Question 1

The area of a rectangle is thrice that of a square. If the length of the rectangle is 40 cm and its breadth is $\frac{3}{2}$ times that of the side of the square, then the side of the square is

Accepted Answer

Concept & Algebraic Formulation of Areas: By defining variables for the unknown side lengths, we can translate the relationships between the areas and dimensions of the square and rectangle into an algebraic equation. Area of Rectangle = 3 × Area of Square Step-by-Step Solution: • Let the side of the square be x cm. Its area is x^2 cm^2. • The length of the rectangle is given as l = 40 cm. • The breadth of the rectangle is (3/2) times the side of the square: b = (3/2)x cm. • The area of the rectangle is l × b = 40 × (3/2)x = 60x cm^2. • According to the problem, the area of the rectangle is thrice the area of the square: 60x = 3(x^2). • Simplify the equation: 60x = 3x^2 \Rightarrow 20x = x^2. • Since x representing a length cannot be 0, divide both sides by x: x = 20 cm. Exam Strategy & Shortcut: Instead of writing out full algebraic equations, you can quickly test the given options. If the side is 20 (Option b), the square area is 400. The rectangle breadth is (3/2)(20) = 30. The rectangle area is 40 × 30 = 1200. Since 1200 is exactly 3 × 400, Option (b) is verified instantly. Common Pitfall: Forgetting to square the side of the square when representing its area, thereby writing 3x instead of 3x^2, is a frequent and fatal calculation error. Final Answer: Therefore, the correct answer is 20 cm.

Question 2

The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq. m. The length of the rectangle is

Accepted Answer

Concept & Quadratic Equations in Geometry: When given the sum and product of two variables (like length and breadth), it naturally forms a quadratic equation. The sum is derived from the perimeter, and the product from the area. l + b = \frac{Perimeter}{2} l × b = Area Step-by-Step Solution: • The perimeter of the square is 160 m. Its side is 160 / 4 = 40 m. • The area of the square is 40^2 = 1600 sq m. • The perimeter of the rectangle is also 160 m: 2(l + b) = 160 \Rightarrow l + b = 80 m. • The area of the rectangle is 100 sq m less than the square's area: 1600 - 100 = 1500 sq m. Thus, l × b = 1500. • We need two numbers that add up to 80 and multiply to 1500. By inspecting the factors, these are clearly 50 and 30. • Since the length is typically considered the longer dimension, l = 50 m and b = 30 m. Exam Strategy & Shortcut: Once you know l + b = 80 and l × b = 1500, look at the options. You need an option that, when subtracted from 80, gives a number that multiplies with the option to yield 1500. 50 × (80 - 50) = 50 × 30 = 1500. Option (c) is immediately correct. Common Pitfall: A common mistake is misreading the phrase "less than that of the square by 100" and instead adding 100 to the square's area, which leads to incorrect factors and a confusing equation. Final Answer: Therefore, the correct answer is 50 m.

Question 3

The area of a rectangle is four times the area of a square. The length of the rectangle is 90 cm and the breadth of the rectangle is $\frac{2}{3}	ext{rd}$ the side of the square. What is the side of the square? (Bank Recruitment, 2008)

Accepted Answer

Concept & Setting up Algebraic Ratios: Translate the word problem into mathematical expressions by assigning a variable to the side of the square and expressing all other given values in terms of this variable to form an equation. Area of Rectangle = 4 × Area of Square Step-by-Step Solution: • Let the side of the square be x cm. The area of the square is x^2 cm^2. • The length of the rectangle is l = 90 cm. • The breadth of the rectangle is (2/3) the side of the square: b = (2/3)x cm. • The area of the rectangle is l × b = 90 × (2/3)x = 60x cm^2. • We are given that the area of the rectangle is four times the area of the square: 60x = 4x^2. • Simplify the equation by dividing both sides by 4x (since length x 
eq 0): 15 = x. • The side of the square is 15 cm. • Checking the given options: (a) 9 cm, (b) 10 cm, (c) 20 cm, (d) Cannot be determined. None of these match the calculated side of 15 cm. Exam Strategy & Shortcut: You can quickly solve the mental math: 90 × (2/3)x = 4x^2 \Rightarrow 60x = 4x^2 \Rightarrow x = 15. Scanning the options, 15 is evidently missing, making "None of these" the clear and obvious choice. Do not doubt your calculation just because the explicit number isn't listed. Common Pitfall: Students often second-guess themselves when their calculated answer is not among the explicit numeric options and might pick a closely related incorrect option or mistakenly choose "Cannot be determined" out of confusion. Final Answer: Therefore, the correct answer is None of these.

