/2) * 8x * 24. Step 5: Simplify the right-hand side. (1/2) * 8x = 4x. So, 1440 = 4x * 24 = 96x. Step 6: Solve for x. x = 1440 / 96 = 15. Step 7: Compute the longer parallel side. Longer side = 5x = 5 * 15 = 75 m. Verification / Alternative check:We can verify by computing the shorter side and checking if the original area is obtained. Shorter side = 3x = 3 * 15 = 45 m. Then sum of parallel sides = 75 + 45 = 120 m. Using the formula: Area = (1/2) * 120 * 24 = 60 * 24 = 1440 sq m, which matches the given area. This confirms that the value x = 15 and longer side = 75 m are correct. Why Other Options Are Wrong: Option 85 m: If the longer side were 85 m, with the same ratio, the shorter side would be 51 m, which would not produce an area of exactly 1440 sq m with height 24 m. Option 95 m and 105 m: These values would require correspondingly larger shorter sides based on the 5 : 3 ratio, resulting in areas much greater than 1440 sq m. Option 65 m: This would be too small compared to the required side lengths that satisfy the area and height conditions. Common Pitfalls:A typical mistake is to forget to apply the factor (1/2) in the trapezium area formula, leading to an incorrect equation and a wrong value for x. Many students also mis-handle the ratio, for example using 5x and x instead of 5x and 3x. Another common error is in arithmetic while solving 1440 = 96x, especially in division. Writing out each step carefully reduces the chance of error. Final Answer:The length of the longer parallel side of the trapezium is 75 metres.

Question 15

A rectangular sheet of paper is folded into two congruent parts along one set of sides, and each of these folded sheets has a perimeter of 34 cm. When the same original sheet is folded into two congruent parts along the other set of sides, the perim…

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Introduction:This question is a clever application of rectangle perimeter and the effect of folding a rectangle in half. The same sheet is folded along two different directions and the perimeters of the resulting congruent parts are given. Using this information, we must determine the original dimensions of the sheet and then compute its area. This tests understanding of perimeter formulas, simple algebra, and how folding affects side lengths. Given Data / Assumptions: • The original sheet is a rectangle with length L and breadth B. • When folded along one set of sides, each half has perimeter 34 cm. • When folded along the other set of sides, each half has perimeter 38 cm. • Folding divides one side into two equal parts while keeping the other dimension unchanged. • We must find the area L * B of the original rectangle. Concept / Approach:When a rectangle is folded along its length, the length becomes L/2 and the breadth remains B for the folded part. When folded along its breadth, the breadth becomes B/2 and the length remains L. The perimeter of a rectangle is 2 * (length + breadth). By writing perimeter equations for both folding cases, we get two linear equations in L and B. Solving these gives us L and B, and we then multiply them to obtain the area. Step-by-Step Solution: Step 1: Fold along the length. The folded part has dimensions L/2 and B. Perimeter in this case is 2 * (L/2 + B) = 34. Simplify: L + 2B = 34. Call this Equation (1). Step 2: Fold along the breadth. The folded part now has dimensions L and B/2. Perimeter = 2 * (L + B/2) = 38. Simplify: 2L + B = 38. Call this Equation (2). Step 3: Solve the system of equations. Equation (1): L + 2B = 34. Equation (2): 2L + B = 38. Step 4: Multiply Equation (1) by 2 if needed or use elimination directly. From Equation (1): L = 34 - 2B. Substitute into Equation (2): 2(34 - 2B) + B = 38. 68 - 4B + B = 38. 68 - 3B = 38. 3B = 68 - 38 = 30. B = 10 cm. Step 5: Substitute B back to find L. L + 2 * 10 = 34 ⇒ L + 20 = 34 ⇒ L = 14 cm. Step 6: Compute the area of the original rectangle. Area = L * B = 14 * 10 = 140 sq cm. Verification / Alternative check:Check both folding scenarios. First scenario (fold along length): dimensions 14/2 and 10 give 7 cm and 10 cm, perimeter = 2 * (7 + 10) = 2 * 17 = 34 cm, which matches. Second scenario (fold along breadth): dimensions 14 and 10/2 give 14 cm and 5 cm, perimeter = 2 * (14 + 5) = 2 * 19 = 38 cm, also matching. Therefore, the dimensions 14 cm by 10 cm are consistent, and the area 140 sq cm is correct. Why Other Options Are Wrong: Option 150 sq cm, 160 sq cm, 170 sq cm, and 180 sq cm: These would correspond to different pairs of length and breadth. None of those pairs will simultaneously satisfy both perimeter conditions for the two different foldings. Only 14 cm by 10 cm meets both conditions exactly. Common Pitfalls:Students sometimes confuse which dimension gets halved in which folding scenario, leading to incorrect perimeter equations. Another source of error is algebraic manipulation when solving the two simultaneous equations. Mixing up L and B or substituting incorrectly can produce wrong dimensions and therefore wrong area. Carefully defining folding direction and writing explicit equations for each case helps prevent such errors. Final Answer:The area of the original rectangular sheet of paper is 140 square centimetres.