Question 4

The cost of fencing a square field @ ₹ 20 per metre is ₹ 10,080. How much will it cost to lay a three metre wide pavement along the fencing inside the field @ ₹ 50 per sq. metre?

Accepted Answer

Concept & Inner Path Area: To find the cost of a pavement inside a square field, we first determine the side length of the outer square using the total fencing cost. Then, we find the area of the pavement by subtracting the inner square's area from the outer square's area. Perimeter = \frac{Total Cost of Fencing}{Cost per Metre} Path Area = Area of Outer Square - Area of Inner Square Step-by-Step Solution: • Find the perimeter of the square field: 10080 / 20 = 504 metres. • Find the side of the outer square field (a): 504 / 4 = 126 metres. • A 3-metre wide pavement is laid *inside* the field. The side of the inner square (b) will be 126 - 2(3) = 120 metres. • Calculate the area of the pavement: Outer Area - Inner Area = 126^2 - 120^2. • Use the algebraic identity a^2 - b^2 = (a-b)(a+b) for easier calculation: (126 - 120)(126 + 120) = 6 × 246 = 1476 sq. metres. • Total cost of laying the pavement = 1476 × 50 = 73,800. Exam Strategy & Shortcut: Instead of squaring 126 and 120 manually, the difference of squares shortcut (a-b)(a+b) is essential here. Multiplying 6 × 246 × 50 can be further simplified by grouping: 246 × (6 × 50) = 246 × 300 = 73800, saving valuable calculation time. Common Pitfall: A frequent error is subtracting the path width only once instead of twice to find the inner dimension (i.e., calculating 126 - 3 instead of 126 - 6). Final Answer: Therefore, the correct answer is ₹ 73,800.

Question 5

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

Accepted Answer

Concept & Path Optimization in Geometry: Walking along the edges involves traveling two sides of the square, whereas walking diagonally involves traversing the hypotenuse of the right-angled triangle formed by the sides. Diagonal = a\sqrt{2} Step-by-Step Solution: • Let the side of the square lot be a. • If the man walks along the edges from one corner to the opposite, he covers two sides. Distance = a + a = 2a. • If he walks diagonally, the distance is a\sqrt{2}. Since \sqrt{2} \approx 1.414, the diagonal distance is approximately 1.414a. • Distance saved = 2a - 1.414a = 0.586a. • Percentage saved = (\frac{Distance Saved}{Original Distance}) × 100. • Percentage saved = ((0.586a/2a)) × 100 = 0.293 × 100 = 29.3%. • Approximately, this is 30%. Exam Strategy & Shortcut: Memorize that the ratio of the diagonal to the sum of two sides of a square is \sqrt{2} : 2, or about 1.41 : 2. The saving is (2 - 1.41) / 2 = 0.59 / 2 \approx 0.30 or 30%. Knowing this standard result saves you from doing calculations on the spot. Common Pitfall: A common mistake is calculating the percentage relative to the new diagonal path (dividing by 1.414a) rather than the original longer path (dividing by 2a). Final Answer: Therefore, the correct answer is 30.

Question 6

If the length of diagonal AC of a square ABCD is 5.2 cm, then the area of the square is (L.I.C.A.D.O., 2008)

Accepted Answer

Concept & Area from Diagonal: When the diagonal of a square is given, its area can be calculated directly without finding the side length first. Area of Square = (d^2/2) Step-by-Step Solution: • We are given the diagonal d = 5.2 cm. • Substitute the diagonal into the area formula: Area = ((5.2)^2/2). • First, square 5.2: 5.2 × 5.2 = 27.04. • Divide by 2: (27.04/2) = 13.52 sq.cm. Exam Strategy & Shortcut: To quickly square 5.2, you can think of it as (5 + 0.2)^2 = 25 + 2(5)(0.2) + 0.04 = 25 + 2 + 0.04 = 27.04. Dividing 27.04 by 2 yields 13.52 seamlessly. Common Pitfall: A frequent mistake is forgetting to divide by 2, yielding 27.04, or mistakenly thinking the given diagonal is the side length. Final Answer: Therefore, the correct answer is 13.52 sq.cm.

Question 7

A man walking at the speed of 4 kmph crosses a square field diagonally in 3 minutes. The area of the field is

Accepted Answer

Concept & Distance and Area Relation: First, use the speed and time to find the total distance walked, which corresponds to the length of the diagonal. Then, use the diagonal to find the area. Distance = Speed × Time Area of Square = (1/2)d^2 Step-by-Step Solution: • Convert speed from kmph to m/s: 4 kmph = 4 × (5/18) = (10/9) m/s. • Convert time from minutes to seconds: 3 minutes = 3 × 60 = 180 seconds. • Calculate the diagonal length (distance): d = (10/9) × 180 = 200 metres. • Calculate the area using the diagonal: Area = (1/2) × (200)^2. • Area = (1/2) × 40000 = 20000 m^2. Exam Strategy & Shortcut: You can keep units simpler: 4 km/hr = 4000 m / 60 min. In 3 minutes, he covers 4000 × (3/60) = (4000/20) = 200 metres. Area = (200^2/2) = 20000. This avoids messy fractional m/s conversions. Common Pitfall: Failing to convert units consistently (e.g., mixing kmph with minutes directly) will lead to vastly incorrect diagonal lengths. Final Answer: Therefore, the correct answer is 20000 m^2.

Question 8

If the length of the diagonal of a square is 20 cm, then its perimeter must be

Accepted Answer

Concept & Diagonal to Perimeter Conversion: To find the perimeter from the diagonal, establish the side length first using the diagonal-side relationship, then multiply by 4. d = a\sqrt{2} Perimeter = 4a Step-by-Step Solution: • Given the diagonal d = 20 cm. • Set up the equation for the side length a: a\sqrt{2} = 20. • Solve for a: a = (20/\sqrt{2)}. • Rationalize the denominator: a = \frac{20\sqrt{2}}{2} = 10\sqrt{2} cm. • Calculate the perimeter: 4a = 4 × 10\sqrt{2} = 40\sqrt{2} cm. Exam Strategy & Shortcut: You can combine the formulas to create a direct relation: Perimeter = 4 × ((d/\sqrt{2)}) = 2d\sqrt{2}. Plugging in d = 20 immediately gives 2(20)\sqrt{2} = 40\sqrt{2} cm. Common Pitfall: Students might calculate the side length 10\sqrt{2} and mistakenly pick option (a) without proceeding to calculate the final perimeter. Final Answer: Therefore, the correct answer is 40\sqrt{2} cm.

Question 9

The area of a square field is 0.5 hectare. Its diagonal would be

Accepted Answer

Concept & Formula: The key to solving this problem is knowing the conversion factor between hectares and square meters, and the relationship between a square's area and its diagonal. Area of a square = (1/2) × (Diagonal)^2 Step-by-Step Solution: • Step 1: Convert the given area into square meters. We know that 1 hectare = 10,000 m^2. • Step 2: Area = 0.5 hectare = 0.5 × 10,000 m^2 = 5,000 m^2. • Step 3: Use the area-diagonal formula: 5,000 = (1/2) × d^2. • Step 4: Multiply both sides by 2: d^2 = 10,000. • Step 5: Take the square root: d = 100 m. Exam Strategy & Shortcut: Memorize standard land area conversions like hectares to square meters (1 ha = 10^4 m^2). Instead of finding the side length first (\sqrt{5000}) and then multiplying by \sqrt{2}, directly use the Area = (d^2/2) formula to save time. Common Pitfall: A frequent mistake is not knowing the hectare-to-meter conversion, or incorrectly assuming 1 hectare = 100 m^2 or 1,000 m^2, which leads to incorrect orders of magnitude in the final answer. Final Answer: Therefore, the correct answer is 100 m.