Question 16

The perimeter of a square is 40 metres. What is the length of one side of the square?

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Introduction:This is a very basic geometry problem that tests understanding of the relationship between the perimeter and side length of a square. The perimeter of a square is four times the length of one side because all four sides are equal. Once the perimeter is given, finding the side length is a simple division problem. Given Data / Assumptions: • The figure is a square. • Perimeter of the square = 40 metres. • All four sides of a square are equal in length. • Let the length of one side be s metres. Concept / Approach:The perimeter P of a square is given by P = 4 * s, where s is the side length. To find s from a known perimeter, we simply rearrange the formula to s = P / 4. This direct relationship makes these problems quick to solve once the formula is remembered. Step-by-Step Solution: Step 1: Write the formula for the perimeter of a square. P = 4 * s. Step 2: Substitute the given perimeter P = 40 m. 40 = 4 * s. Step 3: Solve for s. s = 40 / 4 = 10. Step 4: Interpret the result. The length of one side of the square is 10 metres. Verification / Alternative check:We can check quickly by multiplying the side back by 4. If side s = 10 m, then perimeter = 4 * 10 = 40 m, which is exactly the given value. This confirms that the side calculation is correct and consistent with the perimeter definition. Why Other Options Are Wrong: Option 20 m: This would give a perimeter of 4 * 20 = 80 m, which is double the given perimeter. Option 30 m: This would produce a perimeter of 4 * 30 = 120 m, far larger than required. Option 40 m: This would make the perimeter 4 * 40 = 160 m, four times the given value. Option 5 m: This would give a perimeter of 4 * 5 = 20 m, only half of the given perimeter. Common Pitfalls:The most common mistake is to divide by 2 instead of 4, confusing the formula for the perimeter of a square with that of some other figure. Another minor error can be misreading the units, but here the units are consistent. Remembering that a square has four equal sides and that perimeter is the sum of all sides helps avoid confusion. Final Answer:The length of one side of the square is 10 metres.

Question 17

The perimeter of a rectangle is 160 cm. It is given that the rectangle is four times as long as it is wide. Using this information, determine the exact length and width of the rectangle in centimetres.

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Introduction / Context: This question tests basic understanding of rectangle geometry and perimeter relations. When a rectangle’s perimeter and the ratio between its length and width are known, the actual dimensions can be found by forming a simple algebraic equation. Given Data / Assumptions: Perimeter = 160 cm Length is four times the width Let width = b cm Then length = 4b cm Concept / Approach: The perimeter of a rectangle is given by 2*(length + breadth). Substitute the given relationship between length and width into the perimeter formula and solve for the unknown dimension. Step-by-Step Solution: Perimeter = 2*(l + b) 160 = 2*(4b + b) 160 = 2*5b = 10b b = 16 cm l = 4b = 4*16 = 64 cm Verification / Alternative check: Substituting l = 64 and b = 16 into the perimeter formula gives 2*(64 + 16) = 2*80 = 160 cm, which matches the given perimeter exactly. Why Other Options Are Wrong: 48 cm and 24 cm: length is only twice the width, not four times. 65 cm and 16 cm: perimeter exceeds 160 cm. 16 cm and 64 cm: reverses length and width. 40 cm and 20 cm: does not satisfy the four-times condition. Common Pitfalls: Forgetting to multiply the sum by 2 in the perimeter formula. Mixing up which dimension is four times the other. Arithmetic errors while solving for b. Final Answer: Length = 64 cm and Width = 16 cm

Question 18

A rectangle of dimensions 3 units by 4 units is inscribed in a circle. Assuming the rectangle’s vertices lie on the circle, find the circumference of the circle in terms of pi.