Question 10

Area of a square natural lake is 50 sq. kms. A diver wishing to cross the lake diagonally, will have to swim a distance of

Accepted Answer

Concept & Geometry: This question tests the area-to-diagonal relationship of a square and tests unit conversion awareness (kilometers to miles). Area = (1/2) d^2 Step-by-Step Solution: • Step 1: The area of the square lake is given as 50 km^2. • Step 2: Using the formula relating area and diagonal (d): (1/2) d^2 = 50. • Step 3: d^2 = 100, so the diagonal d = 10 km. • Step 4: The options are given in miles. Convert kilometers to miles (approximate conversion factor: 1 mile \approx 1.609 km). • Step 5: Distance in miles = (10/1.609) \approx 6.21 miles. • Step 6: Since 6.21 miles does not match any of the given numeric options (10, 12, 15), the correct choice is "None of these". Exam Strategy & Shortcut: As soon as you calculate d = 10 km, immediately check the units in the options. Realize that 10 km is definitely less than 10 miles (since a mile is longer than a kilometer). Therefore, options a, b, and c are impossible without any complex calculations. Common Pitfall: The most common mistake is ignoring the unit change and carelessly selecting "10 miles" after calculating a value of 10, forgetting that the calculation yielded kilometers. Final Answer: Therefore, the correct answer is None of these.

Question 11

A square and a rectangle have equal areas. If their perimeters are $p_1$ and $p_2$ respectively, then

Accepted Answer

Concept & Optimization Principle: This is a standard geometric property: for any given, fixed area, the quadrilateral with the minimum perimeter is a square. As a shape gets more elongated (like a long rectangle), its perimeter increases for the same area. Step-by-Step Solution: • Step 1: Let the area of both the square and the rectangle be A. • Step 2: For the square, side a = \sqrt{A}. Its perimeter p_1 = 4\sqrt{A}. • Step 3: For the rectangle, let dimensions be l and b such that l × b = A. Its perimeter is p_2 = 2(l + b). • Step 4: By the AM-GM inequality, for positive numbers l and b: (l + b/2) \ge \sqrt{lb}. • Step 5: Substituting A: (l + b/2) \ge \sqrt{A} \implies l + b \ge 2\sqrt{A} \implies 2(l + b) \ge 4\sqrt{A}. • Step 6: So, p_2 \ge p_1. Since a rectangle (not a square) has l 
eq b, the inequality is strict: p_2 > p_1, or p_1 < p_2. Exam Strategy & Shortcut: Use a counter-example. Let Area = 16. Square: side = 4, Perimeter p_1 = 4 × 4 = 16. Rectangle: sides = 8 and 2. Perimeter p_2 = 2(8 + 2) = 20. Clearly, 16 < 20, so p_1 < p_2. Common Pitfall: Students often confuse this with the opposite rule: "For a fixed perimeter, the square maximizes area." Memorizing these rules backwards leads to picking the wrong inequality sign. Final Answer: Therefore, the correct answer is p_1 < p_2.

Question 12

If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is

Accepted Answer

Concept & Linear Ratios: The ratio of the perimeters of two squares is exactly the same as the ratio of their sides. The ratio of their sides is the square root of the ratio of their areas. Step-by-Step Solution: • Step 1: Let the areas of the two squares be A_1 and A_2. We are given (A_1/A_2) = (225/256). • Step 2: The area of a square is the side squared (s^2). So, (s_1^2/s_2^2) = (225/256). • Step 3: Take the square root of both sides to find the ratio of their sides: (s_1/s_2) = \frac{\sqrt{225}}{\sqrt{256}} = (15/16). • Step 4: The perimeter of a square is 4s. The ratio of their perimeters is (P_1/P_2) = (4s_1/4s_2) = (s_1/s_2). • Step 5: Therefore, the ratio of the perimeters is also (15/16), or 15 : 16. Exam Strategy & Shortcut: Recognize immediately that perimeter is a 1D measure and area is a 2D measure. Simply take the square root of the area ratio (\sqrt{225} : \sqrt{256}) to get the 1D ratio directly. Common Pitfall: A common mistake is forgetting to take the square root and selecting 225 : 256, or reversing the order and selecting 16 : 15. Final Answer: Therefore, the correct answer is 15 : 16.

Question 13

Of the two square fields, the area of one is 1 hectare while the other one is broader by 1%. The difference in their areas is

Accepted Answer

Concept & Unit Conversion and Percentage Increase: This problem combines metric area conversions with percentage changes in linear dimensions. 1 hectare (ha) = 10,000 m^2 Step-by-Step Solution: • Step 1: The area of the first square field is 1 hectare, which equals 10,000 m^2. • Step 2: Find the side length of the first field. Side s_1 = \sqrt{10000} = 100 m. • Step 3: The second field is "broader by 1%". This means its side length is 1% larger. • Step 4: Calculate the new side length: s_2 = 100 + (1% of 100) = 100 + 1 = 101 m. • Step 5: Calculate the area of the second field: A_2 = 101^2 = 101 × 101 = 10,201 m^2. • Step 6: Find the difference in their areas: 10,201 m^2 - 10,000 m^2 = 201 m^2. Exam Strategy & Shortcut: Use the successive percentage formula x + y + (xy/100) for area changes when sides change by a percentage. With a 1% increase in side, the area increases by 1 + 1 + (1 × 1/100) = 2.01%. 2.01% of 10,000 m^2 is 201 m^2. This skips calculating the new side length and squaring it entirely. Common Pitfall: A frequent mistake is calculating 1% of the *area* directly instead of the side, resulting in an incorrect difference of 100 m^2. Final Answer: Therefore, the correct answer is 201 m^2.

Question 14

If each side of a square is increased by 50%, the ratio of the area of the resulting square to that of the given square is

Accepted Answer

Concept & Proportional Area Scaling: If the side of a square is multiplied by a certain factor, the area of the square is multiplied by the square of that factor. Step-by-Step Solution: • Step 1: Let the original side of the square be s. The original area is A_1 = s^2. • Step 2: The side is increased by 50%. The new side is s + 0.5s = 1.5s. • Step 3: Calculate the area of the resulting square: A_2 = (1.5s)^2 = 2.25s^2. • Step 4: Find the requested ratio (Resulting Area to Given Area): (A_2/A_1) = (2.25s^2/s^2) = 2.25. • Step 5: Convert the decimal to a fraction: 2.25 = (225/100). • Step 6: Simplify the fraction by dividing numerator and denominator by 25: (9/4). • Step 7: The ratio is 9 : 4. Exam Strategy & Shortcut: Assume the original side is 2. Original area = 4. A 50% increase on 2 adds 1, so the new side is 3. New area = 3^2 = 9. The ratio of the new area to the old area is immediately 9 : 4. Common Pitfall: Choosing 4 : 9 because of reading the ratio order backward. The question asks for the "resulting square" first, then the "given square". Final Answer: Therefore, the correct answer is 9 : 4.