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Introduction / Context: This problem tests the relationship between a rectangle and its circumcircle. When a rectangle is inscribed in a circle, the diagonal of the rectangle becomes the diameter of the circle. Using the Pythagoras theorem, the diagonal can be found easily. Given Data / Assumptions: Rectangle dimensions = 3 units and 4 units Diagonal of rectangle = diameter of the circle Circumference of circle = pi * diameter Concept / Approach: Use Pythagoras theorem to find the diagonal of the rectangle. Treat this diagonal as the diameter of the circle, then compute the circumference using the standard formula. Step-by-Step Solution: Diagonal = sqrt(3^2 + 4^2) Diagonal = sqrt(9 + 16) = sqrt(25) = 5 Diameter of circle = 5 Circumference = pi * diameter = 5π Verification / Alternative check: A 3–4–5 triangle is a well-known right triangle. Since the diagonal is exactly 5 units, the resulting circumference of 5π is consistent and exact. Why Other Options Are Wrong: 2.5π and 3π: correspond to smaller diameters than the actual diagonal. 4π: assumes diameter equals the longer side, not the diagonal. 6π: overestimates the diagonal length. Common Pitfalls: Using one side of the rectangle as the diameter. Forgetting that the diagonal is the diameter, not the radius. Arithmetic errors while applying Pythagoras theorem. Final Answer: Circumference of the circle = 5π

Question 19

A triangle has a perimeter of 13 units. The two shorter sides are consecutive integers with lengths x and x + 1. Based on triangle inequality conditions, determine which of the following could be the length of the third side.

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Introduction / Context: This question tests understanding of triangle inequality rules. For any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Given the perimeter and two sides in terms of variables, the possible value of the remaining side can be deduced logically. Given Data / Assumptions: Total perimeter = 13 Two sides are x and x + 1 Third side = 13 - (x + x + 1) = 12 - 2x Concept / Approach: Use triangle inequality: the sum of the two shorter sides must be greater than the third side. Check integer values of x that satisfy all triangle inequalities and see which option matches the possible third side length. Step-by-Step Solution: Let sides be x, x + 1, and 12 - 2x For a valid triangle: x + (x + 1) > (12 - 2x) 2x + 1 > 12 - 2x 4x > 11 => x > 2.75 Try x = 3: sides are 3, 4, and 6 (valid triangle) Thus third side = 6 But 6 is not shorter than x and x+1, violating the given condition Try x = 4: sides are 4, 5, and 4 (valid) Verification / Alternative check: For x = 4, all triangle inequalities are satisfied and the perimeter is 13. The third side equals 4, which is consistent with the problem statement. Why Other Options Are Wrong: 6 and 8: violate triangle inequality or ordering of sides. None of the above: incorrect because 4 is valid. 5: does not match the computed third side. Common Pitfalls: Ignoring the triangle inequality conditions. Forgetting that x and x+1 are the shorter sides. Arithmetic errors while forming the third side. Final Answer: The possible length of the third side is 4

Question 20

A pathway of uniform width 2 m is constructed inside a rectangular plot. The area of the original plot is 96 square metres. If the cost of construction is ₹50 per square metre, determine the total cost of constructing the pathway.

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Introduction / Context: This problem checks whether sufficient information is available to compute the area of a pathway constructed inside a rectangle. While the width of the pathway and the total area of the plot are given, the individual dimensions of the rectangle are not specified. Given Data / Assumptions: Area of rectangular plot = 96 m^2 Width of pathway = 2 m (inside the plot) Cost = ₹50 per m^2 Concept / Approach: To find the area of the pathway, the dimensions of the rectangle (length and breadth) are required. Knowing only the area does not uniquely determine the shape of the rectangle, so the inner dimensions after subtracting the pathway width cannot be calculated. Step-by-Step Solution: Let original dimensions be l and b such that l*b = 96 Inner rectangle dimensions become (l - 4) and (b - 4) Pathway area = l*b - (l - 4)(b - 4) Without knowing l and b separately, pathway area cannot be determined uniquely Verification / Alternative check: Different rectangles with area 96 (for example 12x8 or 16x6) give different pathway areas, proving that the answer is not unique. Why Other Options Are Wrong: ₹2000, ₹3000, ₹4000, ₹4800: each assumes specific dimensions not provided in the question. Common Pitfalls: Assuming the rectangle is a square without justification. Proceeding with calculations despite missing essential data. Ignoring that multiple rectangles can have the same area. Final Answer: The given data is insufficient to determine the cost

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