Question 15

What happens to the area of a square when its side is halved? Its area will

Accepted Answer

Concept & Quadratic Relationship: The area of a square (A) is equal to its side (s) squared (A = s^2). Therefore, any fractional change applied to the side is squared when determining the impact on the area. Step-by-Step Solution: • Step 1: Let the original side of the square be s. The initial area is A_1 = s^2. • Step 2: The new side is halved, so s_{new} = (s/2). • Step 3: The new area is A_2 = ((s/2))^2. • Step 4: Expanding this gives A_2 = (s^2/4). • Step 5: Substituting the original area A_1 for s^2, we get A_2 = (1/4) A_1. • Step 6: This means the area becomes one-fourth of its original size. Exam Strategy & Shortcut: Use a simple base number. Let the original side be 4, so Area = 16. Halve the side to 2, the new Area = 4. Since 4 is one-fourth of 16, the area becomes one-fourth. Common Pitfall: The most common mistake is assuming area scales linearly and picking "become half". Always square the scaling factor for 2D measurements. Final Answer: Therefore, the correct answer is become one-fourth.

Question 16

An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is

Accepted Answer

Concept & Error Propagation in Area: When there is a small percentage error in measuring the linear dimension of a square, the error propagates in two dimensions. We can use the successive percentage change formula to find the net error in area: Net Error = x + y + (xy/100) Here, x and y represent the percentage error in length and width, which are both +2%. Step-by-Step Solution: • Step 1: The percentage error in measuring the side is 2% in excess, meaning it is a +2% change. • Step 2: Let x = 2 and y = 2. • Step 3: Plug these values into the successive percentage formula: Error in Area = 2 + 2 + (2 × 2/100). • Step 4: Calculate the terms: 4 + (4/100) = 4 + 0.04. • Step 5: The total percentage error is 4.04%. Exam Strategy & Shortcut: Assume the true side is 100. The true area is 100 × 100 = 10,000. With a 2% excess error, the measured side is 102. The calculated area is 102^2 = 10,404. The excess area is 404. Percentage error = ((404/10000)) × 100 = 4.04%. Common Pitfall: A common mistake is simply adding the errors (2% + 2% = 4%) or assuming the area error is identical to the side error (2%). This ignores the compounding factor ((xy/100)). Final Answer: Therefore, the correct answer is 4.04%.

Question 17

If the diagonal of a square is made 1.5 times, then the ratio of the areas of two squares is

Accepted Answer

Concept & Formula: The area of a square can be calculated directly from its diagonal using the formula: Area = (d^2/2) where d is the length of the diagonal. Step-by-Step Solution: 1. Let the diagonal of the original square be d. 2. The area of the original square is A_1 = (d^2/2). 3. The diagonal of the new square is made 1.5 times, so it is 1.5d = (3/2)d. 4. The area of the new square is A_2 = ( (3d/2)^2 / 2 ) = ( \frac{9d^2/4) }{ 2 } = (9d^2/8). 5. The ratio of the areas of the original square to the new square is A_1 : A_2 = (d^2/2) : (9d^2/8). 6. Simplify the ratio by multiplying both sides by (8/d^2): 4 : 9. Exam Strategy & Shortcut: Since Area is proportional to the square of any linear dimension (like the diagonal), if the diagonal is multiplied by a factor k, the area is multiplied by k^2. Here, k = 1.5 = (3/2). Therefore, the area factor is ((3/2))^2 = (9/4). The ratio of original area to new area is 1 : (9/4), which simplifies to 4 : 9. Common Pitfall: Confusing the order of the ratio. The question asks for the ratio of the areas of the two squares, implying original to new. The inverse ratio 9 : 4 is not among the options, which helps prevent this error. Final Answer: Therefore, the correct answer is 4 : 9.

Question 18

The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by

Accepted Answer

Concept & Net Percentage Change: When two dimensions of a 2D shape are increased by certain percentages, the net percentage change in area can be found using the effective percentage formula: Net Change = x + y + (xy/100) Step-by-Step Solution: 1. Let the original side of the square be 10 units. 2. The original area is 10 × 10 = 100 sq units. 3. The length is increased by 40%, so the new length is 10 + 4 = 14 units. 4. The breadth is increased by 30%, so the new breadth is 10 + 3 = 13 units. 5. The area of the resulting rectangle is 14 × 13 = 182 sq units. 6. The increase in area is 182 - 100 = 82 sq units. 7. Since the original area is 100, the percentage increase is exactly 82%. Exam Strategy & Shortcut: Apply the net percentage change formula directly: Here, x = +40 and y = +30. Net % Change = 40 + 30 + (40 × 30/100) = 70 + (1200/100) = 70 + 12 = 82% Common Pitfall: Simply adding the two percentage increases (40% + 30% = 70%) without accounting for the compounded effect of both dimensions increasing simultaneously. Final Answer: Therefore, the correct answer is 82%.

Question 19

A rectangle becomes a square when its length is reduced by 10 units and its breadth is increased by 5 units. But by this process the area of the rectangle is reduced by 210 sq. units. The area of the rectangle (in square units) is

Accepted Answer

Concept & Algebraic Formulation: Establish equations for the equality of sides to form a square, and the difference in area between the original rectangle and the new square. Area of Rectangle = l × b Step-by-Step Solution: 1. Let the original length be l and breadth be b. 2. The figure becomes a square when l is reduced by 10 and b is increased by 5: l - 10 = b + 5 \Rightarrow l = b + 15. 3. The new area is (l - 10)(b + 5). The problem states this new area is 210 less than the original area (lb): lb - (l - 10)(b + 5) = 210. 4. Expand the equation: lb - (lb + 5l - 10b - 50) = 210 -5l + 10b + 50 = 210 -5l + 10b = 160 5. Substitute l = b + 15 into the equation: -5(b + 15) + 10b = 160 -5b - 75 + 10b = 160 5b = 235 \Rightarrow b = 47. 6. Calculate length: l = 47 + 15 = 62. 7. The area of the original rectangle A = l × b = 62 × 47 = 2914. 8. Evaluate the options based on A = 2914: (a) 2950 > 2914 2914 > 2875 (False) (c) 2925 2875 (False) (d) 2925 > 2914 > 2900 (True, since 2914 is between 2900 and 2925). Exam Strategy & Shortcut: Write the area equation in terms of the new square's side S: Original length = S + 10, Original breadth = S - 5. Original Area = (S+10)(S-5) = S^2 + 5S - 50. New Area = S^2. Difference = (S^2 + 5S - 50) - S^2 = 210. 5S = 260 \Rightarrow S = 52. Original Area = S^2 + 210 = 52^2 + 210 = 2704 + 210 = 2914. This avoids simultaneous equations. Common Pitfall: Misinterpreting the confusing inequality notation in the options. Option (d) 2925 > A > 2900 correctly signifies that A falls within the range (2900, 2925). Final Answer: Therefore, the correct answer is 2925 > A > 2900.

Question 20

The difference of the areas of two squares drawn on two line segments of different lengths is 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.

Accepted Answer

Concept & Algebraic Formulation: The area of a square is the square of its side. If we have two squares with side lengths a and b, the difference in their areas is a^2 - b^2. We can solve for the unknown side by setting up a quadratic equation using the given constraints. Step-by-Step Solution: 1. Let the length of the smaller line segment (side of the smaller square) be x cm. 2. The length of the greater line segment is given as (x + 2) cm. 3. The area of the smaller square is x^2. 4. The area of the larger square is (x + 2)^2. 5. According to the given condition, the difference in areas is 32 sq. cm: (x + 2)^2 - x^2 = 32 6. Expand the squared term: (x^2 + 4x + 4) - x^2 = 32 7. Simplify the equation: 4x + 4 = 32 4x = 28 x = 7 cm 8. The length of the greater line segment is x + 2 = 7 + 2 = 9 cm. Exam Strategy & Shortcut: Use option elimination directly on the greater segment. If greater is 9: smaller is 7. Areas are 81 and 49. Difference = 81 - 49 = 32. This matches perfectly and takes seconds to compute mentally. Common Pitfall: Solving the equation correctly to find x = 7, but forgetting that the question asks for the greater line segment, thus incorrectly choosing 7 cm as the answer. Final Answer: Therefore, the correct answer is 9 cm.

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